THE LIBRARY 
 
 OF 
 
 THE UNIVERSITY 
 OF CALIFORNIA 
 
 LOS ANGELES
 
 DIOPHANTUS OF ALEXANDRIA
 
 CAMBRIDGE UNIVERSITY PRESS 
 
 : FETTER LANE, E.G. 
 C. F. CLAY, MANAGER 
 
 ffiUinburgi): too, PRINCES STREET 
 
 Berlin: A. ASHER AND CO. 
 
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 atrt Calcutta: MACMILLAN AND CO., LTD. 
 
 All rights reserved
 
 DIOPHANTUS OF ALEXANDRIA 
 
 A STUDY IN THE HISTORY 
 
 OF 
 
 GREEK ALGEBRA 
 
 BY 
 
 SIR THOMAS L. HEATH, K.C.B., 
 
 SC.D., SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE 
 
 SECONT> EDITION 
 
 WITH A SUPPLEMENT CONTAINING AN ACCOUNT OF FERMAT'S 
 
 THEOREMS AND PROBLEMS CONNECTED WITH DIOPHANTINE 
 
 ANALYSIS AND SOME SOLUTIONS OF DIOPHANTINE 
 
 PROBLEMS BY EULER 
 
 Cambridge : 
 
 at the University Press 
 
 1910
 
 (ffamtm&ge: 
 
 PRINTED BY JOHN CLAY, M.A. 
 AT THE UNIVERSITY PRESS
 
 Library 
 
 PREFACE 
 
 r I "'HE first edition of this book, which was the first English 
 * Diophantus, appeared in 1885, and has long been out of 
 print. Inquiries made for it at different times suggested to me 
 that it was a pity that a treatise so unique and in many respects 
 so attractive as the Arithmetica should once more have become 
 practically inaccessible to the English reader. At the same time 
 I could not but recognise that, after twenty-five years in which so 
 much has been done for the history of mathematics, the book 
 needed to be brought up to date, Some matters which in 1885 
 were still subject of controversy, such as the date of Diophantus, 
 may be regarded as settled, and some points which then had to 
 be laboured can now be dismissed more briefly. Practically the 
 whole of the Introduction, except the chapters on the editions of 
 Diophantus, his methods of solution, and the porisms and other 
 assumptions found in his work, has been entirely rewritten and 
 much shortened, while the chapters on the methods and on the 
 porisms etc., have been made fuller than before. The new text of 
 Tannery (Teubner 1893, 1895) has enabled a number of obscure 
 passages, particularly in Books V and VI, to be cleared up and, 
 as a basis for a reproduction of the whole work, is much superior to 
 the text of Bachet. I have taken the opportunity to make my 
 version of the actual treatise somewhat fuller and somewhat closer 
 to the language of the original. In other respects also I thought 
 I could improve upon a youthful work which was my first essay in 
 the history of Greek mathematics. When writing it I was solely 
 concerned to make Diophantus himself known to mathematicians, 
 
 833088
 
 vi PREFACE 
 
 and I did not pay sufficient attention to Fermat's notes on the 
 various problems. It is well known that it is in these notes that 
 many of the great propositions discovered by Fermat in the 
 theory of numbers are enshrined ; but, although the notes are 
 literally translated in Wertheim's edition, they do not seem to 
 have appeared in English ; moreover they need to be supple- 
 mented by passages from the correspondence of Fermat and from 
 the Doctrinae analyticae Inventum Novum of Jacques de Billy. 
 The histories of mathematics furnish only a very inadequate 
 description of Fermat's work, and it seemed desirable to attempt 
 to give as full an account of his theorems and problems in 
 or connected with Diophantine analysis as it is possible to 
 compile from the scattered material available in Tannery and 
 Henry's edition of the Oeuvres de Fermat (1891 1896). So much 
 of this material as could not be conveniently given in the notes 
 to particular problems of Diophantus I have put together in 
 the Supplement, which is thus intended to supply a missing 
 chapter in the history of mathematics. Lastly, in order to make 
 the book more complete, I thought it right to add some of the 
 more remarkable solutions of difficult Diophantine problems given 
 by Euler, for whom such problems had a great fascination ; the last 
 section of the Supplement is therefore devoted to these solutions. 
 
 T. L. H. 
 October, 1910.
 
 CONTENTS 
 
 INTRODUCTION 
 
 PAGES 
 
 CHAP. I. Diophantus and his Works !_i 3 
 
 II. The MSS. of and writers on Diophantus . . 14 31 
 
 III. Notation and definitions of Diophantus . . . 3253 
 
 IV. Diophantus' methods of solution . . - . . 54 98 
 
 V. The Porisms and other assumptions in Diopbantus 99110 
 
 VI. The place of Diophantus m 127 
 
 THE ARITHMETICA 
 
 BOOK L 129-143 
 
 IL 143-156 
 
 IIL 156168 
 
 IV - 168199 
 
 \7 
 
 V * 200 225 
 
 VI. 
 
 . - . 226246 
 
 On Polygonal numbers 247 2 ^ 9 
 
 Conspectus of the Arithmetica 260266 
 
 SUPPLEMENT 
 
 SECTIONS I V. NOTES, THEOREMS AND PROBLEMS BY FERMAT. 
 
 I. On numbers separable into integral squares . . 267277 
 
 II- Equation 3*-Ay i =\ ....... 277292 
 
 III. Theorems and Problems on rational right-angled 
 
 triangles 293 _ 3 ,g 
 
 IV. Other problems by Fermat 318320 
 
 V. Fermat's Triple-equations 321328 
 
 SECTION VI. SOME SOLUTIONS BY EULER 329380 
 
 INDEX: I. GREEK 381382 
 
 II. ENGLISH 382 387
 
 INTRODUCTION 
 CHAPTER I 
 
 DIOPHANTUS AND HIS WORKS 
 
 THE divergences between writers on Diophantus used to begin, 
 as Cossali said 1 , with the last syllable of his name. There is now, 
 however, no longer any doubt that the name was Diophantar, not 
 Diophanter 2 . 
 
 The question of his date is more difficult Abu'lfaraj, the 
 Arabian historian, in his History of tlte Dynasties, places Diophantus 
 under the Emperor Julian (A.D. 361-3), but without giving any 
 authority; and it may be that the statement is due simply to a 
 confusion of our Diophantus with a rhetorician of that name, 
 mentioned in another article of Suidas, who lived in the time of 
 Julian*. On the other hand, Rafael Bombelli in his Algebra, 
 
 1 Cossali, Origine, trasporto in Italia, primi progressi in essa ddr Algebra (Parma, 
 1797-9), I. p. 61 : "Su la desinenza del nome comincia la diversita tra gli scrittori." 
 
 1 Greek authority is overwhelmingly in favour of Diophant<w. The following is the 
 evidence, which is collected in the second volume of Tannery's edition of Diophantus 
 (henceforward to be quoted as "Dioph.," "Dioph. n. p. 36" indicating page 36 of 
 Vol. ii., while "Dioph. II. 20" will mean proposition 20 of Book II.): Suidas s.v. 
 'TraTia (Dioph. n. p. 36), Theon of Alexandria, on Ptolemy's Syntaxis Book I. c. 9 
 (Dioph. II. p. 35), Anthology, Epigram on Diophantus (Ep. xiv. 126; Dioph. II. p. 60), 
 Anonymi prolegomena in Introductionem arithmeticam Nicomachi (Dioph. II. p. 73), 
 Georgii Pachymerae paraphrasis (Dioph. n. p. 122), Scholia of Maximus Planudes 
 (Dioph. n. pp. 148, 177, 178 etc.), Scholium on larublichus / Nicomachi arithm. inirod., 
 ed. PisteUi, p. 127 (Dioph. n. p. 72), a Scholium on Dioph. n. 8 from the MS. "A" 
 (Dioph. n. p. 260), which is otherwise amusing (H ifwxn aw > Ai6#arre, eft; perk TOV 
 Sarai-a frexa Trft Siv/coXi'as TUV re iXXwr ffov tfewpij/uiTwr KCU Si) ecu TOV rttporros ffcwpj- 
 ftATm, " Your soul to perdition, Diophantus, for the difficulty of your problems in general 
 and of this one in particular ") ; John of Jerusalem ( loth c.) alone ( Vita loannis Damas- 
 ceni xi. : Dioph. n. p. 36), if the reading of the MS. Parisinus 1559 is right, wrote, in 
 the plural, wj UvOaydpcu. 1} AIO^OJTOI, where however Aio^oirai is dearly a mistake for 
 
 3 \i3dvios, ffoQurriis 'Arrtox/s, rwr tfl 'louXtoyoO TOV /ScurtAlcds 
 Qeodoffiov TOV -rpfff^vripov Qajryariov xarp6j, /ua^TTjj AIOC><UTOI/. 
 
 H. D.
 
 2 INTRODUCTION 
 
 published in 1572, says dogmatically that Diophantus lived under 
 Antoninus Pius (138-161 A.D.), but there is no confirmation of this 
 date either. 
 
 The positive evidence on the subject can be given very shortly. 
 An upper limit is indicated by the fact that Diophantus, in his 
 book on Polygonal Numbers, quotes from Hypsicles a definition 
 of such a number 1 . Hypsicles was also the writer of the sup- 
 plement to Euclid's Book XIII. on the Regular Solids known as 
 Book XIV. of the Elements ; hence Diophantus must have written 
 later than, say, 150 B.C. A lower limit is furnished by the fact that 
 Diophantus is quoted by Theon of Alexandria 2 ; hence Diophantus 
 wrote before, say, 350 A.D. There is a wide interval between 
 150 B.C. and 350 A.D., but fortunately the limits can be brought 
 closer. We have a letter of Psellus (nth c.) in which Diophantus 
 and Anatolius are mentioned as writers on the Egyptian method 
 of reckoning. " Diophantus," says Psellus 3 , " dealt with it more 
 accurately, but the very learned Anatolius collected the most 
 essential parts of the doctrine as stated by Diophantus in a 
 different way (reading erepwf) and in the most succinct form, 
 dedicating (irpoae^xav^a-e) his work to Diophantus." It would 
 appear, therefore, that Diophantus and Anatolius were contem- 
 poraries, and it is most likely that the former would be to the 
 latter in the relation of master to pupil. Now Anatolius wrote 
 about 278-9 A.D., and was Bishop of Laodicea about 280 A.D. We 
 may therefore safely say that Diophantus flourished about 250 A.D. 
 or not much later. This agrees well with the fact that he is not 
 quoted by Nicomachus (about 100 A.D.), Theon of Smyrna (about 
 130 A.D.) or lamblichus (end of 3rd c.). 
 
 1 Dioph. I. p. 470-2. 
 
 2 Theo Alexandrinus in primum librum Ptolemaei Mathematicae Compositionis (on c. 
 IX.) : see Dioph. II. p. 35, Ka.0' a KOI &i6<pavr6s 0ij<rr TTJS yap parados d/uera^Tou o&o-ijs 
 /cat fffTWffrjs irai/Tore, TO iro\\aTT\a(na'6/j.ei>ov elSos CTT' avrrjv avrb TO eZSoj &TTCU K.r.e. 
 
 3 Dioph. II. p. 38-9 : irepl de rrjs aiyvimaKrjs fj,e668ov Tatirrjs Ai6<f>ai>Tos (j.ev di^\a^i> 
 aKpi^ffTfpov, 6 54 \oyiibraTos 'AvctToXios TCI ffweKTiKurara /J.^pij rrjs KOT' iKfivov (iriffTrnj.?)* 
 airo\el;a/J.evos ertpw (PfWpws or eraifnf) Aio^di'Ty ffwoTTTiKurara Trpoffe<p&vr]ffe. The MSS. 
 read rr^pw, which is apparently a mistake for erfyws or possibly for eralpip. Tannery con- 
 jectures rif fralpif, but this is very doubtful ; if the article had been there, Aio^avTy T 
 fralpq would have been better. On the basis of eratpy Tannery builds the further 
 hypothesis that the Dionysius to whom the Arithmetica is dedicated is none other than 
 Dionysius who was at the head of the Catechist school at Alexandria 232-247 and was 
 Bishop there 248-265 A.D. Tannery conjectures then that Diophantus was a Christian 
 and a pupil of Dionysius (Tannery, "Sur la religion des derniers mathematicians de 
 1'antiquite," Extrait des Annales de Philosophic Chretienne, 1896, p. 13 sqq.). It is 
 however difficult to establish this (Hultsch, art. "Diophantos aus Alexandreia" in Pauly- 
 Wissowa's Real-Encyclopddie der classischen Altertumswissenchaften}.
 
 DIOPHANTUS AND HIS WORKS 3 
 
 The only personal particulars about Diophantus which are 
 known are those contained in the epigram-problem relating to him 
 in the Anthology x . The solution gives 84 as the age at which he 
 died. His boyhood lasted 14 years, his beard grew at 21, he 
 married at 33; a son was born to him five years later and died, at 
 the age of 42, when his father was 80 years old. Diophantus' own 
 death followed four years later 2 . It is clear that the epigram was 
 written, not long after his death, by an intimate personal friend 
 with knowledge of and taste for the science which Diophantus 
 made his life-work 3 . 
 
 The works on which the fame of Diophantus rests are : 
 
 (1) The Arithmetica (originally in thirteen Books). 
 
 (2) A tract On Polygonal Numbers. 
 
 Six Books of the former and part of the latter survive. 
 Allusions in the Aritkmetica imply the existence of 
 
 (3) A collection of propositions under the title of Porisms; 
 in three propositions (3, 5 and 16) of Book V. Diophantus quotes 
 as known certain propositions in the Theory of Numbers, prefixing 
 to the statement of them the words " We have it in the Porisms 
 that " (e^o/iey eV rot? TLopt,afj,acriv ort K.r.e.}. 
 
 A scholium on a passage of lamblichus where he quotes a 
 dictum of certain Pythagoreans about the unit being the dividing 
 line (fj,e06piov) between number and aliquot parts, says "thus 
 
 Diophantus in the Moriastica* for he describes as 'parts' the 
 
 progression without limit in the direction of less than the unit." 
 Tannery thinks the Moptacrrucd may be ancient scholia (now 
 lost) on Diophantus I. Def. 3 sqq. 5 ; but in that case why should 
 Diophantus be supposed to be speaking ? And, as Hultsch 
 
 1 Anthology, Ep. xiv. 126; Dioph. n. pp. 60-1. 
 
 2 The epigram actually says that his boyhood lasted | of his life; his beard grew 
 after T \ more ; after f more he married, and his son was born five years later ; the son 
 lived to half his father's age, and the father died four years after his son. Cantor ( Gesch. 
 d. Math. I 3 , p. 465) quotes a suggestion of Heinrich Weber that a better solution is 
 obtained if we assume that the son died at the time when his father's age was double his, 
 not at an age equal to half the age at which his father died. In that case 
 
 is would substitute lof for 14, i6 for 21, 25$ for 33, 30! for 42, 6i for 80, 
 and 65^ for 84 above. I do not see any advantage in this solution. On the contrary, 
 4 think the fractional results are an objection to it, and it is to be observed that the 
 ^nholiast has the solution 84, derived from the equation 
 \x + T^X + f x + 5 + \x + 4 = x. 
 
 3 Hultsch, art. Diophantos in Pauly-Wissowa's Real-Encyclopadie. 
 
 4 lamblichus In Nicomachi arithm. introd. p. 127 (ed. Pistelli) ; Dioph. II. p. 72. 
 jjs 5 Dioph. n. p. 72 note. 
 
 I 2
 
 4 INTRODUCTION 
 
 remarks, such scholia would more naturally have been quoted 
 as a-yoXia and not by the separate title Mopiavriicd 1 . It may 
 have been a separate work by Diophantus giving rules for reckon- 
 ing with fractions ; but I do not feel clear that the reference 
 may not simply be to the definitions at the beginning of the 
 Arithmetica. 
 
 With reference to the title of the Arithmetica, we may observe 
 that the meaning of the word dpid^TiKd here is slightly different 
 from that assigned to it by more ancient writers. The ancients 
 drew a marked distinction between dpidpijTiKij and \OJIO-TIKIJ, 
 though both were concerned with numbers. Thus Plato states 
 that dpid/j,rjTiKrj is concerned with the abstract properties of 
 numbers (as odd and even, etc.), whereas \oyia-TiKtj deals with the 
 same odd and even, but in relation to one another 2 . Geminus also 
 distinguishes the two terms 3 . According to him dpiO^riK^ deals 
 with numbers in themselves, distinguishing linear, plane and solid 
 numbers, in fact all the forms of number, starting from the unit, 
 and dealing with the generation of plane numbers, similar and 
 dissimilar, and then with numbers of three dimensions, etc. 
 \oyiariKij on the other hand deals, not with the abstract properties 
 of numbers in themselves, but with numbers of concrete things 
 (ala-drjTwv, sensible objects), whence it calls them by the names of 
 the things measured, e.g. it calls some by the names /^XtTi?? and 
 <iaXtT?79 4 . But in Diophantus the calculations take an abstract 
 form (except in V. 30, where the question is to find the number 
 of measures of wine at two given prices respectively), so that the 
 distinction between XO^IO-TIKIJ and dpiQ/jLijTucr/ is lost. 
 
 We findjthe Arithmetica quoted under slightly different titles. 
 Thus the anonymous author of prolegomena to Nicomachus" 
 Introductio Arithmetica speaks of Diophantus' " thirteen Books of 
 Arithmetic 5 ." A scholium on lamblichus refers to " the last 
 theorem of the first Book of Diophantus' Elements of Arithmetic 
 
 1 Hultsch, loc. cit. 
 
 2 Gorgias, 451 B, C : rd fjxv &\\a Kadairep 17 dpiB '^T/TIKT; i] hoyicrTiKri Hx fi ' iff pi TO avrb 
 ydp dffTi, rb re apriov Kal TO irepiTTbv dia<p4pei 5 roffovrov, OTL Kal irpos avra Kal irpds 
 d\\ri\a TTWJ ?x fi TXTjtfous tiriffKoirel TO irepiTrov Kal TO apTiov i) \oyiaTiK7]. 
 
 3 Proclus, Comment, on Euclid I., p. 39, 14-40, 7. 
 
 4 Cf. Plato, Laws 819 B, c, on the advantage of combining amusement with instruction 
 in arithmetical calculation, e.g. by distributing apples or garlands (^Xow T TLVUV 
 diavofMl Kal ffT<pdvuv) and the use of different bowls of silver, gold, or brass etc. (tfudXas 
 a/Mi xpvffov Kal X&XKOU *cai apytipov Kal rotoi/raw TIV&V aXXwc KepawuvTes, ol 5 6'Xas TTWS 
 5ia5t56rcj, oirep elirov, eij iraidiav tvapubrrovTes ras TUV dva 
 
 5 Dioph. II. p. 73, 26.
 
 DIOPHANTUS AND HIS WORKS 5 
 
 ) 1 ." A scholium on one of the epigrams 
 in Metrodorus' collection similarly speaks of the " Elements of 
 Diophantus 2 ." 
 
 None of the MSS. which we possess contain more than the 
 first six Books of the Arithmetic^ the only variation being that 
 some few divide the six Books into seven 3 , while one or two give 
 the fragment on Polygonal Numbers with the number vm. The 
 idea that Regiomontanus saw, or said he saw, a MS. containing 
 the thirteen Books complete is due to a misapprehension. There 
 is no doubt that the missing Books were 'lost at a very early date. 
 Tannery 4 suggests that Hypatia's commentary extended only to 
 the first six Books, and that she left untouched the remaining 
 seven, which accordingly were first forgotten and then lost ; he 
 compares the case of Apollonius' Conies, the first four Books of 
 which were preserved by Eutocius, who wrote a commentary on 
 them, while the rest, which he did not include in his commentary, 
 were lost so far as the Greek text is concerned. While, however, 
 three of the last four Books of the Conies have fortunately reached 
 us through the Arabic, there is no sign that even the Arabians 
 ever possessed the missing Books of Diophantus. Thus the 
 second part of an algebraic treatise called the Fakhrl by Abu 
 Bekr Muh. b. al-Hasan al-Karkhl (d. about 1029) is a collection of 
 problems in determinate and indeterminate analysis which not 
 only show that their author had deeply studied Diophantus, but in 
 many cases are taken direct frcjmjihe Arithmetica, with the change, 
 occasionally, of some of the constants. In the fourth section of 
 this work, which begins and ends with problems corresponding to 
 problems in Diophantus Books II. and in. respectively, are 25 
 problems not found in Diophantus ; but the differences from 
 Diophantus in essential features (e.g. several of the problems lead 
 to equations giving irrational results, which are always avoided 
 by Diophantus), as well as other internal evidence, exclude the 
 hypothesis that we have here a lost Book of Diophantus 5 . Nor is 
 there any sign that more of the work than we possess was known 
 
 1 Dioph. II. p. 72, 17 ; lamblichus (ed. Pistelli), p. 132, 12. 
 
 2 Dioph. II. p. 62, 25. 
 
 3 e.g. Vaticanus gr. 200, Scorialensis 0-1-15, ar d the Broscius MS. in the University 
 Library of Cracow ; the two last divide the first Book into two, the second beginning 
 immediately after the explanation of the sign for minus (Dioph. I. p. 14, i). 
 
 4 Dioph. II. p. xvii, xviii. 
 
 5 See F. Woepcke, Extrait du Fakhrl, traiti cTAlgtbre par Abou Bekr Mohammed 
 ben Alhafan AlkarkhT (tnanuscrit 952, supplement arabe de la bibliotheque fnipdriale), Paris, 
 1853.
 
 6 INTRODUCTION 
 
 to Abu'l Wafa al-Buzjam (940-998 A.D.), who wrote a "commentary 
 (tafslr) on the algebra of Diophantus " as well as a " Book of 
 proofs of the propositions used by Diophantus in his work..." 
 These facts again point to the conclusion that the lost Books were 
 lost before the loth c. 
 
 Tannery's suggestion that Hypatia's commentary was limited 
 to the six Books, and the parallel of Eutocius' commentary on 
 Apollonius' Conies, imply that it is the last seven Books, and the 
 most difficult, which 'are lost. This view is in strong contrast to 
 that which Jiad previously found most acceptance among com- 
 petent authorities. The latter view was most clearly put, and 
 most ably supported, by Nesselmann 1 , though Colebrooke 2 had 
 already put forward a conjecture to the same effect ; and historians 
 of mathematics such as Hankel, Moritz Cantor, and Giinther have 
 accepted Nesselmann's conclusions, which, stated in his own 
 words, are as follows: (i) that much less of Diophantus is wanting 
 than would naturally be supposed on the basis of the numerical 
 proportion of 6 to 13; (2) that the missing portion is not to be 
 looked for at the end but in the middle of the work, and indeed 
 mostly between the first and second Books. Nesselmann's general 
 argument is that, if we carefully read the last four Books, from the 
 third to the sixth, we find that Diophantus moves in a rigidly 
 defined and limited circle of methods and artifices, and that any 
 attempts which he makes to free himself are futile ; " as often as 
 he gives the impression that he wishes to spring over the magic 
 circle drawn round him, he is invariably thrown back by an 
 invisible hand on the old domain already known ; we see, similarly, 
 in half-darkness, behind the clever artifices which he seeks to use 
 in order to free himself, the chains which fetter his genius, we hear 
 their rattling, whenever, in dealing with difficulties only too freely 
 imposed upon himself, he knows of no other means of extricating 
 himself except to cut through the knot instead of untying it." 
 Moreover, the sixth Book forms a natural conclusion to the whole, 
 in that it consists of exemplifications of methods explained and 
 used in the preceding Books. The subject is the finding of right- 
 angled triangles in rational numbers such that the sides and area 
 satisfy given conditions, the geometrical property of the right-angled 
 triangle being introduced as a fresh condition additional to the 
 purely arithmetical conditions which have to be satisfied in the 
 
 1 Algebra der Griechen, pp. 264-273. 
 
 2 Algebra of the Hindus, Note M, p. Ixi.
 
 DIOPHANTUS AND HIS WORKS 7 
 
 problems of the earlier Books. But, assuming that Diophantus' 
 resources are at an end in the sixth Book, Nesselmann has to 
 suggest possible topics which would have formed approximately 
 adequate material for the equivalent of seven Books of the 
 Arithmetical. The first step is to consider what is actually wanting 
 which we should expect to find, either as foreshadowed by the 
 author himself or as necessary for the elucidation or completion of 
 the whole subject. Now the first Book contains problems leading 
 to determinate equations of the first degree ; the remainder of the 
 work is a collection of problems which, with few exceptions, lead 
 to indeterminate equations of the second degree, beginning with 
 simpler cases and advancing step by step to more complicated 
 questions. There would have been room therefore for problems 
 involving (i) determinate equations of the second degree and (2) 
 indeterminate equations of the first. There is indeed nothing to 
 show that (2) formed part of the writer's plan ; but on the other 
 hand the writer's own words in Def. 1 1 at the beginning of the 
 work promise a discussion of the solution of the complete or 
 adfected quadratic, and it is clear that he employed his method of 
 solution in the later Books, where in some cases he simply states 
 the solution without working it out, while in others, where the 
 roots are " irrational," he gives approximations which indicate 
 that he was in possession of a scientific method. Pure quadratics 
 Diophantus regarded as simple equations, taking no account of the 
 negative -root. Indeed it would seem that he adopted as his 
 ground for the classification of quadratics, not the index of the 
 highest power of the unknown quantity contained in it, but the 
 number of terms left in it when reduced to its simplest form. His 
 words are 1 : " If the same powers of the unknown occur on both 
 sides, but with different coefficients (prj 6fjW7r\r)0i) Se), we must 
 take like from like until we have one single expression equal to 
 another. If there are on both sides, or on either side, any terms 
 with negative coefficients (ev eXXefy-eo-t riva eiS?)), the defects must 
 be added on both sides until the terms on both sides have 
 none but positive coefficients (evvTrap^ovra), when we must again 
 take like from like until there remains one term on each side. 
 This should be the object aimed at in framing the hypotheses of 
 propositions, that is to say, to reduce the equations, if possible, 
 until one term is left equated to one term. But afterwards I will 
 
 1 Dioph. I. Def. u, p. 14.
 
 8 INTRODUCTION 
 
 show you also how, when two terms are left equal to one term, 
 such an equation is solved." That is to say, reduce the quadratic, 
 if possible, to one of the forms ax*= bx, ax z =c, or bx = c\ I will 
 show later how to solve the equation when three terms are left of 
 which any two are equal to the third, z.^.'the complete quadratic 
 ax* bx c o, excluding the case ax"- + bx + c = o. The exclusion 
 of the latter case is natural, since it is of the essence of the work 
 to find rational and positive solutions. Nesselmann might have 
 added that Diophantus' requirement that the equation, as finally 
 stated, shall contain only positive terms, of which two are equated 
 to the third, suggests that his solution would deal separately with 
 the three possible cases (just as Euclid makes separate cases of the 
 equations in his propositions VI. 28, 29), so that the exposition 
 might occupy some little space. The suitable place for it would 
 be between the first and second Books. There is no evidence 
 tending to confirm Nesselmann's further argument that the six 
 Books may originally have been divided into even more than 
 seven Books. He argues from the fact that there are often better 
 natural divisions in the middle of the Books (e.g. at II. 19) than 
 between them as they now stand ; thus there is no sign of a 
 marked division between Books I. and II. and between Books II. 
 and in., the first five problems of Book II. and the first four of 
 Book in. recalling similar problems in the preceding Books 
 respectively. But the latter circumstances are better explained, 
 as Tannery explains them, , by the supposition that the first 
 problems of Books II. and III. are interpolated from some ancient 
 commentary. Next Nesselmann points out that there are a 
 number of imperfections in the text, Book V. especially having 
 been " treated by Mother Time in a very stepmotherly fashion " ; 
 thus it seems probable that at V. 19 three problems have dropped 
 out altogether. Still he is far from accounting for seven whole 
 Books; he has therefore to press into the service the lost 
 "Porisms" and the tract on Polygonal Numbers. 
 
 If the phrase which, as we have said, occurs three times in 
 Book V., "We have it in the Porisms that...," indicates that the 
 "Porisms" were a definite collection of propositions concerning 
 the properties of certain numbers, their divisibility into a certain 
 number of squares, and so on, it is possible that it was from the 
 same collection that Diophantus took the numerous other pro- 
 positions which he assumes, either explicitly enunciating them, or 
 implicitly taking them for granted. May we not then, says
 
 DIOPHANTUS AND HIS WORKS 9 
 
 Nesselmann, reasonably suppose the " Porisms " to have formed 
 an introduction to the indeterminate and semi-determinate analy- 
 sis of the second degree which forms the main subject of the 
 Arithmetica, and to have been an integral part of the thirteen 
 Books, intervening, probably, between Books I. and II. ? Schulz* on 
 the other hand, considered this improbable, and in recent years 
 Hultsch 1 has definitely rejected the theory that Diophantus filled 
 one or more Books of his Arithmetica exclusively with Porisms. 
 Schulz's argument is, indeed, not conclusive. It is based on the 
 consideration that " Diophantus expressly says that his work deals 
 with arithmetical problems*" \ but what Diophantus actually says is 
 " Knowing you, O Dionysius, to be anxious to learn the solution 
 (or, nerhaps, ' discovery,' evpeviv) of problems in numbers, I have 
 endeavoured, beginning from the foundations on which the study is 
 bufit up, to expound (v-rrofTrfja-ai = to lay down) the nature and 
 force subsisting in numbers," the last of which words would easily 
 c ver 'propositions in the theory of numbers, while " propositions," 
 ot T: prJems," is the word used at the end of the Preface, where 
 
 he says, "let us now proceed to the propositions (7rpoT<ra<?) 
 
 which have been treated in thirteen Books." 
 
 On reconsideration of the whole matter, I now agree in the 
 view of Hultsch that the Porisms were not a separate portion of 
 the Arithmetica or included in the Arithmetica at all. If they had 
 been, I think the expression " we have it in the Porisms " would 
 have been inappropriate. In the first place, the Greek mathe- 
 maticians do not usually give references in such a form as this 
 to propositions which they cite when they come from the same 
 work as that in which they are cited ; as a rule the propositions 
 are quoted without any references at all. The references in this 
 case would, on the assumption that the Porisms were a portion of 
 the thirteen Books, more naturally have been to particular pro- 
 positions of particular Books (cf. Eucl. XII. 2, " For it was proved 
 
 1 Hultsch, loc. cit, 
 
 2 The whole passage of Schulz is as follows (pref. xxi): " Es ist daher nicht unwahr- 
 scheinlich, dass diese Porismen eine eigene Schrift unseres Diophantus waren, welche 
 vorziiglich die Zusammensetzung der Zahlen aus gewissen Bestandtheilen zu ihrem 
 Gegenstande hatten. Konnte man diese Schrift als einen Bestandtheil des grossen in 
 dreizehn Biichern abgefassten arithmetischen Werkes ansehen, so ware es sehr erklarbar, 
 dass gerade dieser Theil, der den blossen Liebhaber weniger anzog, verloren ging. Da 
 indess Diophantus ausdriicklich sagt, sein Werk behandele arithmetisfhe Probleme, so hat 
 wenigstens die letztere Annahme nur einen geringen Grad von \Vahrscheinlichkeit."
 
 io INTRODUCTION 
 
 in the first theorem of the loth Book that..."). But a still vaguer 
 reference would have been enough, even if Diophantus had chosen 
 to give any at all ; if the propositions quoted had preceded those 
 in which they are used, some expression like TOVTO yap irpo- 
 yeypaTTTai, " for this has already been proved," or SeSeitcrai yap 
 TOVTO, " for this has been shown," would have sufficed, or, if the 
 propositions occurred later, some expression like J>? efr;? Sei^BijareTai 
 or Set%#/7reTat v$> fipwv va-repov, "as will be proved in due course" 
 or "later." The expression "we have it in the Porisms" (in the 
 plural) would have been still more inappropriate if the "Porisms" 
 had been, as Tannery supposes 1 , not collected together as one or 
 more Books of the Arithmetica, but scattered about in the work as 
 corollaries to particular propositions 2 . And, as Hultsch says, it is 
 hard, on Tannery's supposition, to explain why the three partv.iiar 
 theorems quoted from " the Porisms " were lost, while a ','air 
 number of other additions survived, partly under the title iropia-^. 
 (cf. I. 34, I. 38), partly as "lemmas to what follows," Xf///,/*r 
 e^9 (cf. lemmas before IV. 34, 35, 36, V. 7, 8, VI. 12, 15). O> 
 other hand, there is nothing improbable in the supposition that 
 Diophantus was induced by the difficulty of his problems to give 
 place in a separate work to the " porisms " necessary to their 
 solution. 
 
 The hypothesis that the Porisms formed part of the Arithmet- 
 ica being thus given up, we can hardly hold any longer to 
 Nesselmann's view of the contents of the lost Books and their 
 place in the treatise; and I am now much more inclined to the 
 opinion of Tannery that it is the last and the most difficult Books 
 which are lost. Tannery's argument seems to me to be very 
 attractive and to deserve quotation in full, as finally put in the 
 preface to Vol. II. of his Diophantus 3 . He replies first to the 
 assumption that Diophantus could not have proceeded to problems 
 more difficult than those of Book V. " But if the fifth or the sixth 
 Book of the Arithmetica had been lost, who, pray, among us would 
 have believed that such problems had ever been attempted by the 
 Greeks? It would be the greatest error, in any case in which a 
 
 1 Dioph. II. p. xix. 
 
 2 Thus Tannery holds (loc. cit.) that the solution of the complete quadratic was given 
 in the form of corollaries to I. 27, 30; and he refers the three "porisms" quoted in v. 3, 
 5, 16 respectively to a second (lost) solution of III. io, to III. 15, and to iv. i, 2. 
 
 3 Dioph. II. p. xx.
 
 DIOPHANTUS AND HIS WORKS n 
 
 thing cannot clearly be proved to have been unknown to all the 
 ancients, to maintain that it could not have been known to some 
 Greek mathematician. If we do not know to what lengths 
 Archimedes brought the theory of numbers (to say nothing of 
 other things), let us admit our ignorance. But, between the 
 famous problem of the cattle and the most difficult of Diophantus' 
 problems, is there not a sufficient gap to require seven Books to 
 fill it? And, without attributing to the ancients what modern 
 mathematicians have discovered, may not a number of the things 
 attributed to the Indians and Arabs have been drawn from 
 Greek sources ? May not the same be said of a problem solved by 
 Leonardo of Pisa, which is very similar to those of Diophantus but 
 is nc now to be found in the Arithmetical In fact, it may fairly 
 be said that, when Chasles made his reasonably probable restitution 
 of the Porisms of Euclid, he, notwithstanding the fact that he had 
 Pappus' lemmas to help him, undertook a more difficult task than 
 he would have undertaken if he had attempted to fill up seven 
 Diophantine Books with numerical problems which the Greeks 
 may reasonably be supposed to have solved." 
 
 On the assumption that the lost portion came at the end of the 
 existing six Books, Schulz supposed that it contained new methods 
 of solution in addition to those used in Books I. to VI., and in 
 particular extended the method of solution by means of the double 
 equation (Bi7r\rj tcrori;? or SiTrXoia-oTT)?). By means of the double 
 equation Diophantus shows how to find a value of the unknown 
 which will make two expressions (linear or quadratic) containing it 
 simultaneously squares. Schulz then thinks that he went on, in 
 the lost Books, to make three such expressions simultaneously 
 squares, i.e. advanced to a triple equation. But this explanation 
 does not in any case take us very far. 
 
 Bombelli thought that Diophantus went on to solve deter- 
 minate equations of the third and fourth degree 1 ; this view, 
 however, though natural at that date, when the solution of cubic 
 and biquadratic equations filled so large a space in contemporary 
 investigations and in Bombelli's own studies, has nothing to 
 support it. 
 
 Hultsch 2 seems to find the key to the question in the fragment 
 of the treatise on Polygonal Numbers and the developments to 
 
 1 Cossali, I. pp. 75, 76. 2 Hultsch, loc. fit.
 
 12 INTRODUCTION 
 
 which it might have been expected to lead. In this he differs 
 from Tannery, who says that, as Serenus' treatise on the sections 
 of cones and cylinders was added to the mutilated Conies of 
 Apollonius consisting of four Books only, in order to make up a 
 convenient volume, so the tract on Polygonal Numbers was added 
 to the remains of the Arithmetical, though forming no part of the 
 larger work 1 . Thus Tannery would seem to deny the genuineness 
 of the whole tract on Polygonal Numbers, though in his text he 
 only signalises the portion beginning with the enunciation of the 
 problem " Given a number, to find in how many ways it can be 
 a polygonal number " as a " vain attempt by a commentator " to 
 solve this problem. Hultsch, on the other hand, thinks we may 
 conclude that Diophantus really solved the problem. He points 
 out moreover that the beginning of the tract is like the beginning 
 of Book I. of the Arithmetica in containing definitions and pre- 
 liminary propositions. Then came the difficult problem quoted, 
 the discussion of which breaks off in our text after a few pages ; 
 and to this it would be easy to tack on a great variety of other 
 problems. Again, says Hultsch, the supplementary propositions 
 added by Bachet may serve to give an approximate idea of the 
 difficulty of the problems which were probably treated in Books VII. 
 and the following. And between these and the bold combination 
 of a triangular and a square number in the Cattle-Problem 
 stretches, as Tannery says, a wide domain which was certainly 
 not unknown to Diophantus, but was his hunting-ground for the 
 most various problems. Whether Diophantus dealt with plane 
 numbers, and with other figured numbers, such as prisms and 
 tetrahedra, is uncertain. 
 
 The name of Diophantus was used, as were the names of Euclid, 
 Archimedes and Heron in their turn, for the purpose of palming 
 off the compilations of much later authors. Tannery prints in 
 his edition three fragments under the head of "Diophantus 
 Pseudepigraphus." The first 2 , which is not " from the Arithmetic 
 of Diophantus " as its heading states, is worth notice as containing 
 some particulars of one of " two methods of finding the square 
 root of any square number"; we are told to begin by writing the 
 number " according to the arrangement of the Indian method," t.e. 
 according to the Indian numerical notation which reached us 
 through the Arabs. The fragment is taken from a Paris MS. 
 
 1 Dioph. ii. p. xviii. 2 Dioph. n. p. 3, 3-14.
 
 DIOPHANTUS AND HIS WORKS 13 
 
 (Supplem. gr. 387), where it follows a work with the title 'Ap^) 
 7779 fi r yd\r)<; KOI 'Iv8itcfj<; "^r}(pi(f)opia<; (i.e. ifnj<f>o<j>opia<;), written in 
 1252 and raided about half a century later by Maximus Planudes. 
 The second fragment 1 is the work edited by C. Henry in 1879 as 
 Opusculum de multiplicatione et divisione sexagesimalibus Diophanto 
 vel Pappo attribuendum. The third 2 , beginning with Aio(j>dvTov 
 eTri7re8ofj,6TpiKd, is a compilation made in the Byzantine period out 
 of late reproductions of the yecof^erpovfieva and <nepeop,rpovfiva 
 of Heron. The second and third fragments, like the first, have 
 nothing to do with Diophantus. 
 
 1 Dioph. II. p. 3, 15-15, 17. 2 Dioph. n. p. 15, 18-31, 22.
 
 CHAPTER II 
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 
 
 FOR full details of the various MSS. and of their mutual 
 relations, reference should be made to the prefaces to the first and 
 second volumes of Tannery's edition 1 . Tannery's account needs 
 only to be supplemented by a description given by Gollob 2 of 
 another MS. supposed by Tannery to be non-existent, but actually 
 rediscovered in the Library of the University of Cracow (Nr 544). 
 Only the shortest possible summary of the essential facts will be 
 given here. 
 
 After the loss of Egypt the work of Diophantus long remained 
 almost unknown among the Byzantines ; perhaps one copy only 
 survived (of the Hypatian recension), which was seen by Michael 
 Psellus and possibly by the scholiast to lamblichus, but of which 
 no trace can be found after the capture of Constantinople in 1 204. 
 From this one copy (denoted by the letter a in Tannery's table of 
 the MSS.) another MS. (a) was copied in the 8th or 9th century ; 
 this again is lost, but is the true archetype of our MSS. The 
 copyist apparently intended to omit all scholia, but, the distinction 
 between text and scholia being sometimes difficult to draw, he 
 included a good deal which should have been left out. For 
 example, Hypatia, and perhaps scholiasts after her, seem to have 
 added some alternative solutions and a number of new problems ; 
 some of these latter, such as II. 1-7, 17, 18, were admitted into the 
 text as genuine. 
 
 The MSS. fall into two main classes, the ante-Planudes class, 
 as we may call it, and the Planudean. The most ancient and the 
 best of all is Matritensis 48 (Tannery's A\ which was written in 
 the 1 3th century and belongs to the first class; it is evidently a 
 most faithful copy of the lost archetype (a). Maximus Planudes 
 wrote a systematic commentary on Books I. and II., and his scholia, 
 
 1 Dioph. I. pp. iii-v, II. pp. xxii-xxxiv. 
 
 2 Eduard Gollob, "Ein wiedergefundener Diophantuscodex " in Zeitschrift filr Math. 
 u. Physik, XLIV. (1899), hist.-litt. Abtheilung, pp. 137-140.
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 15 
 
 which are edited by Tannery for the first time, are preserved in the 
 oldest representative which we possess of the Planudean class, 
 namely, Marcianus 308 (Tannery's B^, itself apparently copied 
 from an archetype of the I4th century now lost, with the exception 
 of ten leaves which survive in Ambrosianus Et 157 sup. 
 
 Tannery shows the relation of the MSS. in the following 
 diagram : 
 
 (a) Lost copy of the Hypatian recension, 
 (a) Lost copy, of eighth or ninth c. 
 
 (FIRST CLASS) 
 
 1 
 
 (PLANUDEAN CLASS) 
 
 i. Matritensis 48 = A, 
 1 3th c. 
 
 2. Vaticanusgr. 191= V, 
 second half of i5th c. 
 
 3. Vaticanus gr. 304, 
 beginning of i6th c. 
 
 
 9. Lost MS. of the i4th c. of which ten leaves 
 are extant in Ambrosianus Et 157 sup. 
 
 i o. Marcianus 308 = B^ , 
 beginning of i5th c. 
 
 1 1 . Guelferbytanus 
 Gudianus i, i5th c. 
 
 14. Ambrosianus 
 A 91 sup. 
 (1545) 
 
 15. Vaticanus gr. 200 
 (1545) 
 
 4. Parisinus 2379 =C 
 (after first two 
 Books), 
 middle of i6th c. 
 
 5. Parisinus 2 3 78 = />, 
 middle of i6th c. 
 
 6. Neapolitanus 
 IIIC 17, 
 middle of i6th c. 
 
 7. Urbinas gr. 74, 
 end of 1 6th c. 
 
 8. Oxon. Baroccianus 
 1 66 (part of Book I. 
 only) 
 
 12. Palatinus gr. 391, 
 end of 1 6th c. 
 
 13. Reginensis 128, 
 end of 1 6th c. 
 
 
 1 6. Scorialensis T I 1 1 
 (1545) 
 
 17. Parisinus 2485 = A", 
 middle of i6th c. 
 
 1 8. Scorialensis 
 R-III-iS, 
 middle of i6th c. 
 
 19. Ambrosianus 
 Q 121 sup. (part of 
 Book I.), 
 middle of i6th c. 
 
 
 4. Parisinus 2379 = C 
 (first two Books) 
 
 
 20. Taurinensis C III 16 
 
 21. Parisinus Ars. 8406 
 = X 
 
 22. Scorialensis fl-I-i 5, 
 middle of 6th c. 
 
 23. ScorialensisR-II-3, 
 end of 1 6th c. 
 
 
 ? 
 24. Oxon. Savilianus, 
 end of 1 6th c. 
 
 -^ 
 
 
 Auria's recension made up out of MSS. 2, 3, 15 above and Xylander's 
 translation: 25. Parisinus 2380 = Z>. 
 
 26. Ambrosianus E 5 sup. 
 
 27. MS. (Patavinus) of Broscius (Brozek) now at 
 
 28. Lost MS. of Cardinal du Perron. 
 
 Cracow.
 
 1 6 INTRODUCTION 
 
 The addition of a few notes as regards the most important and 
 interesting of the MSS., in the order of their numbers in Tannery's 
 arrangement, will now sufficiently complete the story. 
 
 1. The best and most ancient MS., that of Madrid (Tannery's 
 A), was unfortunately spoiled at a late date by corrections made, 
 especially in the first two Books, from some MS. of the Planudean 
 class, in such a way that the original reading is sometimes entirely 
 erased or made quite illegible. In these cases recourse must be 
 had to the Vatican MS. 191. 
 
 2. The MS. Vaticanus graecus 191 was copied from A before 
 it had suffered the general alteration by means of a MS. of the 
 other class, though not before various other corrections had been 
 made in different hands not easily distinguished ; thus V some- 
 times has readings which Tannery found to have arisen from some 
 correction in A. A appears to have been at Rome for a con- 
 siderable period at the time when V was copied ; for the librarian 
 who wrote the old table of contents 1 at the beginning of V inserted 
 in the margin in one place 2 the word a/ofa/^evo?, which had been 
 omitted, direct from the original (A). 
 
 3. Vat. gr. 304 was copied from V, not from A ; Tannery 
 inferred this mainly from a collation of the scholia, and he notes 
 that the word apa/iei/o<? above mentioned is here brought into the 
 text by the erasure of some letters. This MS. 304, being very 
 clearly written, was used thenceforward to make copies from. The 
 next five MSS. do not appear to have had any older source. 
 
 4. The MS. Parisinus 2379 (Tannery's C) was that used by 
 Bachet for his edition. It was written by one loannes Hydruntinus 
 after 1545, and has the peculiarity that the first two Books were 
 copied from the MS. Vat. gr. 200 (a MS. of the Planudean class), 
 evidently in order to include the commentary of Planudes, while 
 the MS. Vat. gr. 304 belonging to the pre-Planudes class was 
 followed in the remaining Books, no doubt because it was con- 
 sidered superior. Thus the class of which C is the chief repre- 
 sentative is a sort of mixed class. 
 
 5. 6. Parisinus 2378 = P, and Neapolitans III C 17, were 
 copied by Angelus Vergetius. In the latter Vergetius puts the 
 
 1 The MS. V was made up of various MSS. before separated. The old table of 
 contents has Aio^dWou d/ji^ijTcm}- apfioviicii. didQopa. The appoviKa include the Intro- 
 duction to Harmony by Cleonides, but without any author's name. This fact sufficiently 
 explains the error of Ramus in saying, Schola mathematics Bk I. p. 35, "Scripserat et 
 Diophantus harmonica. " 
 
 2 Dioph. i. p. 2, 5-6.
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 17 
 
 numbers A, B, F, A, E, Z, H at the top of the pages (as we put 
 headlines) corresponding to the different Books, implying that he 
 regarded the tract on Polygonal Numbers as Book VII. 
 
 The other MSS. of the first class call for no notice, and we pass 
 to the Planudean class. 
 
 9. Tannery, as he tells us, congratulated himself upon finding 
 in Ambrosianus Et 157 sup. ten pages of the archetype of the 
 class, and eagerly sought for new readings. So far, however, as he 
 was able to carry his collation, he found no difference from the 
 principal representative of the class (B^) next to be mentioned. 
 
 10. The MS. Marcianus 308 (= B^) of the 1 5th century formerly 
 belonged to Cardinal Bessarion, and was seen by Regiomontanus 
 at Venice in 1464. It contains the recension by Planudes with his 
 commentary. 
 
 11. It seems certain that the Wolfenbuttel MS. Guelferbytanus 
 Gudianus I (i5th c.) was that which Xylander used for his 
 translation ; Tannery shows that, if this was not the MS. lent 
 to Xylander by Andreas Dudicius Sbardellatus, that MS. must 
 have been lost, and there is no evidence in support of the latter 
 hypothesis. It is not possible to say whether the Wolfenbuttel 
 MS. was copied from Marcianus 308 (B^) or from the com- 
 plete MS. of which Ambrosianus Et 157 sup. preserves the ten 
 leaves. 
 
 12. Palatinus gr. 391 (end of i6th c.) has notes in German in 
 the margin which show that it was intended to print from it ; it 
 was written either by Xylander himself or for him. It is this MS. 
 of which Claudius Salmasius (Claude de Saumaise, 1588-1653) 
 told Bachet that it contained nothing more than the six Books, 
 with the tract on Polygonal Numbers. 
 
 13. Reginensis 128 was copied at the end of the i6th century 
 from the Wolfenbuttel MS. 
 
 14. 15. Ambrosianus A 91 sup. and Vaticanus gr. 200 both 
 come from B^ ; as they agree in omitting V. 28 of Diophantus, one 
 was copied from the other, probably the latter from the former. 
 They were both copied by the same copyist for Mendoza in 1545. 
 Vat. gr. 200 has headings which make eight Books ; according to 
 Tannery the first Book is numbered a', the fourth 8 OV ; before V. 20 
 (in Bachet's numbering) should this be IV. 20 ? is the heading 
 Ato<ai/TOL> 6", before the fifth Book Aio<j>dvTov r", before the sixth 
 At,o(j)dvTov f ", and before the tract on Polygonal Numbers 
 &.io<f>dvTov 77" ; this wrong division occurs in the next three MSS. 
 
 H. D. 2
 
 i8 INTRODUCTION 
 
 (16, 17, 1 8 in the diagram), all of which seem to be copied from 
 Vat. 200. 
 
 The MSS. numbered 20, 21, 22, 23 in the diagram are of the 
 hybrid class derived from Parisinus 2379 (C). Scorialensis ft-I-15 
 and Scorialensis R-II-3, the latter copied from the former, have 
 the first Book divided into two (cf. p. 5 above), and so make 
 seven Books of the Arithmetica and an eighth Book of the 
 Polygonal Numbers. 
 
 27. The Cracow MS. has the same division into Books as the 
 MSS. last mentioned. According to Gollob, the collation of this 
 MS., so far as it was carried in 1899, showed that it agrees in the 
 main with A (the best MS.), B^ (Marcianus 308) and C (Parisinus 
 2379) ; but, as it contains passages not found in the two latter, it 
 cannot have been copied from either of them. 
 
 25. Parisinus 2380 appears to be the copy of Auria's 
 Diophantus mentioned by Schulz as having been in the library of 
 Carl von Montchall and bearing the title " Diophanti libri sex, cum 
 scholiis graecis Maximi Planudae, atque liber de numeris poly- 
 gonis, collati cum Vaticanis codicibus, et latine versi a Josepho 
 Auria 1 ." 
 
 The first commentator on Diophantus of whom we hear is 
 Hypatia, the daughter of Theon of Alexandria ; she was murdered 
 by Christian fanatics in 415 A.D. According to Suidas she wrote 
 commentaries on Diophantus, on the Astronomical Canon (sc. of 
 Ptolemy) and on the Conies of Apollonius 2 . Tannery suggests 
 that the remarks of Michael Psellus (nth c.) at the beginning of 
 his letter about Diophantus, Anatolius, and the Egyptian method 
 of arithmetical reckoning were taken bodily from some MS. of 
 Diophantus containing an ancient and systematic commentary ; 
 and he believes this commentary to have been that of Hypatia. I 
 have already mentioned the attractive hypothesis of Tannery that 
 Hypatia's commentary extended only to our six Books, and that 
 this accounts for the loss of the rest. 
 
 Georgius Pachymeres (1240 to about 1310) wrote in Greek a 
 paraphrase of at least a portion of Diophantus. Sections 25-44 of 
 
 1 Schulz, Diophantus, pref. xliii. 
 
 2 Suidas s.v. 'firaria: typa\f/ev virbnvwa els Aufyavrov, <ets> rbv dffrpovofUKbv Kavdva, 
 efc TO. KuviKa 'AiroXXowfou vir&fu>7ifM. So Tannery reads, following the best MSS. ; he 
 gives ample reasons for rejecting Kuster's conjecture ets bioQdvTov rbv dffrpovofj.tKi>v Kavova, 
 viz. (i) that the order of words would have been TOV Aiotpavrov derpovofUKov Kavova, 
 (i) that there is nothing connecting Diophantus with astronomy, while Suidas mentions, 
 s.v. Qiuv, a commentary et's rbv UroXe^aiov -irpoxeipov Kavova.
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 19 
 
 this survive and are published by Tannery in his edition of 
 Diophantus 1 . The chapters lost at the beginning may have con- 
 tained general observations and introductions to the first two 
 paragraphs of Book I. ; section 25 begins with the third paragraph 
 (Def. i), and the rest of the fragment takes us up to the problem 
 in I. ii. 
 
 Soon afterwards Maximus Planudes (about 1260-1310) wrote 
 a systematic commentary on Books I., II. This is also included by 
 Tannery in his edition 2 . 
 
 There are a number of other ancient scholia, very few of which 
 seemed to Tannery to be worth publication 3 . 
 
 But in the meantime, and long before the date of Georgius 
 Pachymeres, the work of Diophantus had become known in Arabia, 
 where it was evidently the subject of careful study. We are told 
 in the Fihrist, the main part of which was written in the year 
 987 A.D., (i) that Diophantus was a Greek of Alexandria who 
 wrote a book "On the art of algebra 4 /' (2) that Abu'l Wafa 
 al-Buzjanl (940-998) wrote (a) a commentary (tafsir) on the 
 algebra of Diophantus and (b} a book of " proofs to the pro- 
 positions used by Diophantus in his book and to that which 
 he himself (Abu'l Wafa) stated in his commentary 5 ," (3) that 
 Qusta. b. Luqa al-Ba'labakkl (died about 912) wrote a "com- 
 mentary on three and a half Books of Diophantus' work on 
 arithmetical problems 6 ." Qusta b. Luqa, physician, philosopher, 
 astronomer, mathematician and translator, was the author of works 
 on Euclid and of an " introduction to geometry " in the form of 
 question and answer, and translator of the so-called Books XIV., XV. 
 of Euclid ; other Arabian authorities credit him with an actual 
 " translation of the book of Diophantus on Algebra 7 ." Lastly, we 
 are told by Ibn abi Usaibi'a of " marginal glosses which Ishaq b. 
 Yunis (died about 1077), the physician of Cairo, after Ibn al- 
 ii aitham, added to the book of Diophantus on algebraic problems." 
 The title is somewhat obscure ; probably Ibn al-Haitham (about 
 965-1039), who wrote several works on Euclid, wrote a commentary 
 on the Arithmetic* and Ishaq b. Yunis added glosses to this 
 commentary 8 . 
 
 Dioph. ii. pp. 78-122. 2 Dioph. II. pp. 125-255. 
 
 The few that he gives are in Vol. II. pp. 256-260; as regards the collection in 
 general cf. Hultsch in Berliner philologisctie Wochenschrift, 1896, p. 615. 
 
 Fihrist, ed. Suter, p. 22. 5 ibid, p. 39. 6 ibid, p. 43. 
 
 Suter, Die Mathematiker und Astroiiomen der Araber, 1900, p. 41. 
 
 Suter, op. cit, pp. 107-8. Cf. Bibliotheca Malhematica iv 3 , 1903-4, p. 296. 
 
 2 2
 
 20 INTRODUCTION 
 
 To Regiomontanus belongs the credit of being the first to call 
 attention to the work of Diophantus as being extant in Greek. 
 We find two notices by him during his sojourn in Italy, whither he 
 journeyed after the death of his teacher Georg von Peurbach, 
 which took place on the 8th April, 1461. In connexion with 
 lectures on the astronomy of Alfraganus which he gave at Padua 
 he delivered an Oratio introductoria in omnes scientias mathe- 
 maticas 1 . In this he observed: "No one has yet translated from 
 the Greek into Latin the fine thirteen Books of Diophantus, in 
 which the very flower of the whole of Arithmetic lies hid, the ars 
 rei et census which to-day they call by the Arabic name of 
 Algebra 2 ." Secondly, he writes to Bianchini, in answer to a letter, 
 dated 5th February, 1464, that he has found at Venice "Diofantus," 
 a Greek arithmetician, who has not yet been translated into Latin ; 
 that in his preface Diophantus defines the various powers up to 
 the sixth ; but whether he followed out all the combinations of 
 these Regiomontanus does not know: "for not more than six 
 Books are found, though in the preface he promises thirteen. If 
 this book, which is really most wonderful and most difficult, could 
 be found entire, I should like to translate it into Latin, for the 
 knowledge of Greek which I have acquired while staying with my 
 most reverend master [Bessarion] would suffice for this...." He 
 goes on to ask Bianchini to try to discover a complete copy and, 
 in the meantime, to advise him whether he should begin to translate 
 the six Books 3 . The exact date of the Oratio is not certain. 
 Regiomontanus made some astronomical observations at Viterbo 
 in the summer and autumn of 1462. He is said to have spent a 
 year at Ferrara, and he seems to have gone thence to Venice. 
 Extant letters of his written at Venice bear dates from 27th July, 
 1463, to 6th July, 1464, and it may have been from Venice 
 that he made his visit to Padua. At all events the Oratio at 
 Padua must have been near in time to the discovery of the 
 MS. at Venice. 
 
 Notwithstanding that attention was thus called to the work, it 
 
 1 Printed in the work Rudimenta astronomica Alfragani, Niirnberg, 1537. 
 
 2 As the ars rei et census, the solution of determinate quadratic equations, is not found 
 in our Diophantus, it would seem that at the time of the Oratio Regiomontanus had only 
 looked at the MS. cursorily, if at all. 
 
 3 The letter to Bianchini is given on p. 135 of Ch. Th. v. Murr's Memorabilia, 
 Norimbergae, 1786, and partly in Doppelmayer's Historische Nachricht von den Niirn- 
 bergischen Mathematicis und Kiinstlern (Niirnberg, 1730), p. 5, note 7.
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 21 
 
 seems to have remained practically a closed book from the date of 
 Maximus Planudes to about 1570. Luca Paciuolo, towards the 
 end of the isth c., Cardano and Tartaglia about the middle of the 
 1 6th, make no mention of it. Only Joachim Camerarius, in a 
 letter published in I556 1 , mentions that there is a MS. of 
 Diophantus in the Vatican which he is anxious to see. Rafael 
 Bombelli was the first to find a MS. in the Vatican and to conceive 
 the idea of publishing the work. This was towards 1570, for in his 
 Algebra 2 published in 1572 Bombelli tells us that he had in the 
 years last past discovered a Greek book on Algebra written by " a 
 certain Diofantes, an Alexandrine Greek author, who lived in the 
 time of Antoninus Pius " ; that, thinking highly of the contents of 
 the work, he and Antonio Maria Pazzi determined to translate it ; 
 that they actually translated five books out of the seven into 
 which the MS. was divided ; but that, before the rest was finished, 
 they were called away from it by other labours. Bombelli did not 
 carry out his plan of publishing Diophantus in a translation, but 
 he took all the problems of the first four Books and some of those 
 of the fifth, and embodied them in his Algebra, interspersing them 
 with his own problems. He took no pains to distinguish 
 Diophantus' problems from his own ; but in the case of the former 
 he adhered pretty closely to the original, so that Bachet admits his 
 obligations to him, remarking that in many cases he found 
 
 1 De Graecis Latinisque numerorum ttotis et praeterea Saracenis seu Indicts, etc. etc., 
 studio Joachimi Camerarii, Papeberg, 1556. 
 
 3 Nesselmann tells us that he has not seen this work but takes his information about 
 it from Cossali. I was fortunate enough to find in the British Museum one of the copies 
 dated 1579 (really the same as the original edition of 1572 except that the title-page and 
 date are new, and a dedicatory letter on pp. 3-8 is reprinted ; there were not two 
 separate editions). The title is L" Algebra, opera di Rafael Bombelli da Bologna diuisa in 
 
 tre Libri In Bologna, Per Giovanni Rossi, MDLXXIX. The original of the passage 
 
 from the preface is : 
 
 "Questi anni passati, essendosi ritrouato una opera greca di questa disciplina nella 
 libraria di Nostro Signore in Vaticano, composta da un certo Diofante Alessandrino Autor 
 Greco, il quale fu a tempo di Antonin Pio, e havendomela fatta vedere Messer Antonio 
 Maria Pazzi Reggiano, publico lettore delle Matematiche in Roma, e giudicatolo con lui 
 Autore assai intelligente de' numeri (ancorche non tratti de' numeri irrational!, ma solo 
 in lui si vede vn perfetto ordine di operare) egli, ed io, per arrichire il mondo di cosl fatta 
 opera, ci dessimo a tradurlo, e cinque libri (delli sette che sono) tradutti ne habbiamo ; lo 
 restante non hauendo potuto finire per gli trauagli auenuti all' uno, e all' altro; e in delta 
 opera habbiamo ritrouato, ch' egli assai volte cita gli Autori Indiani, col che mi ha fatto 
 conoscere, che questa disciplina appo -gl' indiani prima fu, che a gli Arabi." The last 
 words stating that Diophantus often quotes from Indian authors are no doubt due to 
 Bombelli's taking for part of Diophantus the tract of Maximus Planudes about the Indian 
 method of reckoning.
 
 22 INTRODUCTION 
 
 Bombelli's translation better than Xylander's and consequently 
 very useful for the purpose of amending the latter 1 . 
 
 It may be interesting to mention a few points of notation in 
 this work of Bombelli. At the beginning of Book II. he explains 
 that he uses the word "tanto" to denote the unknown quantity, 
 not "cosa" like his predecessors ; and his symbol for it is -i, the 
 square of the unknown (x^) is ,., the cube d; and so on. or plus 
 and -minus (piu and mend) he uses the initial letters / and m. 
 Thus corresponding to x+6 we should find in Bombelli ii/>. 6, 
 and for x* + $x 4, \^ p. 5-1 m. 4. This notation shows, as will be 
 seen later, some advance upon that of Diophantus in one important 
 respect. 
 
 The next writer upon Diophantus was Wilhelm Holzmann who 
 published, under the Graecised form of his name, Xylander, by 
 which he is generally known, a work bearing the title : Diophanti 
 Alexandrini Rerum Arithmeticarum Libri sex, quorum primi duo 
 adiecta habent Scholia Maximi (tit coniectura esf) Planudis. Item 
 Liber de Numeris Polygonis sen Multangulis, Opus incomparabile, 
 uerae Arithmeticae Logisticae perfectionem continens, paucis adhuc 
 uisum. A Gut/. Xylandro Augustano incredibili labore Latinc 
 redditum, et Commentariis explanatum, inque lucem editum, ad 
 Illustriss. Principem Ludovicum Vuirtembergensem. Basileae per 
 Eusebium Episcopium, et Nicolai Fr. haeredes. MDLXX V. Xylander 
 was according to his own statement a " public teacher of Aristotelian 
 philosophy in the school at Heidelberg 2 ." He was a man of almost 
 universal culture 3 , and was so thoroughly imbued with the classical 
 literature, that the extraordinary aptness of his quotations and his 
 wealth of expression give exceptional charm to his writing whenever 
 he is free from the shackles of mathematical formulae and techni- 
 calities. The Epistola Nuncupatoria is addressed to the Prince 
 Ludwig, and Xylander neatly introduces it by the line " Offerimus 
 numeros, numeri sunt principe digni." This preface is very quaint 
 and interesting. He tells us how he first saw the name of 
 Diophantus mentioned in Suidas, and then found that mention 
 
 1 "Sed suas Diophanteis quaestionibus ita immiscuit, ut has ab illis distinguere non 
 sit in promptu, neque vero se fidum satis interpretem praebuit, cum passim verba 
 Diophanti immutet, hisque pleraque addat, pleraque pro arbitrio detrahat. In multis 
 nihilominus interpretationem Bombellii, Xilandriana praestare, et ad hanc emendandam 
 me adjuvisse ingenue fateor." Ad lectorem. 
 
 3 "Publicus philosophiae Aristoteleae in schola Heidelbergensi doctor." 
 3 Even Bachet, who, as we shall see, was no favourable critic, calls him " Vir omnibus 
 disciplinis excultus."
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 23 
 
 had been made of his work by Regiomontanus as being extant 
 in an Italian library and having been seen by him. But, as the 
 book had not been edited, he tried to think no more of it but, 
 instead, to absorb himself in the study of such arithmetical books 
 as he could obtain, and in investigations of his own 1 . Self-taught 
 except in so far as he could learn from published works such as 
 those of Christoff Rudolff (of the "Coss"), Michael Stifel, Cardano, 
 Nunez, he yet progressed so far as to be able to add to, modify 
 and improve what he found in those works. As a result he fell 
 into what Heraclitus called oiija-iv, lepav VOGOV, that is, into the 
 conceit of " being somebody " in the field of Arithmetic and 
 "Logistic"; others too, themselves learned men, thought him an 
 arithmetician of exceptional ability. But when he first became 
 acquainted with the problems of Diophantus (he continues) right 
 reason brought such a reaction that he might well doubt whether he 
 ought previously to have regarded himself as an object of pity or of 
 derision. He considers it therefore worth while to confess publicly his 
 own ignorance at the same time that he tries to interest others in 
 the work of Diophantus, which had so opened his eyes. Before this 
 critical time he was so familiar with methods of dealing with surds 
 that he had actually ventured to add something to the discoveries 
 of others relating to them ; the subject of surds was considered to 
 be of great importance in arithmetical questions, and its difficulty 
 
 1 I cannot refrain from quoting the whole of this passage : " Sed cum ederet nemo : 
 cepi desiderium hoc paulatim in animo consopire, et eonim quos consequi poteram 
 Arithmeticorum librorum cognitione, et meditationibus nostris sepelire. Veritatis porro 
 apud me est autoritas, ut ei coniunctum etiam cum dedecore meo testimonium lubentissime 
 perhibeam. Quod Cossica seu Algebrica (cum his enim reliqua comparata, id sunt quod 
 umbrae Homerice in Necya ad animam Tiresiae) ea ergo quod non assequebar modo, 
 quanquam mutis duntaxat usus preceptoribus caetera at/roSidaKTos, sed et augere, uariare, 
 adeoque corrigere in loco didicissem, quae summi et fidelissimi in docendo uiri Christifer 
 Rodolphus Silesius, Micaelus Stifelius, Cardanus, Nonius, aliique litteris mandauerant : 
 incidi in otyaiv, Upav vbaov, ut scite appellauit Heraclitus sapientior multis aliis philoso- 
 phis, hoc est, in Arithmetica, et uera Logistica, putaui me esse aliquid : itaque de me 
 passim etiam a multis, iisque doctis uiris iudicatum fuit, me non de grege Arithmeticum 
 esse. Verum ubi primum in Diophantea incidi : ita me recta ratio circumegit, ut flendusne 
 mihi ipsi antea, an uero ridendus fuissem, haud iniuria dubitauerim. Operae precium est 
 hoc loco et meam inscitiam inuulgare, et Diophantei operis, quod mihi nebulosam istam 
 caliginem ab oculis detersit, immo eos in coenum barbaricum defossos eleuauit et repur- 
 gauit, gustum aliquem exhibere. Surdorum ego numerorum tractationem ita tenebatn, 
 ut etiam addere aliorum inuentis aliquid non poenitendum auderem, atque id quidem in 
 rebus arithmeticis magnum habetur, et difficultas istarum rerum multos a rnathematibus 
 deterret. Quanto autem hoc est praeclarius, in iis problematis, quae surdis etiam 
 numeris uix posse uidentur explicari, rem eo deducere, ut quasi solum arithmeticum 
 uertere iussi obsurdescant illi plane, et ne mentio quidem eorum in tractatione ingenio- 
 sissimarum quaestionum admittatur. "
 
 2 4 INTRODUCTION 
 
 was even such as to deter many from the study of mathematics. 
 "But how much more splendid," says Xylander, "in the case of 
 problems which seem to be hardly capable of solution even with 
 the help of surds, to bring the matter to the point that, while the 
 surds, when bidden (so to speak) to plough the arithmetic soil, 
 become true to their name and deaf to entreaty, they are not so 
 much as mentioned in these most ingenious solutions ! " He then 
 describes the enormous difficulties which beset his work owing 
 to the corruptions in his text. In dealing, however, with the 
 mistakes and carelessness of copyists he was, as he says, no novice; 
 for proof of which he appeals to his editions of Plutarch, Stephanus 
 and Strabo. This passage, which is good reading, but too long 
 to reproduce here, I give in full in the note 1 . Next Xylander 
 tells us how he came to get possession of a manuscript of Dio- 
 phantus. In October of the year 1571 he made a journey to 
 Wittenberg ; while there he had conversations on mathematical 
 subjects with two professors, Sebastian Theodoric and Wolfgang 
 Schuler by name, who showed him a few pages of a Greek 
 
 1 " Id uero mihi accidit durum et uix superabile incommodum, quod mirifice deprauata 
 omnia inueni, cum neque problematum expositio interdum integra esset, ac passim numeri 
 (in quibus sita omnia esse in hoc argumento, quis ignorat?) tarn problematum quam 
 solutionum siue explicationum corruptissimi. Non pudebit me ingenue fateri, qualem me 
 heic gesserim. Audacter, et summo cum feruore potius quam alacritate animi opus ipsum 
 initio sum aggressus, laborque mihi omnis uoluptati fuit, tantus est meus rerum arithmeti- 
 carum amor, quin et gratiam magnam me apud omnes liberalium scientiarum amatores ac 
 patronos initurum, et praeclare de rep. litteraria meriturum intelligebam, eamque rem 
 mihi laudi (quam a bonis profectam nemo prudens aspernatur) gloriaeque fortasse etiam 
 emolumento fore sperabam. Progressus aliquantulum, in salebras incidi : quae tantum 
 abest ut alacritatem meam retuderint, ut etiam animos mihi addiderint, neque enim mihi 
 novum aut insolens est aduersus librariorum incuriam certamen, et hac in re militaui, (ut 
 Horatii nostri uerbis utar) non sine gloria, quod me non arroganter dicere, Dio, 
 Plutarchus, Strabo, Stephanusque nostri testantur. Sed cum mox in ipsum pelagus 
 monstris scatens me cursus abripuit : non despondi equidem animum, neque manus dedi, 
 sed tamen saepius ad oram unde soluissem respexi, quam portum in quern esset euadendum 
 cogitando prospicerem, depraehendique non minus uere quam eleganter ea cecinisse 
 Alcaeum, quae (si possum) Latine in hac quasi uotiua mea tabula scribam. 
 
 Qui uela uentis uult dare, dum licet, 
 
 Cautus futuri praeuideat modum 
 
 Cursus. mare ingressus, marine 
 
 Nauiget arbitrio necesse est. 
 
 Sane quod de Echeneide pisce fertur, eum nauim cui se adplicet remorari, poene credibile 
 fecit mihi mea cymba tot mendorum remoris retardata. Expediui tamen me ita, ut facile 
 omnes mediocri de his rebus iudicio praediti, intellecturi sint incredibilem me laborem et 
 aerumnas difficilimas superasse : pudore etiam stimulatum oneris quod ultro mihi impos- 
 uissem, non perferendi. Paucula quaedam non plane explicata, studio et certis de causis 
 in alium locum reiecimus. Opus quidem ipsum ita absoluimus ut neque eius nos pudere 
 debeat, et Arithmeticae Logisticesque studiosi nobis se plurimum debere sint haud dubie 
 professuri."
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 25 
 
 manuscript of Diophantus and informed him that it belonged to 
 Andreas Dudicius whom Xylander describes as "Andreas Dudicius 
 Sbardellatus, hoc tempore Imperatoris Romanorum apud Polonos 
 orator." On his departure from Wittenberg Xylander wrote out 
 and took with him the solution of a single problem of Diophantus, 
 to amuse himself with on his journey. This he showed at Leipzig 
 to Simon Simonius Lucensis, a professor at that place, who wrote to 
 Dudicius on his behalf. A few months afterwards Dudicius sent 
 the MS. to Xylander and encouraged him to persevere in his 
 undertaking to translate the Arithmetica into Latin. Accordingly 
 Xylander insists that the glory of the whole achievement belongs 
 in no less but rather in a greater degree to Dudicius than to 
 himself. Finally he commends the work to the favour of Prince 
 Ludwig, extolling the pursuit of arithmetical and algebraical 
 science and dwelling in enthusiastic anticipation on the influence 
 which the Prince's patronage would have in helping and advancing 
 the study of Arithmetic 1 . This Epistola Nuncupatoria bears the 
 date I4th August, I574 3 . Xylander died on the loth of February 
 in the year following that of the publication, 1 576. 
 
 Tannery has shown that the MS. used by Xylander was 
 Guelferbytanus Gudianus I. Bachet observes that he has not been 
 able to find out whether Xylander ever published the Greek text, 
 though parts of his commentary seem to imply that he had, or at 
 least intended to do so. It is now clear that he intended to bring 
 out the text, but did not carry out his intention. Tannery observes 
 that the MS. Palatinus gr. 391 seems to have been written either by 
 Xylander himself or for him, and there are German notes in the 
 margin showing that it was intended to print from it. 
 
 Xylander's achievement has been, as a rule, quite inadequately 
 appreciated. Very few writers on Diophantus seem to have studied 
 the book itself: a fact which may be partly accounted for by its 
 rarity. Even Nesselmann, whose book appeared in 1842, says that 
 he has never been able to find a copy. Nesselmann however seems 
 to have come nearest to a proper appreciation of the value of the 
 work : he says " Xylander's work remains, in spite of the various 
 
 1 " Hoc non modo tibi, Princeps Illustrissime, honorificum erit, atque gloriosum ; sed 
 te labores nostros approbante, arithmeticae studium cum alibi, turn in tua Academia et 
 Gymnasiis, excitabitur, confirmabitur, prouehetur, et ad perfectam eiusscientiam multi tuis 
 auspiciis, nostro labore perducti, magnam hac re tuis in remp. beneficiis accessionem 
 factam esse gratissima commemoratione praedicabunt." 
 
 2 " Heidelberga. postrid. Eidus Sextiles CID ID LXXIV."
 
 26 INTRODUCTION 
 
 defects which are unavoidable in a first edition of so difficult an 
 author, especially when based on only one MS. and that full of 
 errors, a highly meritorious achievement, and does not deserve 
 the severe strictures which it has sometimes had passed upon it. 
 It is true that Xylander has in many places not understood his 
 author, and has misrepresented him in others ; his translation is 
 often rough and un-Latin, this being due to a too conscientious 
 adherence to the actual wording of the original ; but the result 
 was none the less brilliant on that account. The mathematical 
 public was put in possession of Diophantus' work, and the 
 appearance of the translation had an immediate and enormous 
 influence on the development and shaping of Algebra 1 ." As a 
 rule, the accounts of Xylander's work seem to have been based 
 on what Bachet says about it and about his obligations to it. 
 When I came to read Bachet myself and saw how disparaging, 
 as a rule, his remarks upon Xylander were, I could not but suspect 
 that they were unfair. His repeated and almost violent repudiation 
 of obligation to Xylander suggested to me the very thing which he 
 disclaimed, that he was under too great obligation to his predecessor 
 to acknowledge it duly. I was therefore delighted at my good 
 fortune in finding in the Library of Trinity College, Cambridge, 
 a copy of Xylander, and so being able to judge for myself of 
 the relation of the later to the earlier work. The result was to 
 confirm entirely what I had suspected as to the unfair attitude 
 taken up by Bachet towards his predecessor. I found it every- 
 where ; even where it is obvious that Xylander's mistakes or 
 difficulties are due only to the hopeless state of his solitary MS. 
 Bachet seems to make no allowance for the fact. The truth is that 
 Bachet's work could not have been as good as it was but for the 
 pioneer work of Xylander; and it is the great blot in Bachet's 
 otherwise excellent edition that he did not see fit to acknowledge 
 the fact. 
 
 I must now pass to Bachet's work itself. It was the first 
 edition published which contained the Greek text, and appeared 
 in 1621 bearing the title: Diophanti Alexandrini Arithmeticorum 
 libri sex, et de numeris multangulis liber unus. Nunc primiim 
 Graece et Latine editi, atque absolutissimis Commentariis illustrati. 
 Auctore Claudia Gaspare Baclieto Meziriaco Sebusiano, V.C. Lutetiae 
 Parisiorum, Sumptibus Hieronymi Drovart" 1 , via Jacobaea, sub Scuto 
 
 1 Nesselmann, p. 279-80. 
 
 2 For " sumptibus Hieronymi Drovart etc. " some copies have " sumptibus Sebastiani
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 27 
 
 Solari. MDCXXI. Bachet's Greek text is based, as he tells us, 
 upon a MS. which he calls "codex Regius," now in the Bibliotheque 
 Nationale at Paris (Parisinus 2379) ; this MS. is his sole authority, 
 except that Jacobus Sirmondus had part of a Vatican MS. (Vat. 
 gr. 304) .transcribed for him. He professes to have produced a 
 good Greek text, having spent incalculable labour upon its emenda- 
 tion, to have inserted in brackets all additions which he made to it, 
 and to have given notice of all corrections, except those of an 
 obvious or trifling nature ; a few passages he has left asterisked, in 
 cases where correction could not be safely ventured upon. He 
 is careful to tell us what previous works relating to the subject he 
 had been able to consult. First he mentions Xylander (he spells 
 the name as X/lander throughout), who had translated the whole of 
 Diophantus, and commented upon him throughout, "except that 
 he scarcely touched a considerable part of the fifth book, the whole 
 of the sixth and the treatise on multangular numbers, and even 
 the rest of his work was not very successful, as he himself admits 
 that he did not thoroughly understand a number of points." Then 
 he speaks of Bombelli (as already mentioned) and of the Zetetica of 
 Vieta (in which the author treats in his own way a large number 
 of Diophantus' problems : Bachet thinks that he so treated them 
 because he despaired of restoring the book completely). Neither 
 Bombelli nor Vieta (says Bachet) made any attempt to demonstrate 
 the difficult porisms and abstruse theorems in numbers which 
 Diophantus assumes as known in many places, or sufficiently 
 explained the causes of his operations and artifices. All these 
 omissions on the part of his predecessors he thinks he has supplied 
 in his notes to the various problems and in the three books of 
 "Porisms" which he prefixed to the work 1 . As regards his Latin 
 translation, he says that he gives us Diophantus in Latin from the 
 version of Xylander most carefully corrected, in which he would 
 have us know that he has done two things in particular, first, 
 
 Cramoisy, via Jacobaea, sub Ciconiis." The copy (from the Library of Trinity College, 
 Cambridge) which I used in preparing my first edition has the former words ; a copy in 
 the Library of the Athenaeum Club has the latter. 
 
 1 On the nature of some of Bachet's proofs Nicholas Saunderson (formerly Lucasian 
 Professor) remarks in Elements of Algebra, 1740, apropos of Dioph. ill. 15 : " M. Bachet 
 indeed in the i6th and i7th props, of his second book of Porisms has given us demonstra- 
 tions, such as they are, of the theorems in the problem : but in the first place he 
 demonstrates but one single case of those theorems, and in the next place the demonstra- 
 tions he gives are only synthetical, and so abominably perplexed withal, that in each 
 demonstration he makes use of all the letters in the alphabet except I and O, singly to 
 represent the quantities he has there occasion for. *
 
 28 INTRODUCTION 
 
 corrected what was wrong and filled the numerous lacunae, 
 secondly, explained more clearly what Xylander had given in 
 obscure or ambiguous language ; " I confess however," he says, 
 " that this made so much change necessary, that it is almost 
 fairer to attribute the translation to me than to Xilander. But if 
 anyone prefers to consider it as his, because I have held fast, tooth 
 and nail, to his words when they do not misrepresent Diophantus, 
 I have no objection 1 ." Such sentences as these, which are no 
 rarity in Bachet's book, are certainly not calculated to increase 
 our respect for the author. According to Montucla 2 , "the historian 
 of the French Academy tells us " that Bachet worked at this edition 
 during the course of a quartan fever, and that he himself said that, 
 disheartened as he was by the difficulty of the work, he would never 
 have completed it, had it not been for the stubbornness which his 
 malady generated in him. 
 
 As the first edition of the Greek text of Diophantus, this work, 
 in spite of any imperfections we may find in it, does its author all 
 honour. 
 
 The same edition was reprinted and published with the addition 
 of Fermat's notes in 1670: Diophanti Alexandrini Arithmeticortim 
 libri sex, et de numeris multangiilis liber unus. Cum commentariis 
 C. G. Bacheti V.C. et obseruationibus D. P. de Fermat Senator is 
 Tolosani. Accessit Doctrinae Analyticae inuentum nouum, collectum 
 ex variis eiusdem D. de Fermat Epistolis. Tolosae, Excudebat 
 Bemardus Bosc, e Regione Collegii Societatis Jesu. MDCLXX. 
 This edition was not published by Fermat himself, but by his 
 son after his death. S. Fermat tells us in the preface that this 
 publication of Fermat's notes to Diophantus 3 was part of an 
 attempt to collect together from his letters and elsewhere his 
 contributions to mathematics. The "Doctrinae Analyticae In- 
 uentum nouum" is a collection made by Jacobus de Billy 4 
 
 1 Deinde Latinum damus tibi Diophantum ex Xilandri versione accuratissime castigata, 
 in qua duo potissimum nos praestitisse scias velim, nam et deprauata correximus, hiantesque 
 passim lacunas repleuimus : et quae subobscure, vel ambigue fuerat interpretatus Xilander, 
 dilucidius exposuimus; fateor tamen, inde tantam inductam esse mutationem, vt prope- 
 modum aequius sit versionem istam nobis quam Xilandro tribuere. Si quis autem potius 
 ad eum pertinere contendat, quod eius verba, quatenus Diophanto fraudi non erant, 
 mordicus retinuimus, per me licet.'' - i. 323. 
 
 3 Now published in CEuvres de Fermat by P. Tannery and C. Henry, Vol. I. (1891), 
 pp. 289-342 (the Latin original), and Vol. in. (1896), pp. 241-274 (French translation). 
 
 4 Now published in CEuvres de Fermat, in. 323-398 (French translation). De 
 Billy had already published in 1660 a book under the title Diophantus geometra sive 
 opus contextum ex arithmetica et geometria.
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 29 
 
 from various letters which Fermat sent to him at different times. 
 The notes upon Diophantus' problems, which his son hopes will 
 prove of value very much more than commensurate with their 
 bulk, were (he says) collected from the margin of his copy of 
 Diophantus. From their brevity they were obviously intended 
 for the benefit of experts 1 , or even perhaps solely for Fermat's 
 own, he being a man who preferred the pleasure which he had 
 in the work itself to any reputation which it might bring 
 him. Fermat never cared to publish his investigations, but was 
 always perfectly ready, as we see from his letters, to acquaint 
 his friends and contemporaries with his results. Of the notes 
 themselves this is not the place to speak in detail. This edition 
 of Diophantus is rendered valuable only by the additions in it 
 due to Fermat; for the rest it is a mere reprint of that of 1621. 
 So far as the Greek text is concerned, it is very much inferior 
 to the first edition. There is a far greater number of misprints, 
 omissions of words, confusions of numerals ; and, most serious of 
 all, the brackets which Bachet inserted in the edition of 1621 to 
 mark the insertion of words in the text are in this later edition 
 altogether omitted. These imperfections have been already noticed 
 by Nesselmann 2 . Thus the reprinted edition of 1670 is untrust- 
 worthy as regards the text. 
 
 In 1585 Simon Stevin published a French version of the first 
 four books of Diophantus 3 . It was based on Xylander and was 
 a free reproduction, not a translation, Stevin himself observing that 
 the MS. used by Xylander was so full of mistakes that the text of 
 
 1 Lectori Seneuolo, p. iii : " Doctis tantum quibus pauca sufficiunt, harum obserua- 
 tionum auctor scribebat, vel potius ipse sibi scribens, his studiis exerceri malebat quam 
 gloriari; adeo autem ille ab omni ostentatione alienus erat, vt nee lucubrationes suas 
 typis mandari curauerit, et suorum quandoque responsorum autographa nullo seruato 
 exemplari petentibus vitro miserit ; norunt scilicet plerique celeberrimorum huius saeculi 
 Geometrarum, quam libenter ille et quanta humanitate, sua iis inuenta patefecerit." 
 
 2 "Was dieser Abdruck an ausserer Eleganz gewonnen hat (denn die Bachet'sche 
 Ausgabe ist mit ausserst unangenehmen, namentlich Griechischen Lettern gedruckt), das 
 hat sie an innerm Werthe in Bezug auf den Text verloren. Sie ist nicht bloss voller 
 Druckfehler in einzelnen Worten und Zeichen (z. B. durchgehends JT statt ~^, 900) 
 sondern auch ganze Zeilen sind ausgelassen oder doppelt gedruckt (z. B. ill. 12 eine 
 Zeile doppelt, iv. 25 eine doppelt und gleich hinterher eine ausgelassen, IV. 52 eine 
 doppelt, v. ii eine ausgelassen, desgleichen v. 14, 15, 33, VI. 8, 13 und so weiter), die 
 Zahlen verstiimmelt, was aber das Aergste ist, die Bachet'schen kritischen Zeichen sind 
 fast uberall, die Klammer durchgangig weggefallen, so dass diese Ausgabe als Text des 
 Diophant vollig unbrauchbar geworden ist," p. 283. 
 
 3 Included in L' Arithmelique de Simon Stevin de Bruges. ..A Leyde, De I'lmprimerie 
 de Christophle Plantin, cio . ID . LXXXV.
 
 3 o INTRODUCTION 
 
 Diophantus could not be given word for word 1 . Albert Girard 
 added the fifth and sixth books to the four, and this complete 
 version appeared in i625 2 . 
 
 In 1810 was published an excellent translation (with additions) 
 of the fragment upon Polygonal Numbers by Poselger : Diophantus 
 von Alexandrien iiber die Polygonal-Zahlen. Uebersetzt mit Zusdtzen 
 von F. Th. Poselger. Leipzig, 1810. 
 
 In 1822 Otto Schulz, professor in Berlin, published a very 
 meritorious German translation with notes : Diophantus von 
 Alexandria arithmetische Aufgaben nebst dessen Schrift iiber die 
 Polygon-Zahlen. Aus dent Griechischen ilbersetzt und mit An- 
 merkungen begleitet von Otto Schulz, Professor am Berlinisch- 
 Colnischen Gymnasium zum grauen Kloster. Berlin, 1822. In der 
 Schlesingerschen Buck- tind Mtisikhandlung. The work of Poselger 
 just mentioned was with the consent of its author incorporated in 
 Schulz's edition along with his own translation and notes upon 
 the larger treatise, the Arithmetica. According to Nesselmann 
 Schulz was not a mathematician by profession ; he produced, 
 however, a thoroughly useful edition, with notes chiefly upon 
 the matter of Diophantus and not on the text (with the exception 
 of a very few emendations) : notes which, almost invariably correct, 
 help much to understand the author. Schulz's translation is based 
 upon the edition of Bachet's text published in 1670. 
 
 Another German translation was published by G. Wertheim 
 in 1 890 : Die A rithmetik und die Schrift fiber Polygonalzahlen des 
 Diophantus von Alexandria. Ubersetzt tmd mit Anmerkungen 
 begleitet von G. Wertheim (Teubner). Though it appeared before 
 the issue of Tannery's definitive text, it is an excellent translation, 
 the translator being thoroughly equipped for his task ; it is valuable 
 also as containing Fermat's notes, also translated into German, with 
 a large number of other notes by the translator elucidating both 
 Diophantus and Fermat, and generalising a number of the problems 
 which, with very few exceptions, receive only particular solutions 
 from Diophantus himself. Wertheim has also included 46 epigram- 
 problems from the Greek anthology and the enunciation of the 
 famous Cattle-Problem attributed to Archimedes. 
 
 1 See Bibliotheca Mathematica Vii 3 , 1906-7, p. 59. 
 
 2 U Arithmetiqiu de Simon Stevin de Brvges, Reiteue, corrigee & augmenlee de phisieurs 
 traictez et annotation par Albert Girard Samielois Mathematicien. A Leide, de 
 I'lmprimerie des Elzeviers cio . 10 . cxxv. Reproduced in the edition of Les (Euvres 
 Mathematiques de Simon Stevin de Bruges. Par Albert Girard. Leyde, cio . 10 . cxxxiv.
 
 THE MSS. OF AND WRITERS ON DIOPHANTUS 31 
 
 No description is necessary of the latest edition, by Tannery, 
 in which we at last have a definitive Greek text of Diophantus 
 with the ancient commentaries, etc., Diophanti Alexandrini opera 
 omnia cum Graecis commentaries. Edidit et Latine interpretatus 
 est Paulus Tannery (Teubner). The first volume (1893) contains 
 the text of Diophantus, the second (1895) the Pseudepigrapha, 
 Testimonia veterum, Pachymeres' paraphrase, Planudes' com- 
 mentary, various ancient scholia, etc., and 38 arithmetical epigrams 
 in the original Greek with scholia. Any further edition will neces- 
 sarily be based on Tannery, who has added all that is required in 
 the shape of introductions, etc. 
 
 Lastly we hear of other works on Diophantus which, if they 
 were ever written, are lost or remain unpublished. First, we find 
 it asserted by Vossius (as some have understood him) that the 
 Englishman John Pell wrote an unpublished Commentary upon 
 Diophantus. John Pell (1611-1685) was at one time professor 
 of mathematics at Amsterdam and gave lectures there on Dio- 
 phantus, but what Vossius says about his commentary may 
 well be only a recommendation to undertake a commentary, 
 rather than a historical assertion of its completion. Secondly, 
 Schulz states in his preface that he had lately found a note in 
 Schmeisser's Orthodidaktik der Mathematik that Hofrath Kausler 
 by command of the Russian Academy prepared an edition of 
 Diophantus 1 . This seems however to be a misapprehension on the 
 part of Schulz. Kausler is probably referring, not to a translation 
 of Diophantus, but to his memoir of 1798 published in Nova Acta 
 Acad. PetropoL XI. p. 125, which might easily be described as an 
 Ausarbeitung of Diophantus' work. 
 
 I find a statement in the New American Cyclopaedia (New York, 
 D. Appleton and Company), Vol. vi., that " a complete translation 
 of his (Diophantus') works into English was made by the late 
 Miss Abigail Lousada, but has not been published." 
 
 1 The whole passage of Schmeisser is : "Die mechanische, geistlose Behandlung der 
 Algebra ist ins besondere von Herrn Hofrath Kausler stark geriigt worden. In der 
 Vorrede zu seiner Ausgabe des Uflakerschen Exempelbuths beginnt er so : ' Seit mehreren 
 Jahren arbeitete ich fur die Russisch-Kaiserliche Akademie der Wissenschaften Diophants 
 unsterbliches Werk iiber die Arithmetik ans, und fand darin einen solchen Schatz von 
 den feinsten, scharfsinnigsten algebraischen Auflosungen, dass mir die mechanische, 
 geistlose Methode der neuen Algebra rait jedem Tage mehr ekelte u.s.w.' " (p. 33).
 
 CHAPTER III 
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 
 
 As it is my intention, for the sake of brevity and per- 
 spicuity, to make use of the modern algebraical notation in giving 
 my account of Diophantus' problems and general methods, it is 
 necessary to describe once for all the machinery which our author 
 uses for working out the solutions of his problems, or the notation 
 by which he expresses such relations as would be represented in 
 our time by algebraical equations, and, in particular, to illustrate 
 the extent to which he is able to manipulate unknown quantities. 
 Apart, however, from the necessity of such a description for the 
 proper and adequate comprehension of Diophantus, the general 
 question of the historical development of algebraical notation 
 possesses great intrinsic interest. Into the general history of this 
 subject I cannot enter in this essay, my object being the elucidation 
 of Diophantus ; I shall accordingly in general confine myself to an 
 account of his notation solely, except in so far as it is interesting 
 to compare it with the corresponding notation of his editors and 
 (in certain cases) that of other writers, as, for example, certain of 
 the early Arabian algebraists. - 
 
 First, as to the representation of an unknown quantity. The 
 unknown quantity, which Diophantus defines as containing TrXrjQos 
 /jiovdBcov dopio-rov, i.e. an undefined number of units (def. 2), is 
 denoted throughout by what was printed in the editions before 
 Tannery's as the Greek letter 9 with an accent, thus <?', or in the 
 form 9. This symbol in verbal description he calls o dpi&fj,6<>, " the 
 number," i.e. t by implication, the number par excellence of the problem 
 in question. In the cases where the symbol is used to denote in- 
 flected forms, e.g., the accusative singular or the dative plural, the 
 terminations which would have been added to the stem of the full 
 word dpiOpos were printed above the symbol 9 in the manner of an
 
 NOTATION AND DEFINITIONS OF DlOPHANTUS 33 
 
 exponent, thus ? xv (for apidpov, as T VV for TOJ/), s 5 , the symbol being 
 in addition doubled in the plural cases, thus 55*," w *** ??""> 5?**, f r 
 dpiO/jLOi K.r.e. When the symbol is used in practice, the coefficient 
 is expressed by putting the required Greek numeral immediately 
 after it, thus s<?' ta corresponds to I ix, 9' a to x and so on. 
 
 Tannery discusses the question whether in the archetype (a) of 
 the MSS. this duplication of the sign for the plural and this 
 addition of the terminations of the various cases really occurred 1 . 
 He observes that any one accustomed to reading Greek MSS. will 
 admit that the marks of cases are common in the later MSS. but 
 are very frequently omitted in the more ancient. Further, the 
 practice of duplicating a sign to express the plural is more ancient 
 than that of adding the case-terminations. Tannery concludes that 
 the case-terminations (like the final syllables of abbreviations used 
 for other words) were very generally, if not always, wanting in the 
 archetype (a). If this seems inconsistent with the regularity with 
 which they appear in our MSS., it has to be remembered that A 
 and B l do not represent the archetype (a) but the readings of a, the 
 copyist of which probably took it upon himself to substitute the 
 full word for the sign or to add the case-terminations. Tannery's 
 main argument is the frequent occurrence of instances where the 
 wrong case-ending has been added, e.g., the nominative for the 
 genitive ; the conclusion is also confirmed by instances in which 
 different cases of the word dpi0/j,6<;, e.g. dpiffpov, dptffpov, and even 
 aptBpAv written in full are put by mistake for xai owing to the 
 resemblance between the common abbreviation for icai and the 
 sign for dpiOpo*;, and of course in such cases the abbreviation would 
 not have had the endings. As regards the duplication of the sign 
 for the plural, Tannery admits that this was the practice of the 
 Byzantines ; but he considers that the evidence is against sup- 
 posing that Diophantus duplicated the sign ; he does not do so 
 with any other of his technical abbreviations, those for p.ovds, 
 Svvapis, etc. Accordingly in his text of Diophantus Tannery has 
 omitted the case-endings and written the single sign for dpi0/j,6<; 
 whether in the singular or in the plural ; in his second volume, 
 however, containing the scholia, etc., he has retained the duplicated 
 sign. 
 
 On the assumption that the sign was the Greek final sigma, it 
 was natural that Nesselmann should explain it by the supposition 
 
 1 Dioph. II. pp. xxxiv-xxxix. 
 H. D. 3
 
 34 INTRODUCTION 
 
 that Diophantus, in search of a convenient symbol for his unknown 
 quantity, would select the only letter of the Greek alphabet which 
 was not already appropriated as a numeral 1 . But he made the 
 acute observation 2 that, as the symbol occurred in many places (of 
 course in Bachet's text) for dpi0/j,6<t used in the ordinary un- 
 technical sense, and was therefore, as it appeared, not exclusively 
 used to designate the unknown quantity, the technical apropos, it 
 must after all be more of the nature of an abbreviation than an 
 algebraical symbol like our x. It is true that this uncertainty in 
 the use of the sign in the MSS. is put an end to by Tannery, who 
 uses it for the technical dpiO/j>6f alone and writes the untechnical 
 dpiQjjLos in full ; but, even if Diophantus' practice was as strict as 
 this, I do not think this argues any difference in the nature of the 
 abbreviation. There is also a doubt whether the final sigma, ?, 
 was developed as distinct from the form a- so early as the date of 
 the MSS. of Diophantus, or rather so early as the first copy of his 
 work, if the author himself really gave the explanation of the sign 
 as found in our text of his second definition. These considerations 
 suggested to me that the sign was not the final sigma at all, but 
 must be explained in some other way. I had to look for con- 
 firmation of this to the precise shape of the sign as found in extant 
 MSS. The only MS. which I had the opportunity of inspecting 
 personally was the MS. of the first ten problems of Diophantus in 
 the Bodleian ; but here I found strong confirmation of my view in 
 the fact that the sign appeared as '^, quite different in shape from, 
 and much larger than, the final sigma at the end of words in the 
 same MS. (There is in the Oxford MS. the same irregularity as 
 was pointed out by Nesselmann in the use of the sign sometimes 
 for the technical, and sometimes for the untechnical, aptfyio? 3 .) 
 But I found evidence that the sign appeared elsewhere in some- 
 what different forms. Thus Rodet in the Journal Asiatique of 
 January, 1878, quoted certain passages from Diophantus for the 
 purpose of comparison with the algebra of Muhammad b. Musa 
 al-Khuwarazml. Rodet says he copied these passages exactly 
 from Bachet's MS. ; but, while he generally gives the sign as the 
 final sigma, he has in one case iji] 01 for dpiffftoi. In this last case 
 
 1 Nesselmann, pp. 290-1. 2 ibid. pp. 300-1. 
 
 3 An extreme case is lraa TO rov devrtpov'^o api8/j,ovtt>6$, where the sign (contrary to 
 what would be expected) means the untechnical <i/H0/u6s, and the technical is written in 
 full. Also in the definition 6 5e nydef rotiruv T&V Idiw/A&TUv KTriffd/j.evos...apit)[j.bs KaXerrai 
 the word dpify^s is itself denoted by the symbol, showing that the word and the symbol 
 are absolutely convertible.
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 35 
 
 Bachet himself reads 9? ot . But the same form ijip which Rodet 
 gives is actually found in three places in Bachet's own edition, 
 (i) In his note to IV. 3 he gives a reading from his own MS. which 
 he has corrected in his own text and in which the signs ija and 
 LJLjr) occur, evidently meaning dpi0/j,o<; a and dpiO^ol rj, though the 
 sign should have been that for dpid^oarov (= \\x). (2) In the text of 
 IV. 1 3 there is a sentence (marked by Bachet as interpolated) which 
 contains the expression LjLjr, where the context again shows that 
 i]q is for dpiOfjuoL (3) At the beginning of V. 9 there is a difficulty 
 in the text, and Bachet notes that his MS. has prjre 6 8nr\a<ria)v 
 avrov LJ where a Vatican MS. reads dpid^ov (Xylander notes that 
 his MS. had in this place fjurjre 6 BnrXaalcov avrov dp po a ...). 
 It is thus clear that the MS. (Paris. 2379) which Bachet used 
 sometimes has the sign for dpiQ/jios in a form which is at least 
 sufficiently like i| to be taken for it. Tannery states that the form 
 of the sign found in the Madrid MS. (A) is tj, while B 1 has it in a 
 form ($) nearly approaching Bachet's reproduction of it. 
 
 It appeared also that the use of the sign, or something like 
 it, was not confined to MSS. of Diophantus ; on reference to 
 Gardthausen, Griechische Palaeographie, I found under the head 
 " hieroglyphisch-conventionell " an abbreviation 9, 99 for aptfytd?, 
 -oL, which is given as occurring in the Bodleian MS. of Euclid 
 (D'Orville 301) of the 9th century. Similarly Lehmann 1 notes as 
 a sign for dpi0/j,6<; found in that MS. a curved line similar to that 
 which was used as an abbreviation for icai. He adds that the 
 ending is placed above it and the sign is doubled for the plural. 
 Lehmann's facsimile is like the form given by Gardthausen, but has 
 the angle a little more rounded. The form iji| ot above mentioned 
 is also given by Lehmann, with the remark that it seems to be 
 only a modification of the other. Again, from the critical notes to 
 Heiberg's texts of the Arenarius of Archimedes it is clear that the 
 sign for dpidpos occurred several times in the MSS. in a form 
 approximating to that of the final sigma, and that there was the 
 usual confusion caused by the similarity of the signs for dpidpbs 
 and teal 2 . In Hultsch's edition of Heron, similarly, the critical 
 notes to the Geodaesia show that one MS. had an abbreviation for 
 
 1 Lehmann, Die tachygraphischen Abkiirzungen der griechischen Handschriften, 1880, 
 p. 107 : " Von Sigeln, welchen ich auch anderwarts begegnet bin, sincl zu nennen dpi6/j.6s, 
 das in der Oxforder Euclidhandschrift mit einer der Note xal ahnlichen Schlangenlinie 
 bezeichnet wird." 
 
 2 Cf. Heiberg, Quaestiones Archimedean, pp. 172, 174, 187, 188, 191, 192; Archimedis 
 opera omnia, II., pp. 268 sqq. 
 
 32
 
 36 INTRODUCTION 
 
 oi in various forms with the case-endings superposed ; some- 
 times they resembled the letter , sometimes p, sometimes O and 
 once J . Lastly, the sign for dpiBfjios resembling the final sigma 
 evidently appeared in a MS. of Theon of Smyrna 2 . 
 
 All these facts strongly support the assumption that the sign 
 was a mere tachygraphic abbreviation and not an algebraical 
 symbol like our x, though discharging much the same function. 
 The next question is, what is its origin ? The facts (i) that the 
 sign has the breathing prefixed in the Bodleian MS., which writes 
 '^-7 for dptOfjios, and (2) that in one place Xylander's MS. read dp 
 tor the full word, suggested to me the question whether it could 
 be a contraction of the first two letters of dpidpos ; and, on con- 
 sideration, this seemed to me quite possible when I found a 
 contraction for ap given by Gardthausen, namely cf. It is easy to 
 see that a simplification of this in different ways would readily 
 produce signs like the different forms shown above. This then 
 was the hypothesis which I put forward twenty-five years ago, and 
 which I still hold to be the easiest and best explanation. Two 
 alternatives are possible, (i) Diophantus may not have made the 
 contraction himself. In that case I suppose the sign to be a cur- 
 sive contraction made by scribes ; and I conceive it to have come 
 about through the intermediate form < p. The loss of the downward 
 stroke, or of the loop, would produce a close approximation to 
 the forms which we know. (2) Diophantus may have used a sign 
 approximately, if not exactly, like that which we find in the MSS. 
 For it is from a papyrus of 154 A.D., in writing of the class which 
 Gardthausen calls the "Majuskelcursive," that the contraction c?f> for 
 the two letters is taken. The great advantage of my hypothesis is 
 that it makes the sign for aptfytov exactly parallel to those for the 
 powers of the unknown, e.g., J r for Svvafjus and K Y for /cu/8o?, and 
 
 o 
 
 to that for the unit fjwvds which is denoted by M, with the sole 
 difference that the letters coalesce into one instead of being 
 written separately. 
 
 Tannery's views on the subject are, I think, not very con- 
 sistent, and certainly they do not commend themselves to me. He 
 seems to suggest that the sign is the ancient letter Koppa, perhaps 
 slightly modified ; he first says that the sign in Diophantus is 
 peculiar to him and that, although the word dpi6p,b<i is very often 
 
 1 Heron, ed. Hultsch, pp. 146, 148, 149, 150. 
 
 2 Theon of Smyrna, ed. Hiller, p. 56, critical notes.
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 37 
 
 represented in mathematical MSS. by an abbreviation, it has much 
 oftener the form $ or something similar, closely resembling the 
 ancient Koppa. In the next sentence he seems to say that " on 
 the contrary the Diophantine abbreviation is an inverted di- 
 gamma " ; yet lower down he says that the copyist of a (copied 
 from the archetype a) got the form i] by simplifying the more 
 complicated Koppa. And, just before the last remark, he has 
 stated that in the archetype a the form must have been 5 or very 
 like it, as is shown by the confusion with the sign for ical. (If this 
 is so, it can hardly have been peculiar to Diophantus, seeing that 
 the same confusion occurs fairly often in the MSS. of other 
 authors, as above shown.) I think the last consideration (the con- 
 fusion with ical} is very much against the Koppa-hypothesis ; and, 
 in any case, it seems to me very unlikely that a sign would be 
 used by Diophantus for the unknown which was already appro- 
 priated to the number 90. And I confess I am unable to see in 
 the sign any resemblance to an inverted digamma. 
 
 Hultsch 1 regards it as not impossible that Diophantus may 
 have adopted one of the signs used by the Egyptians for their 
 unknown quantity hau, which, if turned round from left to right, 
 would give V; but here again I see no particular resemblance. 
 Prof. D'Arcy Thompson 2 has a suggestion that the sign might be 
 the first letter of o-p6<?, a heap. But, apart from the fact that the 
 final sigma (<?) is not that first letter, there is no trace whatever 
 in Diophantus of such a use of the word o-twpo? ; and, when 
 Pachymeres 3 speaks of a number being crwpeia /j,ovdSa>v, he means 
 no more than the 7r\fj0os fAovdSwv which he is explaining : his 
 words have no connexion with the Egyptian hau. 
 
 Notwithstanding that the sign is not the final sigma, I shall 
 not hesitate to use 9 for it in the sequel, for convenience of 
 printing. Tannery prints it rather differently as =>. 
 
 We pass to the notation which Diophantus used to express the 
 different powers of the unknown quantity, corresponding to x*, x 3 , 
 and so on. He calls the square of the unknown quantity 8vz>a/u<?, 
 and denotes it by the abbreviation J r . The word Svvapts, 
 literally " power," is constantly used in Greek mathematics for 
 
 1 Art. Diophantus in Pauly-Wissowa's Keal-Encyclopadie der classischen Altertums- 
 wis sense h aften . 
 
 2 Transactions of the Royal Society of Edinburgh, Vol. xxxvm. (1896), pp. 607-9. 
 
 3 Dioph. n. p. 78, 4. Cf. lamblichus, ed. Pistelli, p. 7, 7 ; 34, 3 ; 81, 14, where <rw/>efa 
 is similarly used to elucidate TrXTjflos.
 
 38 INTRODUCTION 
 
 square*. With Diophantus, however, it is not any square, but only 
 the square of the unknown ; where he speaks of any particular 
 square number, it is Terpdytovos dpid/Mos. The higher powers of the 
 unknown quantity which Diophantus makes use of he calls Kvftos, 
 &vva,f*,o8vva/j,t<;, Swa/jLotcvfios, KvftoKvftos, corresponding respectively 
 to X s , x*, X s , x 6 . Beyond the sixth power he does not go, having 
 no occasion for higher powers in the solutions of his problems. For 
 these powers he uses the abbreviations K Y , A Y A, AK Y , K Y K re- 
 spectively. There is a difference between Diophantus' use of the word * 
 8vva/j,i<; and of the complete words for the third and higher powers, 
 namely that the latter are not always restricted like 8iW/u<? to powers 
 of the unknown, but may denote powers of ordinary known num- 
 bers as well. This is no doubt owing to the fact that, while there 
 are two words Bvva/j,i<; and rerpdyaivo^ which both signify " square," 
 there is only one word for a third power, namely /cu/3o?. It is 
 important, however, to observe that the abbreviations K Y , A Y A, 
 AK Y , K Y K, are, like Svvafw and A Y , only used to denote powers 
 of the unknown. The coefficients of the different powers of the 
 unknown, like that of the unknown itself, are expressed by the 
 addition of the Greek letters denoting numerals, e.g., AK V *r cor- 
 responds to 26x*. Thus in Diophantus' system of notation the signs 
 A Y and the rest represent not merely the exponent of a power like 
 the 2 in x z , but the whole expression x*. There is no obvious 
 connexion between the symbol A Y and the symbol 9 of which it is 
 the square, as there is between x* and x, and in this lies the great 
 inconvenience of the notation. But upon this notation no advance 
 was made by Xylander, or even by Bachet and Fermat. They wrote 
 A 7 (which was short for Numerus) for the 9 of Diophantus, Q (Quad- 
 ratus) for J r , C (Cubus) for K Y , so that we find, for example, 
 I Q + 5^= 24, corresponding to x"- + $x = 24. Other symbols were 
 however used even before the publication of Xylander's Diophantus, 
 e.g. in Bombelli's Algebra. Bombelli denotes the unknown and its 
 powers by the symbols -i 1, , and so on. But it is certain that 
 up to this time (1572) the common symbols had been R (Radix 
 or Res), Z (Zensus, i.e. square), C (Cubus). Apparently the first 
 important step towards x*, x*, etc., was taken by Vieta (1540 
 
 1 In Plato we have SiW/uj used for a square number (Titnaeus, 31) and also 
 (TTieaetetus , 147 D) for a square root of a number which is not a complete square, i.e. for 
 a surd ; but the commonest use is in geometry, in the form 8vvA.fj.tL, " in square," e.g. "AB 
 is dw6/j.ei double of C" means " AB 2 = 2.5C 2 ."
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 39 
 
 1603), w ho wrote Aq, Ac, Aqq, etc. (abbreviated for A quadratus 
 and so on) for the powers of A. This system, besides showing the 
 con .exion between the different powers, has the infinite advantage 
 that by means of it we can use in one and the same solution any 
 number of unknown quantities. This is absolutely impossible with 
 the notation used by Diophantus and the earlier algebraists. 
 Diophantus in fact never uses more than one unknown quantity in 
 the solution of a problem, namely the dpi6/j.6<; or <?. 
 
 Diophantus has no symbol for the operation of multiplication ; 
 it is rendered unnecessary by the fact that his coefficients are all 
 definite numbers or fractions, and the results are simply put down 
 without any preliminary step which would call for the use of a 
 symbol. On the ground that Diophantus uses only numerical 
 expressions for coefficients instead of general symbols, it might 
 occur to a superficial observer that there must be a great want 
 of generality in his methods, and that his problems, being solved 
 with reference to particular numbers only, would possess the 
 attraction of a clever puzzle rather than any more general interest. 
 The answer to this is that, in the first place, it was absolutely 
 impossible that Diophantus should have used any other than 
 numerical coefficients, for the reason that the available symbols of 
 notation were already employed, the letters of the Greek alphabet 
 always doing duty as numerals, with the exception of the final 9. 
 In the second place, it is not the case that the use of none but 
 numerical coefficients makes his solutions any the less general. 
 This will be clearly seen when I come to give an account of his 
 problems and methods. 
 
 Next as to Diophantus' expressions for the operations of 
 addition and subtraction. For the former no symbol at all is 
 used: it is expressed by mere juxtaposition, thus K Y a^ Y ^y^ 
 corresponds to x* + \yc- + $x. In this expression, however, there 
 is no absolute term, and the addition of a simple numeral, as 
 for instance /8, directly after e, the coefficient of ?, would cause 
 confusion. This fact makes it necessary to have some expression 
 to distinguish the absolute term from the variable terms. For this 
 purpose Diophantus uses the word /u,ovaSe<?, or units, and denotes 
 
 o 
 
 them after his usual manner by the abbreviation M. The number 
 of units is expressed as a coefficient. Thus corresponding to 
 the expression X s + \y? + $x+2 we should find in Diophantus 
 
 K Y a J r t79eJ/. As Bachet uses the sign + for addition, he
 
 40 INTRODUCTION 
 
 has no occasion for a distinct symbol to mark an absolute term. 
 He accordingly writes iC+ i^Q+^N+2. It is worth observing, 
 however, that the Italians do use a symbol in this case, namely N 
 (Numero), the first power of the unknown being with them R 
 (Radice). Cossali 1 makes an interesting comparison between the 
 terms used by Diophantus for the successive powers of the unknown 
 and those employed by the Italians after their instructors, the 
 Arabians. He observes that Fra Luca (Paciuolo), Tartaglia, and 
 Cardano begin their scale of powers from the power o, not from the 
 power i, as does Diophantus, and he compares the scales thus : 
 
 Scala Diofantea. Scala Araba. 
 
 i. Numero.. .il Noto. 
 
 x \. Numero... 1' Ignoto. i. Cosa, Radice, Lato. 
 
 X* i. Podesta. 3. Censo. 
 
 x 3 3. Cubo. 4. Cubo. 
 
 x* 4. Podesta-Podesta. 5. Censo di Censo. 
 
 x s 5. Podesta-Cubo. 6. Relate i. 
 
 x 6 6. Cubo-Cubo. 7. Censo di Cubo, o Cubo di Censo. 
 
 xi 7 8. Relate 2. 
 
 X s 8 9. Censo di Censo di Censo. 
 
 x 9 9 10. Cubo di Cubo. 
 
 and so on.* So far, however, as this is meant to be a comparison 
 between Diophantus and the early Arabian algebraists themselves 
 (as the title " Scala Araba" would seem to imply), there appears to 
 be no reason why Cossali should not have placed some term to 
 express Diophantus' /ioi>Se? in the same line with Numero in the 
 Arabian scale, and moved the numbers i, 2, 3, etc. one place 
 upward^ in the first scale, or downwards in the second. As 
 Diophantus does not go beyond the sixth power, the last three 
 places in the first scale are left blank. An examination of these 
 two scales will show also that the evolution of the successive 
 powers differs in the two systems. The Diophantine terms for 
 them are based on the addition of exponents, the Arabic on 
 
 1 Upon Wallis' comparison of the Diophantine with the Arabian scale Cossali 
 remarks: "ma egli non ha riflettuto a due altre differenze tra le scale medesime. La 
 prima si e, che laddove Diofanto denomina con singolarita Numero il nuinero ignoto, 
 denominando Monade il numero dato di comparazione : gli antichi italiani degli arabi 
 seguaci denominano questo il Numero ; e Radice, o Lato, o Cosa il numero sconosciuto. 
 La seconda e, che Diofanto comincia la scala dal numero ignoto ; e Fra Luca, Tartaglia, 
 Cardano la incominciano dal numero noto. Ecco le due scale di rincontro, onde meglio 
 risaltino all' occhio le differenze loro ", I. p. 195.
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 41 
 
 their multiplication^. Thus the "cube-cube" means in Diophantus 
 x?, while the Italian and Arabian system uses the expression " cube 
 of cube " and applies it to x 9 . The first system may (says Cossali) 
 be described as the method of representing each power by the 
 product of the two lesser powers which are the nearest to it, the 
 method of multiplication; the second the metliod of elevation, i.e. the 
 method which forms by the process of squaring and cubing all 
 powers which can be so formed, as the 4th, 6th, 8th, 9th, etc. 
 The intermediate powers which cannot be so formed are called 
 in Italian Relati. Thus the fifth power is Relate i, x 1 is 
 Relato 2, x 10 is Censo di Relato i, x" is Relato 3, and so on. 
 Another name for the Relati in use among European algebraists in 
 the 1 6th and I7th centuries was sursolida, with the variants super- 
 solida and surdesolida. 
 
 It is interesting to compare with these systems the Egyptian 
 method described by Psellus 2 . The next power after the fourth 
 (8vi>a/ioSiW/u9), i.e. x 6 , the Egyptians called " the first undescribed " 
 ((1X0709 here apparently meaning that of which no account can 
 be given), because it is neither a square nor a cube ; alternatively 
 they called it " the fifth number," corresponding to the fifth power 
 of x. The sixth power they apparently called " cube-cube " ; but 
 the seventh was " the second undescribed " (0X0709 Sevrepos), as 
 being the product of the square and the " first undescribed," or, 
 alternatively, the "seventh number." The eighth power was the 
 "quadruple-square" (rerpaTrXfj 8ui>a/u9), the ninth the "extended 
 cube " (/eu/8o9 eeXt*T09). Thus the " first undescribed " and the 
 "second undescribed" correspond to "Relato i" and "Relato 2" 
 respectively, but the "quadruple-square" exhibits the additive 
 principle. 
 
 For subtraction Diophantus uses a symbol. His full term for 
 negation or wanting is Xenjrt9, corresponding to inrap%i<; which 
 denotes the opposite. The symbol used to denote it in the MSS., 
 and corresponding to our for minus, is (Def. 9 KOI 7-^9 Xetye<9 
 eXXi7T9 Kara) vevov, A) " an inverted with the top 
 
 1 This statement of Cossali's needs qualification however. There is at least one Arabian 
 algebraist, al-Karkhi (died probably about 1029), the author of the Fakkri, who uses the 
 Diophantine system of powers of the unknown depending on the addition of exponents. 
 Al-Karkhl, namely, expresses all powers of the unknown above the third by means of 
 mat, his term for the square, and kab, his term for the cube of the unknown, as follows. 
 The fourth power is with him mdl mal, the fifth mal kab, the sixth kab ka'b, the seventh 
 mdl mal kab, the eighth mdl kab kab, the ninth kab kab kab, and so on. Among the 
 Italians too there was an exception, Leonardo of Pisa, who proceeded on the additive 
 principle (Bibliotheea Mathematica, vi s , 1905-6, p. 310). 2 Dioph. H. p. 37-38.
 
 42 INTRODUCTION 
 
 shortened, /ft." As Diophantus uses no distinct sign for +, it 
 is clearly necessary, in order to avoid confusion, that all the 
 negative terms in an expression should be placed together 
 after all the positive terms. And so in fact he does place them. 
 Thus corresponding to x 3 $*? + &tr i, Diophantus would write 
 
 K Y a 9 T; A^l 1 e Ma. With respect to this curious sign, given in 
 the MSS. as T and described as an inverted truncated M*", I believe 
 that I was the first to suggest that it could not be what it is 
 represented as being. Even when, as in Bachet's edition, the 
 sign was printed as ^ I could not believe that Diophantus used 
 so fantastic a sign for minus as an inverted truncated ty. In 
 the first place, an inverted ^ seems too far-fetched ; to one who 
 was looking for a symbol to express minus many others more 
 natural and less fantastic than jp> must have suggested themselves. 
 Secondly, given that Diophantus used an inverted M/", why should he 
 truncate it ? Surely that must have been unnecessary ; we could 
 hardly have expected it unless, without it, confusion was likely 
 to arise; but ./p. could not well have been confused with anything. 
 This very truncation itself appears to throw doubt on the description 
 of the symbol as we find it in the MS. I concluded that the con- 
 ception of this symbol as an inverted truncated M* was a mistake, 
 and that the description of it as such is not Diophantus' description, 
 but an explanation by a scribe of a symbol which he did not 
 understand 1 . I believe that the true explanation is the following. 
 Diophantus here took the same course as in the case of the other 
 symbols which we have discussed (those for apiOftos, Svvafjus, etc.). 
 As in those cases he took for his abbreviation the first letter of the 
 word with such an addition as would make confusion with numbers 
 impossible (namely the second letter of the word, which in each of 
 the cases happens to come later in the alphabet than the corre- 
 sponding first letter), so, in seeking an abbreviation for A,et-\Ja<? 
 and cognate inflected forms developed from \ITT, he began by 
 taking the initial letter of the word. The uncial 2 form is A. 
 Clearly A by itself would not serve his purpose, since it denotes 
 a number. Therefore an addition is necessary. The second letter 
 is E, but AE is equally a number. The second letter of the stem 
 
 1 I am not even sure that the description can be made to mean all that it is intended 
 to mean. AXiTr^s scarcely seems to be sufficiently precise. Might it not be applied to 
 /]\ with any part cut off, and not only the top ? 
 
 2 I adhere to the uncial form above for clearness' sake. If Diophantus used the 
 " Majuskelcursive " form, the explanation will equally apply, the difference of form being 
 for our purpose negligible.
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 43 
 
 \ITT is I, but Al is open to objection when so written. Hence 
 Diophantus placed the I inside the A, thus, A. Of the possibility 
 of this I entertain no doubt, because there are undoubted cases 
 of combination, even in uncial writing, of two letters into one sign. 
 I would refer in particular to X, which is an uncial abbreviation for 
 TAAANTON. Now this sign, A, is an inverted and truncated W 
 (written in the uncial form, y) ; and we can, on this assumption, 
 easily account for the explanation of the sign for minus which is 
 given in the text. 
 
 The above suggestion, made by me twenty-five years ago, 
 seems to be distinctly supported by what Tannery says of the form 
 in which the sign appears in the MSS. 1 Thus he remarks (i) that 
 the sign in the MSS. is often made to lean to the right so that it 
 resembles the letter Lambda, (2) that Planudes certainly wrote fc as 
 if he meant to write the first letter of Xen/ret, and (3) that the 
 letter A appears twice in A where it seems to mean XotTro?. Yet 
 in his edition of Diophantus Tannery did not adopt my explanation 
 or even mention it, but explained the sign as being in reality 
 adapted from the old letter Sampi (~>>), the objection to which 
 suggestion is the same as that to which the identification of <? with 
 Koppa is open, namely that ~^ represented the number 900, as ? 
 represented 90. Tannery however afterwards 2 saw reason to 
 abandon his suggestion that the symbol was originally an archaic 
 form of the Greek Sampi rather than "un monogramme se 
 rattachant a la racine de Xen/rt?." The occasion for this change 
 of view was furnished by the appearance of the same sign in the 
 critical notes to Schone's edition of the Metrica of Heron 3 , which 
 led Tannery to re-examine the evidence of the MSS. of Diophantus 
 as to the sign and as to the exact word or words which it re- 
 presented in different places, as well as to search for any similar 
 expressions denoting subtraction which might occur in the works 
 of other Greek mathematicians. In the MSS. of Diophantus, 
 when the sign is resolved by writing a full word instead of it, 
 it is generally resolved into Xen/ret, the dative of Xen/rt<? ; in such 
 cases the only grammatical possibility is to construct it with the 
 genitive case of the quantity subtracted, the meaning then being 
 "with the wanting, or deduction, of ...". But the best MS. (A) 
 
 1 Dioph. n. p. xli. 
 
 - Bibliotheca Mathematica v 3 , 1904-5, pp. 5-8. 
 
 3 Heronis Alexandrini opera, Vol. in., 1903, pp, 156, 8, 10. The MS. reading is 
 t'd', the meaning of which is 74 - T V
 
 44 INTRODUCTION 
 
 has in some places the nominative Xetyv 9, while in others it has the 
 symbol instead of parts of the verb \ei-7reiv, namely \nr(i)v or 
 XeiS/ra? and once even XITTOJO-I ; hence we may conclude that in the 
 cases where A and B^ have \etyei followed by the accusative (which 
 is impossible grammatically) the sign was wrongly resolved, and 
 the full word should have been a participle or other part of the 
 verb \eliretv governing the accusative. The question therefore 
 arises whether Diophantus himself used the dative \efyet at all 
 or whether it was introduced into the MSS. later. Certain it is 
 that the use is foreign to Classical Greek ; but, even if it began 
 with Diophantus, it did not finally hold the field before the time of 
 Planudes. No evidence for it can be found in Greek mathe- 
 maticians before Diophantus. Ptolemy has in two places Xenjrai/ 
 and \ei7rovaav respectively, followed by the accusative, and in 
 one case TO airo rrjs FA \ei<l>9ev VTTO TOV aVo -n)9 Zf (where the 
 meaning is ZP- FA 2 ). Consequently we cannot suppose that the 
 sign where it occurs in the Metrica of Heron represents the dative 
 Xenjr; it must rather stand for a participle, active or passive. 
 Tannery suggests that the full expression in that passage was 
 fjiovdSwv oB \ei<f)0evTo$ re(r<TapaKaiSKdrov, the participle being 
 passive and the construction being the genitive absolute ; but I 
 think a perhaps better alternative would be povaSmv 08 \en/rao-<wz/ 
 reo-a-apaKaiSeicaTov, where the active participle would govern 
 the accusative case of the term subtracted. From all this we 
 may infer that the sign had no exclusive reference to the sub- 
 stantive \enjrt9, still less to the dative case of that substantive, but 
 was a conventional abbreviation associated with the root of the 
 verb \ei7reiv. In these circumstances I think I may now fairly 
 claim Tannery as, substantially, a convert to my view of the 
 nature of the sfgn. 
 
 For division it often happens that no symbol is necessary, 
 i.e. in the cases where the divisor divides the dividend without 
 a remainder. In other cases the quotient has to be expressed 
 as a fraction, whether the divisor is a specific number or contains 
 the variable. The case of division comes then under that of 
 fractions. 
 
 Fractions are represented in different ways according as they are 
 submultiples (fractions with unity as numerator) or not. In the 
 case of submultiples the Greeks did not write the numerator, but 
 only the denominator, distinguishing the submultiple from the 
 cardinal number itself by affixing a certain sign. In more recent
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 45 
 
 MSS. a double accent was used for this purpose: thus 7" = ^. 
 Diophantus follows this plan in the hypothesis and analysis of his 
 problems, though in the solutions he seems to have written the 
 numerator a and assimilated the notation to that used for other 
 fractions. The sign, however, added to the cardinal number to 
 express the submultiple takes somewhat different forms in A : 
 sometimes it is a simple accent, sometimes more elaborate, as /" 
 above the letter and to the right, or actually forming a continuation 
 of the numeral sign, e.g. fr' = ^. Tannery adopts as the genuine 
 mark in Diophantus the affix x in place of the accent : thus 7 X = 3. 
 For he writes L ' as being most suitable for the time of Diophantus, 
 though A has <~*/, sometimes without the dot. 
 
 Of the other class of fractions (numerator not unity) f stands by 
 itself, having a peculiar sign of its own ; curiously enough it occurs 
 only four times in Diophantus. A has a sign for it which was 
 confused with that for dpidpos in one place ; Tannery judges from 
 the Greek mathematical papyrus of Achmlm 1 that its original form 
 was <y ; he himself writes in his text the common form iff'. In the 
 rare cases where the first hand in the oldest MS. (A) has fractions 
 as such with numerator and denominator written in full, the 
 denominator is written above the numerator. Tannery therefore 
 adopts, in his text, this way of writing fractions, separating the 
 
 numerator and denominator by a horizontal line: thus pica = ^. 
 
 PACT? 
 
 It is however better to omit the horizontal line (cf. p in Kenyon 
 Papyri II. No. cclxv. 40; also the fractions in Schone's edition 
 of Heron's Metrica). Once we find in the same MS. (A) in the first 
 hand the form ie s = 1 f-. In this latter method of writing fractions 
 the denominator is written as we write exponents ; and this is the 
 method adopted by Planudes and by Bachet in his edition. 
 Another alternative is to write the numerator first, and then the 
 denominator after it in the same line, marking the denominator with 
 the submultiple sign in some form ; thus jB' would mean ; this is 
 the most convenient method for purposes of printing. Or the de- 
 nominator may be written as an abbreviation for the ordinal number, 
 and the case-termination may be added higher up ; e.g. v K>f v = 50 
 twenty-thirds. But the denominators are nearly always omitted 
 
 1 Published by Baillet in Memoire s publih par Its Membres de la Mission archeologique 
 franfaise au Caire, T. Ix, Fascicule i, pp. 1-88. Paris, 1892.
 
 46 INTRODUCTION 
 
 altogether in the first hand of A ; in the first two Books B^ and the 
 second hand of A give the denominator in the place in which we write 
 an exponent, following the method of Planudes ; in the last four 
 Books both MSS. almost invariably omit the denominator. In 
 some cases the omission is not unnatural, i.e. where the denominator 
 has once been given, and it is almost superfluous to repeat it 
 in other fractions immediately following which have the same 
 denominator ; in other cases it was probably omitted because the 
 superposed denominator was taken by the copyist to be an inter- 
 linear scholium. A few examples of fractions from Diophantus 
 may be added : 
 
 (v. 9) ; 
 
 A ff> 
 
 ' 
 
 = (IV. 16) ; pKa-XS = (IV. 39) ; 
 
 (V . 2) . 
 
 152 V 
 
 Diophantus however often expresses fractions by putting eV 
 or popiov between the numerator and denominator, i.e. he 
 
 r 
 
 says one number divided by another. Cf. Mpiv . g^irb poplov 
 _ _ r 
 
 _ 
 
 *r . ,/8/3/iS = 1507984/262144 (IV. 28), where of course M = 
 
 (tens of thousands); (3 . ,e% ev ^opiw picfS . / aice= 25600/1221025 
 (v. 22). As we said, the most orthodox way of writing a sub- 
 multiple was to omit the numerator (unity) and use the denominator 
 with a distinguishing sign attached, e.g. r x or r' = . But in his 
 solutions Diophantus often uses the form applicable to fractions 
 
 W i 
 
 other than submultiples ; e.g. he writes a for - (IV. 28). 
 
 Numbers partly integral and partly fractional, where the 
 fraction is a submultiple or the sum of submultiples, are written 
 much as we write them, the fraction simply following the integer ; 
 e.g. a 7 X = i^ ; in the Lemma to V. 8 we have ft L ' r' = 2 \ % or 2, 
 where f is decomposed into submultiples as in Heron. Cf. also 
 (m. ii)roZ.V x =37oi T V. 
 
 Before leaving the subject of numerical notation, it may be 
 convenient to refer to the method of writing large numbers. 
 
 r 
 Myriads (tens of thousands) are expressed by M, myriads to the
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 47 
 
 second power by MM or, in words, Bevrepa fj.vpia<f. The de- 
 nominator 1 87474560 in V. 8 would thus be written /j,opiov 
 
 a teal [ivpi(i8(i)v TrpcaTfav fij^n^ ical M ,<>, and the fraction 
 131299224/1629586560 would be written Bevrepa pvpias a 
 
 M 0(TK?) fioptov Bevrepfav pup id Sow if 
 M~r&\ 
 
 But there is another kind of fraction, besides the purely 
 numerical one, which is continually occurring in the Arithmetica, 
 such fractions namely as involve the unknown quantity in some 
 form or other in their denominators. The simplest case is that in 
 which the denominator is merely a power of the unknown, 9. 
 Concerning fractions of this kind Diophantus says (Def. 3) : " As 
 fractions named after numbers have similar names to those of the 
 numbers themselves (thus a third is named from three, a fourth 
 from four), so the fractions homonymous with the numbers just 
 defined are called after them ; thus from a/>ifyto<? we name 
 the fraction dpidfioarov \t. ijx from x\, TO Swapoarov from 
 is, TO KvftoffTov from >ti>/8o?, TO BwafioBwapoa-rov from 
 , TO Bwa/AOKV/Soa'Tov from SvvafjLoicvftos, and TO 
 from KV&OHV&OS. And every such fraction shall 
 have, above the sign for the homonymous number, a line to 
 indicate the species." Thus we find, for example, IV. 3, ? x 17 cor- 
 responding to 8/x and, IV. 15, ? x Xe for 35/4:. Cf. J rx or for 250^. 
 
 Where the denominator is a compound expression involving the 
 unknown and its powers, Diophantus uses the expedient which he 
 often adopts with numerical fractions when the numerators and 
 denominators are large numbers, namely the insertion of ev popiy 
 or /jiopiov between the expressions for the numerator and de- 
 nominator. Thus in VI. 12 we have 
 
 = (6ar 2 + 
 and in VI. 14 
 
 For to-o?, equal, connecting the two sides of an equation, the 
 sign in the archetype seems to have been i ff but copyists intro- 
 
 1 Hultsch, he. tit.
 
 48 INTRODUCTION 
 
 duced a sign which was sometimes confused with the sign i| for 
 dpidpos ; this was no doubt the same abbreviation Lj as that shown 
 (with terminations of cases added above) in the list given at 
 the end of Codex Parisinus 2360 (Archimedes) of contractions 
 found in the " very ancient " MS. from which it was copied and 
 which was at one time the property of Georgius Valla 1 . 
 
 Diophantus evidently put down his equations in the ordinary 
 course of writing, i.e. they were written straight on, as are the 
 steps in the propositions of Euclid, and not put in separate lines for 
 each step in the process of simplification. In the scholia of 
 Maximus Planudes however we find conspectuses of the problems 
 with steps in separate lines which, except for the slightly more 
 cumbrous notation, make the work scarcely more difficult to follow 
 than it is in our notation 2 . Though in the MSS. we have the 
 abbreviation t " to denote equality, Bachet makes no use of any 
 symbol for the purpose in his Latin translation. He uses 
 throughout the full Latin word. It is interesting however to observe 
 that in the notes to his earlier translation (1575) Xylander had 
 already used a symbol to denote equality, namely ||, two short 
 vertical parallel lines. Thus we find, for example (p. 76), 
 
 which we should express by x* 1 + 12 =x*+ 6x+ 9. 
 
 Now that we have described in detail Diophantus' method of 
 expressing algebraical quantities and relations, it is clear that it is 
 essentially different in its character from the modern notation. 
 While in modern times signs and symbols have been developed 
 
 1 Heiberg, Quaestiones Archimedeae, p. 115. 
 
 2 One instance will suffice. On the left Planudes has abbreviations for the words 
 showing the nature of the steps or the operations they involve, e.g. &r0. = ftcflecns (setting - 
 out), rerp. = rer/scry wpwy^s (squaring), wuvO. ff^vOecrts (adding), a<f>. = d.<j>alpfffis (subtrac- 
 tion), pep. = fj.epiffiJ.6s (division), OTT. = virap% is (resulting fact). 
 
 Dioph. I. 28. 
 Planudes. " Modern equivalent. 
 
 (K&. 
 rerp. 
 
 atvd. A Y p(J.ff I" M ^? 
 
 A Y a 
 
 [Given numbers] 20, 208 
 Put for the numbers .*+ 10, 10 - * 
 Squaring, we have * 2 +2a*+ 100, 
 # 2 + 100-20*. 
 Adding, 2jr 2 +2OO = 2o8. 
 Subtracting, 2.* 2 =8. 
 Dividing, Jf 2 = 4. 
 
 Result: [the numbers are] 12, 8.
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 49 
 
 which have no intrinsic relationship to the things which they 
 represent, but depend for their use upon convention, the case 
 is quite different in Diophantus, where algebraic notation takes 
 the form of mere abbreviation of words which are considered as 
 pronounced or implied. 
 
 In order to show in what place, in respect of systems of 
 algebraic notation, Diophantus stands, Nesselmann observes that 
 we can, as regards the form of exposition of algebraic operations 
 and equations, distinguish three historical stages of development, 
 well marked and easily discernible, (i) The first stage Nessel- 
 mann represents by the name Rhetorical Algebra or "reckoning by 
 complete words." The characteristic of this stage is the absolute 
 want of all symbols, the whole of the calculation being carried on 
 by means of complete words, and forming in fact continuous prose. 
 As representatives of this first stage Nesselmann mentions lambli- 
 chus (of whose algebraical work he quotes a specimen in his fifth 
 chapter) "and all Arabian and Persian algebraists who are at 
 present known." In their works we find no vestige of algebraic 
 symbols; the same may be said of the oldest Italian algebraists 
 and their followers, and among them Regiomontanus. (2) The 
 second stage Nesselmann proposes to call the Syncopated A Igebra. 
 This stage is essentially rJielorical, and therein like the first in 
 its treatment of questions ; but we now find for often-recurring 
 operations and quantities certain abbreviational symbols. To 
 this stage belong Diophantus and, after him, all the later 
 Europeans until about the middle of the seventeenth century 
 (with the exception of Vieta, who was the first to establish, 
 under the name of Logistica speciosa, as distinct from Logistica 
 numerosa, a regular system of reckoning with letters denoting 
 magnitudes and not numbers only). (3) To the third stage 
 Nesselmann gives the name Symbolic Algebra, which uses a com- 
 plete system of notation by signs having no visible connexion 
 with the words or things which they represent, a complete language 
 of symbols, which supplants entirely the r/ietorical system, it being 
 possible to work out a solution without using a single word of the 
 ordinary written language, with the exception (for clearness' sake) 
 of a connecting word or two here and there, and so on 1 . Neither 
 
 1 It may be convenient to note here the beginnings of some of our ordinary algebraical 
 symbols. The signs + and - first appeared in print in Johann Widman's arithmetic 
 (1489), where however they are scarcely used as regular symbols of operation ; next they 
 are found in the Rechenbuch of Henricus Grammateus (Schreiber), written in 1518 but 
 perhaps not published till 15.21, and then regularly in Stifel's Arithmetica Integra (1544)
 
 5 
 
 INTRODUCTION 
 
 is it the Europeans from the middle of the seventeenth century 
 onwards who were the first to use symbolic forms of Algebra. 
 In this they were anticipated by the Indians. 
 
 Nesselmann illustrates these three stages by three examples, 
 quoting word for word the solution of a quadratic equation 
 by Muhammad b. Musa as an example of the first stage, and 
 the solution of a problem from Diophantus as representing the 
 second. 
 
 First Stage, Example from Muhammad b. Musa (ed. Rosen, 
 p. 5). "A square and ten of its roots are equal to nine and thirty 
 dirhems, that is, if you add ten roots to one square, the sum is equal 
 to nine and thirty. The solution is as follows. Take half the number 
 of roots, that is in this case five; then multiply this by itself, and 
 the result is five and twenty. Add this to the nine and thirty, 
 which gives sixty-four; take the square root, or eight, and subtract 
 from it half the number of roots, namely five, and there remain 
 three: this is the root of the square which was required, and the 
 square itself is nine 1 ." 
 
 Here we observe that not even are symbols used for numbers, 
 so that this example is even more "rhetorical" than the work of 
 lamblichus who does use the Greek symbols for his numbers. 
 
 as well as in his edition of RudolfFs Coss (1553). Vieta (1540-1603) has, in addition, 
 = for ~. Robert Recorde (1510-1558) had already in his Algebra (The Whetstone of 
 Witte, 1557) used =(but with much longer lines) to denote equality (" bicause noe.2. 
 thynges, can be moare equalle"). Harriot (1560-1621) denoted multiplication by a dot, 
 and also by mere juxtaposition of letters; Stifel (1487-1567) had however already 
 expressed the product of two magnitudes by the juxtaposition of the two letters represent- 
 ing them. Oughtred (1574-1660) used the sign x for multiplication. Harriot also 
 introduced the signs > and < for greater and less respectively, -f- for division is found 
 in Rahn's Algebra (1659). Descartes introduced in his Geometry (1637) our method of 
 writing powers, as a 3 , a* etc. (except a 2 , for which he wrote aa) ; but this notation was 
 practically anticipated by Pierre Herigone (Cours matktmatique, 1634), who wrote ai, a$, 
 04, etc., and the idea is even to be found in the Rechenbuch of Grammateus above 
 mentioned, where the successive powers of the unknown are denoted by pri, se, ter, etc. 
 The use of x for the unknown quantity began with Descartes, who first used 2, then y, and 
 then x for this purpose, showing that he intentionally chose his unknowns from the last 
 letters of the alphabet. ^/ for the square root is traceable to Rudolff, with whom it had 
 only two strokes, the first (down) stroke being short, and the other relatively long. 
 1 Thus Muhammad b. Musa states in words the following solution. 
 
 ^+10^ = 39, 
 x z + 10^+25=64; 
 therefore x + 5 = 8,
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 51 
 
 Second Stage. As an example of Diophantus I give a trans- 
 lation word for word of II. 8. So as to make the symbols correspond 
 exactly I use 5 (Square) for J r (Svvafus), N (Number) for 9, U 
 (Units) for M (^o^aSe?). 
 
 "To divide the proposed square into two squares. Let it be 
 proposed then to divide 16 into two squares. And let the first be 
 supposed to be \S\ therefore the second will be 16 U 1 5. Thus 
 16 U i 5 must be equal to a square. I form the square from any 
 number of N's minus as many (7's as there are in the side of 
 1 6 /'s. Suppose this to be 2/V ^U. Thus the square itself will 
 be 4S i6U- \6N. These are equal to i6U- I S. Add to each 
 the negative term (77 Xen/rt?, the deficiency) and take likes from 
 likes. Thus 56" are equal to i6N, and the N is 16 fifths. One 
 [square] will be ^, and the other l -gg, and the sum of the two 
 makes up 4 ^, or \6U, and each of the two is a square." 
 
 Of the third stage any exemplification is unnecessary. 
 
 To the form of Diophantus' notation is due the fact that he 
 is unable to introduce into his solutions more than one unknown 
 quantity. This limitation has made his procedure often very dif- 
 ferent from our modern work. In the first place we can begin 
 with any number of unknown quantities denoted by different 
 symbols, and eliminate all of them but one by gradual steps in the 
 course of the work ; Diophantus on the other hand has to perform 
 all his eliminations beforehand, as a preliminary to the actual 
 work, by expressing every quantity which occurs in the problem 
 in terms of only one unknown. This is the case in the great 
 majority of questions of the first Book, which involve the solu- 
 tion of determinate simultaneous equations of the first degree 
 with two, three, or four variables; all these Diophantus expresses 
 in terms of one unknown, and then proceeds to find it from a 
 simple equation. Secondly, however, this limitation affects much of 
 Diophantus' work injuriously; for, when he handles problems which 
 are by nature indeterminate and would lead with our notation to an 
 indeterminate equation containing two or three unknowns, he is 
 compelled by limitation of notation to assume for one or other of 
 these some particular number arbitrarily chosen, the effect of the 
 assumption being to make the problem a determinate one. How- 
 ever, it is but fair to say that Diophantus, in assigning an arbitrary 
 value to a quantity, is careful to tell us so, saying, "for such and 
 such a quantity we put any number whatever, say such and such a 
 
 42
 
 52 INTRODUCTION 
 
 number." Thus it can hardly be said that there is (as a rule) any 
 loss of generality. We may say, then, that in general Diophantus is 
 obliged to express all his unknowns in terms, or as functions, of 
 one variable. He compels our admiration by the clever devices 
 by which he contrives so to express them in terms of his single 
 unknown, 9, as to satisfy by that very expression of them all 
 conditions of the problem except one, which then enables us to 
 complete the solution by determining the value of 9. Another 
 consequence of Diophantus' want of other symbols besides 9 to 
 express more variables than one is that, when (as often happens) 
 it is necessary in the course of a problem to work out a subsidiary 
 problem in order to obtain the coefficients etc. in the functions of 9 
 which express the numbers to be found, the unknown quantity 
 which it is the object of the new subsidiary problem to find is also 
 in its turn denoted by the same symbol 9 ; hence we often have 
 in the same problem the same variable 9 used with two different 
 meanings. This is an obvious inconvenience and might lead to 
 confusion in the mind of a careless reader. Again we find two 
 cases, II. 28 and 29, where for the proper working-out of the 
 problem two unknowns are imperatively necessary. We should of 
 course use x andjp; but Diophantus calls the first 9 as usual; the 
 second, for want of a term, he agrees to call "one unit" i.e. I. 
 Then, later, having completed the part of the solution necessary to 
 find 9, he substitutes its value, and uses 9 over again to denote 
 what he had originally called " I " the second variable and so 
 finds it. This is the most curious case of all, and the way in which 
 Diophantus, after having worked with this " I " along with other 
 numerals, is yet able to put his finger upon the particular place 
 where it has passed to, so as to substitute 9 for it, is very remark- 
 able. This could only be possible in particular cases such as those 
 which I have mentioned; but, even here; it seems scarcely possible 
 now to work out the problem by using x and I for the variables 
 as originally taken by Diophantus without falling into confusion. 
 Perhaps, however, in working out the problems before writing them 
 down as we have them Diophantus may have given the " I " which 
 stood for a variable some mark by which he could recognise it 
 and distinguish it from other numbers. 
 
 Diophantus will have in his solutions no numbers whatever 
 except "rational" numbers; and in pursuance of this restriction he 
 excludes not only surds and imaginary quantities, but also negative 
 quantities. Of a negative quantity per se, i.e. without some positive
 
 NOTATION AND DEFINITIONS OF DIOPHANTUS 53 
 
 quantity to subtract it from, Diophantus had apparently no con- 
 ception. Such equations then as lead to surd, imaginary, or 
 negative roots he regards as useless for his purpose: the solution 
 is in these cases aSui/aro?, impossible. So we find him (v. 2) 
 describing the equation 4 = 4^+20 as aroTros, absurd, because it 
 would give x = ^. Diophantus makes it his object throughout 
 to obtain solutions in rational numbers, and we find him frequently 
 giving, as a preliminary, the conditions which must be satisfied in 
 order to secure a result rational in his sense of the word. In the 
 great majority of cases, when Diophantus arrives in the course of 
 a solution at an equation which would give an irrational result, he 
 retraces his steps and finds out how his equation has arisen, and 
 how he may, by altering the previous work, substitute for it 
 another which shall give a rational result. This gives rise, in 
 general, to a subsidiary problem the solution of which ensures 
 a rational result for the problem itself. Though, however, Dio- 
 phantus has no notation for a surd, and does not admit surd 
 results, it is scarcely true to say that he makes no use of quadratic 
 equations which lead to such results. Thus, for example, in V. 30 
 he solves such an equation so far as to be able to see to what 
 integers the solution would approximate most nearly.
 
 CHAPTER IV 
 
 DIOPHANTUS' METHODS OF SOLUTION 
 
 BEFORE I give an account in detail of the different methods 
 which Diophantus employs for the solution of his problems, so far 
 as they can be classified, it is worth while to quote some remarks 
 which Hankel has made in his account of Diophantus 1 . Hankel, 
 writing with his usual brilliancy, says in the place referred to, "The 
 reader will now be desirous to become acquainted with the classes 
 of indeterminate problems which Diophantus treats of, and with 
 his methods of solution. As regards the first point, we must observe 
 that included in the 130 (or so) indeterminate problems, of which 
 Diophantus treats in his great work, there are over 50 different 
 classes of problems, strung together on no recognisable principle 
 of grouping, except that the solution of the earlier problems facili- 
 tates that of the later. The first Book is confined to determinate 
 algebraic equations; Books II. to v. contain for the most part 
 indeterminate problems, in which expressions involving in the first 
 or second degree two or more variables are to be made squares or 
 cubes. Lastly, Book VI. is concerned with right-angled triangles 
 regarded purely arithmetically, in which some linear or quadratic 
 function of the sides is to be made a square or a cube. That is all 
 that we can pronounce about this varied series of problems without 
 exhibiting singly each of the fifty classes. Almost more different 
 in kind than the problems are their solutions, and we are completely 
 unable to give an even tolerably exhaustive review of the different 
 turns which his procedure takes. Of more general comprehensive 
 methods there is in our author no trace discoverable : every 
 question requires a quite special method, which often will not 
 serve even for the most closely allied problems. It is on that 
 
 1 Zur Geschichte der Mathematik in Alterthum und Mittelaller, Leipzig, 1874, 
 pp. 164-5.
 
 DIOPHANTUS' METHODS OF SOLUTION 55 
 
 account difficult for a modern mathematician even after studying 
 100 Diophantine solutions to solve the icist problem; and if we 
 have made the attempt, and after some vain endeavours read 
 Diophantus' own solution, we shall be astonished to see how 
 suddenly he leaves the broad high-road, dashes into a side-path 
 and with a quick turn reaches the goal, often enough a goal with 
 reaching which we should not be content; we expected to have 
 to climb a toilsome path, but to be rewarded at the end by an 
 extensive view; instead of which our guide leads by narrow, 
 strange, but smooth ways to a small eminence; he has finished! 
 He lacks the calm and concentrated energy for a deep plunge 
 into a single important problem ; and in this way the reader also 
 hurries with inward unrest from problem to problem, as in a 
 game of riddles, without being able to enjoy the individual one. 
 Diophantus dazzles more than he delights. He is in a wonderful 
 measure shrewd, clever, quick-sighted, indefatigable, but does not 
 penetrate thoroughly or deeply into the root of the matter. As 
 his problems seem framed in obedience to no obvious scientific 
 necessity, but often only for the sake of the solution, the solution 
 itself also lacks completeness and deeper signification. He is a 
 brilliant performer in the art of indeterminate analysis invented by 
 him, but the science has nevertheless been indebted, at least directly, 
 to this brilliant genius for few methods, because he was deficient 
 in the speculative thought which sees in the True more than the 
 Correct. That is the general impression which I have derived from 
 a thorough and repeated study of Diophantus' arithmetic." 
 
 It might be inferred from these remarks of Hankel that 
 Diophantus' object was less to teach methods than to obtain a 
 multitude of mere results. On the other hand Nesselmann 
 observes 1 that Diophantus, while using (as he must) specific 
 numbers for numbers which are " given " or have to be arbitrarily 
 assumed, always makes it clear how by varying our initial as- 
 sumptions we can obtain any number of particular solutions of 
 the problem, showing "that his whole attention is directed to 
 the explanation of the method, to which end numerical examples 
 only serve as means"; this is proved by his frequently stopping 
 short, when the method has been made sufficiently clear, and 
 the remainder of the work is mere straightforward calculation. 
 The truth seems to be that there is as much in the shape of general 
 
 1 Algebra der Griechen, pp. 308-9.
 
 56 INTRODUCTION 
 
 methods to be found in Diophantus as his notation and the nature 
 of the subject admitted of. On this point I can quote no better 
 authority than Euler, who says 1 : " Diophantus himself, it is true, 
 gives only the most special solutions of all the questions which he 
 treats, and he is generally content with indicating numbers which 
 furnish one single solution. But it must not be supposed that his 
 method was restricted to these very special solutions. In his 
 time the use of letters to denote undetermined numbers was not 
 yet established, and consequently the more general solutions which 
 we are now enabled to give by means of such notation could not 
 be expected from him. Nevertheless, the actual methods which he 
 uses for solving any of his problems are as general as those which 
 are in use today; nay, we are obliged to admit that there is 
 hardly any method yet invented in this kind of analysis of which 
 there are not sufficiently distinct traces to be discovered in Dio- 
 phantus." 
 
 In his 8th chapter, entitled "Diophantus' treatment of equations 2 ," 
 Nesselmann gives an account of Diophantus' solutions of (i) Deter- 
 minate, (2) Indeterminate equations, classified according to their 
 kind. In chapter 9, entitled "Diophantus' methods of solution 3 ," 
 he classifies these " methods " as follows 4 : (i) " The adroit assump- 
 tion of unknowns," (2) "Method of reckoning backwards and 
 auxiliary questions," (3) "Use of the symbol for the unknown in 
 different significations," (4) "Method of Limits," (5) "Solution by 
 mere reflection," (6) "Solution in general expressions," (7) "Arbi- 
 trary determinations and assumptions," (8) "Use of the right- 
 angled triangle." 
 
 At the end of chapter 8 Nesselmann observes that it is not 
 his solutions of equations that we have to wonder at, but the art, 
 amounting to virtuosity, which enabled Diophantus to avoid such 
 equations as he could not technically solve. We look (says Nessel- 
 mann) with astonishment at his operations, when he reduces the 
 most difficult problems by some surprising turn to a quite simple 
 
 1 Nffvi Commentarii Academiae Petropolitanae, 1756-7, Vol. VI. (1761), p. 1 55 = Com- 
 mentationes arithmtticae collectae (ed. Fuss), 1849, I. p. 193. 
 
 2 " Diophant's Behandlung der Gleichungen." 
 
 3 " Diophant's Auflbsungsmethoden." 
 
 4 (r) "Die geschickte Annahme der Unbekannten," (i) " Methode der Zuriick- 
 rechnung und Nebenaufgabe," (3) " Gebrauch des Symbols fur die Unbekannte in 
 verschiedenen Bedeutungen," (4) "Methode der Grenzen," (5) " Auflosung durch blosse 
 Reflexion," (6) "Auflosung in allgemeinen Ausdriicken," (7) " Willkiihrliche Bestim- 
 mungen und Annahmen," (8) "Gebrauch des rechtwinkligen Dreiecks."
 
 DIOPHANTUS' METHODS OF SOLUTION 57 
 
 equation. Then, when in the pth chapter Nesselmann passes to the 
 "methods," he prefaces it by saying: "To give a complete picture 
 of Diophantus' methods in all their variety would mean nothing else 
 than copying his book outright. The individual characteristics of 
 almost every problem give him occasion to try upon it a peculiar 
 procedure or found upon it an artifice which cannot be applied to 
 any other problem.... Mean while, though it may be impossible to 
 exhibit all his methods in any short space, yet I will try to describe 
 some operations which occur more often or are particularly re- 
 markable for their elegance, and (where possible) to bring out 
 the underlying scientific principle by a general exposition and by 
 a suitable grouping of similar cases under common aspects or 
 characters." Now the possibility of giving a satisfactory account of 
 the methods of Diophantus must depend largely upon the meaning 
 we attach to the word "method." Nesselmann's arrangement seems 
 to me to be faulty inasmuch as (i) he has treated Diophantus' 
 solutions of equations, which certainly proceeded on fixed rules, 
 and therefore by "method" separately from what he calls "methods 
 of solution," thereby making it appear as though he did not 
 look upon the "treatment of equations" as "methods"; (2) the 
 classification of the "Methods of solution" seems unsatisfactory. 
 Some of the latter can hardly be said to be methods of solution at 
 all; thus the third, " Use of the symbol for the unknown in different 
 significations," might be more justly described as a "hindrance to 
 the solution"; it is an inconvenience to which Diophantus was 
 subjected owing to the want of notation. -Indeed, on the as- 
 sumption of the eight "methods," as Nesselmann describes them, 
 it is really not surprising that no complete account of them 
 could be given without copying the whole book. To take the 
 first, "the adroit assumption of unknowns." Supposing that a 
 number of essentially different problems are proposed, the differences 
 make a different choice of an unknown in each case absolutely 
 necessary. That being so, how could a rule be given for all cases? 
 The best that can be done is to give a number of typical instances. 
 Precisely the same remark applies to "methods" (2), (5), (6), (7). 
 The case of (4), " Method of Limits," is different ; here we have 
 a " method " in the true sense of the term, Le. in the sense of an 
 instrument for solution. And accordingly in this case the method 
 can be exhibited, as I hope to show later on; (8) also deserves 
 to some extent the name of a " method."
 
 58 INTRODUCTION 
 
 In one particular case, Diophantus formally states a method or 
 rule ; this is his rule for solving what he calls a " double-equation," 
 and will be found in II. 1 1, where such an equation appears for the 
 first time. Apart from this, we do not find in Diophantus' work 
 statements of method put generally as book-work to be applied to 
 examples. Thus we do not find the separate rules and limitations 
 for the solution of different kinds of equations systematically 
 arranged, but we have to seek them out laboriously from the 
 whole of his work, gathering scattered indications here and there, 
 and to formulate them in the best way that we can. 
 
 I shall now attempt to give a short account of those methods 
 running through Diophantus which admit of general statement. 
 For the reasons which I have stated, my arrangement will be 
 different from that of Nesselmann ; I shall omit some of the heads 
 in his classification of "methods of solution"; and, in accordance 
 with his remark that these "methods" can only be adequately 
 described by a transcription of the entire work, I shall leave them 
 to be gathered from a perusal of my reproduction of Diophantus' 
 book. 
 
 I shall begin my account with 
 
 I. DIOPHANTUS' TREATMENT OF EQUATIONS. 
 
 This subject falls naturally into two divisions: (A) Determinate 
 equations of different degrees, (B) Indeterminate equations. 
 
 (A) Determinate equations. 
 
 Diophantus was able without difficulty to solve determinate 
 equations of the first and second degrees; of a cubic equation we 
 find in his Arithmetica only one example, and that is a very 
 special case. The solution of simple equations we may pass over; 
 we have then to consider Diophantus' methods of solution in the 
 case of (i) Pure equations, (2) Adfected, or mixed, quadratics. 
 
 (i) Pure determinate equations. 
 
 By pure equations I mean those equations which contain only 
 one power of the unknown, whatever the degree. The solution is 
 effected in the same way whatever the exponent of the term in the
 
 DIOPHANTUS' METHODS OF SOLUTION 59 
 
 unknown; and Diophantus treats pure equations of any degree 
 as if they were simple equations of the first degree. 
 
 He gives a general rule for this case without regard to the 
 degree 1 : "If a problem leads to an equation in which any terms 
 are equal to the same terms but have different coefficients, we must 
 take like from like on both sides, until we get one term equal to 
 one term. But, if there are on one side or on both sides any negative 
 terms, the deficient terms must be added on both sides until all the 
 terms on both sides are positive. Then we must take like from like 
 until one term is left on each side." After these operations have 
 been performed, the equation is reduced to the form Ax m = B and 
 is considered solved. The cases which occur in Diophantus are 
 cases in which the value of x is found to be a rational number, 
 integral or fractional. Diophantus only recognises one value of x 
 which satisfies this equation; thus, if m is even, he gives only the 
 positive value, excluding a negative value as "impossible." In the 
 same way, when an equation can be reduced in degree by dividing 
 throughout by any power of x, the possible values, x=o, thus 
 arising are not taken into account. Thus an equation of the form 
 x* = ax, which is of common occurrence in the earlier part of the 
 book, is taken to be merely equivalent to the simple equation x=a. 
 
 It may be observed that the greater proportion of the problems 
 in Book I. are such that more than one unknown quantity is sought. 
 Now, when there are two unknowns and two conditions, both 
 unknowns can easily be expressed in terms of one symbol. But, 
 when there are three or four quantities to be found, this reduction 
 is much more difficult, and Diophantus shows great adroitness in 
 effecting it: the ultimate result being that it is only necessary 
 to solve a simple equation with one unknown quantity. 
 
 (2) Mixed quadratic equations. 
 
 After the remarks in Def. 1 1 upon the reduction of equations 
 until we have one term equal to another term, Diophantus 
 adds 2 : "But we will show you afterwards how, in the case also 
 when two terms are left equal to a single term, such an equation 
 can be solved." That is to say, he promises to explain the 
 solution of a mixed quadratic equation. In the Arithmetica, 
 as we possess the book, this promise is not fulfilled. The first 
 
 1 Def. ii. 
 
 * fcrrepo*' & eroi Sfl&nev /tot TWJ Suo eiSwv tffuv evl KaraXfi^vruv TO TOIOVTOV Xi/erot.
 
 60 INTRODUCTION 
 
 indications we have on the subject are a number of cases in which 
 the equation is given, and the solution written down, or stated to 
 be rational, without any work being shown. Thus 
 
 (IV. 22) x* = 4JF 4, therefore x = 2 ; 
 
 (IV. 31) 325^ = 3^+18, therefore x = $fc or &; 
 
 (vi. 6) 84*2 + 7.^ = 7, whence x=\; 
 
 (vi. 7) 84*2 - 7-*-= 7, hence .* = ; 
 
 (VI. 9) 630.^2 - fix = 6, therefore .ar = -^ ; 
 
 and (vi. 8) 630.^2 + 73^ = 6, and x is rational. 
 
 These examples, though proving that Diophantus had somehow 
 arrived at the result, are not in themselves sufficient to show that 
 he was necessarily acquainted with a regular method for the 
 solution of quadratics ; these solutions might (though their variety 
 makes it somewhat unlikely) have been obtained by mere trial. 
 That, however, Diophantus' solutions of mixed quadratics were not 
 merely empirical is shown by instances in V. 30. In this problem 
 he shows that he could approximate to the root in cases where it is 
 not " rational." As this is an important point, I give the substance 
 of the passage in question: "This is not generally possible unless 
 we contrive to make x. > | (x- 60) and < \ (x^ 60). Let then 
 x* 60 be > 5*, but x* 60 < $>x. Since then x z -6o> $x, let 60 be 
 added to both sides, so that & > $x + 60, or x*=$x + some number 
 > 60; therefore x must not be less than n." In like manner 
 Diophantus concludes that "x* = Sx+ some number less than 60 ; 
 therefore x must be found to be not greater than 1 2." 
 
 Now, solving for the positive roots of these two equations, we 
 have 
 
 x > 2 (5 + V265) and x < 4 + ^76, 
 or x> 10*6394. .. and x< 127177.... 
 
 It is clear that x may be < 1 1 or > 1 2, and therefore Dio- 
 phantus' limits are not strictly accurate. As however it was 
 doubtless his object to find integral limits, the limits u and 12 
 are those which are obviously adapted for his purpose, and are 
 a fortiori safe. 
 
 In the above equations the other roots obtained by prefixing 
 the negative sign to the radical are negative and therefore would 
 be of no use to Diophantus. In other cases of the kind occurring
 
 DIOPHANTUS' METHODS OF SOLUTION 61 
 
 in Book V. the equations have both roots positive, and we have to 
 consider why Diophantus took no account of the smaller roots in 
 those cases. 
 
 We will take first the equations in V. 10 where the inequalities 
 to be satisfied are 
 
 17 ........................ (i). 
 
 19 ........................ (2). 
 
 Now, if a, /8 be the roots of the equation 
 
 x- px + q = o (p, q both positive), 
 and if a > fi, then 
 
 (a) in order that x" 2 - px + q may be > o 
 
 we must have x> a or < /3, 
 and (b) in order that x* px + q may be < o 
 
 we must have x < a and > /9. 
 (i) The roots of the equation 
 
 J2X+ 17 =o 
 
 36 + V 1007 , 6773... ,4-26... 
 
 are - -- - ; that is, ' ' J and - ; 
 
 17 17 17 
 
 and, in order that ijx* 72^+ 17 may be.< o, we must have 
 
 (2) The roots of the equation 
 
 19=0 
 
 
 that . 66-577... and 
 
 19 19 19 
 
 and, in order that ic^tr 2 J2X + 19 may be > o, we must have 
 
 x> 66^77^ Qr< 5-422^ 
 19 19 
 
 Diophantus says that x must not be greater than f| or less than 
 ff . These are again doubtless intended to be a fortiori limits ; 
 but ff should have been f, and the more correct way of stating the 
 case would be to say that, if x is not greater than ff and not less 
 than \\, the given conditions are a fortiori satisfied. 
 
 Now consider what alternative (if any) could be obtained, on 
 Diophantus' principles, if we used the lesser positive roots of the
 
 62 INTRODUCTION 
 
 equations. If, like Diophantus, we were to take a fortiori limits, 
 we should have to say 
 
 ^< T 5 gbut>^, 
 
 which is of course an impossibility. Therefore the smaller roots 
 are here useless from his point of view. 
 
 This is, however, not so in the case of another pair of in- 
 equalities, used later in V. 30 for finding an auxiliary x, namely 
 
 X* + 60 > 22X, 
 
 The roots of the equation 
 
 X* 22X + 6O = O 
 
 are ii + V6i ; that is, i8'8i... and 3'i8...; 
 and the roots of the equation x* - 2<\x + 60 = o 
 
 are 12 + \/84; that is, 2ri6... and 2-83.... 
 In order therefore to satisfy the above inequalities we must have 
 
 x> i8'8i ... or < 3'i8..., 
 and x< 2i'i6 ... but > 2 - 83. 
 
 Diophantus, taking a fortiori integral limits furnished by the 
 greater roots, says that x must not be less than 19 but must be 
 less than 21. But he could also have obtained from the smaller 
 roots an integral value of x satisfying the necessary conditions, 
 namely the value x = 3 ; and this would have had the advantage 
 of giving a smaller value for the auxiliary x than that actually 
 taken, namely 2O 1 . Accordingly the question has been raised 2 
 whether we have not here, perhaps, a valid reason for believing 
 that Diophantus only knew of the existence of roots obtained by 
 using the positive sign with the radical, and was unaware of the 
 solution obtained by using the negative sign. But in truth we 
 can derive no certain knowledge on this point from Diophantus' 
 treatment of the particular equations in question. Thus, e.g., if he 
 chose to use the first of the two equations 
 
 17, 
 19, 
 for the purpose of obtaining an upper limit only, and the second 
 
 1 This is remarked by Loria (Le scienze esatte delP antica Grecia, V. p. 128). 
 But in fact, whether we take 20 or 3 as the value of the auxiliary unknown, we get 
 the same value for the original .r of the problem. For the original x has to be found 
 from x*-6o=(x-m)* where m is the auxiliary.*; and we obtain x= n whether we 
 put .r 2 - 60 = (.*- 2 o) 2 or x' 2 -6o = (x-3) 2 . 
 
 2 Loria, op. cit. p. 129.
 
 DIOPHANTUS' METHODS OF SOLUTION 63 
 
 for the purpose of obtaining a lower limit only, he could only use 
 the values obtained by using the positive sign. Similarly, if, with 
 
 the equations 
 
 x 2 + 6o> 22.x, 
 
 x* + 60 < 24*, 
 
 he chose to use the first in order to find a lower limit only, and 
 the second in order to find an upper limit only, it was not open to 
 him to use the values corresponding to the negative sign 1 . 
 
 For my part, I find it difficult or impossible to believe that 
 Diophantus was unaware of the existence of two real roots in 
 such cases. The numerical solution of quadratic equations by the 
 Greeks immediately followed, if it did not precede, their geometrical 
 solution. We find the geometrical equivalent of the solution of 
 a quadratic assumed as early as the fifth century B.C., namely by 
 Hippocrates of Chios in his Quadrature of limes*, the algebraic form 
 of the particular equation being ^ 2 + Vf -ax = a' i . The complete 
 geometrical solution was given by Euclid in VI. 27-29: and the 
 construction of VI. 28 corresponds in fact to the negative sign 
 before the radical in the case of the particular equation there 
 solved, while a quite obvious and slight variation of the con- 
 struction would give the solution corresponding to the positive sign. 
 In VI. 29 the solution corresponds to the positive sign before the 
 radical; in the case of the equation there dealt with the other sign 
 would not give a "real" solution 3 . It is true that we do not find 
 the negative sign taken in Heron any more than in Diophantus, 
 though we find Heron 4 stating an approximate solution of the 
 equation 
 
 x ( 1 4 - x) = 6/2O/ 144, 
 
 without showing how he arrived at it; x is, he says, approximately 
 equal to 8. It is clear however that Heron already possessed 
 a scientific method of solution. Again, the author of the so-called 
 Geometry of Heron 5 practically states the solution of the equation 
 
 212 
 
 r A/ ( 154 X 212 + 841) 20 
 
 in the form x= v - - -, 
 
 ii 
 
 1 Enestrom in Bibliotheca Mathematica IX 3 , 1908-9, pp. 71-2. 
 
 2 Simplicius, Comment, in Aristot. Phys., ed. Diels, p. 64, t8; Rudio, Der Bericht 
 des Simplicius iiber die Quadraluren des Antiphon und des Hippokrates, 1907, p. 58,8-11. 
 
 3 The Thirteen Books of Euclid's Elements, Cambridge, 1908, Vol. II. pp. 257-267. 
 
 4 Heron, Metrica, ed. Schdne, pp. 148-151. The text has 8 as the approximate solu- 
 tion, but the correction is easy, as the inference immediately drawn is that \\-x=\. 
 
 6 Heron, ed. Hultsch, p. 133, 10-23. See M. Cantor, Geschichte der Math. I 3 , p. 405.
 
 64 INTRODUCTION 
 
 showing pretty clearly the rule followed after the equation had 
 been written in the form 
 
 121^ + 638^ = 212 x 154. 
 
 We cannot credit Diophantus with less than a similar uniform 
 method ; and, if he did not trouble to give two roots where both 
 were real, this seems quite explicable when it is remembered that he 
 did not write a text-book of algebra, and that his object through- 
 out is to obtain a single solution of his problems, not to multiply 
 solutions or to show how many can be found in each case. 
 
 In solving such an equation as 
 
 ax 1 bx + c = o, 
 
 it is our modern practice to divide out by a in order to make the 
 first term a square. It does not appear that Diophantus divided 
 out by a\ rather he multiplied by a so as to bring the equation 
 into the form 
 
 cPx? abx + ac = o ; 
 then, solving, he found 
 
 and wrote the solution in the form 
 
 a 
 
 wherein his method corresponds to that of the Pseudo-Heron above 
 referred to. 
 
 From the rule given in Def. 1 1 for removing by means of addition 
 any negative terms on either side of an equation and taking equals 
 from equals (operations called by the Arabians aljabr and almukd- 
 bala) it is clear that, as a preliminary to solution, Diophantus so 
 arranged his equation that all the terms were positive. Thus, 
 from his point of view, there are three cases of mixed quadratic 
 equations. 
 
 Case I. Form mx?+px=q; the root is 
 
 m 
 
 according to Diophantus. An instance is afforded by vi. 6. Dio- 
 phantus namely arrives at the equation 6x* + yc = 7, which, if it is 
 to be of any service to his solution, should give a rational value 
 of x ; whereupon he says " the square of half the coefficient 1 of x 
 
 1 For " coefficient " Diophantus simply uses ir\rj6os, multitude or number ; thus 
 "number of dpiO/ioi " = coeff. of x. The absolute term is described as the "units."
 
 DIOPHANTUS' METHODS OF SOLUTION 65 
 
 together with the product of the absolute term and the coefficient 
 of x? must be a square number ; but it is not," i.e. \jp + mq, or in 
 this case (|) 2 + 42, must be a square in order that the root may be 
 rational, which in this case it is not. 
 
 Case 2. Form mx* =px + q. Diophantus takes 
 
 An example is IV. 39, where 2^ 2 >6>+i8. Diophantus says: 
 "To solve this take the square of half the coefficient of x, i.e. 9, and 
 the product of the absolute term and the coefficient of x*, i.e. 36. 
 Adding, we have 45, the square root 1 of which is not 2 < 7. Add 
 half the coefficient of x, [and divide by the coefficient of x*\ ; whence 
 x is not < 5." Here the form of the root is given completely ; and 
 the whole operation of finding it is revealed. Cf. IV. 31, where 
 Diophantus remarks that the equation 5^r 2 =3^r+ 18 "is not rational. 
 But 5, the coefficient of x*, is a square plus I, and it is necessary 
 that this coefficient multiplied by the 18 units and then added to 
 the square of half the coefficient of x, namely 3, that is to say 2$, 
 shall make a square." 
 
 Case 3. Form mx* + q =px. Diophantus' root is 
 
 Cf. in V. 10 the equation already mentioned, 1 7^r 2 + 1 7 
 Diophantus says: " Multiply half the coefficient of x into itself and 
 we have 1296; subtract the product of the coefficient of x* and the 
 absolute term, or 289. The remainder is 1007, the square root of 
 which is not 3 > 31. Add half the coefficient of x, and the result is 
 not > 67. Divide by the coefficient of ;r 2 , and x is not > 67/17." 
 Here again we have the complete solution given. Cf. VI. 22, where, 
 having arrived at the equation 172^=336^+24, Diophantus 
 remarks that "this is not always possible, unless half the coefficient 
 of x multiplied into itself, minus the product of the coefficient of x* 
 and the units, makes a square." 
 
 For the purpose of comparison with Diophantus' solutions of 
 quadratic equations we may refer to a few of his solutions of 
 
 1 The " square root" is with Diophantus TrXeupd, or "side." 
 
 2 7, though not accurate, is clearly the nearest integral limit which will serve the 
 purpose. 
 
 3 As before, the nearest integral limit. 
 
 H. D. 5
 
 66 INTRODUCTION 
 
 (3) Simultaneous equations involving quadratics. 
 Under this heading come the pairs of equations 
 
 
 I use Greek letters to distinguish the numbers which the 
 problem requires us to find from the one unknown which Dio- 
 phantus uses and which I shall call x. 
 
 In the first two of the above problems, he chooses his x thus. 
 Let, he says, 
 
 Then it follows, by addition and subtraction, that 
 
 = a + x, r) = a x. 
 Consequently, in I. 27, 
 
 7 = (a + x) (a -x) = a>-x* = B, 
 and x is found from this "pure" quadratic equation. 
 
 If we eliminate from the original equations, we have 
 
 if - 2ai) + B = o, 
 which we should solve by completing the square (a T/) 2 , whence 
 
 (a-^^tf-B, 
 which is Diophantus' ultimate equation with a tj for x. 
 
 Thus Diophantus' method corresponds here again to the ordi- 
 nary method of solving a mixed quadratic, by which we make it 
 into a pure quadratic with a different x. 
 
 In I. 30 Diophantus puts + i\ = 2;r, and the solution proceeds 
 in the same way as in I. 27. 
 
 In I. 28 the resulting equation in x is 
 
 x?= 2 (a* +* 2 ) =B. 
 
 (4) Cubic equation. 
 
 There is no ground for supposing that Diophantus was acquainted 
 
 with the algebraical solution of a cubic equation. It is true that there 
 
 is one cubic equation to be found in the Arithmetica, but it is only 
 
 a very particular case. In VI. 17 the problem leads to the equation 
 
 x* + 2x + 3 = x s + ix - 3** - i,
 
 DIOPHANTUS' METHODS OF SOLUTION 67 
 
 and Diophantus says simply "whence x is found to be 4." All 
 that can be said of this is that, if we write the equation in true 
 Diophantine fashion, so that all the terms are positive, 
 
 x 3 + x = 4* 2 + 4. 
 This equation being clearly equivalent to 
 
 Diophantus no doubt detected the presence of the common factor 
 on both sides of the equation. The result of dividing by it 
 is x= 4, which is Diophantus' solution. Of the other two roots 
 x= v'( i) no account is taken, for the reason stated above. 
 
 It is not possible to judge from this example how far Dio- 
 phantus was acquainted with the solution of equations of a degree 
 higher than the second. 
 
 I pass now to the second general division of equations. 
 
 (B) Indeterminate equations. 
 
 As I have already stated, Diophantus does not, in his 
 Arithmetica as we have it, treat of indeterminate equations of the 
 first degree. Those examples in Book I. which would lead to such 
 equations are, by the arbitrary assumption of a specific value for 
 one of the required numbers, converted into determinate equations. 
 Nor is it likely that indeterminate equations of the first degree 
 were treated of in the lost Books. For, as Nesselmann observes, 
 while with indeterminate quadratic equations the object is to obtain 
 a rational result, the whole point in solving indeterminate simple 
 equations is to obtain a result in integral numbers. But Diophantus 
 does not exclude fractional solutions, and he has therefore only to 
 see that his results are positive, which is of course easy. Inde- 
 terminate equations of the first degree would therefore, from 
 Diophantus' point of view, have no particular significance. We 
 take therefore, as our first division, indeterminate equations of 
 the second degree. 
 
 (a) Indeterminate equations of tJte second degree. 
 The form in which these equations occur in Diophantus is 
 invariably this: one or two (but never more) functions of the 
 unknown quantity of the form Ax* + Bx + C or simpler forms 
 are to be made rational square numbers by finding a suitable 
 value for x. Thus the most general case is that of solving one or 
 two equations of the form Ax* + Bx + C=y*. 
 
 52
 
 68 INTRODUCTION 
 
 (i) Single equation. 
 
 The single equation takes special forms when one or more of 
 the coefficients vanish or satisfy certain conditions. It will be well 
 to give in order the different forms as they can be identified in 
 Diophantus, premising that for " = j 2 " Diophantus simply uses the 
 formula ia~ov rerpayoovo), " is equal to a square," or iroiel rerpdytavov, 
 " makes a square." 
 
 I. Equations which can always be solved rationally. This is 
 the case when A or C or both vanish. 
 
 Form Bx=y*. Diophantus puts for j 2 any arbitrary square 
 number, say m*. Then x = m*\B. 
 
 Ex. III. 5: 2x=y z , y is assumed to be 16, and ;r= 8. 
 
 Form Bx+ C=y*. Diophantus puts for y* any square nP, and 
 x=(m*C)IB. He admits fractional values of x, only taking 
 care that they are "rational," i.e. rational and positive. 
 
 Ex. III. 6: 6x+ i =y* = 121, say, and x= 20. 
 
 Form Ax z + Bx=y*. Diophantus substitutes for y any multiple 
 of x, as x; whence Ax + B = ^ x, the factor x disappearing and 
 
 the root x = o being neglected as usual. Thus x - -^. 
 
 Exx. II. 21: 43?+ yc =jj/ 2 = ($xf, say, and x = f . 
 II. 33 : i6x* + *]x =y* = ($x)*, say, and x = $. 
 
 2. Equations which can only be rationally solved if certain 
 conditions are fulfilled. 
 
 The cases occurring in Diophantus are the following. 
 
 Form Ax* + C=y*. This can be rationally solved according to 
 Diophantus 
 
 (a) When A is positive and a square, say 2 . 
 Thus cPx* + C=y i . In this case f is put = (ax mf ; 
 therefore a^x- + C = (ax mf, 
 
 ^- 
 and x= + , 
 
 " 2ma 
 
 (m and the doubtful sign being always assumed so as to give x 
 a positive value). 
 
 (/?) When C is positive and a square number, say c*. 
 Thus Ax z + c* =_/. Here Diophantus puts y = (mx c) ;
 
 DIOPHANTUS' METHODS OF SOLUTION 69 
 
 therefore Ax* + * = (mx cj, 
 
 2mc 
 
 and x = + -; -- , . 
 
 ~ A m* 
 
 (7) When one solution, is known, any number of other 
 solutions can be found. This is enunciated in the Lemma to vi. 1 5 
 thus, though only for the case in which C is negative: "Given two 
 numbers, if, when one is multiplied by some square and the other 
 is subtracted from the product, the result is a square, then another 
 square also can be found, greater than the aforesaid square which 
 has the same property." It is curious that Diophantus does not 
 give a general enunciation of this proposition, inasmuch as not 
 only is it applicable to the cases + Ax* C=y 3 , but also to the 
 general form Ax- + Bx + C=y\ 
 
 Diophantus' method of finding other greater values of x satisfy- 
 ing the equation Ax*- C=y* when one such value is known is as 
 follows. 
 
 Suppose that x is the value already known and that q is the 
 corresponding value of y. 
 
 Put ;r = jF + in the original expression, and equate it to 
 (q k%)-, where k is some integer. 
 
 Since A(x. + &-C=(q-k& t 
 
 it follows (because by hypothesis Ax C = q*) that 
 
 2 (Ax + kq) 
 whence g= #_ 
 
 2(A 
 and *.+ 
 
 In the second Lemma to VI. 12 Diophantus does prove that the 
 equation Ax* + C=y* has an infinite number of solutions when 
 A +C is a square, i>. in the particular case where the value x= I 
 satisfies the equation. But he does not always bear this in mind; 
 for in III. 10 the equation S2x* + I2=y 2 is regarded as impossible of 
 solution although 52+12 = 64, a square, and a rational solution is 
 therefore possible. Again in III. 12 the equation 266r 2 10 =y* is 
 regarded as impossible though x = i satisfies it. 
 
 The method used by Diophantus in the second Lemma to 
 VI. 12 is like that of the Lemma to VI. 15. 
 
 Suppose that A + C = q 11 . 
 
 Put i + f for x in the original expression Ax* + C, and equate it 
 to (q kgf, where k is some integer.
 
 70 INTRODUCTION 
 
 Thus A(i + ?+C=(q-k%}\ 
 
 and it follows that 2 (A + kq} = ? (k- -A), 
 
 so that 
 
 It is of course necessary to choose k? such that & > A. 
 
 It is clear that, if X G satisfies the equation, C is a square, and 
 therefore this case (7) includes the previous case (yS). 
 
 It is to be observed that in VI. 14 Diophantus says that a rational 
 solution of the equation 
 
 Ax* - & = j/ 8 
 
 is impossible unless A is the sum of two squares. 
 
 [In fact, if x-=p\q satisfies the equation, and Ax*- c 2 = &, 
 we have Ap* = c*q* + k-q-, 
 
 Lastly, we have to consider 
 
 Form Ax* + Bx+C=y*. 
 
 This equation can be reduced by means of a change of variable 
 to the preceding form wanting the second term. For, if we put 
 
 D 
 
 x = z -- -j , the transformation gives 
 
 Diophantus, however, treats ' this form of the equation quite 
 separately from the other and less fully. According to him the 
 rational solution is only possible in the following cases. 
 
 (a) When A is positive and a square, or the equation is 
 
 tfx n - + Bx + C=y\ 
 Diophantus then puts y* = (ax mf, whence 
 
 = ?-c ^ Exx 22 ^ 
 
 2am + B 
 (#) When C is positive and a square, or the equation is 
 
 Ax* + Bx + c* = y\ 
 Diophantus puts_y 2 = (c mxf, whence 
 
 (Exx.IV.8, 9 etc.)
 
 DIOPHANTUS' METHODS OF SOLUTION 71 
 
 (7) When ^B* - AC is positive and a square number. Dio- 
 phantus never expressly enunciates the possibility of this case; 
 but it occurs, as it were unawares, in IV. 31. In that problem 
 
 is to be made a square. To solve this Diophantus assumes 
 
 which leads to the quadratic $x+ 18 $x a = o; but "the equation 
 is not rational." Accordingly the assumption 4** will not do; 
 "and we must find a square [to replace 4] such that 18 times 
 (this square + i) +(f) 2 may be a square." Diophantus then solves 
 the auxiliary equation i8(? 2 + 1) + =f y finding m=i8. He 
 then assumes 3*+ 18 -^ a = (i8) 2 ^ 2 , which gives $2$x* 3*- 18 = 0, 
 "and x=-^, that is ^.' Jl 
 
 1 With this solution should be compared the much simpler solution of this case given 
 by Euler (Algebra, tr. Hewlett, 1840, Part n. Arts. 50-53), depending on the separation 
 of the quadratic expression into factors. (Curiously enough Diophantus does not separate 
 quadratic expressions into their factors except in one case, vi. 19, where however his 
 purpose is quite different : he has made the sum of three sides of a right-angled triangle 
 4^ + 6^ + 2, which has to be a cube, and, in order to simplify, he divides throughout by 
 x+ i, which leaves 4^ + 2 to be made a cube.) 
 
 Since \EP-AC is a square, the roots of the quadratic Ax* + Bx + C=o are real, and 
 the expression has two real linear factors. Take the particular case now in question, 
 where Diophantus actually arrives at 3*+ 18- x* as the result of multiplying 6-x and 
 3+jr, but makes no use of the factors. 
 
 We have 3*+ i8-x n - = (6-x) (3+*). 
 
 Assume then (6 - x) (3 + x) =^ (6 - *) 2 , 
 
 and we have / J (6-* 
 
 where/, q may be any numbers subject to the condition that ip*>q*. If ^ = 9, q*= 16, 
 we have Diophantus' solution x = . 
 
 In general, if Ax? + x+C= (f+gx) (A+Jtx), 
 
 we can put (f+gx) (h + kx) = (f+gx)*, 
 
 whence ? 2 (A + >br) =/(/+*), 
 
 
 
 This case, says Euler, leads to a fourth case in which Ax 3 + Bx + C=y* can be solved, 
 though neither A nor C is a square, and though ^-^Cis not a square either. The 
 fourth case is that in which Ax* + Bx + C is the sum of two parts, one of which is a square 
 and the other is the product of two factors linear in x. For suppose 
 
 Ax* + Bx +C=Z* + XY, 
 where Z=dx + e, X=/x+g, Y=hx + k.
 
 7 2 
 
 INTRODUCTION 
 
 It is worth observing that from this example of Diophantus we 
 can deduce the reduction of this general case to the form 
 
 Ax* + C=y* 
 wanting the middle term. 
 
 Assume, with Diophantus, that Ax* + Bx+ C=m 2 x 2 : therefore 
 by solution we have 
 
 A-m 2 
 
 and x is rational provided that IB* AC+ Cm* is a square. This 
 condition can be fulfilled if \B* - AC is a square, by the pre- 
 
 We can then put Z 2 + X Y= ( Z + ^ JfY, 
 
 whence F=2^Z + 4/A', 
 
 q q> 
 
 , 
 
 that is, x (p 2 f+ ipqd -fA) = kq* - ipqe -p 2 g. 
 
 Ex. i. Equation ix 2 - i=y 2 . 
 Put 2X*-i 
 
 * A 
 
 Therefore x- i = it-x+^(x + i), 
 
 and x (f + ipq -g*)=-(p* + q\ 
 
 As x 2 is alone found in our equation, we can take either the positive or negative sign 
 and we may put 
 
 Ex. 2. Equation ix 2 + 'i=y 2 . 
 Here we put 
 
 Equating this to -J2 + - (x+ i)| , 
 
 P P* , 
 we have i(x- i) = 4 <- + -(*+ i), 
 
 1 ? 2V 
 or x (/> - iq 2 ) =-(iqt+4j>q +/), 
 
 It is to be observed that this method enables us to solve the equation 
 
 Ax' 1 -c 2 =y 2 
 
 whenever it can be solved rationally, i.e. whenever A is the sum of two squares (d 2 + e 2 , 
 say). For then 
 
 Ax 2 - 2 = d^x 2 + (ex - c) (ex + c). 
 
 In cases not covered by any of the above rules our only plan is to try to discover one 
 solution empirically. If one solution is thus found, we can find any number of others; if 
 we cannot discover such a solution by trial (even after reducing the equation to the 
 simplest form A'x' 2 + C=y a ), recourse must be had to the method of continued fractions 
 elaborated by Lagrange (cf. Oeuvres, II. pp. 377-535 and pp. 655 726 ; additions to 
 Euler's Algebra).
 
 DIOPHANTUS' METHODS OF SOLUTION 73 
 
 ceding case. If \B' i AC is not a square, we have to solve 
 (putting, for brevity, D for \B* AC) the equation 
 
 D + Cm 2 =y\ 
 and the reduction is effected. 
 
 (2) Double-equation. 
 
 By the name "double-equation" Diophantus denotes the pro- 
 blem of finding one value of the unknown quantity which will make 
 two different functions of it simultaneously rational square numbers. 
 The Greek term for the "double-equation" occurs variously as SwrXoi'- 
 croT?;?, St7rX?7 tVor??? or 8frr\rj to-axrt?. We have then to solve the 
 equations 
 
 mx 2 + a.x + a = u 2 } 
 
 nx 2 + fix + & = w 2 } 
 
 in rational numbers. The necessary preliminary condition is that 
 each of the two expressions can severally be made squares. This 
 is always possible when the first term (in x 2 ) is wanting. This is 
 the simplest case, and we shall accordingly take it first. 
 
 I. Double-equation of the first degree. 
 
 Diophantus has one general method of solving the equations 
 
 ax + a = u 2 ' 
 
 taking slightly different forms according to the nature of the 
 coefficients. 
 
 (a) First method of solution of 
 
 ax + a = u 2 
 
 This method depends upon the identity 
 
 If the difference between the two expressions in x can be separated 
 into two factors p, q t the expressions themselves are equated 
 to {^(p+q)} 2 respectively. Diophantus himself (II. u) states his 
 rule thus. 
 
 "Observing the difference [between the two expressions], seek 
 two numbers such that their product is equal to this difference; 
 then equate either the square of half the difference of the two 
 factors to the lesser of the expressions or the square of half the 
 sum to the greater."
 
 74 INTRODUCTION 
 
 We will take the general case and investigate to what particular 
 classes of cases the method is applicable, from Diophantus' point 
 of view, remembering that his cases are such that the final quadratic 
 equation in x always reduces to a simple equation. 
 
 Take the equations 
 
 Q.X + a = 
 
 Subtracting, we have 
 
 (a - ) x + (a - b) = if - w\ 
 
 We have then to separate (a $)x + (a-b} into two factors; 
 let these be/, {(a - /3)* + (a - b]}lp. 
 We write accordingly 
 
 . 
 
 u w = - 
 
 P 
 
 u + w 
 Thus u 2 
 
 4 
 
 therefore {(a-@)x + a-6 +/ 2 } 2 = 4/ 2 (ax + a), 
 
 or (a - /3) 2 ;tr 2 + 2x {(a - ff) (a - b +/ 2 ) - 2p" a}+(a-b +/ 2 ) 2 - 4/ 2 =o, 
 that is, (a - /S) 2 * 2 + 2x {(a -ft) (a- b} -/ 2 (a + )} 
 
 + (a- by -2^(a + b} +^ = o. 
 
 Now, in order that this equation may reduce to a simple 
 equation, either 
 
 (i) The coefficient of x" must vanish, so that 
 
 -A 
 
 or (2) The absolute term must vanish, that is, 
 
 or /- 
 
 so that ab must be a square number. 
 
 Therefore either a and b are both squares, in which case we 
 may substitute & and d* for them respectively, / being then equal 
 to c d, or the ratio of a to b is the ratio of a square to a square. 
 
 With respect to (i) we observe that on one condition it is not 
 necessary that a /8 should vanish, z>. provided we can, before 
 solving the equations, make the coefficients of x the same in both 
 expressions by multiplying either equation or both equations by 
 some square number, an operation which does not affect the 
 problem, since a square multiplied by a square is still a square.
 
 DIOPHANTUS' METHODS OF SOLUTION 75 
 
 In other words, it is only necessary that the ratio of a to @ should 
 be the ratio of a square to a square 1 . 
 
 Thus, if a/^ = m 2 /n 2 or an?=@m*, the equations can be solved 
 by multiplying them respectively by n z and m*; we can in fact 
 solve the equations 
 
 like the equations 
 
 tax + a = u' 
 a.x + b = w 
 
 in an infinite number of ways. 
 Again, the equations under (2) 
 
 ax + c 2 = u 2 
 
 can be solved in two different ways according as we write them in 
 this form or in the form 
 
 obtained by multiplying them respectively by d*, c 1 *, in order that 
 the absolute terms may be equal. 
 
 I shall now give those of the possible cases which we find solved 
 in Diophantus' own work. These are equations 
 (i) of the form 
 
 b = 
 
 1 Diophantus actually states this condition in the solution of iv. 32 where, on arriving 
 at the equations 
 
 he says : " And this is not rational because the coefficients of x have not to one another 
 the ratio which a square number has to a square number." 
 
 Similarly in the second solution of III. 15 he states the same condition along with an 
 alternative condition, namely that a has to b the ratio of a square to a square, which is 
 the second condition arrived at under (2) above. On obtaining the equations 
 
 Diophantus observes "But, since the coefficients in one expression are respectively greater 
 than those in the other, neither have they (in either case) the ratio which a square number 
 has to a square number, the hypothesis which we took is useless." 
 Cf. also iv. 39 where he says that the equations 
 
 are possible of solution because there is a square number of units in each expression.
 
 76 INTRODUCTION 
 
 a case which includes the more common one where the coefficients 
 of x in both are equal \ 
 
 (2) of the form 
 
 ax + c* = u 2 } 
 
 solved in two different ways according as they are written in this 
 form or in the alternative form 
 
 General solution of Form ( i ) or 
 
 am 2 x + a = u 2 
 a.n 2 x + b = w 2 \' 
 
 Multiply by n 2 , m 2 respectively, and we have to solve the 
 equations 
 
 am 2 n' 2 x+an 2 = tt' 2 \ 
 
 The difference is an 2 dm 2 ; suppose this separated into two 
 factors p, q. 
 
 Let u' w' =/, 
 
 u' 4- w = q ; 
 
 therefore u' 2 = ^(p + qj, w '* i (P #) 2 > 
 
 and a m 2 n 2 x + an 2 = i (/ + q) 2 , 
 
 or a.m 2 n 2 x + bm 2 \(p q)*. 
 
 Either equation gives the same value of x, and 
 
 a.m 2 n 2 
 since pq = an 2 bm 2 . 
 
 Any factors /, q may be chosen provided that the resulting 
 value of x is positive. 
 
 Ex. from Diophantus : 
 
 65 - 
 
 65-24* 
 
 therefore , 260 - 24^ = u' 2 } 
 
 65 - 24^ = w 2 j ' 
 
 The difference = 195 = 15 . 13, say; 
 therefore (15 I3) 2 = 65 24^; that is, 24^=64, and 
 
 =* 2 l. 
 =^}'
 
 DIOPHANTUS' METHODS OF SOLUTION 77 
 
 
 
 General solution (first method) of Form (2), or 
 ax + & = u ' z \ 
 /3x + d* = w' 2 )' 
 
 In order to solve by this method, we multiply by d 2 , c 2 
 respectively and write 
 
 u being supposed to be the greater. 
 
 The difference = (ad* - fic^x. Let the factors of this be/*, q. 
 Therefore u 2 = ( px + qf, 
 
 w> = \(px-qf. 
 Thus x is found from the equation 
 
 This equation gives 
 
 fx* + 2x (pq - 2ad 2 ) + q^-^d- = O, 
 or, since pq = (ad 2 fie 2 ), 
 
 p*x* - 2x (ad 2 + &<*) + f-4c 2 d 2 ^ o. 
 
 In order that this may reduce to a simple equation, as Dio- 
 phantus requires, the absolute term must vanish, 
 
 and q 2cd. 
 
 Thus our method in this case furnishes us with only one solution 
 of the double-equation, q being restricted to the value 2cd, and the 
 solution is 
 
 _ 2 (ad 2 + &") 
 
 Ex. from Diophantus. This method is only used in one par- 
 ticular case (IV. 39), where c* = d* as the equations originally stand, 
 the equations being 
 
 6x + 4 = 
 
 The difference is 2x, and q is necessarily taken to be 2\/4, or 4; 
 the factors are therefore ^x, 4. 
 
 Therefore %x + 4 = \ (%x + 4) 2 , 
 
 and x= 112. 
 
 General solution (second method) of Form (2) or 
 
 ax + c 2 =
 
 78 INTRODUCTION 
 
 The difference = (a - /8) x + (<? - d*). 
 
 Let the factors of this be p, {(a - /3) ^ + c z - 
 
 Then, as before proved (p. 74), / must be equal to (c d). 
 
 Therefore the factors are 
 
 and we have finally 
 
 which equation gives two possible values for x. Thus in this case 
 we can find by our method two values of x, since one of the factors 
 p may be either (c + d) or (c - d). 
 
 Ex. from Diophantus. To solve the equations 
 
 (ill. 15.) 
 
 The difference is here 5^+ 5, and Diophantus chooses as the 
 factors 5, x+ i. This case therefore corresponds to the value 
 c + d of/. The solution is given by 
 
 (\x + 3) 2 = \QX + 9, whence x = 28. 
 
 The other value, c d, of / is in this case excluded, because it 
 would lead to a negative value of x. 
 
 The possibility of deriving any number of solutions of a double- 
 equation when one solution is known does not seem to have been 
 noticed by Diophantus, though he uses the principle in certain 
 cases of the single equation (see above, pp. 69, 70). Fermat was the 
 first, apparently, to discover that this might always be done, if one 
 value a of x were known, by substituting x + a for x in the equa- 
 tions. By this means it is possible to find a positive solution, even 
 if a is negative, by successive applications of the principle. 
 
 But, nevertheless, Diophantus had certain peculiar artifices by 
 which he could arrive at a second value. One of these artifices 
 (which is made necessary in one case by the unsuitableness of the 
 value of x found by the ordinary method) gives a different way of 
 solving a double-equation from that which has been explained, and 
 is used only in one special case (IV. 39).
 
 DIOPHANTUS' METHODS OF SOLUTION 79 
 
 () Second method of solving a double-equation of the first 
 degree. 
 
 Consider only the special case 
 
 Take these expressions, and 2 , and write them in order of 
 magnitude, denoting them for convenience by A, B, C. 
 
 A -B f , A-B 
 Therefore ~ -^ =-j , and _ _ 
 
 Suppose now that Jix + 2 = (y + )* ; 
 therefore hx = y* + 2ny, 
 
 and 
 
 or 
 
 thus it is only necessary to make this expression a square. 
 Assume therefore that 
 
 and any number of values for y, and therefore for x, can be found, 
 by varying/. 
 
 Ex. from Diophantus (the only one), IV. 39. 
 
 In this case there is the additional condition of a limit to the 
 value of x. The double-equation 
 
 6x + 4 = 
 has to be solved in such a manner that x < 2. 
 
 Here -~= , and B is taken 1 to be (y + 2f. 
 
 Therefore A - B = $ (f + 47) ; 
 
 therefore A 
 
 which must be made a square. 
 
 1 Of course Diophantus uses the same variable JT where I have for clearness used y. 
 Then, to express what I have called m later, he says: "I form a square from 3 minus 
 some number of JT'S, and x becomes some number multiplied by 6 and then added to 12, 
 divided by the difference by which the square of the number exceeds 3."
 
 8o INTRODUCTION 
 
 If we multiply by f , we must make 
 
 3j 2 + 1 2y + 9 = a square, 
 where y must be < 2. 
 Diophantus assumes 
 
 6m + 12 
 
 whence y = , 
 
 *'- 3 
 
 and the value of m is then taken such as to make j> < 2. 
 
 It is in a note on this problem that Bachet shows that the 
 double-equation 
 
 ax + a = 
 
 can be rationally solved by a similar method provided that the 
 coefficients satisfy either of two conditions, although none of the 
 coefficients are squares and neither of the ratios a : /3 and a : b is 
 equal to the ratio of a square to a square. Bachet's conditions are : 
 
 (1) That, when the difference between the two expressions 
 is multiplied or divided by a suitably-chosen number, and the 
 expression thus obtained is subtracted from the smaller of the 
 original expressions, the result is a square number, or 
 
 (2) That, when the difference between the two expressions 
 is multiplied or divided by a suitably-chosen number, and the 
 smaller of the two original expressions is subtracted from the 
 expression obtained by the said multiplication or division, the 
 result is a number bearing to the multiplier or divisor the ratio 
 of a square to a square 1 . 
 
 1 Bachet of course does not solve equations in general expressions (his notation does 
 not admit of this), but illustrates his conditions by equations in which the coefficients are 
 specific numbers. I will give one of his illustrations of each condition, and then set the 
 conditions out more generally. 
 
 Case (i). Equations 
 
 difference 2.r + 6 
 
 The suitably-chosen number (to divide by in this case) is 2 ; 
 
 \ (difference) =x + 3, 
 
 and (lesser expression) - | (difference) = x + 7 - (x + 3) = 4, that is, a square. 
 
 We have then to find two squares such that 
 
 their difference 2 (difference between lesser and 4). 
 Assume that the lesser = (y + 2) 2 , 2 being the square root of 4. 
 Therefore (greater square) = 3 (lesser) - 8
 
 DIOPHANTUS' METHODS OF SOLUTION 81 
 
 2. Double-equation of the second degree, 
 or the general form 
 
 These equations are much less thoroughly treated in Diophan- 
 tus than those of the first degree. Only such special instances 
 
 To make 3;c 2 + 127 + 4 a square we put 
 
 3J 2 +I2J|/ + 4 = (2-/^) 2 , 
 
 where p must lie between certain limits which have next to be found. The equation gives 
 
 In order that y may be positive, fp must be > 3 ; and in order that the second of the 
 original expressions, assumed equal to (jy+2) 2 , may be greater than 7 (it is in fact x+j), 
 we must have 0'+2)>2f (an a fortiori limit, since 2 >/7), or_y>|. 
 
 Therefore 4/ + 12 > f (/ 2 - 3) , 
 
 Suppose that 3/ 2 =i6/ + 53|, which gives/ = 7f. 
 
 Therefore / < 7$, but / 2 > 3. 
 
 Put / = 3 in the above equation ; therefore 
 
 3/+t2;/ + 4 = (2-37)2, 
 
 and j/ = 4 . 
 
 Therefore .r=0/ + 2) 2 - 7 = 29. 
 
 Case (2). Equations 6#+25 = 2 | 
 
 wH-j^f ; 
 
 difference 4^+22 
 
 The suitable-number (again to divide by) in this case is 2 ; 
 
 \ (difference) = ix + 1 1, 
 
 and (difference) - (lesser expression) = 8=2.4, 
 
 where 2 is the divisor used, and 4 is the ratio of a square to a square. 
 Hence two squares have to be found such that 
 
 (their difference) = 2 (sum of lesser and 8). 
 If the lesser is j 2 , the greater is 3jj/ 2 + i6 = (4 -pyf 1 , say. 
 Bachet gives, as limits for /, 
 
 /Mil /*>3 
 and puts p = 3. This gives 7 = 4, so that jr= 6 \. 
 
 Let us now state Bachet's conditions generally. 
 Suppose the equations to be 
 
 The difference is (a - /3) * + (a - b). 
 This has to be multiplied by ^ which is the "suitable" factor in this case, and, if 
 we subtract the product from fix + b, we obtain 
 
 ab-afi 
 
 '-^<*-*>. - T^T-
 
 82 INTRODUCTION 
 
 occur as can be easily solved by the methods which we have 
 described for equations of the first degree. 
 
 The following types are found. 
 
 (i) p 
 
 The difference is (a /3) x + (a b\ and, following the usual 
 course, we may, e.g., resolve this into the factors 
 
 (1) The first of Bachet's conditions is that 
 
 ab-aS 
 a _ B = a square =/ 2 /? 2 , say. 
 
 (2) The second condition is that 
 
 aB-ab _p ft 
 
 ; a ratio of a square to a square. 
 
 It is to be observed that the first of these conditions can be obtained by considering 
 the equation 
 
 obtained on page 74 above. 
 
 Diophantus only considers the cases in which this equation reduces to a simple 
 equation ; but the solution of it as a mixed quadratic gives a rational value of x provided 
 that 
 
 { (a - j8) (a - b) -f (a + B) } 2 - (a - 0) 2 { (a - ) 2 - 2/> 2 (a + b) +/ 4 } is a square, 
 that is, if 
 
 ^{(a + ^-(a-^}+^{(a + l>)(a-^-(a 2 -^)(a-/>)} is a square, 
 which reduces to a/3/ 2 + (a-/3) (ab-aB)~ a square ........................ (A). 
 
 This can be solved (cf. p. 68 above), if 
 ab-aB 
 
 is a square. (Bachet's first condition.) 
 
 Again take Bachet's second condition 
 
 aB - ab r* 
 
 p = a square = - z say, 
 
 and substitute fir 2 /* 2 for a/3 - a/> in the equation (A) above. 
 Therefore a/3/ 2 - (a - ft) /3 -^ = a square, 
 
 or aj3/' 2 -- (a - 18) /3 = a square. 
 
 This is satisfied by p'=i; therefore (p. 69) any number of other solutions can 
 be found . 
 
 The second condition can also be obtained directly by eliminating x from the equations 
 ax + a = 2 1 
 
 for the result is ^ w 2 + -*-y- = 2 > 
 
 which can be rationally solved if 
 
 aB - a/'
 
 DIOPHANTUS' METHODS OF SOLUTION 83 
 
 <"- 
 
 as usual, we put 
 
 or 
 
 In order that x may be rational a condition is necessary; thus 
 x is rational if 
 
 = 1 *~ 8 
 
 This is the case in the only instance of the type where a is not 
 equal to b, namely (III. 13) 
 
 the difference is i6x+4, and the resolution of this into the factors 
 4, ^x + i solves the problem. 
 
 In the other cases of the type a = b\ the difference is then 
 (a ft}x, which is resolved into the factors 
 
 ' 
 
 ifa B \ 2 
 
 and we put p 2 x 2 + ax + a = - 4 2px , 
 
 4\ 2p r j 
 
 
 4\ 2p 
 
 whence x = 
 
 Exx. from Diophantus : 
 
 x z - 
 and ^ + 
 
 (2) The second type found in Diophantus 1 is 
 x 2 + ax + a = u- ) 
 
 0x + a = zv*)' 
 where one equation has no term in x*, and p = i, a = b. 
 
 1 It is perhaps worth noting that the method of the "double-equation " has a distinct 
 advantage in this type of cose. The alternative is to solve by the method of Euler (who 
 does not use the " double-equation "), i.e. to put the linear expression equal to/* and then, 
 substituting the value of JT (in terms of/) in the quadratic expression, to solve the 
 
 62
 
 84 INTRODUCTION 
 
 The difference x* + (*$)x is resolved into the factors 
 
 and we put &x + a = | (a - 
 
 which gives x. 
 
 resulting equation in /. But the difficulties would generally be great. Take the case of 
 vi. 6 where 
 
 > have to be made squares. 
 
 If 
 
 (fp |)2 
 
 therefore x' 2 + i = - -- h i has to be made a square, 
 
 or / 4 - 2/ 2 + 1 97 = a square. 
 
 This does not admit of solution unless we could somehow discover empirically one 
 value of / which would satisfy the requirement, and this would be very difficult. 
 
 Let us take an easier case for solution by this method, 
 
 which is solved by Euler (Algebra, Part II. Art. 222), and let us compare the working of 
 the two methods in this case. 
 
 I. Enter's method. Assuming x+ i ==/> 2 and substituting^ 2 i for x in the quadratic 
 expression, we have 
 
 / 4 - 2/J 2 + 2 = a square. 
 
 This can only be solved generally if we can discover one possible value of / by trial ; 
 this however is not difficult in the particular case, for/= i is an obvious solution. 
 
 To find others we put i + q instead of p in the expression to be made a square ; this 
 gives 
 
 i + 4^ 2 + 4^ 3 + ij 4 = a square. 
 
 This can be solved in several ways. 
 
 1. Suppose i + 4^ 2 + 4^ 8 -i-^ 4 =(i + ?*) 2 ; 
 
 thus 4$r" + 4^=2^ 2 , whence q --, p = - and x= ---. 
 
 2. Suppose i+4? 2 + 4 ? 3 + ^=( I -? 2 ) 2 ; 
 
 thus 4^2 + 4^= -2? 2 , and q= -|, /=-- and *=--. 
 
 3. Suppose i+4^ 2 + 4^ 3 + ^ 4 = (i2^^ 2 ) a ; 
 and we find, in either case, that q= - i, so that/= -1,^ = 0. 
 
 4. Suppose i + 4^2 + 4^ 3 + y 4 = ( i + 2^ 2 ) 2 ; 
 
 and we have 4^ + ^ 4 =4^ 4 , whence <? = -, p = ~ and x=(-\ -i=^. 
 
 3 3 \3/ 9 
 
 This value of x satisfies the conditions, for 
 
 The above five suppositions therefore give only two serviceable solutions 
 
 x=-*, ,= *?. 
 4 9 
 
 To find another solution we take one of the values of q already found, say y= , and
 
 DIOPHANTUS' METHODS OF SOLUTION 85 
 
 Exx. from Diophantus : 
 
 3*- 12 = u*) 
 *i ah ( v - *) 
 
 O*;r 12 = T& j 
 
 (vi. 6.) 
 
 -6144*+ 1048576 = M , yl 22> 
 
 jr + 64 = w i r 
 
 substitute r - - for q. This gives / = i + ? = r + - , and we substitute this value for / in the 
 expression /* - 2/ a + 2. 
 
 We have then ^-^r-- i* + 'it 3 +t* to be made a square, or 
 16 2 2 
 
 25 - 24r- 8^+32^+ i6^=a square. 
 
 i. We take S+A*^ for the root, so that the absolute term and the term in r 4 
 may disappear. We can make the term in r disappear also by putting iof= - 24 or 
 
 /= - . We then have 
 
 (a) The upper sign gives 
 
 - 8 + 32r = 40 +/*+ 8A. 
 
 <* r 
 
 and r=( 
 
 thus P= , and jr=/2_ I= ?r. 
 
 20 400 
 
 () The lower sign gives 
 
 -8 + 32r= -40+/ 2 -8A, 
 
 and r=(f 
 
 thus p= - , and or=^i as before. 
 
 ' 20 400 
 
 We have therefore .r+i = (^ , and 
 
 ) . 
 / 
 
 2. Another solution is found by assuming the root to be 5 +/r+g^ and determining 
 /and g so that the absolute term and the terms in r, r 3 may vanish ; the result is 
 ,_ 2 7 2/g-3? = 1550 
 
 /- 5 ' ^^ 125' ' ~ i6-^r 2 861 ' 
 
 /- 
 
 1 1 . Method of " doubU-tqitation. 
 
 The difference =jc a -jr. 
 
 (i) If we take as factors JT, x- i and, as usual, equate the square of half their 
 difference, or - , to x + i, we have 
 
 "=;
 
 86 INTRODUCTION 
 
 The absolute terms in the last case are made equal by multiply- 
 ing the second equation by (i28) 2 or 16384. 
 
 (3) One separate case must be mentioned which cannot be 
 solved, from Diophantus' standpoint, by the foregoing method, 
 but which sometimes occurs and is solved by a special artifice. 
 
 The form of double-equation is 
 
 our 2 4- ax = u z \ (i), 
 
 /3;r 2 + bx = w* \ ( 2 ) . 
 
 Diophantus assumes ?/ 2 = m^x*, 
 whence, by ( I ), x = a/(m z a), 
 
 (2) If we take -x, tx-t, as factors, half the sum of which is -*- i, so that the 
 
 4 
 absolute terms may disappear in the resulting equation, we have 
 
 U^-libA 
 
 16 2 
 
 and *-. 
 
 9 
 
 (3) To find another value by means of the first, namely x= - - , we substitute y - - 
 for x in the original expressions. We then have to solve 
 ,-3 2| = 1<t 
 
 Multiply the latter by so as to make the absolute terms the same, and we must have 
 
 a,+s.** 
 
 4' 16 
 
 Subtract from the first expression, and the difference is y t -^y=y (y- 1 5 then, 
 
 equating the square of half the difference of the factors to the smaller expression, 
 we have 
 
 so that 961=400^+100. 
 
 Therefore 
 
 (4) If we start from the known value and put j+ for j; in the equations, we 
 
 obtain Euler's fourth value of *, namely . 
 
 7 2965284 
 
 Thus all the four values obtained by Euler are more easily obtained by the method of 
 the "double-equation."
 
 DIOPHANTUS' METHODS OF SOLUTION 87 
 
 and, by substitution in (2), we derive that 
 
 ba 
 
 must be a square, 
 
 m 2 a. 
 
 a 2 ft + ba (m 2 - a) 
 or 
 
 We have therefore to solve the equation 
 
 abm z + a (aft a.b) =y 2 , 
 
 and this form can or cannot be solved by the methods already 
 given according to the nature of the coefficients 1 . Thus it can 
 be solved if (a^ a.b}ja is a square or if a\b is a square. 
 
 Exx. from Diophantus : 
 
 * 4*--, (VL I2 .) 
 
 () Indeterminate equations of a degree higher than the second. 
 
 (i) Single equations, 
 
 These are properly divided by Nesselmann into two classes ; 
 the first comprises those problems in which it is required to make 
 a function of x, of a degree higher than the second, a square ; the 
 second comprises those in which a rational value of x has to be 
 found which will make any function of x, not a square, but a higher 
 power of some number. The first class of problems requires the 
 solution in rational numbers of 
 
 Ax n + Bx n ~ l + . . . + Kx + L =/>, 
 the second the solution of 
 
 Ax n + Bx n ~ l + . . . + Kx -f L = y 3 , 
 
 for Diophantus does not go beyond making a certain function of 
 x a cube. In no instance, however, of the first class does the index 
 n exceed 6, while in the second class n does not (except in a 
 special case or two) exceed 3. 
 
 1 Diophantus apparently did not observe that the above form of double-equation can 
 be reduced to one of the first degree by dividing by x 2 and substituting^ for i/jr, when it 
 becomes 
 
 Adapting Sachet's second condition, we see that the equations can be rationally solved 
 if (/3a - ab)la is a square, which is of course the same as one of the conditions under which 
 the above equation abttfi + a (aft a&) can be solved.
 
 88 INTRODUCTION 
 
 First Class. Equation 
 
 Ax n + Bx n ~ l + . . . + Kx + L = y". 
 The forms found in Diophantus are as follows : 
 i . Equation Ax* + Bx* + Cx + d 2 = y\ 
 
 Here, as the absolute term is a square, we might put for y 
 the expression mx + d, and determine m so that the coefficient 
 of x in the resulting equation vanishes. In that case 
 
 2md = C, and m C\2d\ 
 
 and we obtain, in Diophantus' manner, a simple equation for x, 
 giving 
 
 C* -^B 
 
 4tl* A 
 
 Or we might put for y the expression nPx* + nx + d, and deter- 
 mine m, n so that the coefficients of x, x* in the resulting equation 
 both vanish, in which case we should again have a simple equa- 
 tion for x. Diophantus, in the only example of this form of 
 equation which occurs (VI. 18), makes the first supposition. The 
 equation is 
 
 and Diophantus assumes y = \x + i, whence x = *. 
 
 2. Equation Ax* + Bx* + Cx* + Dx + E=y\ 
 In order that this equation may be solved by Diophantus' 
 method, either A or E must be a square. If A is a square and 
 
 D 
 
 equal to a 2 , we may assume y = ax* H -- x + n, determining n so 
 that the term in x* vanishes. If E is a square (= e z ), we may write 
 y = mx* H x + e, determining m so that the term in x* in the 
 
 resulting equation may vanish. We shall then, in either case, 
 obtain a simple equation in x. 
 
 The examples of this form in Diophantus are of the kind 
 
 a y x 4 + Bx 3 + Cx* + Dx + e* =f, 
 
 where we can assume y=ax* + kxe, determining k so that in 
 the resulting equation the coefficient of x 3 or of x may vanish ; 
 when we again have a simple equation. 
 
 Ex. from Diophantus (iv. 28) : 
 
 Diophantus assumes y= yc* 6x+ i,and the equation reduces to 
 2jr 3 6> 2 = o, whence *=.
 
 DIOPHANTUS' METHODS OF SOLUTION 89 
 
 Diophantus is guided in his choice of signs in the expression 
 ax' 1 + kx e by the necessity for obtaining a " rational " result. 
 
 Far more difficult to solve are those equations in which, the 
 left-hand expression being bi-quadratic, the odd powers of x are 
 wanting, i.e. the equations Ax* + Cx* 1 + E =y* and Ax* + E=y*, 
 even when A or E is a square, or both are so. These cases 
 Diophantus treats more imperfectly. 
 
 3. Equation Ax* + Cx* + E =y*, 
 
 Only very special cases of this form occur. The type is 
 
 a*x* - c*x 2 + e 2 =y\ 
 which is written 
 
 a*x*-(? + ?L=y2. 
 
 Here y is assumed to be ax or ejx, and in either case we have 
 a rational value for x. 
 
 Exx. from Diophantus : 
 
 This is assumed to be equal to 
 
 where y*- is assumed to be equal to 
 
 4. Equation Ax* + E =y\ 
 
 The case occurring in Diophantus is x* + 97 =y z (V. 29). Dio- 
 phantus tries one assumption, y = x z 10, and finds that this gives 
 X* = -Q, which leads to no rational result. Instead, however, of 
 investigating in what cases this equation can be solved, he simply 
 drops the equation x* + 97 =j 2 and seeks, by altering his original 
 assumptions, to obtain, in place of it, another equation of the same 
 type which can be solved in rational numbers. In this case, by 
 altering his assumptions, he is able to replace the refractory equa- 
 tion by a new one, ^ 4 + 337=j 2 , and at the same time to find a 
 suitable substitution for j, namely y = x* 25, which brings out 
 a rational result, namely x = ^-. This is a good example of his 
 characteristic artifice of " Back-reckoning 1 /' as Nesselmann calls it. 
 
 5. Equation of sixth degree in the special form 
 
 x 6 Ax* + J3x + c 2 =y-. 
 
 " Methode der Zuriickrechnung und Nebenaufgabe."
 
 90 INTRODUCTION 
 
 It is only necessary to put j> = x 3 + c, and we have 
 -Ax* + B = 2cx\ 
 
 -, 
 
 A + 2C* 
 
 which gives a rational solution if B/(A + 2c) is a square. 
 
 6. If, however, this last condition does not hold, as in the case 
 occurring iv. 18, x 6 i6x s + x + 64 =j/ 2 , Diophantus employs his 
 usual artifice of "back-reckoning," which enables him to replace 
 the equation by another, X s 128^ + ^ + 4096 =/ 2 , where the 
 condition is satisfied, and, by assuming y = x s + 64, x is found to 
 be^. 
 
 Second Class, Equation of the form 
 
 Ax n + Bx n ~ l + . . . + Kx + L = j 3 . 
 
 Except for such simple cases as Ax 2 =y 3 , Ax* = y 3 t where it is only 
 necessary to assume y = mx, the only cases occurring in Diophantus 
 are of the forms 
 
 i. Equation Ax* + Bx + C=y*. 
 
 There are only two examples of this form. First, in VI. I the 
 expression x* ^x + 4 is to be made a cube, being already a square. 
 Diophantus naturally assumes x 2 = a cube number, say 8, and 
 x= 10. 
 
 Secondly, in VI. 17 a peculiar case occurs. A cube is to be 
 found which exceeds a square by 2. Diophantus assumes (x i) 3 
 for the cube and (x+ i) 2 for the square, and thus obtains the 
 equation 
 
 ** ~ 3^ 2 + Z x ~ i =* 2 + 2 * + 3) 
 or x 3 + x=4x z + 4, 
 
 previously mentioned (pp. 66-7), which is satisfied by ^ = 4. 
 The question arises whether it was accidentally or not that this 
 cubic took so simple a form. Were x-i, x+i assumed with 
 knowledge and intention? Since 27 and 25 are, as Fermat 
 observes 1 , the only integral numbers which satisfy the conditions, 
 it would seem that Diophantus so chose his assumptions as to lead 
 back to a known result, while apparently making them arbitrarily 
 with no particular reference to the end desired. Had this not 
 
 1 Note on vi. 17, Oeuvi-es, I. pp. 333-4, II. p. 434. The fact was proved by Euler 
 (Algebra, Part II. Arts. 188, 193). See note on vi. 17 infra for the proof.
 
 DIOPHANTUS' METHODS OF SOLUTION 91 
 
 been so, we should probably have found him, here as elsewhere 
 in the work, first leading us on a false tack and then showing us 
 how we can correct our assumptions. The fact that he here 
 makes the right assumptions to begin with makes us suspect that 
 the solution is not based on a general principle but is empirical 
 merely. 
 
 2. Equation Ax 3 + Ex* + Cx+D =f. 
 
 If A or D is a cube number, this equation is easy of solution. 
 
 jy 
 
 For, first, if A = a?, we have only to write y = ax + a , and we 
 obtain a simple equation in x. 
 
 Secondly, if D = d 3 , we put y = ^ x + d. 
 
 If the equation is a 3 x s + JBx 2 + Cx+ d s =y 3 , we can use either 
 assumption, or we may put y = ax + d, obtaining a simple equation 
 as before. 
 
 Apparently Diophantus used the last assumption only ; for 
 in IV. 27 he rejects as impossible the equation 
 
 because y = 2x i gives a negative value x= ^, whereas either 
 of the other assumptions gives a rational value 1 . 
 
 ( 2 ) Double-equations. 
 
 There are a few examples in which, of two functions of x, one 
 is to be made a square, and the other a cube, by one and the same 
 rational value of x. The cases are for the most part very simple ; 
 e.g. in vi. 19 we have to solve 
 
 2X+ I = 
 
 thus j/ 3 = 2z*, and z = 2. 
 
 A rather more complicated case is VI. 21, where we have the 
 double-equation 
 
 2X 
 
 Diophantus assumes y = mx, whence x = 2/(m 2 2), and we have 
 
 2 V / 2 \ 2 2 
 2 
 
 2m* 
 
 (m* - 2) 3 
 
 1 There is a special case in which C and D vanish, Ay?+ Bx~y*. Here y is put 
 equal to mx, and x=BI(m z - A). Cf. IV. 6, 28 (i).
 
 92 INTRODUCTION 
 
 To make 2m* a cube, we need only make 2m a cube or put 
 m = 4. This gives x=\. 
 The general case 
 
 Ax 9 + Bx" + Cx = 
 
 would, of course, be much more difficult; for, putting y = mx, we 
 have 
 
 x=cl(m?-b), 
 
 and we have to solve 
 
 or Ccm* + c(Bc- 2bC) m? + be (bC -c) + Ac 3 = u 3 , 
 
 of which equation the above corresponding equation is a very 
 particular case. 
 
 Summary of the preceding investigation, 
 
 1. Diophantus solves completely equations of the first degree, 
 but takes pains to secure beforehand that the solution shall be 
 positive. He shows remarkable address in reducing a number of 
 simultaneous equations of the first degree to a single equation in 
 one variable. 
 
 2. For determinate equations of the second degree he has 
 a general method or rule of solution. He takes, however, in the 
 Arithmetica^ no account of more than one root, even where both 
 roots are positive rational numbers. But, his object being simply 
 to obtain some solution in rational numbers, we need not be 
 surprised at his ignoring one of two roots, even though he knew 
 of its existence. 
 
 3. No equations of a degree higher than the second are solved 
 in the book except a particular case of a cubic. 
 
 4. Indeterminate equations of the first degree are not treated 
 of in the work. Of indeterminate equations of the second degree, 
 as Ax* + Bx + C=y*> only those cases are fully dealt with in which 
 A or C vanishes, while the methods employed only enable us to 
 solve equations of the more general forms 
 
 Ax* + C=y> and Ax* + Bx+ C=f 
 
 when A, or C, or \B? -AC is positive and a square number, or (in 
 the case of Ax* C=y 2 ) when one solution is already known.
 
 DIOPHANTUS' METHODS OF SOLUTION 93 
 
 5. For double-equations of the second degree Diophantus has 
 a definite method when the coefficient of x* in both expressions 
 vanishes ; the applicability of this method is, however, subject to 
 conditions, and it has to be supplemented in one or two cases by 
 another artifice. Of more complicated cases we find only a few 
 examples under conditions favourable for solution by his method. 
 
 6. Diophantus' treatment of indeterminate equations of degrees 
 higher than the second depends upon the particular conditions of 
 the problems, and his methods lack generality. 
 
 7. More wonderful than his actual treatment 9f equations are 
 the clever artifices by which he contrives to avoid such equations 
 as he cannot theoretically solve, e.g. by his device of "back- 
 reckoning," instances of which would have been out of place in 
 this chapter and can only be studied in the problems themselves. 
 
 I shall not attempt to class as "methods" certain headings 
 in Nesselmann's classification of the problems, such as (a) " Solution 
 by mere reflection," () " Solution in general expressions," of which 
 there are few instances definitely so described by Diophantus, or 
 (c) "Arbitrary determinations and assumptions." The most that 
 can be done by way of describing these " methods " is to quote 
 a few characteristic instances. This is what Nesselmann has 
 done, and he regrets at the end of his chapter on " Methods of 
 Solution" that it must of necessity be so incomplete. To under- 
 stand and appreciate the various artifices of Diophantus it is in 
 fact necessary to read the problems themselves in their entirety. 
 
 With regard to the " Use of the right-angled triangle," all that 
 can be said of a general character is that only " rational " right- 
 angled triangles (those namely in which the three sides can all be 
 represented by rational numbers) are used in Diophantus, and 
 accordingly the introduction of the " right-angled triangle " is 
 merely a convenient way of indicating the problem of finding 
 two square numbers, the sum of which is also a square number. 
 The general form used by Diophantus (except in one case, VI. 19, 
 q.v.) for the sides of a right-angled triangle is (at + i?), (a'P), 
 2ab, which expressions clearly satisfy the condition 
 
 The expression of the sides of a right-angled triangle in this form 
 Diophantus calls "forming a right-angled triangle from the 
 numbers a and b" His right-angled triangles are of course 
 formed from particular numbers. " Forming a right-angled
 
 94 INTRODUCTION 
 
 triangle from 7, 2 " means taking a right-angled triangle with sides 
 (7 s + 2"), (7 s ~ 2 2 ), 2 . 7 . 2, or 53, 45, 28. 
 
 II. METHOD OF LIMITS. 
 
 As Diophantus often has to find a series of numbers in 
 ascending or descending order of magnitude, and as he does not 
 admit negative solutions, it is often necessary for him to reject 
 a solution which he has found by a straightforward method 
 because it does not satisfy the necessary condition ; he is then 
 very frequently obliged to find solutions which lie within certain 
 limits in place of those rejected. 
 
 1. A very simple case is the following : Required to find 
 a value of x such that some power of it, x n , shall lie between two 
 given numbers. Let the given numbers be a, b. Then Diophantus' 
 method is to multiply both a and b by 2 n , 3", and so on, successively, 
 until some wth power is seen which lies between the two products. 
 Thus suppose that c n lies between ap n and bp n ; then we can put 
 x = c\p, in which case the condition is satisfied, for (<://) n lies 
 between a and b. 
 
 Exx. In IV. 31 (2) Diophantus has to find a square between 
 i \ and 2. He multiplies both by a square, 64 ; this gives 80 and 
 128, and 100 is clearly a square which lies between them; there- 
 fore (gf or f| satisfies the prescribed condition. 
 
 Here, of course, Diophantus might have multiplied by any 
 other square, as 16. In that case the limits would have become 
 20 and 32 , between these lies the square 25, which gives the same 
 square ff as that before found. 
 
 In VI. 21 a sixth power ("cube-cube") is sought which shall 
 lie between 8 and 16. The sixth powers of the first four natural 
 numbers are I, 64, 729, 4096. Multiply 8 and 16 by 2 6 or 64, and 
 we have as limits 512 and 1024, between which 729 lies. There- 
 fore - 7 ^j 9 - is a sixth power satisfying the given condition. To 
 multiply by 729 in this case would not give us a solution. 
 
 2. Sometimes a value of x has to be found which will give 
 some function of x a value intermediate between the values of two 
 other functions of x. 
 
 Ex. i. In IV. 25 it is necessary to find a value of x such that 
 8/(;r 2 +.*) shall lie between x and x + i. 
 The first condition gives 8 >x 3 + x*.
 
 DIOPHANTUS' METHODS OF SOLUTION 95 
 
 Diophantus accordingly assumes that 
 
 which is greater than x s +x*. 
 
 Thus # = f satisfies one condition. It is also seen to satisfy 
 
 o 
 
 the second condition, or - <x+ I. Diophantus. however, says 
 x* + x 
 
 nothing about the second condition being satisfied ; his method is, 
 therefore, here imperfect. 
 
 Ex. 2. In V. 30 a value of x has to be found which shall make 
 * > | (* 2 - 60) but <l(;tr 2 -6o), 
 
 that is, x* 60 > 
 
 x* - 60 < &r 
 
 Hence, says Diophantus, x is not less than 1 1 and not greater 
 than 12. We have already spoken (p. 60 sqq.) of his treatment 
 of such cases. 
 
 Next, the problem in question requires that x* 6o shall be 
 a square. Assume then that 
 
 x* 60 = (x mf, 
 and we have x (m* + 6o)/2m. 
 
 Since, says Diophantus,^ is greater than 11 and less than 12, 
 it follows that 
 
 m 2 + 60 > 22m but < 247/2 ; 
 
 and m must therefore lie between 19 and 21 (cf. p. 62 above). 
 He puts m = 20, and so finds x=n^. 
 
 III. METHOD OF APPROXIMATION TO LIMITS. 
 
 We come, lastly, to a very distinctive method called by 
 Diophantus TrapHrorrjs or Trapia-orijTos dytoyrj. The object of this 
 is to solve such problems as that of finding two, or three, square 
 numbers the sum of which is a given number, while each of them 
 approximates as closely as possible to one and the same number. 
 
 This method can be best exhibited by giving Diophantus' two 
 instances, in the first of which two such squares, and in the second 
 three, are required. In cases like this the principles cannot be 
 so well indicated with general symbols as with concrete numbers, 
 which have the advantage that their properties are immediately
 
 96 INTRODUCTION 
 
 obvious, and the separate expression of conditions is rendered 
 unnecessary. 
 
 Ex. I. Divide 13 into two squares each of which >6 (v. 9). 
 Take half of 13, or 6^, and find what small fraction ijx^ added 
 to it will make it a square : thus 
 
 6^ + , or 26 H - , must be a square. 
 Diophantus assumes 
 
 26+ -3 Y$ + i) i or 26j 2 + i = ($y -f i) a , 
 
 whence y= 10 and 1/7* = ^, or i/^ = ^; and 
 
 6^+^= a square, (W- 
 
 [The assumption of (i^+i) 4 is not arbitrary, for assume 
 26?*+ i (Py-\- i) s > and_y is then 2//(26 /"); since ijy should be 
 a small proper fraction, $ is the most suitable and the smallest 
 possible value for/, inasmuch as 26 -f < 2p or/ 8 + 2p + i > 27.] 
 
 It is now necessary, says Diophantus, to divide 13 into two 
 squares the sides of which are both as near as possible to f^. 
 
 Now the sides of the two squares into which 13 is naturally 
 decomposed are 3 and 2, and 
 
 3 is > ft by &, 
 2 is < ft by ^ 
 
 But, if 3 -fa , 2 + ^ were taken as the sides of two squares, the 
 sum of the squares would be 
 
 which is > 13. 
 
 Accordingly Diophantus puts 3 gx, 2 + i ix for the sides of 
 the required squares, where therefore x is not exactly ^ but 
 near it. 
 
 Thus (3 - 9*)' + (2 + i \xj = 1 3, 
 
 and Diophantus obtains ^r=_^ T . 
 
 The sides of the required squares are ff, f$f. 
 
 [It is of course a necessary condition that the original number, 
 here 13, shall be a number capable of being expressed as the sum 
 of two squares.]
 
 DIOPHANTUS' METHODS OF SOLUTION 97 
 
 Ex. 2. Divide 10 into three squares such that each of them 
 is > 3 (v. n). 
 
 [The original number, here 10, must of course be expressible 
 as the sum of three squares.] 
 
 Take one-third of 10, or 3^, and find what small fraction of the 
 form i/x* added to 3^ will make a square; i*. we have to make 
 
 30 + , a square, or 30^+ i a square, where 3/r = i/y. 
 
 Diophantus assumes 
 
 307*+ i =(57+ i)*, 
 whence .7 = 2 and therefore i/* = & ; and 3^ + ^ ff = W a square. 
 
 [As before, if we assume y*)P = (py+ i) J ,^' = 2//(3O /*); and, 
 since ijy must be a small proper fraction, 30 p* should be < 2p, 
 or /* + 2/-f i >3i. Accordingly Diophantus chooses 5 for/ as 
 being the smallest possible integral value.] 
 
 We have now, says Diophantus, to make each of the sides 
 of our required squares as near as may be to ty. 
 
 Now 10 
 
 and 3, f, are the sides of three squares the sum of which is 10. 
 Bringing (3, f, ) and -^ to a common denominator, we have 
 
 And 
 
 If now we took 3 f$, f + , f + f for sides of squares, the 
 sum of the squares would be 3 (*f or ^^, which is > 10. 
 
 Accordingly Diophantus assumes as the sides of the three 
 required squares 
 
 3-35*. f + 37*. 1 + 31*. 
 where x must therefore be not exactly ^ but near it 
 
 Solving (3 - 3^r) + (f + 37^ + (| + 31*)* = 10, 
 or 10-116*+ 3555**= 10, 
 
 we find x = j^ ; 
 
 the required sides are therefore 
 
 and the required squares 
 
 1745041 lesigys 1658944 
 505681 505521 50B6X1 *
 
 98 INTRODUCTION 
 
 Other instances of the application of the method will be found 
 in V. 10, 12, 13, 14, where, however, the squares are not required to 
 be nearly equal, but each of them is subject to limits which may 
 be the same or different ; e.g. sometimes each square is merely 
 required to be less than a given number (10, say), sometimes the 
 squares lie respectively between different pairs of numbers, some- 
 times they are respectively greater than different numbers, while 
 they are always subject to the condition that their sum is a given 
 number. 
 
 As it only lies within the scope of this Introduction to explain 
 what we actually find in Diophantus, I cannot do more than give 
 a reference to such investigations as those of Poselger in his 
 " Beitrage zur unbestimmten Analysis" published in the Abhand- 
 lungen der Koniglichen Akademie der Wissenscliaften zu Berlin aus 
 dem Jahre 1832, Berlin, 1834. One section of this paper Poselger 
 entitles " Annaherungs-methoden nach Diophantus," and obtains 
 in it, on Diophantus' principles, a method of approximation to the 
 value of a surd which will furnish the same results as the method 
 of continued fractions, with the difference that the " Diophantine 
 method " is actually quicker than the method of continued frac- 
 tions, so that it may serve to expedite the latter.
 
 CHAPTER V 
 
 THE PORISMS AND OTHER ASSUMPTIONS IN DIOPHANTUS 
 
 I HAVE already mentioned (in Chapter I.) the three explicit refer- 
 ences made by Diophantus to " The Porisms " and the possibility 
 that, if these formed a separate work, it may have been from that 
 work that Diophantus took a number of other propositions relating 
 to properties of numbers which he enunciates or tacitly takes for 
 granted in the Arithmetica. 
 
 I begin with the three propositions for which he expressly 
 refers to " The Porisms." 
 
 Porism i. In V. 3 he says, "We have it in the Porisms that, 
 ' If each of two numbers and their product when severally added to 
 the same given number produce squares, the squares with which 
 they are so connected are squares of two consecutive numbers 1 .'" 
 
 That is to say, if x + a = m z , y + a = n*, and if xy + a is also a 
 square, then m~n=i. 
 
 The theorem is not correctly enunciated, for it would appear 
 that m ~ n = i is not the only condition under which the three 
 expressions may be simultaneously squares. 
 
 For suppose 
 
 x + a = m*, y + a = n\ xy + a =/ 2 . 
 
 By means of the first two equations we have 
 xy + a = m*ri* a (m 2 + n 2 - i) + a\ 
 
 In order that 
 
 nfif- - a (m* + 2 i ) + a* 
 
 may be a square certain conditions must be satisfied. One suffi- 
 cient condition is 
 
 or m ~ n = i , 
 
 which is Diophantus' condition. 
 
 1 Literally "(the numbers) arise from two consecutive squares" (yeybvaffiv avb 8i5o 
 r(av Kara TO (%?}*) 
 
 72
 
 INTRODUCTION 
 But we may also regard 
 
 as an indeterminate equation in m of which we know one solution, 
 namely m = n i . 
 
 Other solutions are then found by substituting z + (n i) for 
 m, whence we obtain the equation 
 
 (n z -a)z*+2{n*(n i)-a(ni)}z 
 
 + O 2 - a)(n i ) 2 - a (n z - i ) + 2 =/ 2 , 
 
 or (n- -a) z 2 + 2 (n* -a)(n i)z + {n(n i)-a} 2 =p\ 
 
 which is easy to solve in Diophantus' manner, since the absolute 
 term is a square. 
 
 But in the problem V. 3 three numbers are required, such that 
 each of them, and the product of each pair, when severally added 
 to a given number, produce squares. Thus if the third number be 
 z, three additional conditions have to be satisfied, namely 
 
 z 4- a = it 2 , zx + a = v 2 , zy + a = w*. 
 The two last conditions are satisfied, if m+i=n, by putting 
 
 z = 2 (x +y) i = 4.m* -f 40? + I - 4a, 
 when xz + a = {m(2m + i) - 2a} 2 
 
 and 
 
 and perhaps this means of satisfying the conditions may have 
 affected the formulation of the Porism 1 . 
 
 The problem V. 4 immediately following assumes the truth of 
 the same Porism with a substituted for -f a. 
 
 Porism 2. In V. 5 Diophantus says, " Again we have it in the 
 Porisms that, ' Given any two consecutive squares, we can find in 
 addition a third number, namely the number greater by 2 than the 
 double of the sum of the two squares, which makes, the greatest of 
 three numbers such that the product of any pair of them added to 
 either the sum of that pair or the remaining number gives a square."' 
 
 That is, the three numbers 
 
 1 Euler has a paper describing and illustrating a general method of finding such 
 "porisms" the effect of which is to secure that, when some conditions are satisfied, the 
 rest are simultaneously satisfied ("De problematibus indeterminatis quae videntur plus 
 quam determinata" in Novi Commentarii Acad. Petropol. 1756-57, Vol. vi. (1761), 
 p. 85 sqq. = Commentationes arithmeticae collectae, I. pp. 245 259). This particular 
 porism of Diophantus appears as a particular case in 13 of the paper.
 
 THE PORISMS AND OTHER ASSUMPTIONS 101 
 
 have the property that the product of any two plus either the sum 
 of those two or the remaining number gives a square. In fact, if 
 X, Y, Z denote the numbers respectively, 
 
 XY+X+ Y=( m* + m+i)*, XY+Z=(m* + m + 2)*, 
 
 \T 7 i y i 7 f ^ jj/- J- "2ff* -I- 2^* V '7 -I- y /?/* _L 24** _l_ ^\* 
 
 ZJf + Z + ^ = (2/ s + / + 2) 2 , Z^+ K=(2* s + +!). 
 
 Porism 3 occurs in v. 16. Unfortunately the text is defective 
 and Tannery has had to supply three words 1 ; but there can be no 
 doubt that the correct statement of the Porism here in question is 
 " The difference of any two cubes is also the sum of two cubes," 
 i.e. can be transformed into the sum of two cubes, or two cubes can 
 be found the sum of which is equal to the difference between any 
 two given cubes. Diophantus contents himself with the enuncia- 
 tion of the proposition and does not show how to prove it or how 
 he effected the transformation in practice. The subject of the 
 transformation of sums and differences of cubes was investigated 
 by Vieta, Bachet and Fermat 
 
 Vieta (Zetetica, IV. 18-20) has three problems on the subject 
 
 (i) Given two cubes, to find in rational numbers two other 
 cubes such that their sum is equal to the difference of the given 
 cubes 8 . 
 
 As a solution of a* fc=x*+y*, he finds 
 a(a?-2P) 
 
 a* + P ' y ~ 
 
 rotj Uoplff fount on " Tarrwr Suo Kvfiur i) irrepaxh Kvfkai> < 
 
 3 The solution given by Vieta is obtainable thus. The given cubes being a 3 , A 3 , where 
 a> b, we assume x - b, a - kx as the sides of the required cubes. 
 
 Thus 
 whence 
 
 This reduces to a simple equation if we assume 
 
 lP-a?Jk=o, or Jk = 
 in which case 
 
 and the sides of the cubes are therefore 
 
 Vieta's second problem is similarly solved by taking a+x, kx-b as the sides of the 
 required cubes, and the third problem by taking x - 6, kx - a as the sides of the required 
 cubes respectively.
 
 102 INTRODUCTION 
 
 (2) Given two cubes, to find in rational numbers two 
 others such that their difference is equal to the sum of the given 
 cubes. 
 
 Solving a 3 + & = x 3 }> 3 , we find that 
 
 (3) Given two cubes, to find in rational numbers two cubes 
 such that their difference is equal to the difference of the given 
 cubes. 
 
 For the equation a 3 - d 3 = x 3 j 3 , Vieta finds 
 
 _ b(2a*-P) _ a (26 s -a 9 ) 
 
 ; at + b 3 ' y= tf + b 3 
 as a solution 1 . 
 
 In the solution of (1} x is clearly negative if 2& >a 3 ; therefore, 
 in order that the result may be " rational," a 3 must be > 2^. But 
 for a " rational " result in (3) we must, on the contrary, have a 3 < 2b 3 . 
 Fermat was apparently the first to notice that, in consequence, the 
 processes in (i) and (3) exactly supplement each other, so that by 
 employing them successively we can effect the transformation 
 required in (i) even when a 3 is not > 2b 3 . 
 
 The process (2) is always possible ; therefore, by a suitable 
 combination of the three processes, the transformation of a sum 
 of two cubes into a difference of two cubes, or of a difference of 
 two cubes into a sum or a difference of two other cubes is always 
 
 1 Vieta's formulae for these transformations give any number of very special solutions 
 (in integers and fractions)of the indeterminate equation jc s +y 3 + z? = v 3 , including solutions 
 in which one of the first three cubes is negative. These special solutions are based on 
 the assumption that the values of two of the unknowns are given to begin with. Euler 
 observed, however, that the method does not give all the possible values of the other two 
 even in that case. Given the cubes 3 3 and 4 3 , the method furnishes the solution 
 
 3 3 + 4 3 +( Y=( Y, but not the simpler solution 3 3 + 4 3 +5 3 = 6 3 . Euler ac- 
 
 cordingly attacked the problem of solving the equation x?+y 3 + z s = v 3 more generally. 
 He began with assuming only one, instead of two, of the cubes to be given, and, on that 
 assumption, found a solution much more general than that of Vieta. Next he gave a 
 more general solution still, on the assumption that none of the cubes are given to begin 
 with. Lastly he proceeded to the problem To find all the sets of three integral cubes the 
 sum of which is a cube and showed how to obtain a very large number of such sets 
 including sets in which one of the cubes is negative (Novi Commentarii Acad. Petropol- 
 175657, Vol. VI. (1761), p. 155 sq. = C omni enta done s arithmeticae, I. pp. 193 207). 
 The problem of solving x s +y 3 =z 3 + 7p in integers in any number of ways had occupied 
 Frenicle, who gave a number of solutions (Oeuvres de Fermat, ill. pp. 420, 535) ; but the 
 method by which he discovered them does not appear,
 
 THE PORISMS AND OTHER ASSUMPTIONS 103 
 
 practicable 1 . Fermat showed also how, by a repeated use of the 
 several processes as required, we can transform a sum of two cubes 
 into a sum of two other cubes, the latter sum into the sum of two 
 others and so on ad infinitum*. 
 
 Besides the " Porisms " there are many other propositions 
 assumed or implied by Diophantus which are not definitely called 
 
 1 Fermat (note on IV. 2) illustrates by the following case : 
 
 Given two cubes 125 and 64, to transform their difference into the sum of two other 
 cubes. 
 
 Here a=s, 6=4, and so 26 s > a 3 ', therefore we must first apply the third process 
 by which we obtain 
 
 '-"(*)' -(*)' 
 
 As f -7 J > 2 / -~- J , we can, by the first process, turn the difference between the 
 
 cubes ( -- ) and ( ?- ) into the sum of two cubes. 
 \<>3/ \63/ 
 
 " In fact," says Fermat, "if the three processes are used in turn and continued 
 ad infinitum, we shall get a succession ad infinitum of two cubes satisfying the same 
 condition ; for from the two cubes last found, the sum of which is equal to the difference 
 of the two given cubes, we can, by the second process, find two more cubes the difference 
 of which is equal to the sum of the two cubes last found, that is, to the difference 
 between the two original cubes; from the new difference between two cubes we can 
 obtain a new sum of two cubes, and so on ad infinitum" 
 
 As a last illustration, to show how a difference between cubes can be transformed into 
 the difference between two other cubes even where the condition for process (3) is not 
 satisfied, Fermat takes the case of 8- i, i.e. the case where 
 a = 2, d=i and o?>ilP. 
 
 First use process (i) and we have 
 
 --='+ 
 
 Then use process (2), and 
 
 2 Suppose it required to solve the fourth problem of transforming the sum of two cubes 
 into the sum of two other cubes. 
 
 Let it be required so to transform 2 3 + i 3 or 9. 
 
 First transform the sum into a difference of two cubes by process (2). This gives 
 
 The latter two cubes satisfy the condition for process (3) and, applying that process, 
 we get 
 
 /2oV_ /i7\ /I88479V _ f 365*oV 
 \ II \7 / \9039 1 / \939 1 / 
 
 The cubes last found satisfy the condition for process ( i), and accordingly the difference 
 between the said last cubes, and therefore the sum of the original cubes, is at last trans- 
 formed into the sum of two other cubes.
 
 io 4 INTRODUCTION 
 
 porisms, though some of them are of the same character as the 
 three porisms above described. 
 
 Of these we may distinguish two classes. 
 
 I. The first class of theorems or facts assumed without ex- 
 planation by Diophantus are more or less of the nature of identical 
 formulae. Some are quite simple, e.g. the facts that the expressions 
 [%(a+b}} z ab and a* (a + i) 2 + a 2 + (a + i) 2 are respectively 
 squares, that the expression a (a 2 a) + a + (a* - a) is always a 
 cube, and the like. 
 
 Others are more difficult and betoken a certain facility in work- 
 ing with quasi-algebraical expressions. Examples of this kind are 
 the following : 
 
 (a) If X=a 2 x+2a, Y=(a + i)*x + 2(a + i ), or, in other words, 
 if xX + i =(ax+ i) 2 , xY+ i=[(a+ i)x+ i} 2 , then XY + i is a 
 square [IV. 20]. As a matter of fact, 
 
 (/3) 8 times a triangular number plus \ gives a square [IV. 38]. 
 In fact, 8 . (1) + j = ( 2 x + i) 2 . 
 
 (7) If Xa = m\ Ya = (m + i) 2 , and Z=2(X+Y)-i, 
 then the expressions YZ a, ZX a, X Y a are all squares. 
 (The upper signs refer to the assumption in V. 3, the lower to that 
 in V. 4.) 
 
 In fact, YZ a = {(m + i)(2m + i) + 2a}\ 
 
 ZX a= [m(2m + i)+ 2a} 2 , 
 XYa={m(m+ i) + a}\ 
 
 then the six expressions 
 
 FZ-( F+Z), ZX-(Z+X), XY-(X+ F) 
 YZ-X, ZX-Y, XY-Z 
 
 are all squares [v. 6]. 
 In fact, 
 YZ - ( F+ Z) = (2m* + $m + 3) 2 , FZ - X = (2m* + $m + 4) 2 , etc. 
 
 2. The second class is much more important, consisting of a 
 number of propositions in the Theory of Numbers which we find
 
 THE PORISMS AND OTHER ASSUMPTIONS 105 
 
 first stated or assumed in the Arithmetica. It was, in general, in 
 explanation or extension of these that Fermat's most famous notes 
 were written. How far Diophantus possessed scientific proofs of 
 the propositions which he assumes, as distinct from a merely 
 empirical knowledge of them, must remain to a great extent 
 matter of speculation. 
 
 (a) Theorems in DiopJtantus respecting the composition of num- 
 bers as the sum of two squares. 
 
 (1) Any square number can be resolved into two squares in 
 any number of ways, II. 8. 
 
 (2) Any number which is the sum of two squares can be 
 resolved into two other squares in any number of ways, II. 9, 
 
 N.B. It is implied throughout that the squares may be frac- 
 tional as well as integral. 
 
 (3) If there are two whole numbers each of which is the sum of 
 tivo squares, their product can be resolved into the sum of two squares 
 in two ways, III. 19. 
 
 The object of III. 19 is to find four rational right-angled triangles 
 having the same hypotenuse. The method is this. Form two 
 right-angled triangles from (a, b) and (c, d) respectively, i.e. let 
 the sides of the triangles be respectively 
 
 and c- + d\ ? - d*, 2cd. 
 
 Multiplying all the sides in each by the hypotenuse of the other, 
 we have two triangles with the same hypotenuse, namely 
 
 (a* + &}((* + d z ), (a 2 - P)(<* + d*\ 2ab (t* + d*\ 
 and (* + P)(e? + d-\ (a 2 + P)((* - d*), 2cd (a* + #). 
 
 Two other triangles having the same hypotenuse are obtained 
 by using the theorem enunciated. In fact, 
 
 (a 2 + b*)(c* + d*) = (ac bd^ + (ad + be? 
 
 and the triangles are formed from ac bd, ad + be, being the 
 triangles 
 
 (a 2 + P)(c- + d*\ \abcd + (a 2 - b*)(c- - d*), 2(ac-\- bd)(ad - bc\ 
 (a* + &)((? + d"-\ ^abcd - (a 2 - &*)(<;* - d*\ 2 (ac - bd)(ad + be).
 
 106 INTRODUCTION 
 
 In the case taken by Diophantus 
 
 and the four triangles are respectively 
 
 (65, 52, 39), (65, 60, 25), (65, 63, 16), (65, 56, 33). 
 
 (If certain relations 1 hold between a, b, c, d, this method fails. 
 Diophantus has provided against them by taking two triangles " in 
 the smallest numbers" (VTTO eXa%to-T&>i> dpid/j,oov), namely 3,4, 5 and 
 5, 12, 13.) 
 
 Upon this problem III. 19 Fermat has a long and important 
 note which begins as follows 2 : 
 
 " [i] A prime number of the form 4^+1 is the hypotenuse of 
 a right-angled triangle in one way only, its square is so in two 
 ways, its cube in three, its biquadrate in four ways, and so on ad 
 infinitum. 
 
 "[2] The same prime number 4 + I and its square are the 
 sum of two squares in one way only, its cube and its biquadrate 
 in two ways, its fifth and sixth powers in three ways, and so on ad 
 infinitum. 
 
 "[3] If a prime number which is the sum of two squares be 
 multiplied into another prime number which is also the sum of 
 two squares, the product will be the sum of two squares in two 
 ways ; if the first prime be multiplied into the square of the second 
 
 1 (i) We must not have a\b c\d or a\b = d\c, for in either case one of the perpendiculars 
 of one of the resulting triangles vanishes, making that triangle illusory. Nor (2) must 
 c\d be equal to (a + b)l(a-b) or to (a-l>)j(a + l>), for in the first case ac-bd=ad+bc, 
 and in the second case ac + bd=ad-bc, so that one of the sums of squares equal to 
 (a? + l> 2 ) (c 2 + cf 2 ) is the sum of two equal squares, and consequently the triangle formed 
 from the sides of these equal squares is illusory, one of its perpendicular sides vanishing. 
 
 3 G. Vacca (in Bibliotheca Mat hematic a, H 3 . 1901, pp. 358-9) points out that Fermat 
 seems to have been anticipated, in the matter of these theorems, by Albert Girard, who 
 has the following note on Diophantus v. 9 (Oeuvres mathhnatiques de Simon Stevin par 
 Albert Girard, 1634, p. 156, col. i): 
 
 " ALB. GlR. Determinaison d'un nombre qui se peut diviser en deux quarrez entiers. 
 
 I. Tout nombre quarre. 
 
 II. Tout nombre premier qui excede un nombre quaternaire de 1'unite. 
 
 III. Le produict de ceux qui sont tels. 
 
 IV. Et le double d'un chacun d'iceux. 
 
 Laquelle determinaison n'estant faicte n'y de PAutheur n'y des interpretes, servira tant 
 en la presente et suivante comme en plusieurs autres. " 
 
 Now Girard died on 9 December, 1632 ; and the Theorems of Fermat above 
 quoted are apparently mentioned by him for the first time in his letter to Mersenne of 
 25 December, 1640 (Oeuvres de Fermat^ II. p. 213). Was the passage of Girard known 
 to Fermat ?
 
 THE PORISMS AND OTHER ASSUMPTIONS 107 
 
 prime, the product will be the sum of two squares in three ways ; 
 if the first prime be multiplied into the cube of the second, the 
 product will be the sum of two squares in four ways, and so on 
 ad infinitum*." 
 
 It is not probable that Diophantus was aware that prime num- 
 bers of the form 4# + I and numbers arising from the multiplication 
 of such numbers are the only classes of numbers which are always 
 the sum of two squares ; this was first proved by Euler 2 . But it 
 is remarkable that Diophantus should have selected the first two 
 prime numbers of the form 4+ I, namely 5 and 13, which are 
 both sums of two squares, as the hypotenuses of his first two right- 
 angled triangles and then made their product, 65, the hypotenuse 
 of other right-angled triangles, that product having precisely the 
 property of being, as in Fermat's [3], the sum of two squares in 
 two ways. Diophantus may therefore have had an inkling, whether 
 obtained empirically or otherwise, of some of the properties enunci- 
 ated by Fermat. 
 
 (4) Still more remarkable is a condition of possibility of solution 
 prefixed to the problem V. 9. The object of this problem is " to 
 divide I into two parts such that, if a given number is added to 
 either part, the result will be a square." Unfortunately, the text 
 of the added condition is uncertain. There is no doubt about the 
 first few words, " The given number must not be odd" />. No number 
 of tlie form 4 + 3 [or 4*1 i] can be tJte sum of two squares. 
 
 The text, however, of the latter half of the condition is corrupt. 
 The true condition is given by Fermat thus : " The given number 
 must not be odd, and the double of it increased by one, when divided 
 by tJte greatest square which measures it, must not be divisible by a 
 prime number of the form %n I." (Note upon V. 9; also in a 
 letter to Roberval 3 .) There is room for any number of conjectures 
 as to what may have been Diophantus' words 4 . 
 
 1 For a fuller account of this note see the Supplement, section I. 
 
 8 Novi Commentarii Acad. Petrofol. 1751-3, Vol. IV. (1758), pp. 3-40, and 1754-5, 
 Vol. v. (1760), pp. 3-58= Commentationes arithmcticae, I. pp. 155-173 and pp. 110233 '> 
 cf. Legendre, Zahlentheorie, tr. Maser, I. p. 108; Weber and Wellstein's EncyclopdtSe 
 der Ekmentar-Mathematik, I 2 . pp. 285 sqq. 
 
 3 Ofttvres de Fermat, II. pp. 203-4, See the Supplement, section I. 
 
 4 Bachet's text has 5e? 5^ rbv di86tJKvov /n^re re/xerffdr flrcu, fir/re 6 SwXcurfuH' oirrou 
 q' n a. neifova. txi P*(KK 5. 17 fifrpeiTai irrb rov a ". t*. 
 
 He also says that a Vatican MS. reads nijre 6 5ir\affiw afoov aptOnov /xortWa a. 
 fjLti^ova txV M^pos TfTaprov, rj fjLerpeirai intb rov ffxarov apiffpov. 
 
 Neither does Xylander help us much. He frankly tells us that he cannot understand
 
 io8 INTRODUCTION 
 
 There would seem to be no doubt that in Diophantus' condition 
 there was something about " double the number " (i.e. a number of 
 the form \n\ also about " greater by unity " and " a prime number." 
 It seems, then, not unlikely that, if Diophantus did not succeed in 
 giving the complete sufficient and necessary condition stated by 
 Fermat, he made an approximation to it ; and he certainly knew 
 that no number of the form 4# + 3 could be the sum of two 
 squares l . 
 
 (b) On members which are the sum of three squares. 
 
 In the problem v. 1 1 a condition is stated by Diophantus re- 
 specting the form of a number which, added to three parts of unity, 
 makes each of them square. If a be this number, clearly 30+1 
 must be divisible into three squares. 
 
 Respecting the number a Diophantus says, " It must not be 2 
 or any multiple of 8 increased by 2." 
 
 That is, a number of the form 2^n + 7 cannot be the sum of three 
 squares. Now the factor 3 of 24 is irrelevant here, for with respect 
 to three this number is of the form ^m + I, and this, so far as 3 is 
 concerned, might be a square or the sum of two or three squares. 
 Hence we may neglect the factor 3 in 24^. 
 
 We must therefore credit Diophantus with the knowledge of 
 
 the passage. " Imitari statueram bonos grammaticos hoc loco, quorum (ut aiunt) est 
 multa nescire. Ego vero nescio heic non multa, sed paene omnia. Quid enim (ut 
 reliqua taceam) est ju^re 6 diirXafflwv O.VTOV op fju> a etc., quae causae huius irpoffSiopifffj,ov, 
 quae processus? immo qui processus, quae operatic, quae solutio?" 
 
 Nesselmann discusses an attempt made by Schulz to correct the text, and himself 
 suggests Atiyre TOV SurXaaiova avTov &pi9/j.bv fj.ovdSi /j.eiova ^x et ") & perpetrat vvo TIVOS 
 irp&Tov dpi6fwv. But this ignores frfpos rtraprov and is not satisfactory. 
 
 Hankel, however (Gesch. d. Math. p. 169), says: " Ich zweifele nicht, dass die 
 von den Msscr. arg entstellte Determination so zu lesen ist : Aei 5^ rbv diS6/j.evov n^re 
 wepiffffbv elva.i, /tijre ^bv oiirXafftova. O.VTOV api8fj.bv fiovASi d (Jifl^ova /j.fTptiff6ai vir6 TOV 
 irp&Tov dpiOfiov, 6s &v fi.ovA.dt. d ndfav fyv V-tpos rtrapTov." This correction seems a 
 decidedly probable one. Here the words ptpos T^raprov find a place ; and, secondly, 
 the repetition of fj.ov&di d fj.d$wv might well confuse a copyist. TOV for TOV is of course 
 natural enough ; Nesselmann reads rtvos for TOV. 
 
 Tannery, improving on Hankel, reads A 5r; TOV $i56u.fvov yu^re wepurffov elvai, //.ijre 
 trip SurXturioj' avrov Kal fj.ov6.8i M'? /xeiforo fjierpeiffOai viro TOV irp&TOv dpiB/J.ov <ov 6 
 fiovddt, fdq. fj.eifav> ?x?7 M^pos T^Taprov t, " the given number must not be odd, and twice 
 it plus i must not be measured by any prime number which, when increased by i, is 
 divisible by 4." 
 
 1 A discussion of the text and a suggestion as to the considerations which may have 
 led to the formulation of the condition will be found in Jacobi, " Ueber die Kenntnisse 
 des Diophantus von der Zusammensetzung der Zahlen" (Berliner Monatsberichte, 1847; 
 Gesammelte Werke, vil., 1891, pp. 332-344).
 
 THE PORISMS AND OTHER ASSUMPTIONS 109 
 
 the fact that no nttmber of the form 8 + 7 can be the sum of 
 three squares*. 
 
 This condition is true, but does not include all the numbers 
 which cannot be the sum of three squares, for it is not true that 
 all numbers which are not of the form 8# + 7 are made up of three 
 squares. Even Bachet remarked that the number a might not be 
 of the form 32^ + 9, or a number of the form 96^ + 28 cannot be 
 the sum of three squares. 
 
 Fermat gives the conditions to which a must be subject thus 2 . 
 
 Write down two geometrical series (common ratio of each 4), 
 the first and second series beginning respectively with i, 8, 
 
 i 4 16 64 256 1024 4096 
 8 32 128 512 2048 8192 32768; 
 
 then a must not be (i) any number obtained by taking twice any 
 term of the upper series and adding all the preceding terms, or 
 (2) the number found by adding to the numbers so obtained any 
 multiple of the corresponding term of the second series. 
 Thus a must not be 
 
 128/^+2.16 + 4+1 =128/^ + 37, 
 512^+ 2.64+ 16 + 4+ i = 512^+ 149, 
 and so on, where k = o or any integer. 
 
 That is, since i + 4 + 4 2 + . . . + 4 n-1 = (4" i ), a cannot be either 
 
 therefore 30 + I cannot be of the form 4" (24^ + 7) or 4 n (8 + 7). 
 
 Again, there are other problems, e.g. v. 10 and V. 20, in which, 
 though conditions are necessary for the possibility of solution, none 
 are mentioned ; but suitable assumptions are tacitly made, without 
 explanation. It does not follow, from the omission to state the 
 conditions, that Diophantus was ignorant of even the minutest 
 points connected with them ; as, however, we have no definite 
 statements, we must be content to remain in doubt. 
 
 1 Legendre proved (Zahlentheorie, tr. Maser, I. p. 386), that numbers of this form are 
 the only odd numbers which are not divisible into three squares. 
 3 Note on Diophantus v. 1 1 .
 
 no INTRODUCTION 
 
 (c) Composition of numbers as the sum of four squares. 
 
 Every number is either a square or the sum of two, three or four 
 squares. This well-known theorem, enunciated by Fermat 1 , and 
 proved by Lagrange 2 (who followed up results obtained by Euler) 
 shows at once that any number can be divided into four squares 
 either integral or fractional, since any square number can be divided 
 into two other squares, integral or fractional. We have now to look 
 for indications in the Arithmetica as to how far Diophantus was 
 acquainted with the properties of numbers as the sum of four squares. 
 Unfortunately, it is impossible to decide this question with anything 
 like certainty. There are three problems, iv. 29, 30 and V. 14, in 
 which it is required to divide a number into four squares, and from 
 the absence of mention of any condition to which the number must 
 conform, considering that in both cases where a number is to 
 be divided into three or two squares, v. 1 1 and V. 9, he does 
 state a condition, we should probably be right in inferring that 
 Diophantus was aware, at least empirically, that any number could 
 be divided into four squares. That he was able to prove the 
 theorem scientifically it would be rash to assert. But we may 
 at least be certain that Diophantus came as near to the proof of 
 it as did Bachet, who takes all the natural numbers up to 120 
 and finds by trial that all of them can actually be expressed 
 as squares, or as the sum of two, three or four squares in whole 
 numbers. So much we maybe sure that Diophantus could do, and 
 hence he might have empirically satisfied himself that it is possible 
 to divide any number into four squares, integral or fractional, even 
 if he could not give a rigorous mathematical demonstration of the 
 general theorem. 
 
 1 See note on Diophantus IV. 29 ; cf. also section I. of the Supplement. 
 
 z " Demonstration d'un theoreme d'arithmetique " in Nouveaux Memoires de PAcad. 
 royale des sciences de Berlin, annee 1770, Berlin 1772, pp. 123-133= Oeuvres de Lagrange, 
 in. pp. 187-201 ; cf. Wertheim's account of the proof in his Diophantus, pp. 324-330.
 
 CHAPTER VI 
 
 THE PLACE OF DIOPHANTUS 
 
 IN algebra, as in geometry, the Greeks learnt the beginnings 
 from the Egyptians. Familiarity on the part of the Greeks with 
 Egyptian methods of calculation is well attested. Thus (i) Psellus 
 in the letter published by Tannery 1 speaks of "the method 
 of arithmetical calculations used by the Egyptians, by which 
 problems in analysis are handled " (77 /car' Alyvrrrtovf rtov 
 aptdfjbwv /j,edoBo<i, 81 779 oiKovofj^eirat ra Kara rrjv dva\VTiicr)v 
 7rpo/3A,?7/4aTa) ; the details which he goes on to give respecting 
 the technical terms for different kinds of numbers, including the 
 powers of the unknown quantity, in use among the Egyptians 
 are doubtless taken from Anatolius. (2) The scholiast to Plato's 
 Charmides 165 E may be drawing on the same source when he 
 says that " parts of \oyia-TtKij (the science of calculation) are the 
 so-called Greek and Egyptian methods in multiplications and 
 divisions, and the additions and subtractions of fractions.... The 
 aim of it all is the service of common life and utility for contracts, 
 though it seems to deal with things of sense as if they were 
 perfect or abstract." (3) Plato himself, in the Laws 2 , says that 
 free-born boys should, as is the practice in Egypt, learn, side by 
 side with reading, simple mathematical calculations adapted to their 
 age, which should be put into a form such as to give amusement 
 and pleasure as well as instruction ; e.g. there should be different 
 distributions of such things as apples, garlands, etc., different 
 arrangements of numbers of boys in contests of boxing or wrestling, 
 illustrations by bowls of different metals, gold, copper, silver, etc., 
 and simple problems of calculation of mixtures ; all of which are 
 useful in military and civil life and " in any case make men more 
 useful to themselves and more wide-awake." 
 
 1 Dioph. II. pp. 37-42. 2 Laws, VII. 819 A-c.
 
 ii2 INTRODUCTION 
 
 The Egyptian calculations here in point (apart from their 
 method of writing and calculating in fractions, which differed 
 from that of the Greeks in that the Greeks worked with ordinary 
 fractions, whereas the Egyptians separated fractions into sums 
 of submultiples, with the exception of | which was not decomposed) 
 are the ^^-calculations. Hau, meaning a heap, is the term used 
 to denote the unknown quantity, and the calculations in terms 
 of it are equivalent to the solutions of simple equations with one 
 unknown quantity 1 . Examples from the Papyrus Rhind 2 corre- 
 spond to the following equations: 
 
 = 19, 
 
 (jr+f*)-i (* + !*) =10- 
 
 Before leaving the Egyptians, it is right to mention their 
 anticipation, though in an elementary form, of a favourite method 
 of Diophantus, that of the " false supposition " or " regula falsi " 
 as it is sometimes called. An arbitrary assumption is made as to 
 the value of the unknown, and the value is afterwards corrected 
 by a comparison of the result of substituting the wrong value in 
 the original expression with the actual fact. Two instances 
 mentioned by Cantor 3 may be given. The first, taken from the 
 Papyrus Rhind, is the problem of dividing 100 loaves among five 
 persons in numbers forming an arithmetical progression and such 
 that one- seventh of the sum of the first three shares is equal to 
 the sum of the other two. If a + qd, a + ^d, a+2d, a + d, a 
 are the shares, we have 
 
 or d = $^a. 
 
 Ahmes merely says, without explanation, " make the difference, 
 as it is, 5^," and then, assuming a=i, writes the series 23, 17^, 
 12, 6^, i. The addition of these gives 60, and 100 is if times 60. 
 Ahmes says simply "multiply if times" and thus gets the correct 
 values 38^, 294, 20, iof , if. The second instance (taken from 
 the Berlin Papyrus 6619) is the solution of the equations 
 
 x*+y*=ioo, 
 x \y = i : f , or y = x. 
 
 1 For a complete account of the subject the reader is referred to Moritz Cantor's 
 Geschichte der Mathematik, I 3 . Chapter II., especially pp. 74-81. 
 
 2 Eisenlohr, Ein mathematisches Handbuch der alten Agypter {Papyrus Rhind des 
 British Museum) , Leipzig, 1877. 
 
 3 Geschichte der Math. I 3 . pp. 78-9 and p. 95.
 
 THE PLACE OF DIOPHANTUS 113 
 
 x is first assumed to be I, and x*+jP is then found to be 25/16. 
 In order to make 100, 25/16 has to be multiplied by 64 or S 2 . The 
 true value of x is therefore 8 times I, or 8. 
 
 The simple equations solved in the Papyrus Rhind are just the 
 kind of equations of which we find numerous examples in the 
 arithmetical epigrams included in the Greek Anthology. Most 
 of these appear under the name of Metrodorus, a grammarian, 
 who probably lived about the time of the Emperors Anastasius I. 
 (491-518 A.D.) and Justin I. (518-527 A.D.). They were obviously 
 only collected by Metrodorus, from ancient as well as more recent 
 sources ; none of them can with certainty be attributed to Metro- 
 dorus himself. Many of the epigrams (46 in number) lead to 
 simple equations, with one unknown, of the type of the hau- 
 equations, and several of them are problems of dividing a number 
 of apples or nuts among a certain number of persons, that is 
 to say, the very type of problem alluded to by Plato. For 
 example, a number of apples has to be determined such that, if 
 four persons out of six receive one-third, one-eighth, one-fourth 
 and one-fifth respectively of the total number of apples, while the 
 fifth person receives ten apples, there remains one apple as the 
 share of the sixth person, i.e. 
 
 We are reminded of Plato's allusion to problems about bowls 
 (<f>id\at) of different metals by two problems (Antliol. Palat. XIV. 
 12 and 50) in which the weights of bowls have to be found. We 
 can now understand the allusions of Proclus 1 and the scholiast 
 on Charmides 165 E to p,rj\lTac and (f>ia\iTai dpt0(j,oi, the adjectives 
 being respectively formed from p.fi\ov y an apple, and </>taXi;, a 
 bowl. It is clear from Plato's allusions that the origin of such 
 simple algebraical problems dates back, at least, to the fifth 
 century B.C. 
 
 I have not thought it worth while to reproduce at length the 
 problems contained in the Anthology 2 , but the following is a 
 classification of them, (i) Twenty-three are simple equations 
 containing one unknown and of the type shown above ; one of 
 these is the epigram on the age of Diophantus and incidents 
 in his life (XIV. 126). (2) Twelve more are easy simultaneous 
 
 1 Proclus, Comment, on Eucl. /., ed. Friedlein, p. 40, 5. 
 
 2 They are printed in Greek, with the scholia, in Tannery's edition of Diophantus 
 (il. pp. 43-72 and x), and they are included in Wertheim's German translation of 
 Diophantus, pp. 331-343. 
 
 H. D. 8
 
 ii4 INTRODUCTION 
 
 equations containing two unknowns, and of the same sort as 
 Diophantus I. 1-6 ; or, of course, they can be reduced to a simple 
 equation with one unknown by means of an easy elimination. 
 One other (XIV. 51) gives simultaneous equations in three un- 
 knowns 
 
 and one (XIV. 49) gives four equations in four unknowns 
 
 With these may be compared Diophantus I. 16-21. (3) Six more 
 are problems of the usual type about the filling of vessels by pipes : 
 e.g. (XIV. 130) one pipe fills the vessel in one day, a second in two, 
 and a third in three ; how long will all three running together 
 take to fill it? Another about brickmakers (XIV. 136) is of the 
 same sort 
 
 The Anthology contains (4) two indeterminate equations of 
 the first degree which can be solved in positive integers in an 
 infinite number of ways (xiv. 48 and 144); the first is a distribution 
 of apples satisfying the equation x ^y =y, where y is not less 
 than 2, and the original number of apples is $x ; the second leads 
 to the following three equations between four unknown quantities : 
 
 the general solution of which is x = 4^, y = , x l = $k, y^ = 2k. These 
 very equations, made however determinate by assuming that 
 x+y = x l +j/i = 100, are solved in Diophantus I. 12. 
 
 We mentioned above the problem in the Anthology (XIV. 49) 
 leading to the following four simultaneous linear equations with 
 four unknown quantities, 
 
 x + y = a, 
 
 x + ti = c, 
 
 The general solution of any number of simultaneous linear 
 equations of this type with the same number of unknown quantities 
 was given by Thymaridas, apparently of Paros, and an early 
 Pythagorean. He gave a rule, e</>oSo<?, or method of attack (as
 
 THE PLACE OF DIOPHANTUS 115 
 
 lamblichus 1 , our informant, calls it) which must have been widely 
 known, inasmuch as it was called by the name of the e-n-avdrffia, 
 " flower" or "bloom," of Thymaridas. The rule is stated in general 
 terms, but, though no symbols are used, the content is pure 
 algebra. Thymaridas, too, distinguishes between what he calls 
 dopia-Tov, the undefined or unknown quantity, and the atpio-pevov, 
 the definite or known, therein anticipating the very phrase of 
 Diophantus, TrXijOos povdSow dopta-rov, "an undefined number of 
 units," by which he describes his dpidfjios or x. Thymaridas' rule, 
 though obscurely expressed, states in effect that, if there are n 
 equations between n unknown quantities ;r, x lt x^...x n _ l of the 
 following form, 
 
 x*+ ... + x 
 then the solution is given by 
 
 2 
 
 lamblichus goes on to show that other types of equations can 
 be reduced to this, so that the rule does not leave us in the lurch 
 (ov Tra/aeX/cet) in those cases either. Thus we can reduce to 
 Thymaridas' form the indeterminate problem represented by the 
 following three linear equations between four unknown quantities : 
 
 b(u + y\ 
 
 From the first equation we obtain 
 
 x + y + z + u = (a + i) (z + u), 
 
 from which it follows that, if x, y, z, u are all to be integers, 
 x+y + z + u must contain a+i as a factor. Similarly it must 
 contain b + I and c + I as factors. 
 
 Suppose now that x+y+z+u = (a+ i)(+ i)fc+ i). There- 
 fore, by means of the first equation, we get 
 
 (x+y} i + =(+ !)(*+!)(*+ I), 
 
 1 lamblichus, / Nicomachi arithmeticam introductionem (ed. Pistelli), pp. 62, 
 18-68, 26. 
 
 82
 
 n6 INTRODUCTION 
 
 or 
 
 Similarly x+ z = b (c + i)(a+ i), 
 
 x + u = c(a+ i)(b+ i), 
 
 and the equations are in the form to which Thymaridas' rule is 
 applicable. 
 
 Hence, by that rule, 
 
 _ a(b+ i)(V+i) +...-(<?+ I) (b 4- 
 
 In order to ensure that x may always be integral, if is only 
 necessary to assume 
 
 x+y+z + u = 2(a+ i)(b + i)(c+ i). 
 
 The factor 2 is of course determined by the number of un- 
 knowns. If there are n unknowns, the factor to be put in place 
 of 2 is n 2. 
 
 lamblichus has the particular case corresponding to a = 2, 
 b = 3, c = 4. He goes on to apply the method to the equations 
 
 ^0"M), 
 
 for the case where //= f , mjn = |, pjq = f . 
 
 Enough has been said to show that Diophantus was not the 
 inventor of Algebra. Nor was he the first to solve indeterminate 
 problems of the second degree. 
 
 Take, first, the problem of dividing a square number into two 
 squares (Diophantus II. 8), or of finding a right-angled triangle 
 with sides in rational numbers. This problem was, as we learn 
 from Proclus 1 , attributed to Pythagoras, who was credited with 
 the discovery of a general formula for finding such triangles which 
 may be shown thus : 
 
 - i\ 2 _ /;? 2 H 
 
 \ 2 
 
 where n is an odd number. Plato again is credited, according 
 to the same authority, with another formula of the same sort, 
 
 Comment, on Euclid, Book /. pp. 428, 7 sqq.
 
 THE PLACE OF DIOPHANTUS 117 
 
 Both these formulae are readily connected with the geometrical 
 proposition in Eucl. II. 5, the algebraical equivalent of which may 
 be stated as 
 
 The content of Euclid Book II. is beyond doubt Pythagorean, and 
 this way of stating the proposition quoted could not have escaped 
 the Pythagoreans. If we put I for b and the square of any odd 
 number for a, we have the Pythagorean formula ; and, if we put 
 a = 211*, b = 2, we obtain Plato's formula. Euclid finds a more 
 general formula in Book X. (Lemma following X. 28). Starting 
 with numbers u = c + b and v = c b, so that 
 
 uv = c z - P, 
 
 Euclid points out that, in order that uv may be a square, u and v 
 must be " similar plane numbers " or numbers of the form mnfP, 
 mngr*. Substituting we have 
 
 But the problem of finding right-angled triangles in rational 
 numbers was not the only indeterminate problem of the second 
 degree solved by the Pythagoreans. They solved the equation 
 
 2x*-y*= i . 
 
 in such a way as to prove that there are an infinite number of 
 solutions of that equation in integral numbers. The Pythagoreans 
 used for this purpose the system of "side-" and "diagonal-" 
 numbers 1 , afterwards fully described by Theon of Smyrna 2 . We 
 begin with unity as both the first "side" and the first " diagonal"; 
 
 thus 
 
 # 1= i, d^= i. 
 
 We then form (a 2 , d 2 ), (a 3 , d 3 ), etc., in accordance with the following 
 law, 
 
 and so on. Theori states, with reference to these numbers, the 
 general proposition that 
 
 d n 2 =2a n * i, 
 and observes that (i) the signs alternate as successive d's and a's 
 
 1 See Proclus, In Platonis rempublicam commtntarii (Teubner, Leipzig), Vol. II. 
 c. 27, p. 27, 11-18. 
 
 2 Theon of Smyrna, ed. Hiller, pp. 43, 44.
 
 n8 INTRODUCTION 
 
 are taken, d? 2a? being equal to i, d?-2a? equal to+i, 
 d? - 2# 3 2 equal to - I and so on, (2) the sum of the squares of all 
 the d's will be double of the sum of the squares of all the a's. For 
 the purpose of (2) the number of successive terms in each series, 
 if finite, must of course be even. The algebraical proof is easy. 
 
 and so on, while df 2a? = I. Proclus tells us that the property 
 was proved by means of the theorems of Eucl. II. 9, 10, which 
 are indeed equivalent to 
 
 (2.X +j) 2 - 2 (X +_X) 2 = 2* 2 -f. 
 
 Diophantus does not particularly mention the indeterminate 
 equation 2x* i =j 2 , still less does he mention "side-" and 
 "diagonal-" numbers. But from the Lemma to VI. 15 (quoted 
 above, p. 69) it is clear that he knew how to find any number of 
 solutions when one is known. Thus, seeing that x= i, /= i is 
 one solution, he would put 
 
 2 (i + xf i = a square 
 
 = (px- i) 2 say, 
 whence # = (4 + 2/)/(/ 2 - 2). 
 
 Take the value p=2, and we have ^ = 4, or ;tr+i=5; and 
 2 . 5 2 i = 49 = 7 2 . Putting x + 5 in place of x, we find a still 
 higher value, and so on. 
 
 In a recent paper Heiberg has published and translated, and 
 Zeuthen has commented on, still further Greek examples of in- 
 determinate analysis 1 . They come from the Constantinople MS. 
 (probably of I2th c.) from which Schone edited the Metrica of 
 Heron. The first two of the thirteen problems had been published 
 before (though in a less complete form) 2 ; the others are new. 
 
 The first bids us find two rectangles such that the perimeter 
 of the second is three times that of the first, and the area of the 
 first is three times that of the second (the first of the two con- 
 ditions is, by some accident, omitted in the text). The number 3 
 
 1 Bibliotheca Mathematica, vm 3 , 1907-8, pp. 118-134. 
 
 2 Hultsch's Heron, Geeponica t 78, 79. The two problems are discussed by Cantor, 
 Agrimensoren, p. 62, and Tannery, Mem. dc la soc. des sc. de Bordeaux, IV 2 , 1882.
 
 THE PLACE OF UIOPHANTUS 119 
 
 is of course only an illustration, and the problem is equivalent to 
 the solution of the equations 
 
 xy = n.uv } 
 
 the solution given in the text is equivalent to 
 x = 2n*-i, y = 2n* ) 
 u = n (4>/ s 2), v = n) 
 
 Zeuthen suggests that this solution may have been arrived at 
 thus. As the problem is indeterminate, it would be natural to 
 make trial of some hypothesis, e.g. to put v = n. It would follow 
 from the first equation in (i) that u is a multiple of n, say nz. We 
 have then 
 
 x +y =1+2, 
 
 xy = n s 2, 
 
 whence xy = n 3 (x +y) n 3 , 
 
 or (x - n 3 ) (y - 3 ) = n 3 (n 3 - i ). 
 
 An obvious solution of this is 
 
 x n 3 = n 3 i, y 1? = n 3 . 
 
 The second problem is equivalent to the solution of the 
 equations 
 
 I ........................ (I); 
 
 xy = n . uv) 
 
 and the solution given in the text is 
 
 i ..................... (2), 
 
 In this case trial may have been made of the assumption 
 
 v = nx, y = 2 #, 
 when the first equation in (i) would give 
 
 (-i);r = ( 2 -i)/,, 
 a solution of which is x= w 2 i, u = n i. 
 
 The fifth problem is of interest in one respect. We are asked 
 to find a right-angled triangle (in rational numbers) with area 
 of 5 feet. We are told to multiply 5 by some square containing 6 
 as a factor, e.g. 36. This makes 180, and this is the area of the 
 triangle (9, ^b, til). Dividing each side by 6, we have the triangle 
 required. The author, then, is aware of the fact that the area 
 of a right-angled triangle with sides in whole numbers is divisible
 
 120 INTRODUCTION 
 
 by 6. If we take the Euclidean formula for a right-angled triangle, 
 thus making the sides 
 
 m 2 # 2 m* 4- # 2 
 
 a . mn. a . - , a . , 
 
 2 2 
 
 where a is any number, and m, n are numbers which are both odd 
 or both even, the area is 
 
 2 mn (m n} (m + n) 
 
 4 
 
 and, as a matter of fact, the numerator mn(m n)(m + n) is 
 divisible by 24, as was proved later (for another purpose) by 
 Leonardo of Pisa 1 . ' There is no sign that Diophantus was aware 
 of the proposition ; this however may be due to the fact that he 
 does not trouble as to whether his solutions are integral, but is 
 satisfied with rational results. 
 
 The last four problems (numbered 10 to 13) are of great 
 interest. They are different particular cases of one problem, that 
 of finding a rational right-angled triangle such that the numerical 
 sum of its area and all its three sides is a given number. The 
 author's solution depends on the following formulae, where a, b 
 are the perpendiculars, and c the hypotenuse, of a right-angled 
 triangle, 6" its area, r the radius of its inscribed circle, and 
 
 S = rs = \ab> r + s = a + b, c = s r. 
 
 (The proof of these formulae by means of the usual figure, that 
 used by Heron to prove his formula for the area of a triangle 
 in terms of its sides, is easy.) 
 
 Solving the first two equations, in order to find a and b, we 
 have 
 
 a] r + s + <J{(r + s) 2 - 8rs] 
 
 which formula is actually used by the author for finding a and b. 
 The method employed is to take the sum of the area and the three 
 sides 5 + 2s, separated into its two obvious factors s (r + 2), to put 
 s(r+2) = A (the given number), and then to separate A into 
 suitable factors to which s and r+ 2 may be equated. They must 
 obviously be such that sr, the area, is divisible by 6. To take the 
 first example where A is equal to 280 : the possible factors are 
 
 1 Scritti, ed. B. Boncompagni, n. (1862), p. 264. Cf. Cantor, Gesch. d. Math. ii lt 
 p. 40.
 
 THE PLACE OF DIOPHANTUS 121 
 
 2x140, 4x70, 5x56, 7x40, 8x35, 10x28, 14x20. The 
 suitable factors in this case are r + 2 = 8, s = 35, because r is then 
 equal to 6, and rs is a multiple of 6. 
 The author then says that 
 
 a = 6 + 35- v1(6+35) 2 -8. 6.35)^41- i ^ 2Q> 
 
 2 2 
 
 and ^=35-6 = 29. 
 
 The triangle is therefore (20, 21, 29) in this case. The 
 triangles found in the other cases, by the same method, are 
 (9, 40, 41), (8, 15, 17) and (9, 12, 15). 
 
 Unfortunately there is no guide to the date of the problems 
 just given. The form, however, cannot be that in which the 
 discoverer or discoverers of the methods indicated originally 
 explained those methods. The probability is that the original 
 formulation of the most important of the problems belongs to 
 the period between Euclid and Diophantus. This supposition best 
 agrees with the fact that the problems include nothing taken from 
 the great collection in the Arithmetica. On the other hand, it is 
 strange that none of the seven problems above mentioned is found 
 in Diophantus. The five of them which relate to rational right- 
 angled triangles might well have been included by him ; thus he 
 finds rational triangles such that the area plus or minus one of the 
 perpendiculars is a given number, but not the rational triangle 
 which has a given area ; and he finds rational triangles such that 
 the area plus or minus the sum of two sides is a given number, 
 but not the rational triangle such that the sum of the area and 
 the three sides is a given number. The omitted problems might, 
 it is true, have come in the lost Books ; but, on the other hand, 
 Book VI. is the place where we should have expected to find 
 them. Nor do we find in the above problems any trace of 
 Diophantus' peculiar methods. 
 
 Lastly, the famous Cattle-Problem attributed to Archimedes 1 
 has to be added to the indeterminate problems propounded before 
 Diophantus' time. According to the heading prefixed to the 
 epigram, it was communicated by Archimedes to the mathe- 
 maticians at Alexandria in a letter to Eratosthenes. The scholiast 
 
 1 Archimedes, ed. Heiberg, Vol. II. p. 450 sqq.
 
 122 INTRODUCTION 
 
 on Charmides 165 E also refers to the problem "called by Archi- 
 medes the Cattle-Problem." Krumbiegel, who discussed the 
 arguments for and against the attribution to Archimedes, con- 
 cluded apparently that, while the epigram can hardly have been 
 written by Archimedes in its present form, it is possible, nay 
 probable, that the problem was in substance originated by 
 Archimedes 1 . Hultsch 2 has a most attractive suggestion as to 
 the occasion of it. It is known that Apollonius in his WKVTOKIOV 
 had calculated an approximation to the value of TT closer than 
 that of Archimedes, and he must therefore, have worked out more 
 difficult multiplications than those contained in the Measurement 
 of a circle. Also the other work of Apollonius on the multipli- 
 cation of large numbers, which is partly preserved in Pappus, was 
 inspired by the Sand-reckoner of Archimedes ; and, though we 
 need not exactly regard the treatise of Apollonius as polemical, 
 yet it did in fact constitute a criticism of the earlier book. That 
 Archimedes should then reply with a problem involving such a 
 manipulation of immense numbers as would be difficult even for 
 Apollonius is not altogether outside the bounds of possibility. And 
 there is an unmistakable vein of satire in the opening words of 
 the epigram, " Compute the number of the oxen of the Sun, giving 
 thy mind thereto, if thou hast a share of wisdom," in the tran- 
 sition from the first part to the second, where it is said that 
 ability to solve the first part would entitle one to be regarded 
 as " not unknowing nor unskilled in numbers, but still not yet 
 to be counted among the wise," and again in the. last lines. 
 Hultsch concludes that in any case the problem is not much 
 later than the time of Archimedes and dates from the beginning 
 of the second century B.C. at the latest. 
 
 I have reproduced elsewhere 3 , from Amthor, details regarding 
 the solution of the problem, and I need do little more than state 
 here its algebraical equivalent. Eight unknown quantities have 
 to be found, namely, the numbers of bulls and cows respectively 
 of each of four colours (I use large letters for the bulls and small 
 letters for the cows). The first part of the problem connects the 
 eight unknowns by seven simple equations ; the second part adds 
 two more conditions. 
 
 1 Zeitschrift fur Math. n. Physik (Hist. lilt. Abtheilung), xxv. (1880), p. 121 sq. 
 Amthor added (p. 1 53 sq. ) a discussion of the problem itself. 
 
 2 Art. Archimedes in Pauly-Wissowa's Real-Encyclopcidie, n. i, pp. 534, 535. 
 
 3 The Works of Archimedes, pp. 319-326.
 
 THE PLACE OF DIOPHANTUS 123 
 
 First part of Problem. 
 
 (I) W=($ + ^X+Y .................. (i), 
 
 .................. (2), 
 
 .................. (3). 
 
 (II) w = (i + i)(.T + *) ........ .......... (4), 
 
 ) .................. (5), 
 
 +0 .................. (7)- 
 
 Second part. 
 
 W+X = a. square ........................... (8), 
 
 F + /? = a triangular number ............ (9). 
 
 The solution of the first part gives 
 
 W = 10366482 n, w = 7206360 n, 
 X = 74605 1 4 n, x = 4893246 n, 
 Y= 4149387 w, y= 5439213 , 
 Z= 7358060 , -s: = 35 1 5820 , 
 
 where is an integer. The solution given by the scholiast 1 corre- 
 sponds to n = 80. 
 
 The complete problem would not be unmanageable but for the 
 condition (8). If for this were substituted the requirement that 
 W + X shall be merely a product of two unequal factors (" Wurm's 
 problem "), the solution in the least possible numbers is 
 W= 1217263415886, w = 846 1 924 1 0280, 
 
 x= 876035935422, *= 574579625058, 
 
 F= 487233469701, 7 = 638688708099, 
 Z= 864005479380, # = 412838131860. 
 
 But, if we include condition (8) and first of all find a solution 
 satisfying the conditions (i) to (8), we have then, in order to 
 satisfy condition (9), to solve the equation 
 
 q(q+\)J2 = 5 1285802909803 . f 
 
 If we multiply by 8, and put 
 
 2q + I = /, 2 . 4657 | = it, 
 we have the equation 
 
 ^-1 = 2.3.7. 1 1. 29. 353. 2 , 
 or f-- 4729494 2 = i. 
 
 1 Archimedes, ed. Heiberg, Vol. II. pp. 454, 455.
 
 i2 4 INTRODUCTION 
 
 The value of W would be a number containing 206545 
 digits. 
 
 Such are the very few and scattered particulars which we 
 possess of problems similar to those of Diophantus solved or 
 propounded before his time. They show indeed that the kind of 
 problem was not invented by him, but on the other hand they 
 show little or no trace of anything like his characteristic alge- 
 braical methods. In the circumstances, and in default of discovery 
 of fresh documents, the question how much of his work represents 
 original contributions of his own to the subject must remain a 
 matter of pure speculation. It is pretty obvious that one man 
 could not have been the author of all the problems contained in 
 the six Books. There are also inequalities in the work ; some 
 problems are very inferior in interest to the remainder, and some 
 solutions may be assumed to be reproduced from other writers 
 of less calibre, since they reveal none of the mastery of the subject 
 which Diophantus possessed. Again, it seems probable that the 
 problem V. 30, which is exceptionally in epigrammatic form, was 
 taken from someone else. The Arithmetica was no doubt a 
 collection, much in the same sense as Euclid's Elements were. And 
 this may be one reason why so little trace remains of earlier 
 labours in the same field. It is well known that Euclid's Elements 
 so entirely superseded the works of the earlier writers of Elements 
 (Hippocrates of Chios, Leon and Theudius) and of the great 
 contributors to the body of the Elements, Theaetetus and Eudoxus, 
 that those works have disappeared almost entirely. So no doubt 
 would Diophantus' work supersede, and have the effect of con- 
 signing to oblivion, any earlier collections of problems of the 
 same kind. But, if it was a compilation, we cannot doubt that 
 it was a compilation in the best sense, therein resembling Euclid's 
 Elements; it was a compilation by one who was a master of the 
 subject, who took account of and assimilated all the best that had 
 been written upon it, arranged the whole of the available material 
 in due and progressive order, but also added much of his own, not 
 only in the form of new problems but also (and even more) in the 
 mode of treatment, the development of more general methods, and 
 so on. 
 
 It is perhaps desirable to add a few words on the previous 
 history of the theory of polygonal numbers. The theory certainly 
 goes back to Pythagoras and the earliest Pythagoreans. The 
 triangle came first, being obtained by first taking I, then adding
 
 THE PLACE OF DIOPHANTUS 125 
 
 2 to it, then 3 to the sum ; each successive number would be 
 represented by the proper number of dots, and, when each number 
 was represented by that number of dots arranged symmetrically 
 under the row representing the preceding number, the triangular 
 form would be apparent to the eye, thus : 
 
 etc. 
 
 Next came the Pythagorean discovery of the fact that a similar 
 successive addition of odd numbers produced 
 successive square numbers, the odd numbers 
 being on that account called gnomons, and again 
 the process was shown by dots arranged to re- 
 present squares. The accompanying figure shows 
 the successive squares and gnomons. 
 
 Following triangles and squares came the figured numbers 
 in which the "gnomons," or the numbers added to make one 
 number of a given form into the next larger of the same form, 
 were numbers in arithmetical progression starting from I, but with 
 common difference 3, 4, 5, etc., instead of I, 2. Thus, if the 
 common difference is 3, so that the successive numbers added to 
 i are 4, 7, 10, etc., the number is a pentagonal number, if the 
 common difference is 4 and the gnomons 5, 9, 1 3, etc., the number 
 is a hexagonal number, and so on. Hence the law that the 
 common difference of the gnomons in the case of a -gon is 
 11 2. 
 
 Perhaps these facts had already been arrived at by Philippus 
 of Opus (4th c. B.C.), who is said to have written a work on 
 polygonal numbers 1 . Next Speusippus, nephew and successor of 
 Plato, wrote on Pythagorean Numbers, and a fragment of his 
 book survives 2 , in which linear numbers, polygonal numbers, 
 triangles and pyramids are spoken of: a fact which leaves no 
 room for doubt as to the Pythagorean origin of all these con- 
 ceptions 3 . 
 
 Hypsicles, who wrote about 170 B.C., is twice mentioned by 
 Diophantus as the author of a " definition " of a polygonal number, 
 
 1 Bioypd(poi, Vitarum scriptores Graeci minores, ed. Westermann, 1845, p. 448. 
 
 2 Theologumena arithmcticae (ed. Ast), 1817, pp. 61, 62 ; the passage is translated with 
 notes by Tannery, Pour Phistoire de la science hellene, pp. 386-390. 
 
 3 Cantor, Geschichte der Mathematik, I 3 , p. 249.
 
 i 2 6 INTRODUCTION 
 
 which is even quoted verbatim 1 . The definition does not mention 
 any polygonal number beyond the pentagonal ; but indeed this 
 was unnecessary : the facts about the triangle, the square and the 
 pentagon were sufficient to enable Hypsicles to pass to a general 
 conclusion. The definition amounts to saying that the nth #-gon 
 (i counting as the first) is 
 
 * (2 +(-!) (a -2)}. 
 
 Theon of Smyrna 2 , Nicomachus 3 and lamblichus 4 all devote 
 some space to polygonal numbers. The first two, who flourished 
 about 100 A.D., were earlier than Diophantus, and are accordingly 
 of interest here. Besides a description of the successive polygonal 
 numbers, Theon gives the theorem that two successive triangular 
 numbers added together give a square. That is, 
 (n-i)n n(n+ i)_ 
 
 ~ 7~. 
 
 2 2 
 
 The fact is of course clear if we divide a square 
 into two triangles as in the figure. 
 
 Nicomachus gave various rules for transforming triangles into 
 squares, squares into pentagons, etc. 
 
 1. If we put two consecutive triangles together we get a square 
 (as in Theon's theorem). 
 
 2. A pentagon is obtained from a square by adding to it a 
 triangle the side of which is i less than that of the square; 
 similarly a hexagon from a pentagon by adding a triangle the side 
 of which is i less than that of the pentagon ; and so on. 
 
 In fact, 
 
 \n {2 + ( n -i)(a- 2)} +i (-i) = J [2 + ( n -i){(a + i) -2}]. 
 Next Nicomachus sets out the first triangles, squares, pentagons, 
 hexagons and heptagons in a diagram thus : 
 
 Triangles 
 
 ... i 
 
 3 
 
 6 
 
 10 
 
 15 
 
 2 i 
 
 28 
 
 36 
 
 45 
 
 55 
 
 Squares 
 
 ... i 
 
 4 
 
 9 
 
 16 
 
 25 
 
 36 
 
 49 
 
 64 
 
 81 
 
 100 
 
 Pentagons 
 
 i 
 
 5 
 
 12 
 
 22 
 
 35 
 
 51 
 
 70 
 
 92 
 
 117 
 
 H5 
 
 Hexagons 
 
 ... i 
 
 6 
 
 15 
 
 28 
 
 45 
 
 66 
 
 91 
 
 120 
 
 153 
 
 190 
 
 Heptagons 
 
 ... i 
 
 7 
 
 18 
 
 34 
 
 55 
 
 81 
 
 112 
 
 148 
 
 189 
 
 235 
 
 and observes 
 
 that 
 
 
 
 
 
 
 
 
 
 
 1 Dioph. i. pp. 470-472- 
 
 2 Expositio rerum mathematicanim ad legendum Platonem ut ilium, ed. Hiller, 
 pp. 31-40. 
 
 8 Introductio arithmetica, ed. Hoche, II. 8-12, pp. 87-99. 
 
 4 In Nicomachi arithmeticam introd., ed. Pistelli, pp. 58-61, 68-72.
 
 THE PLACE OF DIOPHANTUS 127 
 
 3. Each polygon is equal to the polygon immediately above 
 it in the diagram plus the triangle with i less in its side, i.e. the 
 triangle in the preceding column. 
 
 4. The vertical columns are arithmetical progressions, the 
 common difference of which is the triangle in the preceding 
 column. 
 
 But Plutarch, a contemporary of Nicomachus, mentioned 
 another method of transforming triangles into squares: Every 
 triangular number taken eight times and then increased by I gives 
 a square*. That is, 
 
 Diophantus generalised this proposition into his theorem for 
 transforming any polygonal number into a square. 
 If P be a polygonal number, a the number of angles, 
 
 8P (a - 2) + (a - 4) 2 = a square. 
 
 He deduces rules for finding a polygonal number when the 
 side and the number of angles are given, and for finding the side 
 when the number and the number of its angles are given. These 
 fine results and the fragment of the difficult problem of finding 
 the number of ways in which any given number can be a polygonal 
 number no doubt represent part of the original contributions by 
 Diophantus to the theory of that class of numbers. 
 
 1 Plat, quaest. V. 2, 4, 1003 F.
 
 THE ARITHMETICA 
 BOOK I 
 
 PRELIMINARY 
 Dedication. 
 
 " Knowing, my most esteemed friend Dionysius, that you are 
 anxious to learn how to investigate problems in numbers, I have 
 tried, beginning from the foundations on which the science is 
 built up, to set forth to you the nature and power subsisting in 
 numbers. 
 
 " Perhaps the subject will appear rather difficult, inasmuch as 
 it is not yet familiar (beginners are, as a rule, too ready to despair 
 of success) ; but you, with the impulse of your enthusiasm and 
 the benefit of my teaching, will find it easy to master ; for 
 eagerness to learn, when seconded by instruction, ensures rapid 
 progress." 
 
 After the remark that " all numbers are made up of some 
 multitude of units, so that it is manifest that their formation is 
 subject to no limit," Diophantus proceeds to define what he calls 
 the different "species" of numbers, and to describe the abbreviative 
 signs used to denote them. These " species " are, in the first 
 place, the various powers of the unknown quantity from the second 
 to the sixth inclusive, the unknown quantity itself, and units. 
 
 Definitions. 
 
 A square (=x-) is StW/u? (" power "), and its sign is a J with Y 
 
 superposed, thus J r . 
 A cube (=x 3 ) is icvfios, and its sign K Y . 
 
 A square-square (=#*) is 8vvafj,oBvva^i<; 1 , and its sign is J r J. 
 A square-cube (=x 5 ) is 8vva/j,6Kvj3o<;, and its sign JAT r . 
 A cube-cube (= x 6 ) is tcvftoicvfios, and its sign K Y K. 
 
 1 The term Swa/uodiVa/us was already used by Heron (Metrica, ed. Schone, p. 48, 
 n, 19) for the fourth power of a side of a triangle. 
 
 H. D. Q
 
 1 30 THE ARITHMETICA 
 
 " It is," Diophantus observes, " from the addition, subtraction 
 or multiplication of these numbers or from the ratios which they 
 bear to one another or to their own sides respectively that most 
 arithmetical problems are formed" ; and "each of these numbers... 
 is recognised as an element in arithmetical inquiry." 
 
 " But the number which has none of these characteristics, but 
 merely has in it an indeterminate multitude of units (7rX?}$o<? 
 fi.ovd8ci)v dopia-Tov) is called dpiOpos, ' number I and its sign is 
 9 [=*]." 
 
 "And there is also another sign denoting that which is in- 
 variable in determinate numbers, namely the unit, the sign being 
 
 M with o superposed, thus M." 
 
 Next follow the definitions of the reciprocals, the names of 
 which are derived from the names of the corresponding species 
 themselves. 
 
 Thus 
 
 from dpid/jLos [x~\ we derive the term dpidpoarov [= I/*] 
 
 [= i/^ 2 ] 
 [= I/* 3 ] 
 
 8vvafj,o8vvafj,i<; \x*\ SwaftoSwa/jLoa-Tov [= I/*" 4 ] 
 
 Suva/jioicvfios \X 5 ~\ Suva/Motcvftoo-TOv [= I/^ 5 ] 
 
 /eu/3o/ei/3o<? [x 6 ] KvftoKvftoaTov [= I/*" 6 ], 
 
 and each of these has the same sign as the corresponding original 
 species, but with a distinguishing mark which Tannery writes in 
 the form x above the line to the right. 
 Thus JJ'X = l j x \ j u st as 7 * = . 
 
 Sign of Subtraction (minus}. 
 
 " A minus multiplied by a minus makes a plus^ ; a minus 
 multiplied by a plus makes a minus ; and the sign of a minus is a 
 truncated ^f turned upside down, thus fa." 
 
 Diophantus proceeds : " It is well that one who is beginning 
 this study should have acquired practice in the addition, subtraction 
 and multiplication of the various species. He should know how 
 to add positive and negative terms with different coefficients to 
 
 1 The literal rendering would be "A wanting multiplied by a wanting makes a 
 forthcoming." The word corresponding to minus is Xetfis ("wanting"): when it is 
 used exactly as our minus is, it is in the dative Xetyei, but there is some doubt whether 
 Diophantus himself used this form (cf. p. 44 above). For the probable explanation of 
 the sign, see pp. 42-44. The word for "forthcoming" is Uirap!;is, from VTT&PXU, to exist. 
 Negative terms are XeiTrovra eldrj, and positive virdpxoi>ra.
 
 BOOK I ' 131 
 
 other terms \ themselves either positive or likewise partly positive 
 and partly negative, and how to subtract from a combination of 
 positive and negative terms other terms either positive or likewise 
 partly positive and partly negative. 
 
 " Next, if a problem leads to an equation in which certain 
 terms are equal to terms of the same species but with different 
 coefficients, it will be necessary to subtract like from like on both 
 sides, until one term is found equal to one term. If by chance 
 there are on either side or on both sides any negative terms, it will 
 be necessary to add the negative terms on both sides, until the 
 terms on both sides are positive, and then again to subtract like 
 from like until one term only is left on each side. 
 
 " This should be the object aimed at in framing the hypotheses 
 of propositions, that is to say, to reduce the equations, if possible, 
 until one term is left equal to one term ; but I will show you later 
 how, in the case also where two terms are left equal to one term, such 
 a problem is solved" 
 
 Diophantus concludes by explaining that, in arranging the 
 mass of material at his disposal, he tried to distinguish, so far as 
 possible, the different types of problems, and, especially in the 
 elementary portion at the beginning, to make the more simple lead 
 up to the more complex, in due order, such an arrangement being 
 calculated to make the beginner's course easier and to fix what 
 he learns in his memory. The treatise, he adds, has been divided 
 into thirteen Books. 
 
 PROBLEMS 
 
 1. To divide a given number into two having a given 
 difference. 
 
 Given number 100, given difference 40. 
 Lesser number required x. Therefore 
 2^ + 4.0= loo, 
 ^=30. 
 The required numbers are 70, 30. 
 
 2. To divide a given number into two having a given ratio. 
 
 Given number 60, given ratio 3:1. 
 Two numbers x, ^x. Therefore ;r= 15. 
 The numbers are 45, 15. 
 
 1 eTSos, "species," is the word used by Diophantus throughout. 
 
 92
 
 i 3 2 THE ARITHMETTCA 
 
 3. To divide a given number into two numbers such that one 
 is a given ratio of the other plus a given difference 1 . 
 
 Given number 80, ratio 3:1, difference 4. 
 Lesser number x. Therefore the larger is "$x f 4, and 
 4* -f 4 = 80, so that x = 19. 
 The numbers are 61, 19. 
 
 4. To find two numbers in a given ratio and such that their 
 difference is also given. 
 
 Given ratio 5:1, given difference 20. 
 Numbers $x, x. Therefore ^x= 20, x= 5, and 
 the numbers are 25, 5. 
 
 5. To divide a given number into two numbers such that given 
 fractions (not the same) of each number when added together 
 produce a given number. 
 
 Necessary condition. The latter given number must be such 
 that it lies between the numbers arising when the given fractions 
 respectively are taken of the first given number. 
 
 First given number 100, given fractions \ and ^, given 
 
 sum of fractions 30. 
 
 Second part $x. Therefore first part = 3 (30 x). 
 Hence 90 + 2x 100, and x = 5. 
 
 The required parts are 75, 25. 
 
 6. To divide a given number into two numbers such that a 
 given fraction of the first exceeds a given fraction of the other 
 by a given number. 
 
 Necessary condition. The latter number must be less than that 
 which arises when that fraction of the first number is taken which 
 exceeds the other fraction. 
 
 Given number 100, given fractions and respectively, 
 
 given excess 20. 
 
 Second part 6x. Therefore the first part is 4 (;r + 20). 
 Hence lox + 80 = 100, x = 2, and 
 the parts are 88, 12. 
 
 1 Literally "to divide an assigned number into two in a given ratio and difference (ev 
 Xctycf) Kal inrepoxs r-ij 5o0ei<rij)." The phrase means the same, though it is not so clear, as 
 Euclid's expression (Data, Def. 1 1 and passim) boOivrt pelfav 17 ev \kryif. According to 
 Euclid's definition a magnitude is greater than a magnitude "by a given amount (more) 
 than in a (certain) ratio" when the remainder of the first magnitude, after subtracting 
 tlie given amount, has the said ratio to the second magnitude. This means that, if x, y 
 are the magnitudes, d the given amount, and k the ratio, x-d=ky or
 
 BOOK I 133 
 
 7. From the same (required) number to subtract two given 
 numbers so as to make the remainders have to one another a 
 given ratio. 
 
 Given numbers 100, 20, given ratio 3:1. 
 Required number x. Therefore x 20 = 3 (x 100), and 
 x = 140. 
 
 8. To two given numbers to add the same (required) number so 
 as to make the resulting numbers have to one another a given ratio. 
 
 Necessary condition. The given ratio must be less than the 
 ratio which the greater of the given numbers has to the lesser. 
 Given numbers 100, 20, given ratio 3:1. 
 Required number x. Therefore ~$x + 60 =x + 100, and 
 
 9. From two given numbers to subtract the same (required) 
 number so as to make the remainders have to one another a given 
 ratio. 
 
 Necessary condition. The given ratio must be greater than the 
 ratio which the greater of the given numbers has to the lesser. 
 Given numbers 20, 100, given ratio 6 : I. 
 Required number x. Therefore 120 6x = 100 x, and 
 
 10. Given two numbers, to add to the lesser and to subtract 
 from the greater the same (required) number so as to make the 
 sum in the first case have to the difference in the second case 
 a given ratio. 
 
 Given numbers 20, 100, given ratio 4:1. 
 Required numbers. Therefore (20 + x} = 4(100-.*'), and 
 ;r= 7 6. 
 
 11. Given two numbers, to add the first to, and subtract the 
 second from, the same (required) number, so as to make the 
 resulting numbers have to one another a given ratio. 
 
 Given numbers 20, 100, given ratio 3:1. 
 Required number x. Therefore yc 30x3 =x+ 20, and 
 x = 160. 
 
 12. To divide a given number twice into two numbers such 
 that the first of the first pair may have to the first of the second 
 pair a given ratio, and also the second of the second pair to the 
 second of the first pair another given ratio.
 
 134 THE ARITHMETIC A 
 
 Given number 100, ratio of greater of first parts to lesser 
 of second 2:1, and ratio of greater of second parts 
 to lesser of first parts 3:1. 
 
 x lesser of second parts. 
 
 The parts then are 
 
 -<5r) 
 
 and 
 
 IOO2X) X \ 
 
 Therefore 300 - $x= 100, x= 40, and 
 the parts are (80, 20), (60, 40). 
 
 1 3. To divide a given number thrice into two numbers such that 
 one of the first pair has to one of the second pair a. given ratio, 
 the second of the second pair to one of the third pair another 
 given ratio, and the second of the third pair to the second of the 
 first pair another given ratio. 
 
 Given number 100, ratio of greater of first parts to lesser 
 of second 3:1, of greater of second to lesser of 
 third 2:1, and of greater of third to lesser of 
 first 4: i. 
 
 x lesser of third parts. 
 Therefore greater of second parts = 2x, lesser of second 
 
 = loo- 2x, greater of first = 300 - 6>. 
 Hence lesser of first = 6x 200, so that greater of third 
 
 = 24*- - 800. 
 Therefore 2$x 800 = 100, x = 36, and 
 
 the respective divisions are (84, 16), (72, 28), (64, 36). 
 
 14. To find two numbers such that their product has to their 
 sum a given ratio. [One is arbitrarily assumed.] 
 
 Necessary condition. The assumed value of one of the two 
 must be greater than the number representing the ratio 1 . 
 
 Ratio $ : i, x one of the numbers, 12 the other (> 3). 
 Therefore 1 2x $x + 36, x = 4, and 
 the numbers are 4, 12. 
 
 15. To find two numbers such that each after receiving from 
 the other a given number may bear to the remainder a given 
 ratio. 
 
 Let the first receive 30 from the second, the ratio being 
 then 2:1, and the second 50 from the first, the ratio 
 being then 3:1; take x + 30 for the second. 
 
 1 Literally "the number homonymous with the given ratio."
 
 BOOK I 135 
 
 Therefore the first = 2x 30, and 
 
 (x + 80) = 3 (2x - 80). 
 Thus x = 64, and 
 
 the numbers are 98, 94. 
 
 16. To find three numbers such that the sums of pairs are 
 given numbers. 
 
 Necessary condition. Half the sum of the three given numbers 
 must be greater than any one of them singly. 
 
 Let (i) + (2) = 20, (2) + (3) = 30, (3) + (i) = 40. 
 
 x the sum of the three. Therefore the numbers are 
 
 x 30, x 40, x 20. 
 The sum x = *x go, and x = 45. 
 The numbers are 15, 5, 25. 
 
 17. To find four numbers such that the sums of all sets of three 
 are given numbers. 
 
 Necessary condition. One-third of the sum of the four must be 
 greater than any one singly. 
 
 Sums of threes 22, 24, 27, 20 respectively. 
 
 x the sum of all four. Therefore the numbers are 
 
 X22, * 24, X-27, X2O. 
 
 Therefore ^x 93 x, x = 3 1, and 
 the numbers are 9, 7, 4, n. 
 
 1 8. To find three numbers such that the sum of any pair 
 exceeds the third by a given number. 
 
 Given excesses 20, 30, 40. 
 
 2.x the sum of all three. 
 
 We have (i) + (2) = (3) + 20. 
 
 Adding (3) to each side, we have : twice (3) + 20= 2x, and 
 
 (3) = *- 10. 
 Similarly the numbers (i) and (2) are ^15, x - 20 
 
 respectively. 
 
 Therefore yc 45 = 2x, ^=45, and 
 the numbers are 30, 25, 35. 
 \OtJierwise thus 1 . As before, if the third number (3) is x, 
 
 (l) + (2)=;r+20. 
 
 Next, if we add the equations 
 
 (I) + (2) -( 3 ) = 20) 
 (2) + (3)-(l)= 3 OJ' 
 
 1 Tannery attributes the alternative solution of I. 18 (as of I. 19) to an old scholiast.
 
 136 THE ARITHMETICS 
 
 we have (2) = | (20 + 30) = 25. 
 Hence (i)=.r-5. 
 
 Lastly .u) + (i)-(2) = 40, 
 
 or 2r-5-25 = 4 o. 
 
 Therefore x= 35. 
 
 The numbers are 30, 25, 35.] 
 
 19* To find four numbers such that the sum of any three 
 exceeds the fourth by a given number. 
 
 Necessary condition. Half the sum of the four given differences 
 must be greater than any one of them. 
 Given differences 20, 30, 40, 50. 
 
 2r the sum of the required numbers. Therefore the 
 numbers are 
 
 -r 15, x 20, x 25, x 10. 
 Therefore 4-r 70 = 2jr, and x 35. 
 
 The numbers are 20, 15, 10, 25. 
 
 \Otkeronse tk*s\ If the fourth number (4) is x, 
 
 (I) + (2) + (3) = X + 20. 
 
 Put (2) + (3) equal to half the sum of the two excesses 20 
 and 30, i.e. 25 [this is equivalent to adding the two 
 equations 
 
 It follows by subtraction that (I) = JT- 5. 
 Next we add the equations beginning with (2) and (3) 
 respectively, and we obtain 
 
 (3) + (4) = 1 (30 -I- 40) = 35, 
 so that (3) = 35~^ 
 
 It follows that (2) = jr- ia ' 
 Lastly, since ( 4 ) + (i) + (2)-(3) = 50, 
 
 y- I 5-(35-- r >=50, and ^ = 25. 
 The numbers are accordingly 20, 15, 10, 25.] 
 
 2Oi To divide a given number into three numbers such that the 
 sum of each extreme and the mean has to the other extreme a 
 given ratio. 
 
 Given number ioo; and let (i) + (2)= 3 .(3) and (2) + (3) 
 = 4-(0- 
 
 19 (as of I. 1 8) to an old scholiast.
 
 BOOK I 137 
 
 x the third number. Thus the sum of the first and second 
 
 = 3^r, and the sum of the three =^x= 100. 
 Hence x = 25, and the sum of the first two = 75. 
 Let y be the first 1 . Therefore sum of second and third 
 
 The required parts are 20, 55, 25. 
 
 21. To find three numbers such that the greatest exceeds the 
 middle number by a given fraction of the least, the middle exceeds 
 the least by a given fraction of the greatest, but the least exceeds 
 a given fraction of the middle number by a given number. 
 
 Necessary condition. The middle number must exceed the 
 least by such a fraction of the greatest that, if its denominator 2 be 
 multiplied into the excess of the middle number over the least, the 
 coefficient of x in the product is greater than the coefficient of 
 x in the expression for the middle number resulting from the 
 assumptions made 3 . 
 
 Suppose greatest exceeds middle by \ of least, middle 
 exceeds least by \ of greatest, and least exceeds 
 \ of middle by 10. [Diophantus assumes the three 
 given fractions or submultiples to be one and the 
 same.] 
 x + 10 the least. Therefore middle = 3-r, and greatest 
 
 = 6x - 30. 
 
 Hence, lastly, 6x - 30 yc = (x + 10), 
 or x+ 10 = 93- 90, and x= 12^. 
 
 The numbers are 45, yj\, 22^. 
 
 1 As already remarked (p. 52), Diophantus does not use a second symbol for the 
 second unknown, but makes d/M0/xos do duty for the second as well as for the first. 
 
 2 "Denominator," literally the "number homonymous with the fraction," i.e. the 
 denominator on the assumption that the fraction is, or is expressed as, a submultiple. 
 
 3 Wertheim points out that this condition has reference, not to the general solution of 
 the problem, but to the general applicability of the particular procedure which Diophantus 
 adopts in his solution. Suppose X, Y, Z required such that X- Y=Z\m, Y-Z=X\n, 
 Z-a=Y\p. Diophantus assumes Z=x + a, whence Y=px, X=n(j>x-x-a). The 
 condition states that np n >p. If we solve for x by substituting the values of X, Y, Z 
 in the first equation, we in fact obtain 
 
 m {(np-n-#)x-na}=x + a, 
 or x (mnp - mn - mp - i) = a (tnn+ i). 
 
 In order that the value of x may be positive, we must have mnp>mn + mp -f- 1, 
 that is, 
 
 np>n+p + 
 or (if m, , / are positive integers) np>n +/.
 
 138 THE ARITHMETICA 
 
 [Another solution 1 . 
 
 Necessary condition. The given fraction of the greatest must 
 be such that, when it is added to the least, the coefficient of x in 
 the sum is less than the coefficient of x in the expression for the 
 middle number resulting from the assumptions made 2 . 
 
 Let the least number be x + 10, as before, and the given 
 
 fraction \ ; the middle number is therefore $x. 
 Next, greatest = middle + (least) = $%x + 3$. 
 Lastly, 3* = x + i o + (fa + 3i) 
 
 Therefore x = \2\, and 
 
 the numbers are, as before, 45, 37^, 22^.] 
 
 22. To find three numbers such that, if each give to the next 
 following a given fraction of itself, in order, the results after each 
 has given and taken may be equal. 
 
 Let first give ^ of itself to second, second \ of itself to 
 
 third, third of itself to first. 
 Assume first to be a number of x's divisible by 3, say 
 
 $x, and second to be a number of units divisible by 
 
 4, say 4. 
 
 Therefore second after giving and taking becomes -*"+ 3. 
 Hence the first also after giving and taking must become 
 
 ;tr+3; it must therefore have taken x + $2x, or 
 
 3-;r; ^-x must therefore be of third, or third 
 
 = I$-5* 
 
 Lastly, 15 -5*- (3 -#)+!=*+ 3, 
 or 134^ = ^ + 3, and x=2. 
 
 The numbers are 6, 4, 5. 
 
 23. To find four numbers such that, if each give to the next 
 following a given fraction of itself, the results may all be equal. 
 
 Let first give \ of itself to second, second \ of itself 
 
 to third, third of itself to fourth, and fourth of 
 
 itself to first. 
 Assume first to be a number of ;r's divisible by 3, say yc, 
 
 and second to be a number of units divisible by 4, 
 
 say 4. 
 
 1 Tannery attributes this alternative solution, like the others of the same kind, to an 
 ancient scholiast. 
 
 2 Wertheim observes that the scholiast's necessary condition comes to the same thing 
 as Diophantus' own.
 
 BOOK I 
 
 139 
 
 The second after giving and taking becomes x + 3. 
 Therefore first after giving x to second and receiving 
 of fourth =x+ 3 ; therefore fourth 
 
 = 6(^+3 2x}= i8-6.tr. 
 But fourth after giving $x to first and receiving \ of 
 
 third =-*- + 3 ; therefore third =301: -60. 
 Lastly, third after giving 6x i? to fourth and receiving 
 
 I from second = .*+ 3. 
 That is, 24^ 47=;tr+3, and * = f$. 
 
 The numbers are therefore J^, 4, J^f, J^; 
 or, after multiplying by the common de- 
 nominator, 150, 92, 120, 114. 
 
 24. To find three numbers such that, if each receives a given 
 fraction of the sum of the other two, the results are all equal. 
 
 Let first receive of (second + third), second \ of 
 
 (third + first), and third i of (first + second). 
 Assume first =x, and for convenience' sake (rov irpo^Lpov 
 eveicev) take for sum of second and third a number of 
 units divisible by 3, say 3. 
 Then sum of the three = x-\- 3, 
 
 and first -I- 1 (second + third) = x + I . 
 
 Therefore second + ^ (third + first) = x+ I ; 
 hence 3 times second + sum of all = 4^ + 4, 
 and therefore second = x + %. 
 
 Lastly, third + { (first + second) = x + I, 
 or 4 times third + sum of all = 5^+5, 
 
 and third =x + ^. 
 
 Therefore x + (x + ) + (x + ) = x + 3, 
 and ^ = T!- 
 
 The numbers, after multiplying by the common 
 denominator, are 13, 17, 19. 
 
 25. To find four numbers such that, if each receives a given 
 fraction of the sum of the remaining three, the four results are 
 equal. 
 
 Let first receive of the rest, second 1 of the rest, 
 
 third i of rest, and fourth of rest. 
 Assume first to be x and sum of rest a number of units 
 
 divisible by 3, say 3. 
 Then sum of all =x+ 3. 
 
 Now first + (second -f third 4 fourth) = x + I .
 
 i4o THE ARITHMETICA 
 
 Therefore second + | (third + fourth + first) =^+ I, 
 
 whence 3 times second + sum of all = 4.^ + 4, 
 
 and therefore second = x + \. 
 
 Similarly third =.* + , 
 
 and fourth = ;r+f. 
 
 Adding, we have 4^r + f = ^+3, 
 
 and x== $- 
 
 The numbers, after multiplying by a common 
 denominator, are 47, 77, 92, 101. 
 
 26. Given two numbers, to find a third number which, when 
 multiplied into the given numbers respectively, makes one product 
 a square and the other the side of that square. 
 
 Given numbers 200, 5 ; required number x. 
 
 Therefore 2OOtr = ^r 2 , and 
 
 .27. To find two numbers such that their sum and product are 
 given numbers. 
 
 Necessary condition. The square of half the sum must exceed 
 the product by a square number, ecm Se roOro irXao-fiaTiicbv 1 . 
 Given sum 20, given product 96. 
 2x the difference of the required numbers. 
 Therefore the numbers are io+x, 10 x. 
 Hence ioo-x z = g6. 
 Therefore x=2, and 
 
 the required numbers are 12, 8. 
 
 28. To find two numbers such that their sum and the sum of 
 their squares are given numbers. 
 
 Necessary condition. Double the sum of their squares must 
 exceed the square of their sum by a square, ecrrt Se KOI TOVTO 
 
 1 There has been controversy as to the meaning of this difficult phrase. Xylander, 
 Bachet, Cossali, Schulz, Nesselmann, all discuss it. Xylander translated it by "effictum 
 aliunde." Bachet of course rejects this, and, while leaving the word untranslated, 
 maintains that it has an active rather than a passive signification ; it is, he says, not 
 something "made up" (effictum) but something "a quo aliud quippiam effingi et 
 plasmari potest," " from which something else can be made up," and this he interprets as 
 meaning that from the conditions to which the term is applied, combined with the 
 solutions of the respective problems in which it occurs, the rules for solving mixed 
 quadratics can be evolved. Of the two views I think Xylander's is nearer the mark. 
 Tr\afffj.aTiK6v should apparently mean "of the nature of a ir\dfffj.a," just as 5pa/uariK<5j> 
 means something connected with or suitable for a drama ; and ir\d<r/j.a means something
 
 BOOK I 141 
 
 Given sum 20, given sum of squares 208. 
 Difference 2x. 
 
 Therefore the numbers are lo+x, IQ-X. 
 Thus 200 + 2x* = 208, and x = 2. 
 
 The required numbers are 12, 8. 
 
 29. To find two numbers such that their sum and the difference 
 of their squares are given numbers. 
 
 Given sum 20, given difference of squares 80. 
 Difference 2x. 
 
 The numbers are therefore lO + x, 10 x. 
 Hence (lo + xf- (io-xf= 80, 
 
 or 401: = 80, and x = 2. 
 The required numbers are 12, 8. 
 
 30. To find two numbers such that their difference and product 
 are given numbers. 
 
 Necessary condition. Four times the product together with 
 the square of the difference must give a square, eo-ri Se real rovro 
 
 Given difference 4, given product 96. 
 2x the sum of the required numbers. 
 
 Therefore the numbers are x+2, x 2\ accordingly 
 x* 4 = 96, and x= 10. 
 
 The required numbers are 12, 8. 
 
 31. To find two numbers in a given ratio and such that the 
 sum of their squares also has to their sum a given ratio. 
 
 Given ratios 3 : I and 5 : I respectively. 
 Lesser number x. 
 
 Therefore \ox- = 5 . 4-r, whence x = 2, and 
 the numbers are 2, 6. 
 
 32. To find two numbers in a given ratio and such that the 
 sum of their squares also has to their difference a given ratio. 
 
 Given ratios 3 : I and 10 : I. 
 
 Lesser number x, which is then found from the equation 
 
 ior 2 = 10. ix. 
 Hence* =2, and 
 
 the numbers are 2, 6. 
 
 "formed" or "moulded." Hence the expression would seem to mean "this is of the 
 nature of a formula," with the implication that the formula is not difficult to make up 
 or discover. Nesselmann, like Xylancler, gives it much this meaning, translating it "das 
 lasst sich aber bewerkstelligen." Tannery translates irXaoytart/roi' by "formativum."
 
 i 4 2 THE ARITHMETICA 
 
 33. To find two numbers in a given ratio and such that the 
 difference of their squares also has to their sum a given ratio. 
 
 Given ratios 3 : i and 6 : I. 
 Lesser number x, which is found to be 3. 
 The numbers are 3, 9. 
 
 34. To find two numbers in a given ratio and such that the 
 difference of their squares also has to their difference a given 
 ratio. 
 
 Given ratios 3 : I and 12 : I. 
 
 Lesser number x, which is found to be 3. 
 
 The numbers are 3, 9. 
 
 Similarly by the same method can be found two numbers in 
 a given ratio and (i) such that their product is to their sum in a 
 given ratio, or (2) such that their product is to their difference in a 
 given ratio. 
 
 35. To find two numbers in a given ratio and such that the 
 square of the lesser also has to the greater a given ratio. 
 
 Given ratios 3 : i and 6 : i respectively. 
 Lesser numbers, which is found to be 18. 
 The numbers are 18, 54. 
 
 36. To find two numbers in a given ratio and such that the 
 square of the lesser also has to the lesser itself a given ratio. 
 
 Given ratios 3 : i and 6 : i . 
 Lesser number x, which is found to be 6. 
 The numbers are 6, 18. 
 
 37. To find two numbers in a given ratio and such that the 
 square of the lesser also has to the sum of both a given ratio. 
 
 Given ratios 3 : i and 2:1. 
 Lesser number x, which is found to be 8. 
 The numbers are 8, 24. 
 
 38. To find two numbers in a given ratio and such that the 
 square of the lesser also has to the difference between them a 
 given ratio. 
 
 Given ratios 3 : i and 6 : i. 
 Lesser number x, which is found to be 1 2. 
 The numbers are 12, 36.
 
 BOOK II 143 
 
 'Similarly can be found two numbers in a given ratio and 
 
 (1) such that the square of the greater also has to the 
 
 lesser a given ratio, or 
 
 (2) such that the square of the greater also has to the 
 
 greater itself a given ratio, or 
 
 (3) such that the square of the greater also has to the sum 
 
 or difference of the two a given ratio. 
 
 39. Given two numbers, to find a third such that the sums of 
 the several pairs multiplied by the corresponding third number 
 give three numbers in arithmetical progression. 
 Given numbers 3, 5. 
 Required number x. 
 
 The three products are therefore 3^ + 15, 5^+15, &r. 
 Now 3-r + 1 5 must be either the middle or the least of 
 the three, and $x+i$ either the greatest or the 
 middle. 
 
 (i) 5^+15 greatest, 3^+15 least. 
 
 Therefore *>x + 1 5 + 3-*" + 1 5 = 2 . &tr, and 
 
 (2) 5^+15 greatest, 3.^+15 middle. 
 
 Therefore (5*+ 15) -(3^+ 15) = 3*+ IS -&r, and 
 
 *-? 
 
 (3) % x greatest, yc + 1 5 least. 
 
 Therefore &r + 3* + 1 5 = 2 ($x + 1 5), and 
 
 BOOK II 
 
 [The first five problems of this Book are mere repetitions of problems in 
 Book I. They probably found their way into the text from some ancient 
 commentary. In each case the ratio of one required number to the other 
 is assumed to be 2 : i. The enunciations only are here given.] 
 
 1. To find two numbers such that their sum is to the sum of 
 their squares in a given ratio [cf. i. 31]. 
 
 2. To find two numbers such that their difference is to the 
 difference of their squares in a given ratio [cf. I. 34].
 
 144 THE ARITHMETICA 
 
 3. To find two numbers such that their product is to their sum 
 or their difference in a given ratio [cf. I. 34]. 
 
 4. To find two numbers such that the sum of their squares is to 
 their difference in a given ratio [cf. I. 32]. 
 
 5. To find two numbers such that the difference of their squares 
 is to their sum in a given ratio [cf. I. 33]. 
 
 6 1 . To find two numbers having a given difference and such 
 that the difference of their squares exceeds their difference by a 
 given number. 
 
 Necessary condition. The square of their difference must be 
 less than the sum of the said difference and the given excess 
 of the difference of the squares over the difference of the 
 numbers. 
 
 Difference of numbers 2, the other given number 20. 
 Lesser number x. Therefore x + 2 is the greater, and 
 
 4^+4 = 22. 
 Therefore x = 4^, and 
 
 the numbers are 4^, 6. 
 
 7 1 . To find two numbers such that the difference of their 
 squares is greater by a given number than a given ratio of 
 their difference-. [Difference assumed.] 
 
 Necessary condition. The given ratio being 3:1, the square of 
 the difference of the numbers must be less than the sum of three 
 times that difference and the given number. 
 
 Given number 10, difference of required numbers 2. 
 Lesser number x. Therefore the greater is x+ 2, and 
 
 4^ + 4 = 3.2+ 10. 
 Therefore x = 3, and 
 
 the numbers are 3, 5. 
 
 8. To divide a given square number into two squares 3 . 
 
 1 The problems n. 6, 7 also are considered by Tannery to be interpolated from some 
 ancient commentary. 
 
 2 Here we have the identical phrase used in Euclid's Data (cf. note on p. 132 above) : 
 the difference of the squares is rfjs vTrepoxw avr&v doOtvTi. apid/nf /j.flfai> rj ev \6yif, 
 literally "greater than their difference by a given number (more) than in a (given) ratio," 
 by which is meant "greater by a given number than a given proportion or fraction 
 of their difference." 
 
 3 It is to this proposition that Fermat appended his famous note in which he 
 enunciates what is known as. the "great theorem" of Fermat. The text of the note is 
 as follows : 
 
 "On the other hand it is impossible to separate a cube into two cubes, or a
 
 BOOK II 145 
 
 Given square number 16. 
 
 x* one of the required squares. Therefore \6-x* must 
 be equal to a square. 
 
 Take a square of the form 1 (inx 4)*, m being any 
 integer and 4 the number which is the square root 
 of 1 6, e.g. take (2^ 4)*, and equate it to 16 x*. 
 
 Therefore 4x*\6x+i6=\(:>x\ 
 
 or 5** = \6x, and x = g-. 
 
 The required squares are therefore y-, ^. 
 
 9. To divide a given number which is the sum of two squares 
 into two other squares 2 . 
 
 biquadrate into two biquadrates, or generally any power except a square into two pcnvers 
 with the same exponent. I have discovered a truly marvellous proof of this, which 
 however the margin is not large enough to contain." 
 
 Did Fermat really possess a proof of the general proposition that x m +y m = z l * cannot 
 be solved in rational numbers where m is any number >2? As Wertheim says, one 
 is tempted to doubt this, seeing that, in spite of the labours of Euler, Lejeune-Dirichlet, 
 Kummer and others, a general proof has not even yet been discovered. Euler proved 
 the theorem for m = $ and / = 4, Dirichlet for *w = 5, and Kummer, by means of the 
 higher theory of numbers, produced a proof which only excludes certain particular 
 values of m, which values are rare, at all events among the smaller values of m ; thus 
 there is no value of m below 100 for which Kummer's proof does not serve. (I take 
 these facts from Weber and Wellstein's Encyclopddie der Elementar-Mathematik, I 2 , 
 p. 284, where a proof of the formula for m = + is given.) 
 
 It appears that the Gottingen Academy of Sciences has recently awarded a prize 
 to Dr A. Wieferich, of Miinster, for a proof that the equation x p +y p = g p cannot be 
 solved in terms of positive integers not multiples of p, if 2 P - 2 is not divisible by p*. 
 " This surprisingly simple result represents the first advance, since the time of Kummer, 
 in the proof of the last Fermat theorem " (Bulletin of the American Mathematical Society, 
 February 1910). 
 
 Fermat says ("Relation des nouvelles decouvertes en la science des nombres," 
 August 1659, Oeuvres, II. p. 433) that he proved that no cube is divisible into two cutesby 
 a variety of his method of infinite diminution (descente infinie or indefinie) different from 
 that which he employed for other negative or positive theorems ; as to the other cases, see 
 Supplement, sections I., n. 
 
 1 Diophantus' words are: "I form the square from any number of dp<.0/j.oi minus 
 as many units as there are in the side of 16." It is implied throughout that m must 
 be so chosen that the result may be rational in Diophantus' sense, i.e. rational and 
 positive. 
 
 2 Diophantus' solution is substantially the same as Euler's (Algebra, tr. Hewlett, 
 Part n. Art. 219), though the latter is expressed more generally. 
 
 Required to find x, y such that 
 
 If x /, then y $ g. 
 
 Put therefore -r=/+/te, y=g-qz. 
 
 H. D.
 
 146 THE ARITHMETICA 
 
 Given number 13 = 2 2 + 3 2 . 
 
 As the roots of these squares are 2, 3, take (x + 2)* as the 
 
 first square and (mx 3) 2 as the second (where m is 
 
 an integer), say (2.x 3)*. 
 
 Therefore (x* + 4-*" + 4) + (^x* + 9-1 2x} = 1 3, 
 or 5-r 2 + 13 -8x= 13. 
 
 Therefore x = f , and 
 
 the required squares are ~ t -. 
 
 10. To find two square numbers having a given difference. 
 
 Given difference 60. 
 
 Side of one number x, side of the other x plus any 
 
 number the square of which is not greater than 60, 
 
 say 3. 
 
 Therefore (* + 3) 2 - ;r 2 = 60 ; 
 
 x=%\, and 
 
 the required squares are 72^, 132^. 
 
 1 1. To add the same (required) number to two given numbers 
 so as to make each of them a square. 
 
 (i) Given numbers 2, 3 ; required numbers. 
 
 X -4- 2 1 
 
 Therefore \ must both be squares. 
 
 This is called a double-equation (StTrXoiVo-n;?). 
 To solve it, take the difference between the two expressions 
 and resolve it into two factors 1 ; in this case let us say 
 
 4, i- 
 Then take either 
 
 (a) the square of half the difference between these factors 
 
 and equate it to the lesser expression, 
 or (b) the square of half the sum and equate it to the 
 greater. 
 
 hence iffz +/V - igqz + </ 2 s 2 = o, 
 
 and =3az2# 
 
 so that *=^ l j:\f', y= 
 
 in which we may substitute all possible numbers for/, q. 
 
 1 Here, as always, the factors chosen must be suitable factors, i.e. such as will lead to 
 a "rational " result, in Diophantus' sense.
 
 BOOK II t 4 7 
 
 In this case (a) the square of half the difference is %-. 
 Therefore x + 2 = - 2 ^, and x = g, the squares being ^ 5 -, Jj/. 
 
 Taking (b) the square of half the sum, we have x+ 3 = -^, 
 
 which gives the same result. 
 (2) To avoid a double-equation 1 , 
 
 first find a number which when added to 2, or to 3, 
 
 gives a square. 
 Take e.g. the number x~ 2, which when added to 2 gives 
 
 a square. 
 
 Therefore, since this same number added to 3 gives a 
 square, 
 
 x* + i = a square = (x 4)", say, 
 the number of units in the expression (in this case 4) 
 
 being so taken that the solution may give x- > 2. 
 Therefore jr = - 1 g 5 -, and 
 
 the required number is ^, as before. 
 
 12. To subtract the same (required) number from two given 
 numbers so as to make both remainders squares. 
 Given numbers 9, 21. 
 
 Assuming 9 x* as the required number, we satisfy one 
 condition, and the other requires that 12 +x* shall be 
 a square. 
 Assume as the side of this square x minus some number 
 
 the square of which > 12, say 4. 
 Therefore (x 4)-=\2 + x z , 
 
 and x=\. 
 
 The required number is then 8f . 
 
 [Diophantus does not reduce to lowest terms, but says 
 x = | and then subtracts |-|~from 9 or ^^.] 
 
 1 This is the same procedure as that of Euler, who does not use double-equations. 
 Euler (Algebra, tr. Hewlett, Part II. Art. 214) solves the problem 
 
 Suppose x + 4 = 
 
 therefore jr=/ 2 ~4, and x + 7= 
 
 Suppose that / 
 
 therefore / = (3 - 
 
 Thus X (Q-22 
 
 or, if we take a fraction r[s instead of q, 
 x = (gs* -
 
 148 THE ARITHMETICA 
 
 13. From the same (required) number to subtract two given 
 numbers so as to make both remainders squares. 
 Given numbers 6, 7. 
 
 (i) Let x be the required number. 
 Therefore x ~ I are both squares. 
 
 The difference is I, which is the product of, say, 2 and 
 and, by the rule for solving a double equation, 
 
 (2) To avoid a double-equation, seek a number which exceeds 
 
 a square by 6, say x* + 6. 
 
 Therefore x* I must also be a square = (x 2) 2 , say. 
 Therefore x = f , and 
 
 the required number is ^. 
 
 14. To divide a given number into two parts and to find a 
 square which when added to each of the two parts gives a square 
 number. 
 
 Given number 20. 
 
 Take two numbers 1 such that the sum of their squares 
 < 20, say 2, 3. 
 
 1 Diophantus implies here that the two numbers chosen must be such that the sum of 
 their squares <2O. Tannery pointed out (Bibliotheca Matkematua, 1887, p. 103) that 
 this is not so and that the condition actually necessary to ensure a real solution in 
 Diophantus' sense is something different. We have to solve the equations 
 x+y=a, 2 2 -t-jc = 2 , 2 2 +_y = z> 2 . 
 
 We assume u = z + m, z> = 2 + , and, eliminating x, y, we obtain 
 
 i (m + n) 
 
 In order that z may be positive, we must have n& + 2 < a ; but z need not be positive 
 in order to satisfy the above equations. What is really required is that x, y shall both be 
 positive. 
 
 Now from the above we derive 
 
 Solving for x, y, we have 
 
 _ (m - n) (a + 2tnn) 
 
 m (a + mn - 2 ) _ n (a + mn - m 2 ) 
 m+n ' m+n 
 
 If, of the two assumed numbers, m>n, the condition necessary to secure that x, y shall 
 both be positive is a + mn > m*.
 
 BOOK II 149 
 
 Add x to each and square. 
 We then have 
 
 and, if , [ are respectively subtracted, the remainders 
 
 are the same square. 
 Let then x* be the required square, and we have only to 
 
 make . I tne required parts of 20. 
 
 Thus 10* +13 = 20, 
 
 and X = IQ. 
 
 The required parts are then (?, ^Y and 
 the required square is . 
 
 15. To divide a given number into two parts and to find a 
 square which, when each part is respectively subtracted from it, 
 gives a square. 
 
 Given number 20. 
 
 Take (x -I- m? for the required square \ where m* is not 
 
 greater than 20, 
 e.g. take (x + 2)*. 
 
 This leaves a square if either AX + 4 } . , 
 
 > is subtracted. 
 or 2;r + 3J 
 
 Let these then be the parts of 20. 
 
 1 Here again the implied condition, namely that m- is not greater than 10, is not 
 necessary ; the condition necessary for a real solution is something different. 
 
 The equations to be solved are x+y=a, z^ x u^, z*y=iP. 
 
 Diophantus here puts ( + m)* for z z , so that, if x=im$ + m?, the second equation is 
 satisfied. Now ( + w) 2 -y must also be a square, and if this square is equal to ( + m - )*, 
 say, we must have 
 
 Therefore, since x+y=a, 
 
 i (m + n) + m z + *mn -n*=a, 
 
 a m? + w 2 imn 
 
 whence = -- ; - r - , 
 
 2 (/// + ) 
 
 and it follows that 
 
 _ m (a - mn + 8 ) _ (a - mn + m 2 ) 
 m + tt ' y ~ m + n ' 
 
 If m>n, it is necessary, in order that x, y may both be positive, that a + 2 > mn, 
 which is the true condition for a real solution.
 
 150 THE ARITHMETICA 
 
 Therefore 6x + 7 = 20, and x = J^-. 
 
 The required parts are therefore f^-, ^ j , and 
 
 the required square is ~. 
 
 1 6. To find two numbers in a given ratio and such that each 
 when added to an assigned square gives a square. 
 
 Given square 9, given ratio 3:1. 
 
 If we take a square of side (mx + 3) and subtract 9 
 
 from it, the remainder may be taken as one of the 
 
 numbers required. 
 
 Take, e.g., (x + 3)* 9, or x* + 6x, for the lesser number. 
 Therefore 3^ 2 + i8x is the greater number, and 3-r 2 + 18^+9 
 
 must be made a square = (2.x 3) 2 , say. 
 Therefore x = 30, and 
 
 the required numbers are 1080, 3240. 
 
 17. To find three numbers such that, if each give to the next 
 following a given fraction of itself and a given number besides, 
 the results after each has given and taken may be equal 1 . 
 
 First gives to second \ of itself + 6, second to third of 
 
 itself + 7, third to first j of itself + 8. 
 Let first and second be 5^ 6x respectively. 
 When second has taken x + 6 from first it becomes 7^ + 6, 
 
 and when it has given x + J to third it becomes 
 
 6x-\. 
 But first when it has given x + 6 to second becomes 
 
 ^x 6 ; and this too when it has taken \ of third 
 
 + 8 must become 6x i . 
 Therefore f of third + 8 = 2x + 5, and 
 
 third = i^x 21. 
 Next, third after receiving of second + 7 and giving \ of 
 
 itself + 8 must become 6x I. 
 Therefore I ye 19 = 6x I, and x = . 
 
 The required numbers are ^-, - , . 
 
 1 Tannery is of opinion that the problems II. 17 and 18 have crept into the text 
 from an ancient commentary to Book I. to which they would more appropriately belong. 
 Cf. I- 22, 23.
 
 BOOK II 151 
 
 1 8. To divide a given number into three parts satisfying the 
 conditions of the preceding problem 1 . 
 Given number 80. 
 Let first give to second of itself + 6, second to third 
 
 of itself + 7, and third to first f of itself + 8. 
 [What follows in the text is not a solution of the problem 
 but an alternative solution of the preceding. The 
 first two numbers are assumed to be ^x and 12, and 
 
 the numbers found are ^^, - -.1 
 19 19 19 J 
 
 19. To find three squares such that the difference between the 
 greatest and the middle has to the difference between the middle 
 and the least a given ratio. 
 
 Given ratio 3:1. 
 
 Assume the least square = x 3 , the middle =x* + 2x -\- I. 
 Therefore the greatest = x> + 8x + 4 = square = (x + 3> 2 , say. 
 Thus;tr=2, and 
 
 the squares are 30^, 12^, 6. 
 
 20. To find two numbers such that the square of either added 
 to the other gives a square 2 . 
 
 1 Though the solution is not given in the text, it is easily obtained from the general 
 solutjon of the preceding problem, which again, at least with our notation, is easy. 
 
 Let us assume, with Wertheim, that '-the numbers required in n. 17 are 5.*, 6y, "jz. 
 Then by the conditions of the problem 
 
 4x-6 + z+8 = ;,y-7 + x + 6 = 6z-8+y + 7, 
 from which two equations we can find x, z in terms of y. 
 
 In fact *=(20>-i8)/i9 and z=(\-jy- 3)/i9, 
 
 and the general solution is 
 
 In his solution Diophantus assumes x=y, whence y . 
 
 Now, to solve II. 18, we have only to equate the sum of the three expressions to 80, 
 and so findy. 
 
 We have jgaj 
 
 y(i>. 26 + 6. 19 + 7. 17) -5. 18-7. 3 = 80. 19, ? = -^'> 
 
 and the required numbers are 
 
 944Q 9786 9814 
 363 ' 363 ' 363 ' 
 
 2 Euler (Algebra, Part n. Art. 239) solves this problem more generally thus. 
 Required to find x, y such that x 3 +y and jp + x are squares. 
 
 If we begin by supposing x 2 +y=J?, so that y=f^-x^, and then substitute the value 
 of y in terms of x in the second expression, we must have 
 p\ _ 2 / 2 jc 2 + x* + x = square. 
 
 But, as this is difficult to solve, let us suppose instead that 
 * 2 +y=(P~ *) 2 =? ~ tfx -- J: 2 ,
 
 152 THE ARITHMETICA 
 
 Assume for the numbers x, 2x + I, which by their form 
 
 satisfy one condition. 
 The other condition gives 
 
 4^r 2 + $x+ I = square = (2x 2) 2 , say. 
 Therefore x = -^ , and 
 
 the numbers are , . 
 
 21. To find two numbers such that the square of either minus 
 the other number gives a square. 
 
 x+i, 2x+i are assumed, satisfying one condition. 
 The other condition gives 
 
 4^2 + yc = square = gx 2 , say. 
 Therefore x = f , and 
 
 the numbers are -, . 
 
 22. To find two numbers such that the square of either added 
 to the sum of both gives a square. 
 
 Assume x, x + I for the numbers. Thus one condition is 
 
 satisfied. 
 It remains that 
 
 x z + 4*+2 = square = (x 2) 2 , say. 
 Therefore x = ^, and 
 
 the numbers are -, -. 
 [Diophantus has f, Jg-.] 
 
 23. To find two numbers such that the square of either minus 
 the sum of both gives a square. 
 
 Assume x, x + i for the numbers, thus satisfying one 
 
 condition. 
 
 Then x* 2x i = square = (x 3) 2 , say. 
 Therefore x = 2%, and 
 
 the numbers are 2^, 3^. 
 
 and that j/ 2 + x = (q -yf = q* - iqy +y 2 . 
 
 It follows that 
 
 whence 
 
 tfq - i 4/7 - i 
 
 Suppose, for example, /=2, ?=3, and we have * = , y= ; and so on. We 
 
 must of course choose /, q such that x, y are both positive. Diophantus' solution is 
 obtained by putting p= - i, ^=3.
 
 BOOK II 153 
 
 24. To find two numbers such that either added to the square 
 of their sum gives a square. 
 
 Since x 3 + $x*, x 2 + Sx* are both squares, let the numbers 
 
 be $x z , 8x* and their sum x. 
 Therefore I2ix* = x 2 , whence I ix* = x, and x = Jp 
 
 The numbers are therefore , . 
 
 25. To find two numbers such that the square of their sum 
 minus either number gives a square. 
 
 If we subtract 7 or 12 from 16, we get a square. 
 
 Assume then \2x* t jx* for the numbers, and i6x- for the 
 
 square of their sum. 
 Hence i$x* = 4*, and x = ^. 
 
 The numbers are ^ 2 , ** 2 . 
 
 26. To find two numbers such that their product added to 
 either gives a square, and the sides of the two squares added 
 together produce a given number. 
 
 Let the given number be 6. 
 
 Since x (4* i ) + x is a square, let x, 4* - i be the numbers. 
 
 Therefore 4# 2 + 3^1 is a square, and the side of this 
 square must be 6 2x [since 2x is the side of the 
 first square and the sum of the sides of the square 
 is 6]. 
 
 Since 4^ + 3* i = (6 2x)*, 
 
 we have x = fj, and 
 
 the numbers are [. . 
 27 27 
 
 27. To find two numbers such that their product minus either 
 gives a square, and the sides of the two squares so arising when 
 added together produce a given number. 
 
 Let the given number be 5. 
 
 Assume 4x+i,x for the numbers, so that one condition 
 
 is satisfied. 
 
 Also 4* 2 - 3* i = (5 2x)*. 
 
 Therefore * = ff , and 
 
 the numbers are *|, *^.
 
 i 5 4 THE ARITHMETICA 
 
 28. To find two square numbers such that their product added 
 to either gives a square. 
 
 Let the numbers 1 bex*,y 2 . 
 
 X& -1/2 i i/2) 
 
 Therefore * 2 j- are both squares. 
 
 To make the first expression a square we make x* -f i a 
 
 square, putting 
 
 x* + i = (x - 2) 2 , say. 
 Therefore x = f , and x* = ^. 
 We have now to make ^(y* + i) a square [and y must be 
 
 different from x\ 
 
 Put 9^ + 9 = (3y- 4) 2 > say, 
 
 and y = lt. 
 
 Therefore the numbers are -%. Q. 
 io' 576 
 
 29. To find two square numbers such that their product minus 
 either gives a square. 
 
 Let x z ,y* be the numbers. 
 
 ..2 ,,,2 _ \ 
 
 Then * J \ are both squares. 
 x y XT) 
 
 A solution of x* I = (a square) is x" 1 = ff . 
 We have now to solve 
 
 f f ^ - ft = a square. 
 Put ^-i=(j-4) 2 ,say. 
 
 Therefore y = *, and 
 
 the numbers are $9 ??. 
 64 64 
 
 30. To find two numbers such that their product + their sum 
 gives a square. 
 
 Now m z + n? 2mti is a square. 
 
 Put 2, 3, say, for m, n respectively, and of course 
 
 2 2 + 3 2 2 . 2 . 3 is a square. 
 Assume then product of numbers = (2 2 + 3 2 )^ 2 or 1 3^, and 
 
 sum = 2 . 2 . 3^r 2 or 1 2x*. 
 
 The product being 13^, let x, lye be the numbers. 
 Therefore their sum 14^= i2x*, and ^ = ^. 
 The numbers are therefore L . 
 
 1 Diophantus does not use two unknowns, but assumes the numbers to be x z and I 
 until he has found x. Then he uses the same unknown (x) to find what he had first taken 
 to be unity, as explained above, p. 52. The same remark applies to the next problem.
 
 BOOK II 155 
 
 31. To find two numbers such that their sum is a square and 
 their product their sum gives a square. 
 
 2 . 2m .m = a. square, and (2m)* + m* 2 . 2m .m = a. square. 
 
 If m = 2, 4* + 2 2 2 . 4 . 2 = 36 or 4, 
 
 Let then the product of the numbers be (4* + 2 f )x* or 2Or a , 
 
 and their sum 2.4.2JT 2 or idr 1 , and take 2-r, lor for 
 
 the numbers. 
 
 Then \2x= idr 2 , and * = f. 
 
 The numbers are -, . 
 
 32. To find three numbers such that the square of any one of 
 them added to the next following gives a square. 
 
 Let the first be x, the second 2x -f i, and the third 
 2(2*+!)+ i or 4* +3, so that two conditions are 
 satisfied. 
 
 The last condition gives (44: + 3)* + x = square = (44: 4^, 
 say. 
 
 Therefore x=^ y and 
 
 the numbers are ^, g, &. 
 
 33. To find three numbers such that the square of any one of 
 them minus the next following gives a square. 
 
 Assume x + i, 2x+ \, $x+ i for the numbers, so that two 
 
 conditions are satisfied. 
 Lastly, 1 6r 2 + "jx = square = 25*', say, 
 and x = . 
 
 The numbers are ^, ^, ^. 
 999 
 
 34. To find three numbers such that the square of any one 
 added to the sum of all three gives a square. 
 
 \^(m n^-\-mn is a square. Take a number separable 
 into two factors (#/, ) in three ways, say 1 2, which is 
 the product of (i, 12), (2,6) and (3, 4). 
 
 The values then of \ (m n) are 5|, 2, . 
 
 Take ^\x, 2x t ^x for the numbers, and for their sum \2x-. 
 
 Therefore &r = 1 2x*, and x = \. 
 The numbers are , *, -. 
 
 [Diophantus says |, and ^, f , |.]
 
 156 THE ARITHMETICA 
 
 35. To find three numbers such that the square of any one 
 minus the sum of all three gives a square. 
 
 { (m + )} 2 mn is a square, Take, as before, a number 
 
 divisible into factors in three ways, as 12. 
 Let then 6^, 4*, ^\x be the numbers, and their sum \2x*. 
 Therefore \4x=i2x*, and;r = f. 
 
 The numbers are &, , . 
 
 BOOK III 
 
 1. To find three numbers such that, if the square of any one 
 of them be subtracted from the sum of all three, the remainder 
 is a square 1 . 
 
 Take two squares ;r 2 , ^x* ; the sum is 5;r 2 . 
 
 If then we take 5;r 2 as the sum of the three numbers, and 
 
 x, 2x as two of them, we satisfy two conditions. 
 Next divide 5, which is the sum of two squares, into two 
 other squares ^, J^- [ll. 9], and assume \x for the 
 third number. 
 Therefore x + 2x -f \x = 5* 2 , and x = ^. 
 
 The numbers are ^, ^, ~^. 
 [Diophantus writes -^ for x and T %\, ||g, -^ for the numbers.] 
 
 2. To find three numbers such that the square of the sum of 
 all three added to any one of them gives a square. 
 
 Let the square of the sum of all three be x*, and the 
 
 numbers ^, 8x*, i$x\ 
 Hence 26;r 2 = x, x = -%, and 
 
 the numbers are ~y. 7 - i , ,. 
 
 0/0 0/0 OyO 
 
 3. To find three numbers such that the square of the sum of 
 all three minus any one of them gives a square. 
 
 Sum of all three 4*", its square i6.r 2 , the numbers "jx*, 
 
 I2X Z , l$X\ 
 
 Then 34Jtr 2 = 4*, x = &, and 
 
 the numbers are J?-, , . 
 
 1 The fact that the problems III. 1-4 are very like II. 34, 35 makes Tannery suspect 
 that they have found their way into the text from some ancient commentary.
 
 BOOK III 157 
 
 4. To find three numbers such that, if the square of their sum 
 be subtracted from any one of them, the remainder is a square. 
 
 Sum x, numbers 2x z t $x z , lot: 2 . 
 Then 17 x* = x, x = -fa, and 
 
 the numbers are ^, J-, Jj|. 
 
 5. To find three numbers such that their sum is a square and 
 the sum of any pair exceeds the third by a square. 
 
 Let the sum of the three be (x+ i) 2 ; let first + second 
 = third + i, so that third = ^ + x ; let second + third 
 = first + x*, so that first =x + $. 
 Therefore second = ^x* + . 
 
 It remains that first + third = second + a square. 
 Therefore 2.x square = 16, say, and x = 8. 
 
 The numbers are 8|, 32^, 40. 
 Ot/ierwise thus\ 
 
 First find three squares such that their sum is a square. 
 Find e.g. what square number + 4 + 9 gives a square, 
 
 that is, 36 ; 
 
 4> 36, 9 are therefore squares with the required property. 
 Next find three numbers such that the sum of each pair = 
 the third + a given number ; in this case suppose 
 first + second third = 4, 
 second + third first = 9, 
 third + first second = 36. 
 This problem has already been solved [I. 18]. 
 
 The numbers are respectively 20, 6, 22^. 
 
 1 We should naturally suppose that this alternative solution, like others, was inter- 
 polated. But we are reluctant to think so because the solution is so elegant that it 
 can hardly be attributed to a scholiast. If the solution is not genuine, we have here 
 an illustration of the truth that, however ingenious they are, Diophantus' solutions are not 
 always the best imaginable (Loria, Le scienze esatte nelf antica Greda, Libro v. pp. 138-9). 
 In this case the more elegant solution is the alternative one. Generalised, it is as follows. 
 We have to find JT, y, z so that 
 
 -x+y + z = a. square\ 
 x y + 2 = a square Y , 
 x+y-z = a. square) 
 
 and also x +y + z = a square. 
 
 We have only to equate the first three expressions to squares a 2 , P, c* such that 
 square, ^ say, since the sum of the first three expressions is itself 
 
 The solution is then
 
 i 5 8 THE ARITHMETICA 
 
 6. To find three numbers such that their sum is a square and 
 the sum of any pair is a square. 
 
 Let the sum of all three be x* + 2x+ i, sum of first and 
 second x 2 , and therefore the third 2.x + I ; let sum of 
 second and third be (x - i) 2 . 
 
 Therefore the first = ^x, and the second =;r 2 <\x. 
 But first + third = square, 
 that is, 6x + i = square =121, say. 
 Therefore x = 20, and 
 
 the numbers are 80, 320, 41. 
 
 [An alternative solution, obviously interpolated, is practically 
 identical with the above except that it takes the square 36 as 
 the value of 6x+i, so that x ^-, and the numbers are ^ Q - 
 _?4o 385 456-, 
 - 3 6 ' 36 ' 36 <J 
 
 7. To find three numbers in A.P. such that the sum of any 
 pair gives a square. 
 
 First find three square numbers in A.P. and such that half 
 their sum is greater than any one of them. Let 
 ;r 2 , (x+ i) 2 be the first and second of these ; therefore 
 the third is x* + ^x + 2 = (x - 8) 2 , say. 
 
 Therefore x = f $ or f ; 
 
 and we may take as the numbers 961, 1681, 2401. 
 
 We have now to find three numbers such that the sums 
 of pairs are the numbers just found. 
 
 The sum of the three = a^a = 25211, and 
 
 the three numbers are 120^, 840^, 1560^. 
 
 8. Given one number, to find three others such that the sum 
 of any pair of them added to the given number gives a square, and 
 also the sum of the three added to the given number gives a 
 square. 
 
 Given number 3. 
 
 Suppose first required number + second =x"* + 4x+ i, 
 second + third = x* + 6x + 6, 
 sum of all three = x* + Sx + 1 3. 
 Therefore third =^x + 12, second =x*-+2x-6, first = 
 Also first + third + 3 = a square, 
 that is, 6x + 22 = square = 100, suppose. 
 Hence x= 13, and 
 
 the numbers are 33, 189, 64.
 
 BOOK III 159 
 
 9. Given one number, to find three others such that the sum 
 of any pair of them minus the given number gives a square, and 
 also the sum of the three minus the given number gives a square. 
 
 Given number 3. 
 
 Suppose first of required numbers + second = x* + 3, 
 
 second + third =x*+2x + 4, 
 
 sum of the three = x* + ^x + 7. 
 
 Therefore third = 4^+4, second = x* zx, first = 2x + 3. 
 Lastly, first + third 3 = 6;r + 4 = a square = 64, say. 
 Therefore x = 10, and 
 
 (23, 80, 44) is a solution. 
 
 10. To find three numbers such that the product of any pair 
 of them added to a given number gives a square. 
 
 Let the given number be 12. Take a square (say 25) 
 
 and subtract 12. Take the difference (13) for the 
 
 product of the first and second numbers, and let these 
 
 numbers be 13*, \\x respectively. 
 Again. subtract 12 from another square, say 16, and let the 
 
 difference (4) be the product of the second and third 
 
 numbers. 
 
 Therefore the third number = 4*. 
 The third condition gives 52^* 4- 12 = a square; now 
 
 52 = 4. 13, and 13 is not a square; but, if it were a 
 
 square, the equation could easily be solved 1 . 
 Thus \ve must find two numbers to replace 13 and 4 such 
 
 that their product is a square, while either + 12 is 
 
 also a square. 
 Now the product is a square if both are squares ; hence we 
 
 must find two squares such that either + 12 = a square. 
 " This is easy 2 and, as we said, it makes the equation easy 
 
 to solve." 
 The squares 4, \ satisfy the condition. 
 
 1 The equation 52^+ n = 2 can in reality be solved as it stands, by virtue of the fact 
 that it has one obvious solution, namely x = i . Another solution is found by substituting 
 jr+i for jr, and so on. Cf. pp. 69, 70 above. The value JT= i itself gives (13, i, 4) as 
 a solution of the problem. 
 
 2 The method is indicated in II. 34. We have to find two pairs of squares differing 
 by 12. (a) If we put 12 = 6.1, we have 
 
 and 1 6, 4 are squares differing by n, or 4 is a square which when added to n gives a 
 square. () If we put 12 = 4.3, we fid '-(4-3); or - to be a square which when 
 added to 12 gives a square.
 
 160 THE ARITHMETICA 
 
 Retracing our steps, we now put <\x, ijx and x\^ for the 
 numbers, and we have to solve the equation 
 
 x*+i2 = square = (x + 3)", say. 
 Therefore x = %, and 
 
 (2, 2, 1 j is a solution 1 . 
 
 ii. To find three numbers such that the product of any pair 
 minus a given number gives a square. 
 Given number 10. 
 Put product of first and second = a square + 10 =4 -H 10, 
 
 say, and let first = 14*, second = \\x. 
 Let product of second and third = a square + 10= 19, say ; 
 
 therefore third = 19*:. 
 By the third condition, 266x' 2 10 must be a square ; but 
 
 266 is not a square 2 . 
 Therefore, as in the preceding problem, we must find two 
 
 squares each of which exceeds a square by 10. 
 The squares 30^, 12\ satisfy these conditions 3 . 
 Putting now 30^^, \\x, \2-\x for the numbers, we have, 
 
 by the third condition, 37O T 9 5 ;r 2 10 = square [for 
 
 370^ Diophantus writes 370 J^]; 
 therefore 5929*2 - 160 = square = (77* - 2) 2 , say. 
 Therefore x = } |, and 
 
 the numbers are ^, 22, . 
 
 1 Euler (Algebra, Part n. Art. 232) has an elegant solution of this problem in whole 
 numbers. Let it be required to find x, y, z such that xy + a, yz + a, zx + a are all squares. 
 Suppose xy + a=jP, and make z = x+j> + $; 
 
 therefore xz + a = x* + xy + qy. + a = x z + qx + {P, 
 
 and yz + a=xy+y* + qy + a=jP + qy+lP', 
 
 and the right hand expressions are both squares if ^ = 2/, so that z = x+yip. 
 
 We can therefore take any value for p such that / 2 >, split p^-a into factors, 
 take those factors respectively for the values of x and y, and so find 2. 
 
 E.g. suppose a 11 and ^2 = 25, so that xy=\$\ let x=i,y=i^ t and we have 
 2=14=^10=24 or 4, and (i, 13, 4), (i, 13, 24) are solutions. 
 
 2 As a matter of fact, the equation 266^- io = 2 can be solved as it stands, since it 
 has one obvious solution, namely x=i. (Cf. pp. 69, 70 above and note on preceding 
 problem, p. 159.) The value x=i gives (14, i, 19) as a solution of the problem. 
 
 3 Tannery brackets the passage in the text in which these squares are found, on 
 the ground that, as the solution was not given in the corresponding place of in. 10, there 
 was no necessity to give it here. 10 and i being factors of 10, 
 
 thus 3<>i is a square which exceeds a square by 10. Similarly {-(5 + 2) I or 12^ is such 
 a square. The latter is found in the text by putting m z - 10 = square = (m -- 2) 2 , whence 
 * = 34i and a 2 =n|.
 
 BOOK III 161 
 
 12. To find three numbers such that the product of any two 
 added to the third gives a square. 
 
 Take a square and subtract part of it for the third number ; 
 let x- + 6x-\-g be one of the sums, and 9 the third 
 number. 
 Therefore product of first and second =x*+6x; let first 
 
 = x, so that second = x + 6. 
 By the two remaining conditions 
 IOF + 
 
 54] 
 
 ~;r are both squares. 
 6J 
 
 Therefore we have to find two squares differing by 48 ; 
 
 " this is easy and can be done in an infinite number 
 
 of ways." 
 The squares 16, 64 satisfy the condition. Equating these 
 
 squares to the respective expressions, we obtain 
 
 x= i, and 
 
 the numbers are i, 7, g. 
 
 1 3. To find three numbers such that the product of any two 
 minus the third gives a square. 
 
 First x, second x + 4 ; therefore product = x* + ^x, and we 
 
 assume third = 4^. 
 Therefore, by the other conditions, 
 
 AX* + 1 5.*- ) 
 
 \ are both squares. 
 4* 2 -*- 4 j 
 
 The difference = i6x + 4 = 4 (4^ + i), and we put 
 
 Therefore x = f, and 
 
 the numbers are |, *, 
 
 14. To find three numbers such that the product of any two 
 added to the square of the third gives a square 1 . 
 
 1 Wertheim gives a more general solution, as follows. If we take as the required 
 numbers X- ax, V=ajc + 6-, Z=- 2 , two conditions are already satisfied, namely 
 
 A'K+Z 2 = a square, and YZ+ X* = a. square. 
 
 It only remains to satisfy the condition ZX+ F 2 = a square, or 
 
 a 2 * 2 + ^| at?* + # = a square. 
 
 Put a t x 3 + ^a^?x + l>* 
 
 10 
 
 16^(^ 
 and JT = ; 
 
 a (33 - 
 where k remains undetermined. 
 
 H. D.
 
 162 THE ARITHMETICA 
 
 Firsts, second 4^ + 4, third I. Two conditions are thus 
 
 satisfied. 
 The third condition gives 
 
 x + (4* + 4)2 = a square = (4* - 5) 2 > sa y- 
 Therefore x= 7 9 , and 
 
 the numbers (omitting the common denominator) 
 are 9, 328, 73. 
 
 15. To find three numbers such that the product of any two 
 added to the sum of those two gives a square 1 . 
 
 [Lemma.] The product of the squares of any two con- 
 secutive numbers added to the sum of the said 
 squares gives a square 2 . 
 Let 4, 9 be two of the required numbers, x the third. 
 
 Therefore iatr + 'I are both squares. 
 
 The difference = $x + 5 = 5 (x + i ). 
 
 Equating the square of half the sum of the factors to 
 IOF+ 9, we have 
 
 (H* + 6 )} 2 = 10* + 9. 
 Therefore x= 28, and (4, 9, 28) is a solution. 
 
 1 The problem can of course be solved more elegantly, with our notation, thus. (The 
 same remark applies to the next problem, in. 16.) 
 
 If x, y, z are the required numbers, xy + x+y, etc. are to be squares. We may 
 therefore write the conditions in the form 
 
 (y+i) (z + i) = a square + i, 
 (z+i)(x+i) = a square + i, 
 (jr+i)(^+i)=a square + 1. 
 
 Assuming a 2 , 2 , f 2 for the respective squares, and putting = x+i, i)=y+i, f=z+i, 
 we have to solve 
 
 [This is practically the same problem as that in the Lemma to Dioph. v. 8.] 
 Multiplying the second and third equations and dividing by the first, we have 
 
 with similar expressions for r), f. 
 
 x, y, z are these expressions minus i respectively, a 2 , 2 , c 1 must of course be so 
 chosen that the resulting values of , i), f may be rational. Cf. Euler, Commentationes 
 arithmeticae, II. p. 577. 
 
 2 In fact, a 2 (a+i) 2 + a 2 + (a+i) 2 ={a(o+i) + i} 2 .
 
 BOOK III 163 
 
 Otherwise thus 1 . 
 
 Assume first number to be x, second 3. 
 Therefore 4^ + 3 = square = 25 say, whence x = 5^, and 5^, 
 3 satisfy one condition. 
 
 1 This alternative solution would appear to be undoubtedly genuine. 
 Diophantus has solved the equations 
 
 Fermat shows how to solve the corresponding problem with four numbers instead of 
 three. He uses for this purpose Diophantus' solution of V. 5, namely the problem 
 of finding x 2 , y 2 , z 2 , such that 
 
 Diophantus finds I , , J as a solution of the latter problem. Fermat takes 
 
 these as the first three of the four numbers which are to satisfy the condition that the 
 product of any two plus the sum of those two gives a square, and assumes x for the 
 fourth. Three relations out of six are already satisfied, and the other three require 
 
 +*+, or 34* + 25 
 
 9 9 99 
 
 64 64 ij.x 64 
 
 x + x+ , or ta- + 
 9 9 99 
 
 '-a^+'-g, or ~5* + 12? 
 9 9 99 
 
 to be made squares : a "triple-equation" to be solved by Fermat's method. (See the 
 Supplement, section V.) 
 
 Fermat does not give the solution, but I had the curiosity to work it out in order to 
 verify to what enormous numbers the method of the triple-equation leads, even in such 
 comparatively simple cases. 
 
 We may of course neglect the denominator 9 and solve the equations 
 
 73* + 64 = z/2, 
 205*+ 1 96 = ^2. 
 The method gives 
 
 x _ _ 459818598496844787200 
 631629004828419699201 ' 
 the denominator being equal to (25132230399)2. 
 
 Verifying the correctness of the solution, we find that, in fact, 
 
 25 
 
 73 x , l _ / i 
 
 64 ' 
 
 *S X / 12275841601X2 
 
 196 V25i3"3399/ 
 
 Strictly speaking, as the value found for x is negative, we ought to substitute y - A 
 for it (where - A is the value found) in the three equations and start afresh. The 
 portentous numbers which would thus arise must be left to the imagination.
 
 164 THE ARITHMETICA 
 
 Let the third be x, while 5^, 3 are the first two. 
 Therefore ^ ~ I must both be squares; 
 
 but, since the coefficients in one expression are respectively 
 greater than those in the other, but neither of the ratios 
 of corresponding coefficients is that of a square to a 
 square, our suppositions will not serve the purpose \ we 
 cannot solve by our method. 
 
 Hence (to replace 5$, 3) we must find two numbers such 
 that their product + their sum = a square, and the 
 ratio of the numbers increased by I respectively is 
 the ratio of a square to a square. 
 
 Let these be y and 47 + 3, which satisfy the latter con- 
 dition ; and, in order that the other may be satisfied, 
 we must have 
 
 4/ 2 + 87 + 3 = square = (zy - 3) 2 , say. 
 
 Therefore y = T 3 ^. 
 
 Assume now ^, 4\, x for the three numbers. 
 
 Therefore Jf* 4 " 4 /} are both squares, 
 
 or, if we multiply by 25 and 100 respectively, 
 
 130* + 105 ) 
 
 ,30*+ J0 } are both squares. 
 
 The difference is 75 = 3.25, and the usual method of 
 solution gives x = ^. 
 
 The numbers are , g, . 
 
 1 6. To find three numbers such that the product of any two 
 minus the sum of those two gives a square. 
 
 Put x for the first, and any number for the second; we 
 then fall into the same difficulty as in the last 
 problem. 
 We have to find two numbers such that 
 
 (a) their product minus their sum = a square, and 
 
 (b) when each is diminished by I, the remainders 
 
 have the ratio of squares. 
 Now 47+ i,/+ i satisfy the latter condition. 
 The former (a) requires that 
 
 47 s I = square = (2y 2) 2 , say, 
 which gives y . 
 Assume then J, -,x for the numbers.
 
 BOOK III 165 
 
 Therefore , ft [ are both squares, 
 
 * x ~ If J 
 or, if we multiply by 4, 16 respectively, 
 
 lox- 14) 
 
 I0jr _ 26 } * both squares. 
 
 The difference is 12 = 2.6, and the usual method gives 
 *=3- 
 
 The numbers are ^, 3^ = -g-, 3 = y- 
 
 17. To find two numbers such that their product added to 
 both or to either gives a square. 
 
 Assume x,^x\ for the numbers, since 
 
 x (^x i ) + x = 4x 2 , a square. 
 
 A*& _L 2 y _ T \ 
 
 Therefore also , , \ are both squares. 
 
 4* 2 + 4*- - i j 
 
 The difference is x=/^x. \, and we find 
 
 *-4h- 
 
 The numbers are , --. 
 
 1 8. To find two numbers such that their product minus either, 
 or minus the sum of both, gives a square 1 . 
 
 1 With this problem should be compared that in paragraph 42 of Part I. of the 
 Invenlum Novum of Jacobus de Billy (Oeuvres de Fermat, in. pp. 351-2), where three 
 conditions correspond to those of the above problem, and there is a fourth in addition. 
 The problem is to find , -q (!>;) such that 
 
 t 
 
 are all squares. 
 
 Suppose TJ = X, =i-x; the first two conditions are thus satisfied. The other 
 two give 
 
 Separating the difference ix into the factors ix, t, we put, as usual, 
 
 (^y--~.. 
 
 whence .*= | , and the numbers are | , | . 
 
 o so 
 
 To find another value of x by means of the value thus found, we put ^ + 5 * n pl a ce of 
 x in the double-equation, whence 
 
 Multiplying the lower expression by 49, we can solve in the usual way. Our expressions
 
 166 THE ARITHMETICA 
 
 Assume x+ i, 4^ for the numbers, since 
 <\x(x + i) \x = a square. 
 
 Therefore also 4;r 2 _ _ [ are both squares. 
 
 The difference is A [ x = t x . i, and we find 
 
 *-!*. 
 The numbers are 2^, 5. 
 
 19. To find four numbers such that the square of their sum 
 plus or minus any one singly gives a square. 
 Since, in any right-angled triangle, 
 
 (sq. on hypotenuse) + (twice product of perps.) = a square, 
 we must seek four right-angled triangles [in rational 
 
 numbers] having the same hypotenuse, 
 or we must find a square which is divisible into two 
 
 squares in four different ways ; and "we saw how to 
 
 divide a square into two squares in an infinite 
 
 number of ways." [ll. 8.] 
 Take right-angled triangles in the smallest numbers, 
 
 (3, 4, 5) and (5, 12, 13); and multiply the sides of 
 
 are now^ 2 - - y + ~ and 49^ - ^-y + ^ , and the difference between them is ^8y 2 - noy. 
 
 The solution next mentioned by De Billy was clearly obtained by separating this 
 difference into factors such that, when the square of half their difference is equated to 
 
 jP-y+jr\ the absolute terms cancel out. The factors are ^y, Jf--, and we put 
 
 ' 
 
 This gives y= 4 't 5 " , whence x / , and the numbers are 
 
 71362992 7'3 62 99 2 
 
 48647065 ^ 22715927 
 
 71362992' 71362992' 
 
 A solution in smaller numbers is obtained by separating ^8y 2 - noy into factors such 
 that the terms in x 2 in the resulting equation cancel out. The factors are 6y, 8y -- , and 
 we put 
 
 whence y= 479 ' S9 and x= 47959 I 3 = 5l865 . 
 10416' 10416 8 10416' 
 
 This would give a negative value for i x ; but, owing to the symmetry of the 
 
 original double-equation in x, since x=- 1 satisfies it, so does x= 10 *] ; hence the 
 
 10416 51865' 
 
 numbers are |^- and *y : a solution also mentioned by De Billy. 
 Cf. note on iv. 23.
 
 BOOK III 167 
 
 the first by the hypotenuse of the second and vice 
 versa. 
 This gives the triangles (39, 52, 65) and (25, 60, 65); thus 
 
 65 2 is split up into two squares in two ways. 
 Again, 65 is "naturally" divided into two squares in two 
 ways, namely into 7 2 + 4 2 and 8 2 + i 2 , "which is due 
 to the fact that 65 is the product of 13 and 5, each of 
 which numbers is the sum of two squares." 
 Form now a right-angled triangle 1 from 7, 4. The sides 
 
 are (; 2 - 4 2 , 2 . 7 . 4, ; 2 + 4 2 ) or (33, 56, 65). 
 Similarly, forming a right-angled triangle from 8, I, we 
 
 obtain (2 . 8 . i, 8 2 - i 2 , 8 2 + I 2 ) or 16, 63, 65. 
 Thus 65 2 is split into two squares in four ways. 
 Assume now as the sum of the numbers 6$x and 
 as first number 2 . 39 . 52;tr 2 = 4056^, 
 second 2.25. 6ox 2 = 3000^, 
 third 2.33. 56;tr 2 = 3696^, 
 fourth 2. i6.63x*=20i6x*, 
 the coefficients of x* being four times the areas of the 
 four right-angled triangles respectively. 
 The sum 12768^ = 65^, and x 
 The numbers are 
 
 12675000 15615600 J3 
 
 ^ 
 163021824' 163021824' 163021824' 163021824* 
 
 20. To divide a given number into two parts and to find a 
 square which, when either of the parts is subtracted from it, gives 
 a square 2 . 
 
 Given number 10, required square x z + 2x + i. 
 
 Put for one of the parts 2x + i, and for the other ^x. 
 
 The conditions are therefore satisfied if 
 
 i = 10. 
 Therefore x=i 
 
 the parts are (4, 6) and the square 6. 
 
 1 If there are two numbers/, y, to "form a right-angled triangle" from them means 
 to take the numbers / 2 + ^ 2 , /> 2 - q 1 , ipq. These are the sides of a right-angled triangle, 
 
 2 This problem and the next are the same as II. 15, 14 respectively. It may therefore 
 be doubted whether the solutions here given are genuine, especially as interpolations 
 from ancient commentaries occur most at the beginning and end of Books,
 
 168 THE ARITHMETICA 
 
 21. To divide a given number into two parts and to find a 
 square which, when added to either of the parts, gives a square. 
 Given number 20, required square x* + 2x + i. 
 If to the square there be added either 2^+3 or 4*+ 8, 
 
 the result is a square. 
 Take 2x + $, <\x + 8 as the parts of 20, and 6x+ 1 1 = 20, 
 
 whence x = \\. 
 Therefore the parts are (6, 14) and the square 6|. 
 
 BOOK IV 
 
 1. To divide a given number into two cubes such that the sum 
 of their sides is a given number 1 . 
 
 Given number 370, given sum of sides 10. 
 Sides of cubes 5 +x, 5 x, satisfying one condition. 
 Therefore 30^ + 2 50 = 370, x = 2, 
 
 and the cubes are 7 3 , 3 3 , or 343, 27. 
 
 2. To find two numbers such that their difference is a given 
 number, and also the difference of their cubes is a given number. 
 
 Difference 6, difference of cubes 504. 
 
 Numbers X+^XT,. 
 
 Therefore i8;tr 2 + 54 = 504, x z = 25, and x=$. 
 
 The sides of the cubes are 8, 2 and the cubes 512, 8. 
 
 3. To multiply one and the same number into a square and 
 its side respectively so as to make the latter product a cube and 
 the former product the side of the cube. 
 
 Let the square be x z . Its side being x, let the number 
 
 be 8/r. 
 Hence the products are 8x, 8, and 
 
 (8;tr) 3 = 8. 
 
 Therefore 2 = %x, x = \, and the number to be multiplied 
 is 32. 
 The square is ^ and its side -. 
 
 1 It will be observed that Diophantus chooses, as his given numbers, numbers such 
 as will make the resulting "pure" quadratic equation give a " rational " value for x. If 
 the given numbers are ia, ib, respectively, we assume b + x, b-x as the sides of the 
 cubes, and we have 
 
 so that x* = (a-lF)l$b; x is therefore "irrational" unless (a-b 3 )l$b is a square. In 
 Diophantus' hypothesis a is taken as 185, and b as 5, and the condition is satisfied. He 
 shows therefore incidentally that he knew how to find two numbers a, b such that 
 (a-lP)l?,b is a square (Loria, Le scienze esatte nelV antica Grecia, Libro v. pp. 129-30). 
 A similar remark applies to the next problem, iv. i.
 
 BOOK IV 169 
 
 4. To add the same number to a square and its side re- 
 spectively and make them the same 1 \i.e. make the first product a 
 square of which the second product is the side]. 
 Square x*, with side x. 
 Let the number added to x* be such as to make a square 
 
 say 3X*. 
 Therefore yc* + x = side of $x 2 2x, and x \. 
 
 The square is -, its side -, and the number -. 
 
 5. To add the same number to a square and its side and make 
 them the opposite 2 . 
 
 Square x*, the number a square number of times x* 
 
 minus x> say 4^r 2 x. 
 Hence $x 2 x= side of 4^r 2 = 2x, and x = f . 
 
 The square is its side f , and the number ^. 
 25 5 Z 5 
 
 6. To add the same square number to a cube and a square 
 and make them the same. 
 
 Let the cube be x 3 and the square any square number of 
 
 times x*, say gx z . 
 We want now a square which when added to gx* makes 
 
 a square. Take two factors of 9, say 9 and i, sub- 
 
 tract i from 9, take half the difference and square. 
 
 This gives 16. 
 
 Therefore i6x* is the square to be added. 
 Next, x* + i6x*= a cube = Sx 3 , say; and x ^-. 
 
 The cube is therefore ^-, the square ^?4, and 
 343 49 
 
 the added square number ^5-. 
 49 
 
 1 In this and the following enunciations I have kept closely to the Greek partly 
 for the purpose of showing Diophantus' mode of expression and partly for the brevity 
 gained thereby. 
 
 In Prop. 4 to "make them the same" means what I have put in brackets ; to "make 
 them the opposite" in Prop. 5 means to make the first product a side of which the second 
 product is the square. 
 
 2 Nesselmann solves the problem generally, thus (Notes in Zeitschrift fur Math. u. 
 Physik, xxxvii. (1892), Hist. lilt. Abt. p. 162). 
 
 x 2 +y = J(x +>>) ; therefore x* + ix^y +_y 2 = x +y, oiy 2 -(i- ix 2 ) y=x-x*. 
 Solving for y, we obtain, as one of the solutions, 
 
 To make the expression under the radical a square we put -+x x^=(mx I , 
 
 m + i m* + m 3 m i 
 
 whence x= ^ - , _j/= - ^ - TO - Diophantus solution corresponds to m i.
 
 1 70 THE ARITHMETICA 
 
 7. To add the same square number to a cube and a square 
 respectively and make them the opposite. 
 
 For brevity call the cube (i), the second square (2) and 
 
 the added square (3). 
 
 Now, since (2) + (3) = a cube, suppose (2) + (3) = (i). 
 Since a^ + b z 2ab is a square, suppose (i) = (a*+& i ), 
 (3) = 2ab, so that the condition that (i) + (3) = square 
 is satisfied. 
 But (3) is a square, and, in order that 2ab may be a square, 
 
 we put a = x, b = 2x. 
 Suppose then ( i ) = x* + (zxf = $x z , (3) = 2 . x . 2x = ^ ; 
 
 therefore (2) = ^, by subtraction. 
 But $x 2 is a cube; therefore x = 5, 
 
 and the cube (i)=i25, the square (2) = 25, the 
 
 square (3) = 100. 
 Otherwise thus. 
 
 Let .(2)+ (3)= (i). 
 
 Then, since (i ) + (3) = a square, we have to find two squares 
 
 such that their sum + one of them = a square. 
 Let the first of these squares be x*, the second 4. 
 Therefore 2** + 4 = square = (2x 2) 2 , say ; thus x = 4, 
 
 and the squares are 16, 4. 
 Assume now (2) = 4**, (3)= \6x z . 
 Therefore 2ox z is a cube, so that x= 20; 
 
 the cube (i) is 8000, the square (2) is 1600, and the 
 added square (3) is 6400. 
 
 8. To add the same number to a cube and its side and make 
 them the same 1 . 
 
 Added number x, cube Sx 5 , say. Therefore second sum 
 
 = $x, and this must be the side of Sx 3 + x. 
 That is, 8x 3 + x= 27 x*, and igx s = x, or igx 2 = i. 
 
 1 Nesselmann (pp. fit. p. 163) gives a more general solution. 
 We have x 3 +y= (x +y) 3 , whence i = 
 Solving for y, we find 
 
 xndy= n a + m 2 ' If the P ositive sign be taken, then, in order that y may 
 
 always be positive, m\n must be >3 + >/i2; Diophantus' solution corresponds to / = ;,
 
 BOOK IV 171 
 
 But 19 is not a square. Hence we must find, to replace 
 it, some square number. Now igx 3 arises from 
 2JX 3 8x*, where 27 is the cube of 3, and 8 the cube 
 of 2. And the 3* comes from the assumed side 2x, 
 by increasing the coefficient by unity. 
 
 Thus we must find two consecutive numbers such that their 
 cubes differ by a square. 
 
 Let them be 7, 7+ I. 
 
 Therefore y* + -$y + i = square = ( I - 2yf, say, and y = 7. 
 
 Going back to the beginning, we assume added number 
 = *, side of cube = "jx. 
 
 The side of the new cube is then 8*, and 
 343** + *= 512^. 
 
 Therefore * 2 = y^, and x = -^. 
 
 The cube is J^L its side J , and the added 
 number ^-. 
 
 9. To add the same number to a cube and its side and make 
 them the opposite 1 . 
 
 Suppose the cube is S* 3 , its side being 2*, and the added 
 
 number is 27 x 3 2.x. (The coefficients 8, 27 are 
 
 chosen as cube numbers.) 
 
 Therefore 3S* 3 2* = side of cube 27** = 3*, or 35** = 5. 
 This gives no rational value. 
 But 35 = 27 + 8, and 5 = 3+2. 
 Therefore we have to find two cubes such that their sum 
 
 has to the sum of their sides the ratio of a square 
 
 to a square*. 
 Let sum of sides = any number, 2 say, and side of first 
 
 cube = z, so that the side of the other cube is 2 - z. 
 
 1 Nesselmann (op. cit. p. 163) solves as follows. The equation being x+y=(x 3 +y) 3 , 
 putjy^z-j; 3 , and the equation becomes x + z-x 3 = 
 Dividing by x + z, we have -r 2 - xz + z 2 = i . 
 
 Solving for x, we obtain x = - {z >/(4 ~ S 02 )}- 
 
 
 To make 4-322 a square, equate it to f-z-2 } ; 
 
 
 + 
 
 x= 2 ^ n , and y -z - x 3 . If the positive sign be taken, Diophantus' solution 
 
 corresponds to m t, = i. 
 
 2 It will be observed that here and in the next problem Diophantus makes no use of 
 the fact that 
 
 Cf. note on iv. 1 1 below.
 
 1 72 THE ARITHMETICA 
 
 Therefore 8 1 2z -f 6z must be twice a square. 
 
 That is, 4 6z + 32* = square = (2 - 4,3-) 2 , say ; s = i, and 
 
 the sides are |j|, |f . 
 Neglecting the denominator and the factor 2 in the 
 
 numerators, we take 5, 8 for the sides. 
 Starting afresh, we put for the cube 125^ and for the 
 
 number to be added 512^-5^-; we thus get 
 637^ $x = %x, and x = | . 
 
 The cube is ^5, its side 5, and the added number?^. 
 343 7 OTO 
 
 10. To find two cubes the sum of which is equal to the sum 
 of their sides. 
 
 Let the sides be 2x, ^x. 
 
 This gives 35^= $x\ but this equation gives an irrational 
 
 result. 
 We have therefore, as in the last problem, to find two 
 
 cubes the sum of which has to the sum of their sides 
 
 the ratio of a square to a square 1 . 
 These are found, as before, to be 5 3 , 8 3 . 
 Assuming then $x, 8x as the sides of the required cubes, 
 
 we obtain the equation 637^= i^x, and x=\. 
 
 The cubes are g. 
 
 1 Here, as in the last problem, Diophantus could have solved his auxiliary problem 
 of making (xr i +y s )l(x+y) a square by making x^-xy+y 2 a square in the same way as in 
 Lemma I. to V. 7 he makes x^ + xy+y* a square. 
 
 The original problem, however, of solving 
 
 can be more directly and generally solved thus. Dividing out by (x+y), we must have 
 
 This can be solved by the method shown in the note to the preceding problem. 
 Alternatively, we may (with Wertheim) put x 2 - xy+y 2 = (x + ^y)' 2 , and at the same 
 time \ = (x + ky). 
 
 Thus we have to solve the equations 
 
 which give x= I ~ J&2 - 2> J= 
 
 i+k + &' * 
 
 where k remains undetermined. 
 
 Diophantus' solution is obtained by taking the positive sign and putting k = - or by 
 
 taking the negative sign and putting k= - - .
 
 BOOK IV 173 
 
 II. To find two cubes such that their difference is equal to 
 the difference of their sides. 
 
 Assume 2x, ^x as the sides. 
 This gives 193? = x, and x is irrational. 
 We have therefore to find two cubes such that their 
 difference has to the difference of their sides the 
 ratio of a square to a square 1 . Let them be (z+ i) 8 , 
 z 3 , so that the difference of the sides may be a square, 
 namely I. 
 
 Therefore $z* + 3#+ i = square = (i 2#) 2 , say, and z'j. 
 Starting afresh, assume 'jx, %x as the sides; therefore 
 i6$x*=x, and x -^. 
 The sides of the two cubes are therefore ^-, . 
 
 1 Nesselmann (Die Algebra der Griechen> pp. 447-8) comments on the fact that 
 Diophantus makes no use here of the formula (x 3 -y 3 )l(x-y)=x a + x}'+y 2 , although he 
 must of course have known it (it is indeed included in Euclid's much more general 
 summation of a geometrical progression, IX. 35). To solve the auxiliary problem in 
 IV. 1 1 he had only to solve the equation 
 
 x? + xy+y 2 a. square, 
 which equation he does actually solve in his Lemma I. to v. 7. 
 
 The whole problem can be more simply and generally solved thus. We are to have 
 
 Nesselmann's method of solution (cf. note on iv. 9) gives x=- { - 
 
 mn 2/ (w 2 3 2 ) . , . 
 
 and hence y= -? - ?, x= -- ^ ^- - -. Diophantus solution is obtained by 
 2 2 2 2 
 
 putting m= i, n = 2 and taking the lower sign. 
 
 Wertheim's method (see note on preceding problem) gives in this case 
 i-^ ik-i 
 
 where k is undetermined. 
 
 If we take the negative sign and put k = - 3, we obtain Diophantus' solution. 
 Bachet in his notes to IV. 10, n solves the problems represented by 
 
 subject to the condition that m is either a square or the third part of a square. His method 
 corresponds to that of Diophantus. He does not divide out by xy, and he reduces the 
 problem to the subsidiary one of finding , 17 such that the ratio of { 3 =fci^ to 17 is the 
 ratio of a square to a square. His assumptions for the " sides," , 17, are of the same kind 
 as those made by Diophantus; in the first problem he assumes x, 6 x and in the second 
 x, x+2. In fact, it being given that (y?y3)l(xy)=a, Bachet assumes xj>=z and 
 thus obtains 
 
 33? $xz + z* = a, 
 
 which equation can easily be solved by Diophantus' method if a is a square or triple of a 
 square. 
 
 Fermat observes that the Siopt<j>i6j of Bachet is incorrect because not general. It 
 should be added that the number (m) may also be the product of a square number into a 
 prime number of the form 3+ i, as 7, 13, 19, 37 etc. or into any number which has no 
 factors except 3 and prime numbers of the form 3+ i, as 11, 91 etc. "The proof and 
 the solution are to be obtained by my method. "
 
 174 THE ARITHMETICA 
 
 12. To find two numbers such that the cube of the greater 
 + the less = the cube of the less + the greater 1 . 
 
 Assume 2x, ^x for the numbers. 
 
 Therefore 27 x* + 2x = %x 3 + $x, or igx a = x, and x is 
 irrational. 
 
 But 19 is the difference of two cubes, and I the difference 
 of their sides. Therefore, as in the last problem, 
 we have to find two cubes such that their difference 
 has to the difference of their sides the ratio of a 
 square to a square 2 . 
 
 The sides of these cubes are found, as before, to be 7, 8. 
 
 Starting afresh, we assume jx, $x for the numbers; then 
 343-r* + 8^- = 5 1 2x* + jx, and x - -fa. 
 The numbers are , . 
 
 13. To find two numbers such that either, or their sum, or 
 their difference added to unity gives a square. 
 
 Take for the first number any square less I ; let it be, 
 say, o-r 2 + 6^. But the second + I =a square; and 
 first + second + i also = a square. Therefore we must 
 find a square such that the sum of that square and 
 ojr 2 + 6x = a square. 
 
 Take factors of the difference gx* + 6x, say gx + 6, x\ 
 the square of half the difference between these factors 
 
 Therefore, if we put for the second number this expres- 
 sion minus i, or \6x* + 2^x + 8, three conditions are 
 satisfied. 
 
 The remaining condition gives difference + i = square, 
 or 7^ 2 + i8x -f 9 = square = (3 3^r) 2 , say. 
 
 Therefore x= 18, and (3024, 5624) is a solution. 
 
 14. To find three square numbers such that their sum is equal 
 to the sum of their differences. 
 
 Sum of differences = (greatest) (middle) + (middle) 
 (least) +(greatest)least= twice difference of greatest 
 and least. 
 
 This is equal to the sum of all three, by hypothesis. 
 Let the least square be i, the greatest x* + 2x+ i ; 
 
 1 This problem will be seen to be identical with the preceding problem. 
 
 2 See note, p. 173.
 
 BOOK IV 175 
 
 therefore twice difference of greatest and least = sum of 
 
 the three = 2x* + 44:. 
 But least + greatest =x- + 2X+2, so that 
 
 middle = x 2 + 2x 2. 
 Hence X s + 2x - 2 = square = (x - 4)*, say, and x = f . 
 
 The squares are (&, ^, i"\ or (196, 121, 25). 
 \ 25 25 / 
 
 15. To find three numbers such that the sum of any two 
 multiplied into the third is a given number. 
 
 Let (first + second) x third = 35, (second + third) x first 
 
 = 27 and (third + first) x second = 32. 
 Let the third be x. 
 Therefore (first + second) = 3 $/x. 
 Assume first = lo/x, second = 25/4:; then 
 250 , 
 
 These equations are inconsistent ; but they would not be if 
 
 25 10 were equal to 32 27 or 5. 
 Therefore we have to divide 35 into two parts (to replace 
 
 25 and 10) such that their difference is 5. The parts 
 
 are 15, 20. [Cf. I. i.] 
 We take therefore 1 5/ar for the first number, 2ofx for the 
 
 second, and we have 
 
 Therefore x= 5, and (3, 4, 5) is a solution 1 . 
 
 1 As Loria says (Le scienze esatte nelT antica Grecia, Libro V. p. 131), this method of 
 the "false hypothesis," though somewhat indirect, would not be undeserving of a place 
 in a modem textbook. 
 
 Here again, as in IV. r, 2, Diophantus tacitly chooses, for his given numbers, numbers 
 which will make the resulting ' ' pure " quadratic equation give a rational value for jr. 
 
 We may put the solution more generally thus. We have to solve the equations 
 (y + z)x=a, (z + x)y=&, (x+y)i=c. 
 
 Diophantus takes z for his principal unknown and, writing the third equation in the 
 form x+y=e/z, he assumes jr = o/2, y=$\z, where a, /3 have to be determined. One 
 equation connecting a, /3 is a + /3 = <-. Next, substituting the values of JT, y in the first two 
 equations, we have
 
 1 76 THE ARITHMETICA 
 
 1 6. To find three numbers such that their sum is a square, 
 while the sum of the square of each added to the next following 
 number gives a square. 
 
 Let the middle number be any number of x's, say 4^; 
 we have therefore to find what square + ^x gives 
 a square. Split ^x into two factors, say 2x, 2, and 
 take the square of half their difference, (x - I ) 2 . This 
 is the square required. 
 
 Thus the first number is x I. 
 
 Again, i6x* + third number = a square. Therefore, if we 
 subtract \6x* from a square, we shall have the third 
 number. Take as the side of this square the side of 
 \6x*, or ^x, plus i. 
 
 Therefore third number = (^x + i) 2 \6x* = &x + I. 
 
 Now the sum of the three numbers = a square; therefore 
 \$x= a square = i6o,y 2 , say 1 . 
 
 The numbers are then 13^* i, 52jp 2 , 104^+ i. 
 
 Lastly, (third) 2 + first = a square. 
 
 Therefore io8i6y* + 22iy 2 = a. square, 
 
 or io8i6^ 2 + 221 = a square = (1047 + i) 2 , say. 
 
 Therefore j = fg = .|f, 
 
 ' 
 
 17. To find three numbers such that their sum is a square, while 
 the square on any one minus the next following also gives a square. 
 The solution is precisely similar to the last. 
 
 whence it follows that a- fi = a-b. From this condition and a + /3 = c, we obtain 
 a = -( a -6 + c), p = -(-a + i> + c). 
 
 o_ / i( a -6 + c )( a + 6- f )) 8_ /{( 
 
 z V ( i(-a + b + c) /' y ~ ' z~ V * 
 Now x, y, z must all be rational, and this is the case if 
 
 where /, q, r are any integers. 
 
 This gives a=p(tj + r), b = q(r+p), c=r(p + q); 
 
 a fact which can hardly have been unknown to Diophantus, since his values a = ij, b = $i, 
 ^=35 correspond to the values/ = 3, ^ = 4, r=$ (Loria, loc. cit.). 
 
 1 Diophantus uses the same unknown s for y as for x, writing actually Kal ylverai 6 
 s A
 
 BOOK IV I77 
 
 The middle number is assumed to be 4^. The square 
 which exceeds this by a square is (ar+i) 2 , and we 
 therefore take x+ i for the first number. 
 
 For the third number we take i6x* ($x - i) s or %x i. 
 
 The sum of the numbers being a square, 
 1 3^- = a square = i6o.j>> 2 , say. 
 
 The numbers are then i$y z + i, S2^ 2 , lo^/p 2 i. 
 
 Lastly, since (third) 2 first = a square, 
 1 08 i6y 22 iy z = a square, 
 
 or io8i6j 2 221 = a square = (104^ i) 2 , say. 
 
 1 8. To find two numbers such that the cube of the first added 
 to the second gives a cube, and the square of the second added to 
 the first gives a square. 
 
 First number x. Therefore second is a cube number 
 
 minus X s , say 8 x 3 . 
 Therefore x 6 \6x* + 64 + ;r = a square = (x 3 + 8) 2 , say, 
 
 whence ^ > 2x 3 = x, or 32^=1. 
 This gives an irrational result; x would however be 
 
 rational if 32 were a square. 
 But 32 comes from 4 times 8. We must therefore sub- 
 
 stitute for 8 in our assumptions a cube which when 
 
 multiplied by 4 gives a square. If y* is the cube, 
 
 4y = a square = \6y z say; whence y = 4. 
 Thus we must assume x, 64 x 3 for the numbers. 
 Therefore x 6 - 1 2&X 3 + 4096+ x = a square = (x 3 + 64)", say ; 
 
 whence 2^6x 3 = x, and x = -fa. 
 
 The numbers are 4, ~- 
 16' 4090 
 
 19. To find three numbers indeterminately 1 such that the 
 product of any two increased by i is a square. 
 
 Take for the product of first and second some square 
 
 minus i, say x z + 2x\ this satisfies one condition. 
 Let second = x, so that first = x+ 2. 
 Now product of second and third + i =a square; let the 
 
 1 The expression is fr T$ dop/ffry, which is defined at the end of the problem to mean 
 in terms of one unknown (and units), so that the conditions of the problem are satisfied 
 whatever value is given to the unknown. 
 
 H. D. I2
 
 178 THE ARITHMETICA 
 
 square be ($x + i)*, so that product of second and 
 third = 9^ + 6x ; 
 
 therefore third = 9^+6. 
 
 Also product of third and first + I = a square; therefore 
 gx z + 2\x + 1 3 = a square. 
 
 Now, if 13 were a square, and the coefficient of x were 
 twice the product of the side of this square and the 
 side of the coefficient of x 2 , the problem would be 
 solved indeterminately. 
 
 But 13 comes from 2.6+ I, the 2 in this from twice I, 
 and the 6 from twice 3. Therefore we want two 
 coefficients (to replace I, 3) such that the product 
 of their doubles + i = a square, or four times their 
 product + i = a square. 
 
 Now four times the product of any two numbers plus the 
 square of their difference gives a square. Thus the 
 requirement is satisfied by taking as coefficients any 
 two consecutive numbers, since the square of their 
 difference is i. [The assumption of two consecutive 
 numbers for the coefficients simultaneously satisfies 
 the second of the two requirements indicated in the 
 italicised sentence above.] 
 
 Beginning again, we take (x + i) 2 i for the product of 
 first and second and (2x + i) 2 i for the product of 
 second and third. 
 Let the second be x, so that first = x+ 2, third = 4^+4. 
 
 [Then product of first and third + i = 4^ + 12^ + 9, and 
 the third condition is satisfied.] 
 
 Thus the required indeterminate solution 1 is 
 (x + 2, x, 4X + 4)- 
 
 1 The result obtained by Diophantus really amounts to the more general solution 
 
 a?x+ia, x, (a+ i) 2 jr+2 (a+i). 
 
 With this solution should be compared that of Euler (Algebra, Part II. Art. 231). 
 I. To determine x, y, z so that 
 
 xy+i, yz+i, zx+i are all squares. 
 Suppose zx + i - /*, yz+i = <j*, 
 
 so that *=.(/*- i)/, y = (q*-i)lz. 
 
 Therefore xy + 1 = (^Zjl^JLl) + , =a square , 
 
 or (?-i)(f-i)+z* = z square 
 
 = (2 - ry, say ; [Euler has (z + r) 2 ] 
 
 _ 
 
 whence z = - St. - UJC 
 
 2/- 
 where any numbers may be substituted for^, q, r.
 
 BOOK IV i 79 
 
 20. To find four numbers such that the product of any two 
 increased by unity is a square. 
 
 For the product of first and second take a square minus I, 
 
 say (x+ iy*- i =x* + 2x. 
 Let first = x, so that second =x+2. 
 
 For example, if r=pq+ i, we shall have 
 = J/+)L , 2 
 
 II. But, if whole numbers are required, we put ;ry+ 1 =/*, and assume z=x+y+f. 
 We then have xz+ i =* 3 + .ry+^jr + i = 
 
 and ?z+i=xy+y i +2y+i= 
 
 These expressions are both squares if q = ip. 
 Thus a solution is obtained from xy=p*- i combined with either 
 z=x+y+ip, or z=x+y-ip. 
 
 We take a certain value for /*, split /* - i into two factors, take these factors for the 
 values of x, y respectively, and so find z. 
 
 For example, let/=3, so that/ 2 - 1 = 8; if we make x=t, .7=4, we find z=either 11 
 or o ; and in this case x=i, y=\ t z= it is the solution. 
 
 If we put /* = (+ 1) 2 , we have xy=? + if ; and if we put x=t+y,jf={, we have 
 
 or o. 
 
 The solution is then (+ 2, , 4^ + 4), as in Diophantus. 
 
 Fermat in his note on this problem shows how to find three numbers satisfying not 
 only the conditions of the problem but three more also, namely that each of the numbers 
 shall itself when increased by i give a square, i.e. to solve the equations 
 
 Solve, he says, the present problem of Diophantus in such a way that the terms 
 independent of x in the first and third of the numbers obtained by his method shall be 
 such as when increased by i give a square. It is easy to find a value for a such that 
 
 IM + 1 and i (a + 1) + 1 are both squares. Fermat takes the value ia = -j? , which satisfies 
 the conditions, and the general expressions for the three numbers sought are therefore 
 160 13 7225 85 
 
 5 -7r 4 * + 3 -6> - 3-*3- 
 
 Each of these has, when increased by i, to become a square, that is, we have to solve the 
 triple-equation 
 
 rfjL^if 
 
 5184 36 
 
 Fermat does not give the solution ; but it is effected as follows. 
 
 Multiplying the third expression by 36 and the first by ^ . 36 (in order that the 
 absolute terms in the two may be equal), we have to solve 
 
 12 2
 
 i8o THE ARITHMETICA 
 
 For the product of first and third take (2x + i) 2 - i, or 
 4^r 2 + 4, the coefficient of x being the number next 
 following the coefficient (i) taken in the first case, 
 for the reason shown in the last problem ; 
 
 thus third number = 4^ + 4. 
 
 Similarly take (3* + i) 2 - i, or 9x* + 6x, for the product of 
 first and fourth; therefore fourth = 9* +6. 
 
 And product of third and fourth + i 
 
 = (4;r+4)(9;r + 6) + i =36^ + 60^+25, 
 
 which is a square 1 . 
 
 X+I=V 2 
 *+I2I=' 2 
 
 (S) 1 - 
 
 In order to solve by the method of the triple-equation, we make x+ i a square by 
 putting x=y* + iy. 
 
 Substitute this value in the other two expressions, and for convenience multiply each 
 by 144; this gives 
 
 (y)V + V) + ('3*) 8 = a square! 
 (8s) 2 0* + y)+( I3*) 2 = a squareJ 
 The difference = (y* + iy) (s$ + l -^\ (s 5 - *-^\ 
 
 ' 
 
 The square of half the difference of the factors equated to the smaller expression gives 
 
 whence y=~; and we find that 
 
 7239457225 
 It is easily verified that 
 
 so that the value of x satisfies the three equations. 
 
 The numbers satisfying Fermat's six conditions are then 
 
 1&L x + 13 _i 0060498 1 50193144576 7225 85 _ 48191691 
 
 5184 36 171348100' 7239457225' 5 *8 4 J 36~ 4008004 
 
 1 This results from the fact that, if we have three numbers x, y, z such that 
 
 xy+i=(mx+i) z and xz+ i = {(m + i)x+ i} 2 , 
 then yz+i = \i(m+i)x+('2m+i)}^
 
 BOOK IV T 8i 
 
 Lastly, product of second and fourth + i = gx 2 + 24* + 13 ; 
 
 therefore gx* + 24* + 13 = square = (3* - 4)', say ; 
 
 which gives x = -J%. 
 All the conditions are now satisfied 1 , 
 
 and TV TV " * is the solution2 ' 
 
 1 The remaining condition was: product of second and third + i=a square. That this 
 is satisfied also follows from the general property stated in the last note. In fact 
 
 (x + v) (4^ + 4) + i = 4* 2 + iix + g, 
 which is a sqnare. 
 
 2 With this solution should be compared Euler's solution (Algebra, Part n. Art. 233) 
 of the problem of finding x, y, z, v such that the six expressions 
 
 xy + a, yz + a, zx + a, xv + a, yv + a, zv + a 
 
 are all squares. The solution follows the method adopted to solve the corresponding 
 problem with three unknowns x, y, z only. See note on III. 10 above. 
 
 If we begin by supposing xy + a=^P, and take z = x+y+ip, the second and third 
 expressions become squares (vide note on III. 10, p. 160). 
 
 If we further suppose v = x+y-ip, the fourth and fifth expressions also become 
 squares (vide the same note). 
 
 Consequently we have only to secure that the sixth expression zv + a shall be a square ; 
 that is, 
 
 x 2 + 2xy +y z - 4/ 2 + a = a square, 
 
 or (since xy + a =/2) x z - ixy +y* - 30 = a square. 
 
 Suppose that . (x -y) 2 -^a=(x-y- q)* ; 
 
 therefore x- 
 
 Consequently / 2 = xy + a =y* + ^-^ y + a. 
 
 If we put p=y + r, we have 
 
 and y = - 
 
 from which /, x, and therefore z, v also, are found in terms of q, r, where y, r may have 
 any values provided that x, y, z, v are all positive. 
 
 Euler observed that this method is not suited for finding integral solutions, and, 
 pursuing the matter further, he gave the following very elegant solution of Diophantus' 
 actual problem (the case where a=i) in integers ("Miscellanea analytica" in Com- 
 mentationes arithmeticae, n. pp. 45-6)- 
 
 Six conditions have to be satisfied. If x, y, z, v are the required numbers, let x = i, 
 y = n, where m, n are any integers such that mn + i = P. 
 
 Then put z = m + n + il, and three conditions are already satisfied, for 
 xy+i=mn+i= I 2 , by hypothesis, 
 xz + i = in (m + n + il) + i = (/+ /w) 2 , 
 
 The three conditions remaining to be satisfied are 
 
 xv + i mv + i = a square, 
 
 yv + i = nv + i = a square, 
 
 zv + i = (m + n + 2/) v + i = a square. 
 Let us make the continued product of these expressions a square.
 
 182 THE ARITHMETICA 
 
 21. To find three numbers in proportion and such that the 
 difference of any two is a square. 
 
 Assume x for the least, x + 4 for the middle (in order 
 that the difference of middle and least may be a 
 square), x+ 13 for the greatest (in order that differ- 
 ence of greatest and middle may be a square). 
 
 This product will be found to be 
 
 i + v (tK + n + l)v+{(m + n + /) 2 - 1} 2 +#/(w + + 2/)zr 3 . 
 
 Let us equate this to \i+(m + n + l)v--v i i ,in order that the terms in v, z> 2 as well 
 as the absolute term may vanish ; therefore 
 
 whence 
 
 =/(/+)(/+), 
 
 and therefore v = \l (/+ m) (/+). 
 
 It is true that we have only made the product of the three expressions mv+i, nv+i, 
 (m + n + il)v+ i a square ; but, as the value of v has turned out to be an integral number, 
 so that all three formulae are prime to one another, we may conclude that each of the 
 expressions is a square. 
 The solution is therefore 
 
 x=m, y=n, z = m + n + i/, z> = 4/ (/+*)(/+), 
 where mn + i = I 3 . 
 
 In fact, while three of the conditions have been above shown to be satisfied, we find, 
 as regards the other three, that 
 
 xv + i = 4 /w (l+m) (l+n)+i = (*P+'ilm - i) 2 , 
 yv+ i = 4 / (l+m) (/+) + i =(2/ 2 + 2/ - i) 2 , 
 
 2Z>+ I =4/(#Z + + 2/) (/+#*)(/+)+ I = (4/ l +2//W + 2/- i) 2 . 
 
 It is to be observed that / may be either positive or negative. 
 Ex. Let * = 3, = 8, so that /= 5. 
 
 If /= + 5, the solution is 3, 8, 21, 2080 ; if /= - 5, the solution is 3, 8, i, 120. 
 Fermat shows how to solve this problem, alternatively, by means of the "triple-equation. " 
 Take three numbers with the required property, e.g. 3, r, 8. Let x be the fourth, and 
 we have then to satisfy the conditions 
 
 Put x=jfl+ty, so as to make the second expression a square, and then substitute the 
 value of x in the other two. We have then the double-equation 
 
 The difference = 5 (y* + iy) = $y (y + 1 ) . 
 
 We put then (3^+ i) 2 = 
 
 whence y= 10, and xy z + iy= 120, which value satisfies the triple-equation. 
 
 The four numbers are then 3, i, 8, 120, which solution is identical with one of those 
 obtained by Euler as above.
 
 BOOK IV 183 
 
 If now 13 were a square, we should have an indeterminate 
 solution satisfying three of the conditions. 
 
 We must therefore replace 13 by a square which is the 
 sum of two squares. Any rational right-angled 
 triangle will furnish what is wanted, say 3, 4, 5 ; 
 
 we therefore put for the numbers x, x + 9, ;r+ 25. 
 
 The fourth condition gives 
 
 4r + f and x = . 
 
 Thus , ^, -^- is a solution. 
 
 22. To find three numbers such that their solid content 1 added 
 to any one of them gives a square. 
 
 Assume continued product x*+ 2x, first number I, second 
 
 number ^x + 9, so that two conditions are satisfied. 
 The third number is then (x* + 2x)\(&x + 9). 
 This cannot be divided out unless x*\^x=2x\<) or, 
 alternately, x* : 2x = ^x : g ; but it could be done if 4 
 were half of 9. 
 Now ^x comes from 6x 2x, and the 6x in this from 
 
 twice T)X\ the 9 comes from 3 2 . 
 
 Therefore we have to find a number m to replace 3 such 
 that 2m 2 = \nf\ thus m* = tyn - 4, whence 2 m 2. 
 We put therefore for the second number (x+ 2? (x* + 2x\ 
 or 2x + 4 ; the third number is then 
 
 (x* + 2x)l(2x + 4) or \x. 
 Lastly, the third condition requires 
 
 x* + 2x + \x = a square = 4^, say. 
 Therefore * = f, 
 
 and i, is a solution*. 
 
 1 6 e| at/run- ffre/>e6i, " the solid (number formed) from them " = the continued product 
 of the three numbers. 
 
 8 Observe the solution of a mixed quadratic. 
 
 3 Fermat gives a solution which avoids the necessity for the auxiliary problem. 
 
 Let the solid content bejc 3 -**, the first number i, and ihe second number tx\ two 
 conditions are thus satisfied. 
 
 The third number is now A 3 - xr divided by rr . i, or x - i ; and the third condition 
 gives 
 
 x 3 -^x-i=a. square. 
 
 Now x must be greater than i ; we therefore put 
 
 x*-^x-i=(x-m)*, 
 where m is greater than 2.
 
 i8 4 THE ARITHMETIC A 
 
 23. To find three numbers such that their solid content minus 
 any one gives a square 1 . 
 
 First numbers, solid content x* + x; therefore product of 
 
 second and third x-\- 1. 
 
 Let the second be i, so that the third is x + i. 
 The two remaining conditions require that 
 
 2 I shall both be squares. [Double-equation.] 
 
 The difference = x = J . 2x, say ; 
 thus (x + 1) 2 = x 2 + x - i , and x = ty. 
 
 The numbers are (~, i, ^M. 
 
 \ o o / 
 
 24. To divide a given number into two parts such that their 
 product is a cube minus its side. 
 
 Given number 6. First part^r; therefore second = 6 x, 
 
 and 6x x 2 = a cube minus its side. 
 Form a cube from a side of the form mx i, say 2^1, 
 
 and equate 6x x* to this cube minus its side. 
 Therefore Sx 3 1 2x z -f ^x= 6x x z . 
 
 1 A remarkable problem of this kind (in respect of the apparent number of conditions 
 satisfied) is given by De Billy in the Inventum Novum, Part I. paragraph 43 (Oeuvres 
 de Fermat, III. p. 352) : To find three numbers , 77, f (, f, 77 being in ascending order 
 of magnitude) such that the following nine expressions may become squares : 
 
 (0 f-fctf (4) i --!? (7) fr-fcjf, 
 
 (2) 77-f7?f, (5) f-f-fijf, (8) ijf-fijf, 
 
 (3) f-frf, (6) ij-f-frfc (9) V'-M. 
 
 Take x, i, i JT as the values of , 77, f respectively. Then six conditions, namely, 
 (0. (3) (4). (6). (?), (8), are all automatically satisfied. 
 By conditions (2) and (9) alike, 
 
 i -x + x 2 = a. square. 
 And, by (5), i $x + j? = a. square. 
 
 Solving this double-equation in the usual way, we get *=f ,and the numbers are 
 
 o 
 
 * ! I 
 
 8' ' 8 
 
 Another solution can be obtained by putting y + ^ in place of x in the two expressions, 
 
 8 
 
 and so on. Cf. note on III. 18 above. 
 
 It would appear from a letter from Fermat to De Billy of 26 Aug. 1659 (Oeuvres, n. 
 pp. 436-8) that this problem and the above single solution were De Billy's own. De Billy 
 had supposed that this was the only solution, but Fermat observed that there were any 
 number, as the above double-equation has any number of solutions. Fermat gave 
 
 (!O4l6 4l440\ 
 
 ^ , i, H ) as another solution.
 
 BOOK IV 185 
 
 Now, if the coefficient of x were the same on both sides, 
 this would reduce to a simple equation, and x would 
 be rational. 
 
 In order that this may be the case, we must put m for 2 
 in our assumption, where ^m m=6 (the 6 being 
 the given number in the original hypothesis). Thus 
 m = 3. 
 
 We therefore assume 
 
 or 
 and 
 
 The parts are *, ^. 
 
 25. To divide a given number into three parts such that their 
 continued product gives a cube the side of which is equal to the 
 sum of the differences of the parts. 
 Given number 4. 
 Since the product is a cube, let it be Sx 3 , the side of 
 
 which is 2x. 
 Now (second part) - (first) + (third) - (second) + (third) 
 
 (first) = twice difference between third and first. 
 Therefore difference between third and first = half sum of 
 
 differences=;tr. 
 Let the first be any multiple of x, say 2.x; therefore the 
 
 third = $x. 
 Hence second = S-^/dt? = %x\ and, if tJie second had lain 
 
 between the first and third, the problem would have been 
 
 solved. 
 Now the second came from dividing 8 by 2 . 3, and the 
 
 2 and 3 are not two numbers at random but con- 
 
 secutive numbers. 
 Therefore we have to find two consecutive numbers such 
 
 that, when 8 is divided by their product, the quotient 
 
 lies between the numbers. 
 Assume m, m+i; therefore 8/(m* + m) lies between 
 
 m and m + I. 
 
 o 
 
 Therefore -- 1- i > m + I, 
 
 m* + m 
 
 so that m* + m + 8 > m 3 + 2m 3 + m, 
 
 or 8 > m 3 + m*.
 
 186 THE ARITHMETICA 
 
 I form a cube such that it has m 3 , m* as terms, that is, the 
 cube (m -f ^) 3 , which is greater than m 3 + m 2 , and I put 
 
 therefore m + ^ = 2, and m = . 
 
 Assume now for first number %x\ the third is x, and 
 the second is \x. 
 
 Multiplying throughout by 15, we take 2$x, 27^-, 4Ox, 
 and the product of these numbers is a cube the 
 side of which is the sum of their differences. 
 
 The sum = g2x = 4, by hypothesis. 
 
 Therefore x = -^, 
 
 and (g, g, g) are the parts required. 
 
 [N.B. The condition 8/(m* + m) < m + i is ignored in 
 the work, and is incidentally satisfied.] 
 
 26. To find two numbers such that their product added to 
 either gives a cube. 
 
 Let the first number be of the form m*x, say 8x 
 
 Second x* i. Therefore one condition is satisfied, since 
 
 Sx 3 Sx + 8x = a cube. 
 
 Also Sx' Sx + x 2 - i =a cube = (2* i) 3 , say. 
 Therefore 13^ = 14^, and x = \&. 
 The numbers are ^, ^. 
 
 27. a To find two numbers such that their product mimes either 
 gives a cube. 
 
 Let the first be of the form m*x, say 8x, and the second 
 
 x 2 + i (since S^ 3 + 8x - Sx= a cube). 
 Also S^+Sxx 2 i must be a cube, " which is impossible 1 ." 
 
 1 Diophantus means that, if we are to get rid of the third power and the absolute 
 term, we can only put the expression equal to (i^r-i) 3 , which gives a negative and 
 therefore "impossible" value for x. But the equation is not really impossible, for we can 
 get rid of the terms in x 3 and -c 2 by putting 
 
 ^-i='2x-, whence x=-, 
 \ I2 / ! 3/5 2 
 
 or we can make the term in x and the absolute term disappear by putting 
 
 Sjf + Sx-x*- ! = (***- iV, whence * = ^>- 
 \3 / '96 
 
 Diophantus has actually shown us how to do the former in iv. 25 just preceding.
 
 BOOK IV 187 
 
 Accordingly we assume for the first number an expression 
 of the form m*x + i, say Sx+ i, and for the second 
 number x* (since 8* 8 + x* x* = a 'cube). 
 
 Also S.* 8 +x 2 - 8* - i = a cube = (2x- i) 3 , say. 
 
 Therefore x = -J^, 
 
 arid the numbers are ^, ^. 
 
 28. To find two numbers such that their product their 
 sum gives a cube. 
 
 Assume the first cube (product + sum) to be 64, and the 
 
 second (product sum) to be 8. 
 Therefore twice sum of numbers = 64 - 8 = 56, and the 
 
 sum = 28, while the product + the sum = 64; therefore 
 
 the product = 36. 
 Therefore we have to find two numbers such that their 
 
 sum is 28 and their product 36. If 14 + *, 14 -x are 
 
 the numbers 1 , we have 196 ^ = 36, or x 2 = 160; and, 
 
 if 1 60 were a square, we should have a rational 
 
 solution. 
 Now 160 arises from 142-36, and 14 = ^.28 = ^.56 
 
 = i (difference of two cubes); also 36 = (sum of 
 
 the cubes). 
 Therefore we have to find two cubes such that 
 
 (\ of their difference) 2 - ^ their sum = a square. 
 Let the sides of the cubes be (z + i), (z i); 
 therefore of difference = i^ 3 + , and the square of this 
 
 is 2j^+ 1^2 + ; 
 the sum of the cubes is ^ + 32-; 
 therefore 2^2* + \\z z + z* 32 = a square, 
 or 9-s 4 + 6z* + i 40 s 122 = a square =(52* + i 62)*, say; 
 whence 32^ = 36^, and z = |. 
 The sides of the cubes are therefore ^-, |, and the cubes 
 
 Put now product of numbers + their sum = ^^, and pro- 
 duct sum = g-J-j- 
 
 Therefore their sum = %$$, and their product 
 Now let the first number = x + half sum = x + 
 
 and the second = half sum - x = $&- - x; 
 
 t'hprpfoj'p j 50 7 98^4 j*2 = ^ JL4J>7_ 
 
 and 262144^=250000. 
 
 i Cf. i. 27.
 
 i88 THE ARITHMETICA 
 
 Therefore x = \\ , 
 
 and (W' i) is a solution - 
 
 Otherwise thus. 
 
 If any square number is divided into two parts one of 
 
 which is its side, the product of the parts added to 
 
 their sum gives a cube. 
 [That is, x (x* - x) -f x* - x + x = a cube.] 
 Let the square be x z , and be divided into the parts x> x*x. 
 Then, by the second condition of the problem, 
 
 x^ x* x* = x 3 2x* = a cube (less than x 3 ) = (^x) 3 , say. 
 Therefore Sx 3 - i6x 2 = x 9 , so that x = ^ , 
 and f 1 -, J is a solution. 
 
 29. To find four square numbers such that their sum added to 
 the sum of their sides makes a given number 1 . 
 Given number 12. 
 Now x s + x + \ = a. square. 
 
 Therefore the sum of four squares + the sum of their sides 
 
 + i =the sum of four other squares= 1 3, by hypothesis. 
 
 Therefore we have to divide 13 into four squares; then, if 
 
 we subtract from each of their sides, we shall have 
 
 the sides of the required squares. 
 
 1 On this problem Bachet observes that Diophantus appears to assume, here 
 and in some problems of Book v., that any number not itself a square is the sum of 
 two or three or four squares. He adds that he has verified this statement for all 
 numbers up to 325, but would like to see a scientific proof of the theorem. These 
 remarks of Bachet's are the occasion for another of Fermat's famous notes : " I have 
 been the first to discover a most beautiful theorem of the greatest generality, namely this : 
 Every number is either a triangular number or the sum of two or three triangular 
 numbers ; every number is a square or the sum of two, three, or four squares; every 
 number is a pentagonal number or the sum of two, three, four or five pentagonal 
 numbers; and so on ad infinitum, for hexagons, heptagons and any polygons whatever, 
 the enunciation of this general and wonderful theorem being varied according to the 
 number of the angles. The proof of it, which depends on many various and abstruse 
 mysteries of numbers, I cannot give here ; for I have decided to devote a separate and 
 complete work to this matter and thereby to advance arithmetic in this region of inquiry 
 to an extraordinary extent beyond its ancient and known limits." 
 
 Unfortunately the promised separate work did not appear. The theorem so far as it 
 relates to squares was first proved by Lagrange (Nottv. Memoires de F Acad. de Berlin, 
 annee 1770, Berlin 1772, pp. 123-133; Oeuvres, in. pp. 189-201), who followed up 
 results obtained by Euler. Cf. also Legendre, Zahlentheorie, tr. Maser, I. pp. 212 sqq. 
 Lagrange's proof is set out as shortly as possible in Wertheim's Diophantus, pp. 324-330. 
 The theorem of Fermat in all its generality was proved by Cauchy (Oeuvres, il e serie, 
 Vol. VI. pp. 320-353) ; cf. Legendre, Zahlentheorie, tr. Maser, II. pp. 332 sqq.
 
 BOOK IV 189 
 
 Now i 3 = 4 + 9 = (f! + f!) + (W + M)> 
 
 and the sides of the required squares are |, T 7 ^, {$, |$, 
 
 the squares themselves being , &- . &* *Q 
 
 B 100' 100' 100' 100' 
 
 30. To find four squares such that their sum minus the sum of 
 their sides is a given number. 
 
 Given number 4. 
 
 Now x* x + | = a square. 
 
 Therefore (the sum of four squares) (sum of their sides) 
 
 + i =the sum of four other squares = 5, by hypothesis. 
 Divide 5 into four squares, as ^, |f , ff , ff . 
 The sides of these squares plus $ in each case are the sides 
 
 of the required squares. 
 Therefore sides of required squares are }, i, ^, -, . 
 
 and the squares themselves ^, *?, *, *?. 
 
 100' 100' 100' 100 
 
 31. To divide unity into two parts such that, if given numbers 
 are added to them respectively, the product of the two sums gives 
 a square. 
 
 Let 3, 5 be the numbers to be added; x, I # the parts of I. 
 Therefore (x + 3) (6 x) = 1 8 + 3 x x* = a square = 4^, say; 
 thus 1 8 + 3* = 5;r 2 , which does not give a rational result. 
 Now 5 comes from a square + I ; and, in order that the 
 equation may have a rational solution, we must sub- 
 stitute for the square taken (4) a square such that 
 
 (the square + i) . 18 + (|) 2 = a square. 
 Put (m 2 + 1)18 + 2^ = a square, 
 
 or 72 m 2 + 8 1 = a square = (Sm + gf, say, 
 
 and m= 18, w 2 =324. 
 
 Hence we must put 
 
 (x+$)(6-x}= 18 + 3^-^ 2 = 324^r 2 . 
 Therefore 1 325^- 3^-18 = 0, 
 
 * = s 7 A = &> 
 
 and ( , -~\ is a solution. 
 Otherwise thus. 
 
 The numbers to be added being 3, 5, assume the first of 
 
 the two parts to be x 3 ; the second is then 4 x. 
 Therefore x (9 - x} = a square = 4^, say, 
 and *=\- 
 
 But I cannot take 3 from f , and x must be > 3 and < 4. 
 
 1 Observe the solution of a mixed quadratic equation.
 
 190 THE ARITHMETICA 
 
 Now the value of x comes from 9/(a square + i), and, since 
 x> 3, this square + I should be<3, so that the square 
 must be less than 2; but, since x <4, the square 4- I 
 must be > f , so that the square must be > |. 
 
 Therefore I must find a square lying between and 2, or 
 between f$ and ^. 
 
 W or ft satisfies the condition. 
 
 Put now x (9 - x) = $%x* ; 
 
 therefore x = ift, 
 
 and (", *) is a solution. 
 
 32. To divide a given number into three parts such that 
 the product of the first and second the third gives a square. 
 Given number 6. 
 Suppose third part = x, second = any number less than 6, 
 
 say 2; therefore first part = ^ x. 
 
 The two remaining conditions require that 8 2x x a 
 square, 
 
 or ' t [ are both squares. [Double-equation.] 
 
 This does not give a rational result ("is not rational "), since 
 the ratio of the coefficients ofx is not a ratio of a square 
 to a square. 
 
 But the coefficients of x are 2 - I and 2 + I ; therefore we 
 must find a number y to replace 2 such that 
 (y + i)j(y - i) = ratio of square to square = , say. 
 
 Therefore y + i = 4y 4, and y = f . 
 
 Put now second part =f ; therefore first = -^ x. 
 
 Therefore & $- \xx='& square. 
 
 That is, ' X I are both squares, 
 
 260 24^) 
 
 or , \ are both squares. 
 
 65 - 24*] 
 
 The difference = 195 = 15 . 13; 
 
 we put therefore \ (i 5 1 3)* = 65 24*, and x = f . 
 
 Therefore the required parts are f-, -, -V. 
 
 i Fermat observes: "The following is an easier method of solution. Divide the 
 number 6 into two in any manner, e.g. into 5 and i. Divide their product less i, that is 
 
 4, by 6, the given number: the result is -. Subtracting this first from 5 and then from i,
 
 BOOK IV 191 
 
 33. To find two numbers such that the first with a fraction 
 of the second is to the remainder of the second in a given ratio, 
 and also the second with the same fraction of the first is to the 
 remainder of the first in a given ratio. 
 
 Let the first with the fraction of the second = 3 times the 
 
 remainder of the second, and the second with the 
 
 same fraction of the first = 5 times the remainder of 
 
 the first. 
 [The fraction may be either an aliquot part or not, TO 
 
 avro fj.epof or rd avra fiepij as Diophantus says, 
 
 following the ordinary definition of those terms ("the 
 
 same part" or "the same parts")-, cf. Euclid VII. 
 
 Deff. 3, 4-] 
 Let the second =x+ i, and let the part of it received by 
 
 the first = i ; 
 
 therefore the first = 3^1 (since ^x i + I = yc\ 
 Since the second plus the fraction of the first = 5 times 
 
 the remainder of the first, 
 
 the second + the first = 6 times the remainder of the first. 
 And first + second = 4* ; therefore remainder of first 
 
 = f;r, and hence the second receives from the first 
 
 -$x\ \x or \x i. 
 We have therefore to secure that \x\ is the same 
 
 fraction of yc i that I is of x+ i. 
 This requires that (&x i)(^+ i) = (3^r i). I ; 
 therefore \x* + %x - i = 3* i, and x = f . 
 Accordingly the numbers are f, ^; and I is ^ of the 
 
 second. 
 
 we have as remainders and -, which are the first two parts of the number to be 
 
 divided; the third is therefore -." 
 
 3 
 
 That is, if , 77, f be the required parts of the number a, Fermat divides a into two 
 parts x, a - x and then puts 
 
 whence 
 
 The three general expressions in x satisfy the conditions, and x may be given any 
 value <a.
 
 192 THE ARITHMETICA 
 
 Multiply by 7 and the numbers are 8, 12, and the fraction 
 is ^; but 8 is not divisible by 12: so multiply by 3, 
 and (24, 36) is a solution. 
 
 Lemma to the next problem. 
 
 To find two numbers indeterminately such that their product 
 together with their sum is a given number. 
 
 Given number 8. 
 
 Assume the first number to be x, the second 3. 
 
 Therefore $x+x+ 3= given number = 8; * = f, and the 
 numbers are f , 3. 
 
 Now | arises from (8 - 3)/(3 + i), where 3 is the assumed 
 second number. 
 
 We may accordingly put for the second number (instead 
 of 3) any (undetermined) number whatever 1 ; then, 
 substituting this for 3 in the above expression, we 
 have the corresponding first number. 
 
 For example, we may take*- I for the second number; 
 
 the first is then 9 - x divided by x, or - I. 
 
 34. To find three numbers such that the product of any two 
 together with the sum of those two makes a given number 2 . 
 
 1 The Greek phrase is eav dpa rdfw/xo' rbv fi? v s ov oiovS^irore (otonS^Trore j in Lemma 
 to iv. 36), "If we make the second" (literally "put the second at") "any s whatever." 
 But the s is not here, as it is in the Lemma to IV. 36, the actual x of the problem, for 
 Diophantus goes on to say " E.g. let the second be x- i." In the Lemma to iv. 34 the 
 corresponding expression is "any quantity whatever" (offovSjiroTe without s). The 
 present Lemma amounts to saying that, if xy + x +y = a, then x = (a-y)l(y+i). 
 
 2 This determinate set of equations can of course be solved, with our notation, by 
 a simple substitution. 
 
 The equations yz +y + z = a\ 
 
 are equivalent to 
 
 where 
 
 The solution is = x + i 
 
 In order that the result may be rational, it is only necessary that (a+ i) (l>+ i) (c + i) 
 should be a square ; it is not necessary that each of the expressions a + t , b + i , c + i 
 should be a square, as Diophantus says.
 
 BOOK IV I93 
 
 Necessary condition. Each number must be i less than some 
 square 1 . 
 
 Let (product + sum) of first and second = 8. 
 second and third = 15. 
 third and first = 24. 
 By the first equation, if we divide(8 - second)by (second + 1 ), 
 
 we have the first number. 
 Let the second number be x i. 
 
 Therefore the first = 
 
 Similarly the third number = -- i. 
 The third equation remains, which gives 
 
 The numbers are 33, 2 , 
 or, when reduced to a common denominator, @, f* 34?. 
 
 DO DO OO 
 
 Lemma to the following problem. 
 
 To find two numbers indeterminately such that their product 
 minus their sum is a given number. 
 Given number 8. 
 
 First numbers, second 3, suppose; therefore 
 (product) - (sum) = ?>x x*$ = 2x 3 = 8, and x^\. 
 The first number is therefore 5^, the second 3. 
 But 5^ comes from (8 + 3)/(3 - i), and we may put for 3 
 
 any number whatever. 
 E.g. put the second number =x+\; the first is then ^+9 
 
 divided by x, or i +-. 
 
 35. To find three numbers such that the product of any two 
 minus the sum of those two is a given number 2 . 
 
 Necessary condition. Each of the given numbers must be i less 
 than some square 2 . 
 
 Let (product sum) of first and second = 8. 
 
 M ., second and third = 15. 
 
 third and first = 24. 
 
 1 See last paragraph of preceding note. 
 
 2 The notes to iv. 34 above apply, mutatis mutandis, to this problem as well. 
 
 H. D. J 3
 
 1 9 4 THE ARITHMETICA 
 
 By the first equation, if we divide (8 + second) by 
 (second - i), we have the first number. 
 
 Assuming x + i for the second number, we have i + - 
 
 for the first. 
 Similarly i H is the third number, and two conditions 
 
 are satisfied. 
 
 The third gives l ~ - i = 24, and x = - 1 /. 
 The numbers are ^, , , 
 
 or, with a common denominator, ~ s , ^, ^. 
 
 Lemma to the following problem. 
 
 To find two numbers indeterminately such that their product 
 has to their sum a given ratio. 
 
 Let the given ratio be 3:1, the first number x, the 
 
 second 5. 
 Therefore 5^=3(5 + -*'), x = 7\', and the numbers are 
 
 7i 5- 
 But 7^ arises from 15 divided by 2, while the 15 is the 
 
 second number multiplied by the given ratio, and 
 
 the 2 is the excess of the second number over the 
 
 ratio. 
 Putting therefore x (instead of 5) for the second number, 
 
 we have, for the first number, 3 x divided by x - 3. 
 The numbers are therefore ^xl(x 3), x. 
 
 36. To find three numbers such that the product of any two 
 bears to the sum of those two a given ratio. 
 
 Let product of first and second be 3 times their sum. 
 second and third be 4 times their sum. 
 
 third and first be 5 times their sum. 
 
 Let second number be.r; the first is therefore ^xl(x 3), 
 by the Lemma, and similarly the third is <\x\(x 4). 
 
 T>X A.X ( \x A.X \ 
 Lastly -^ . -^ = 5 ( -^ + -^ _ ) 
 
 x-i . *-4 'U-3 *-4/' 
 
 or 1 2 AT 2 = 35^r 2 
 
 Therefore * = J^, 
 
 and the numbers are &
 
 BOOK IV I95 
 
 37. To find three numbers such that the product of any two 
 has to the sum of the three a given ratio 1 . 
 
 Let product of first and second = 3 times sum of the three, 
 of second and third =4 
 of third and first = 5 
 
 First seek three numbers such that the product of any two 
 
 has to an arbitrary number (say 5) the given ratio. 
 Then product of first and second = 15; and, if x be the 
 
 second, the first is i$jx. 
 The product of second and third = 20; therefore third 
 
 It follows that 20. i5/;r' 2 = 25. 
 
 And, if the ratio of 20 . 1 5 to 25 were that of a square to a 
 
 square, the problem would be solved. 
 Now 15 = 3.5, an d 20 is 4. 5, the 3 and 4 being fixed by 
 
 the original hypothesis, but 5 being an arbitrary 
 
 number. 
 We must therefore find a number m (to replace 5) such 
 
 that I2m 2 /$m = ratio of a square to a square. 
 Thus 1 2m 2 . 5z =6om 3 = a square = poow 2 , say ; and m= 1 5. 
 Let then the sum of the three numbers be 15. 
 Product of first and second is therefore 45, and first = 45/^r. 
 Similarly third = 6olx. 
 Therefore 45 . 6o/x 2 = 75, and x = 6. 
 
 Therefore the numbers are 7|, 6, 10, and the sum of 
 
 these = 2 3^. 
 Now, if this sum ^vere 1 5 instead, the problem would be 
 
 solved. 
 
 1 Loria (pp. cit. p. 130) quotes this problem as an instance of Diophantus' ingenious 
 choice of unknowns. Here the equations are, with our notation, 
 
 xy = <(*+)> + z), 
 
 and Diophantus chooses as his principal unknown the sum of the three numbers, 
 x+y + z w, say. 
 
 We may then write x=cw\y, z-aw\y, so that zx = acw i \y 1 = bw, and _j/ 2 = w. 
 
 Putting W = -T 2 , we have 
 
 from which, by eliminating x, y, z, we obtain = 
 Hence x=(6f+fa + a6)la, 
 
 132
 
 196 THE ARITHMETIC A 
 
 Put therefore for the sum of the three numbers 1 5^ 2 , and 
 
 for the numbers themselves J\x, 6x, lox. 
 Therefore 2^x = \$x z , so that x = $ s , 
 
 and 35M, *, 4?o is a solution> 
 30 3" 30 
 
 38. To find three numbers such that their sum multiplied 
 
 into the first gives a triangular number, their sum multiplied into 
 
 the second a square, and their sum multiplied into the third a cube. 
 
 Let the sum be x-, and let the numbers be m/x 2 , n/x*,p/x-, 
 
 where m t n,p are a triangular number, a square and 
 
 a cube respectively ; 
 
 say first number = 6/x*, second 4/*r 2 , third S/x 2 . 
 But the sum is x 2 ; therefore i8/x z = x 2 , or i8=;r 4 . 
 Therefore we must replace 18 by some fourth power. 
 But 1 8 = sum of a triangular number, a square and a cube. 
 Let x* be the required fourth power, which must therefore be 
 
 the sum of a triangular number, a square and a cube. 
 Let the square be x* 2x* + i ; 
 
 therefore the triangular number + the cube = 2x* i. 
 Let the cube be 8; therefore the triangular number is 
 
 2X 2 9. 
 But 8 times a triangular number + i = a square ; therefore 
 
 \6x~- 71 = a square = (4^ i) 2 , say; thus x = g, the 
 
 triangular number is 153, the square 6400 and the 
 
 cube 8. 
 
 Assume then as the numbers I53/-*" 2 , 64<X)/r 2 , S/x 2 . 
 Therefore 6$6i/x* = x*, or ^ = 6561, and x=g. 
 
 1 The procedure may be shown more generally thus. 
 Let , -TI, f be the required numbers; suppose 
 
 It follows that 
 
 Suppose now that /3=j- 2 -3 2 [Diophantus and Bachet assume &= i]. 
 
 Then ^tlU^-^3. 
 
 Eight times the left hand side plus i gives a square (by the property of triangular 
 numbers) ; that is, 
 
 (20 -- 1) 2 = i6sV - 8z 4 - 8-y 3 + i ='a square 
 
 = ( 4 0jr-/&) 2 , say,
 
 BOOK IV I97 
 
 39. To find three numbers such that the difference of the 
 greatest and the middle has to the difference of the middle and the 
 least a given ratio, and also the sum of any two is a square. 
 
 Ratio 3 : i. Since middle number + least = square, let the 
 
 square be 4. 
 
 Therefore middle > 2 ; let it be x + 2, so that least = 2-x. 
 Therefore difference of greatest and middle = 6x, whence 
 the greatest = 7 x + 2. 
 
 Therefore , I are both squares. [Double-equation] 
 
 Take the difference 2.x, split it into factors, say \x, 4, and 
 proceed by the rule ; therefore x= 112. 
 
 But I cannot take 1 12 from 2; therefore x must be found 
 to be < 2, so that 6x + 4< 16. 
 
 Thus there are to be three squares 8^+4, 6^ + 4 and 4 
 (the 4 arising from 2 . 2), and the difference of the 
 greatest and middle is ^ of the difference of the 
 middle and least. 
 
 We have therefore to find three squares having this property 
 and such that the least = 4 and the middle < 16. 
 
 Let side of middle square be z + 2 ; therefore excess of 
 middle over least = & + 42, whence excess of greatest 
 over middle = \z^ + \%z, and therefore the greatest 
 
 This must be a square ; therefore, multiplying by 9, we have 
 1 2z* + 482 + 36 = a square, 
 
 whence ., 
 
 OKZ 
 
 But a ^ must be integral, and therefore a integral, so that -(43^-^-1) must be 
 
 integral ; that is, must be integral. 
 
 Bachet assumes that it is necessary, with Diophantus, to take k=i, observing that 
 trial will show that the problem can hardly be solved otherwise. On this Fermat remarks 
 that Bachet's trial had not been carried far enough. We may, he says, put for -y 3 any 
 cube, for instance, with side of the form 3+i. Suppose, for example, we take 7 3 . 
 Then [2 being i] we have to make 
 
 2.** - 344 a triangle, 
 
 and therefore 16^-2751 a square, and we may take, if we please, 4^-3 as the side of 
 this square [so that k is in this case 3]. 
 
 By varying the cubes we may use an unlimited variety of odd numbers, besides 3, 
 as values for k which will satisfy the required condition. 
 
 Loria (op. cit. p. 138) points out that the problem could have been more simply 
 solved by substituting x fcr * 2 and 2 for 2 2 in the above assumptions. The ultimate 
 expression to be made a square would then have been 162* -8s 2 -873+ r, and we could 
 have equated this to X 2 , thus finding jr.
 
 198 THE ARITHMETICA 
 
 or 3r 2 + 1 2.z + 9 = a square = (mz 3) 2 , say. 
 
 It follows that z (6m + \2)\(in^ 3), which must be < 2. 
 
 Therefore 6m + 12 < 2m 2 6, or 2m z > 6m + 18. 
 
 "When we solve such an equation 1 , we multiply half the 
 coefficient of x into itself this gives 9 then multiply 
 the coefficient of X* into the units 2 . 18 = 36 add 
 this last number to the 9, making 45, and take the 
 side [square root] of 45, which is not less than 7; 
 add half the coefficient of x making a number not 
 less than 10 and divide the result by the coefficient 
 of;tr 2 ; the result is not less than 5." 
 
 [3 2 + 18.2 = 45, and 1^45 +f is not less than f +|.] 
 
 We may therefore put m = f + |, or 5, and we thus have 
 
 Therefore z = \\, and the side of the middle square is 
 
 ^f , the square itself being -^p. 
 Turning to the original problem, we put 6;r+ 4 = 1 f$ L , and 
 
 x = J^R, which is less than 2. 
 
 The greatest of the required numbers = yx + 2 = 
 the middle =^+2=^, 
 
 and the least = 2 - x = & 
 726 
 
 The denominator not being a square, we can make it 
 a square by dividing out by 6; the result is 
 1834^ 
 
 121 ' 121 ' 121 ' 
 
 or again, to avoid the in the numerators, we may 
 multiply numerators, and denominators, by 4; thus 
 
 7 . ' isaso,ution'. 
 
 1 I have quoted Diophantus' exact words here, with the few added by Tannery, 
 "making a number not less than 10... coefficient of jc 2 ," in order to show the precise 
 rule by which Diophantus solved a complete quadratic. 
 
 When he says v /45 is not less than 7, Diophantus is not seeking exact limits. Since 
 ^/45 is between 6 and 7 we cannot take a smaller integral value than 7 in order to 
 satisfy the conditions of the problem (cf. p. 65 above). 
 
 2 A note in the Invmtum Novum (Part n, paragraph 26) remarks upon the prolix and 
 involved character of Diophantus' solution and gives a shorter alternative. The problem 
 is to solve 
 
 -rt = w (>}-{), (!>>?> and ^ = 3, say)
 
 BOOK IV 199 
 
 40. To find three numbers such that the difference of the 
 squares of the greatest and the middle numbers has to the differ- 
 ence of the middle and the least a given ratio, and also the sum of 
 each pair is a square. 
 Ratio 3:1. 
 
 Let greatest + middle number = the square \6x*\ therefore 
 greatest^* 8 : let it be 8^ + 2 ; hence middle=8^- 2. 
 And, since greatest + middle > greatest + least, 
 
 1 6^ > (greatest + least) > 8x* ; 
 
 let greatest + least = ojr 2 , say ; therefore least =x* 2. 
 Now difference of squares of greatest and middle = 64**, 
 
 and difference of middle and least = jx?. 
 But 64 is not equal to 3.7 or 21. 
 Now 64 comes from 32.2; therefore we must find a 
 
 number m (in place of 2) such that 32m =21. 
 Therefore m = f. 
 Assume now greatest number = &x- + f, middle = 8x* , 
 
 least = ;r 2 -f. 
 [And difference of squares of greatest and middle 
 
 The only condition left is: middle + least = square; that is, 
 
 a square = (yc 6Y, say. 
 Therefore x-- 
 and 
 
 Take an arbitrary square number, say 4, for the sum of ij, f; suppose 2 + .1 = 17, 2 - -* - 
 sothatij-i-=wr; therefore - 17 = 3(1;- f)= 6*. whence f= 2 + 7*. 
 The last two conditions require that 
 
 4 + ?*l shall be squares. 
 4 + drf 
 
 Replace x by -)? + -?. This will make 4 + 6x a square. It remains that 
 
 -(*?)*+ 
 
 71105 -K^'iM?" 5 )' 
 
 and y= , so that x =\y*+~ 7=*^ 
 
 The numbers are therefore - , i -
 
 THE ARITHMETICA 
 
 BOOK V 
 
 i. To find three numbers in geometrical progression such that 
 each of them minus a given number gives a square. 
 Given number 12. 
 Find a square which exceeds 12 by a square. "This is 
 
 easy [u. 10]; 42! is such a number." 
 Let the first number be 42^, the third ,r 2 ; therefore the 
 middle number = 6\x. 
 
 Therefore ^ [ are both squares; 
 
 6\x- 12 j 
 
 their difference = x z 6\x = x (x 6) ; half the difference 
 of the factors multiplied into itself =J ^; therefore, 
 putting 6\x- 12=^, we have ^ = ffi, 
 
 2. To find three numbers in geometrical progression such that 
 each of them when added to a given number gives a square. 
 Given number 20. 
 Take a square which when added to 20 gives a square, 
 
 say 1 6. 
 
 Put for one of the extremes 16, and for the other x*, so 
 that the middle term = ^x. 
 
 ,, r *" + 20 ) 
 Therefore \ are both squares. 
 
 4^+20 j 
 
 Their difference is x* \x = x (x 4), and the usual method 
 gives $x + 20 = 4, which is absurd, because the 4 
 ought to be some number greater than 20. 
 
 But the 4 = \ (16), while the 16 is a square which when 
 added to 20 makes a square; therefore, to replace 16, 
 we must find some square greater than 4 . 20 and 
 such that when increased by 20 it makes a square. 
 
 Now 8 1 > 80; therefore, putting (m + 9) 2 for the required 
 square, we have 
 
 (m + p) 2 + 20 = square = (;// 1 1 ) 2 , say ; 
 
 therefore m = , and the square = (9^)" = 90^.
 
 BOOK V 201 
 
 Assume now for the numbers 90^, Q\X, x-, and we have 
 
 The difference =x(x <)$), and we put 
 Therefore x = -^, and 
 
 3. Given one number, to find three others such that any one of 
 them, or the product of any two of them, when added to the given 
 number, gives a square. 
 
 Given number 5. 
 
 " We have it in tlie Porisms that if, of two numbers, each, 
 as well as their product, when added to one and the 
 same given number, severally make squares, the 
 two numbers are obtained from the squares of con- 
 secutive numbers 1 ." 
 
 Take then the squares (x+ 3)' 2 , (x+^f, and, subtracting 
 
 the given number 5 from each, put for the first 
 
 number ;r 2 -j-6;tr+4, and for the second 
 
 and let the third 2 be twice their sum minus i, or 
 
 4_r 2 + 28 x + 29. 
 
 1 On this Porism, see pp. 99, 100 ante. 
 
 2 The Porism states that, if a be the given number, the numbers j^-a, 
 satisfy the conditions. 
 
 In fact, their product + a={x (x + i)} 2 - a ( 3 + wr+i) + a*+a 
 
 Diophantus here adds, without explanation, that, if X, Y denote the above two numbers, 
 we should assume for the third required number Z= i (X+ Y)-i. We want thrte numbers 
 such that any two satisfy the same conditions as X, Y, Diophantus takes for the third 
 Z = i.(X+ Y)- i because, as is easily seen, with this assumption two out of the three 
 additional conditions are thereby satisfied. 
 
 For Z=2(X+Y)-i=i(2Jc l +ix+i)-4a-i 
 
 therefore XZ+a=Jt?(*jc+i)*-a{('ix 
 
 = Jt 2 (2.* + i) s - a . +x (ix -f i) + 4 2 
 
 while 
 
 The only condition remaining is then 
 
 Z+a=a square, 
 
 or (2jr+i) s -3 
 
 and x is found. 
 
 Cf. pp. ioo, 104 above.
 
 THE ARITHMET1CA 
 Therefore 43? + 2%x+ 34 = a square = (2x 6) 2 , say. 
 
 Hence * = ^, and (**, ^, ^) is a solution'. 
 
 4. Given one number, to find three others such that any one 
 of them, or the product of any two, minus the given number gives 
 a square. 
 
 Given number 6. 
 
 Take two consecutive squares x*, x* + 2x + i. 
 
 Adding 6 to each, we assume for the first number x* + 6, 
 
 and for the second x 2 + 2x + 7. 
 For the third 2 we take twice the sum of the first and 
 
 second minus i, or 4^ + 4^+25. 
 Therefore third minus 6 = ^.x- + 4.*+ 19 = square =(2x6) 2 , 
 
 say. 
 Therefore * = |, 
 
 " (?? %, w) is a solution - 
 
 [The same Porism is assumed as in the preceding problem 
 but with a minus instead of a plus. Cf. p. 99 above.] 
 
 5. To find three squares such that the product of any two 
 added to the sum of those two, or to the remaining square, gives 
 a square. 
 
 " We have it in the Porisms" that, if the squares on any two 
 consecutive numbers be taken, and a third number 
 be also taken which exceeds twice the sum of the 
 squares by 2, we have three numbers such that the 
 product of any two added to those two or to the 
 remaining number gives a square 3 . 
 
 1 Diophantus having solved the problem of finding three numbers , 17, f satisfying 
 the six equations 
 
 Fermat observes that we can deduce the solution of the problem 
 
 To find four numbers such that the product of any pair added to a gii-en number 
 produces a square. 
 
 Taking three numbers, as found by Diophantus, satisfying the above six conditions, 
 we take x+ i as the fourth number. We then have three conditions which remain to be 
 satisfied. These give a " triple-equation " to be solved by Fermat's method. 
 
 2 Diophantus makes this assumption for the same reason as in the last problem, v. 3. 
 The second note on p. 201 covers this case if we substitute - a for a throughout. 
 
 3 On this Porism, see pp. 100-1 ante.
 
 BOOK V 203 
 
 Assume as the first square x*+2x+ i, and as the second 
 
 ;r 2 + 4jr+4, so that third number = 4^ + 12*+ 12. 
 Therefore x* + 3* + 3 = a square = (x - $f, say, and x = f . 
 Therefore (|, , ^) is a solution. 
 
 6. To find three numbers such that each minus 2 gives a 
 square, and the product of any two minus the sum of those two, 
 or minus the remaining number, gives a square. 
 
 Add 2 to each of three numbers found as in the Porism 
 
 quoted in the preceding problem. 
 Let the numbers so obtained be 
 
 All the conditions are now satisfied 1 , except one, which 
 gives 
 
 4^? + 44: + 6 2 = a square. 
 
 Divide by 4, and x- + x + I = a square = (x 2f, say. 
 Therefore x = \ t 
 
 anc * ( ' W' ^s) is a s l ut i n - 
 
 Lemma I to tlie following problem. 
 
 To find two numbers such that their product added to the 
 squares of both gives a square. 
 
 Suppose first number x, second any number (;), say I. 
 Therefore x . I +x z + I =x* + x+ I = a square = (x 2) 2 , say. 
 Thus x = \, and 
 
 ($, i) is a solution, or (3, 5). 
 
 Lemma II to the following problem. 
 
 To find three right-angled triangles (t.e. three right-angled 
 triangles in rational numbers-} which have equal areas. 
 
 We must first find two numbers such that their product 
 -f the sum of their squares = a square, e.g. 3, 5, as in 
 the preceding problem. 
 
 1 The numbers are jc 2 + 1, (x+i)*+2, i {x* + (x+ i) 2 + 1} + 2; and if X, Y, Zdenote 
 these numbers respectively, it is easily verified that 
 
 XY-(X+ Y) = (x?+x+i)\ XY-Z=(x*+x?, 
 XZ - (X+ Z) = (2*2 + x + 2) 2 , XZ - Y = (2.x- 2 + x+ a) 2 , 
 and FZ-(F+Z) =(2^2+3^ + 3)2, YZ - X = (2*2 + 3^ + 4)2. 
 
 2 All Diophantus' right-angled triangles must be understood to be right-angled 
 triangles with sides expressible in rational numbers. In future I shall say "right-angled 
 triangle " simply, for brevity.
 
 204 THE ARITHMETICA 
 
 Now form right-angled triangles from the pairs of 
 numbers 1 
 
 (7, 3), (7, 5), (7>3 + 5) 
 
 \t.e. the right-angled triangles (7 2 + 3 2 , 7 2 - 3 2 , 2 . 7 . 3), etc.]. 
 The triangles are (40, 42, 58), (24, 70, 74), (15, 112, 113), 
 the area of each being 840. 
 
 1 Diophantus here tacitly assumes that, if ab + a? + & 2 = c 2 , and right-angled triangles 
 be formed from (c, a), (t, b) and (c, a + b) respectively, their areas are equal. The 
 areas are of course (c 2 - a 2 ) ca, (c 2 - 3 2 ) cb and { (a + 6>)' 2 - c z } (a + b) c, and it is easy to 
 see that each = abc (a + b). 
 
 Nesselmann suggests that Diophantus discovered the property as follows. Let the 
 triangles formed from (n, m), (q, m), (r, m) have their areas equal ; therefore 
 n (m 2 -n*) = q (m 2 - f) = r(r*- m 2 ). 
 
 It follows, first, since nPn - n z = m z q - q 3 , 
 
 that m 2 = ( w 3 - ? 3 )/( -f) = 2 + nq + q*. 
 
 Again, given (q, m, n), to find r. 
 
 We have q (m* - f] = r(r>- m*), 
 
 and m 2 -g*=n z + ng, from above ; 
 
 therefore ?( 3 + ?) = r(r 2 - 2 -?-? 2 ), 
 
 or q(n z + nr) + g*(n + r) = r(r*-n*). 
 
 Dividing by r + , we have qn + q* = r 2 - rn ; 
 
 therefore (q + r) n = r* - q 2 , 
 
 and r=q + n. 
 
 Fermat observes that, given any rational right-angled triangle, say z, />, d, where z 
 is the hypotenuse, it is possible to find an infinite number of other rational right-angled 
 triangles having the same area. Form a right-angled triangle from z 2 , ibd\ this gives 
 the triangle z* + ^b z d' 2 , s 4 -^ 2 ^ 2 , ypbd. Divide each of these sides by iz(!fi-d), 
 
 b being >d ; and we have a triangle with the same area ( ~bd\ as the original triangle. 
 
 Trying this method with Diophantus' first triangle (40, 42, 58), we obtain as the new 
 triangle 1412881 1412880 1681 
 
 1189 ' 1189 ' 1189' 
 
 The method gives ( , , j as a right-angled triangle with area equal 
 to that of (3, 4, 5). 
 
 Another method of finding other rational right-angled triangles having the same area as 
 a given right-angled triangle is explained in the Inventum Novum, Part l, paragraph 38 
 (CEuvres de Fermat, in. p. 348). 
 
 Let the given triangle be 3, 4, 5, so that it is required to find a new rational right- 
 angled triangle with area 6. 
 
 Let 3, or + 4 be the perpendicular sides ; therefore 
 
 the square of the hypotenuse =* 2 +8.*+25 = a square. 
 
 Again, the area is - x + 6 ; and, as this is to be 6, it must be six times a certain square ; 
 
 that is, -x + 6 divided by 6" must be a square, and this again multiplied by 36 must 
 
 be a square ; therefore 
 
 gx -f 36 = a square. 
 Accordingly we have to solve the double-equation 
 
 9* + 36=^1 '
 
 BOOK V 205 
 
 7. To find three numbers such that the square of any one 
 the sum of the three gives a square. 
 
 Since, in a right-angled triangle, 
 
 (hypotenuse) 2 + twice product of perps. = a square, 
 we make the three required numbers hypotenuses and the 
 
 sum of the three four times the area. 
 
 Therefore we must find three right-angled triangles having 
 the same area, e.g. t as in the preceding problem, 
 
 (40, 42, 58), (24, 70, 74), (15, ii2, 113). 
 Reverting to the substantive problem, we put for the 
 numbers 5&r, 744:, \\^x\ their sum 245* = four times 
 the area of any one of the triangles = 3360**. 
 Therefore x = , 
 
 and ( 
 
 f , S* f ) is a solution. 
 
 Leinma to the following problem. 
 
 Given three squares, it is possible to find three numbers such 
 that the products of the three pairs shall be respectively equal to 
 those squares. 
 
 This gives ^ = _ 67,5600 1 
 
 2405601 
 2806804 
 
 whence * +4= .-^6S- 
 
 The triangle is thus found to be 
 
 2896804 7/76485 
 7405601* 2405601' 
 
 The area is 6 times a certain square, namely /242 I , the root of which is . 
 
 2405001 1551 
 
 Dividing each of the above sides by , we obtain a triangle with area 6, namely 
 
 Another solution of the double-equation, r = ~^^ P^ jr+4= 4 ^' leads to 
 the same triangle ( , , -. ] as that obtained by Fermat's rule (see above). 
 
 The method of the Invenium Nvvum has a feature in common with the procedure in 
 the ancient Greek problem reproduced and commented on by Heiberg and Zeuthen 
 (Bibliotheca Mathtmatiea, vm 3 , 1907/8, p. 122), where it is required to find a rational 
 right-angled triangle, having given the area, 5 feet, and where the 5 is multiplied by a 
 square number containing 6 as a factor and such that the product " can form the area of 
 a right-angled triangle." 36 is taken and the area becomes 180, which is the area of 
 (9, 40, 41). The sides of the latter triangle are then divided by 6, and we have the 
 required triangle (cf. p. 119, ante).
 
 2 o6 THE ARITHMETICA 
 
 Squares 4, 9, 16. 
 
 First numbers, so that the others are 4/ar, 9/^5 and 36/^=16. 
 
 Therefore -f =f, and the numbers are (i^, 2|, 6). 
 
 We observe that x \, where 6 is the product of 2 and 3, 
 and 4 is the side of 16. 
 
 Hence the following rule. Take the product of two sides 
 (2, 3), divide by the side of the third square 4 [the 
 result is the first number] ; divide 4, 9 respectively 
 by the result, and we have the second and third 
 numbers. 
 
 8. To find three numbers such that the product of any two + 
 the sum of the three gives a square. 
 
 As in Lemma II to the 7th problem, we find three right- 
 angled triangles with equal areas ; the squares of 
 their hypotenuses are 3364, 5476, 12769. 
 
 Now find, as in the last Lemma, three numbers such that 
 the products of the three pairs are equal to these 
 squares respectively, which we take because each 
 + 4 . (area) or 3360 gives a square ; the three numbers 
 then are 
 
 Affix, z-iy-x ^Ull^x Tannery], 
 
 4 it 1;tr [ & lial &;tr Tannery]. 
 
 It remains that the sum of the three = 3360^. 
 
 Therefore ^ffffflr^ [**$$$** Tannery] = 3360;^. 
 
 TViprpfnrf* r 3282 4 80 6 f 131299224 /-> 781543 Tannprwl 
 
 i iicrciui c ji -ftfTSTj-Gg-fG LT3T958~(>55tf U1 W5W20 -idiiiieryj, 
 
 and the numbers are ly^ , i^gf^ ^ST] 
 
 9. To divide unity into two parts such that, if the same given 
 number be added to either part, the result will be a square. 
 
 Necessary condition. The given number must not be odd and 
 the double of it + i must not be divisible by any prime number 
 which, when increased by i, is divisible by 4 \i.e. any prime number 
 of the form 4 i ] 1 . 
 
 Given number 6. Therefore 13 must be divided into two 
 squares each of which >6. If then we divide 13 into 
 two squares the difference of which < i, we solve the 
 problem. 
 
 1 For a discussion of the text of this condition see pp. 107-8, ante.
 
 BOOK V 207 
 
 Take half of 1 3 or 6, and we have to add to 6 a small 
 fraction which will make it a square, 
 
 or, multiplying by 4, we have to make - + 26 a square, 
 
 X 
 
 i.e. 26x* + I = a square = (5^+ i) 2 , say, whence x= 10. 
 That is, in order to make 26 a square, we must add y^, or, 
 to make 6^ a square, we must add ^^, and 
 
 Therefore we must divide 1 3 into two squares siich that their 
 sides may be as nearly as possible equal to f. [This 
 is the Tra/Ko-oTT/To? dycayij described above, pp. 95-8.] 
 
 Now 1 3 = 2 2 + 3 2 . Therefore we seek two numbers such 
 that 3 minus the first = f, so that the first = ^ , and 
 2 plus the second = f, so that the second = $. 
 
 We write accordingly (ii^r + 2) 2 , (3 gxf for the required 
 squares [substituting x for ^]. 
 
 The sum =2O2^ 2 - io^+ 13 = 13. 
 
 Therefore x= fa, and the sides are fjft, fflf . 
 
 Subtracting 6 from the squares of each, we have, as the 
 parts of unity, 
 
 4843 5358 
 
 I020I ' IO2OI ' 
 
 10. To divide unity into two parts such that, if we add different 
 given numbers to each, the results will be squares. 
 
 Let the numbers 1 be 2, 6 and let them and the unit be 
 represented in the figure, where DA =2, AB=\, 
 BE = 6, and G is a point in AB so chosen that DG, 
 GE are both squares. 
 
 D A GB E 
 
 Now DE = g. Therefore we have to divide 9 into two 
 squares such that one of them lies between 2 and 3. 
 
 Let the latter square be *, so that the other is 9-*', 
 where 3 >^ 2 >2. 
 
 Take two squares, one > 2, the other <3 [the former 
 being the smaller], say fff , ff. 
 
 1 Loria (op. cit. p. isow.), as well as Nesselmann, observes thai Diophantus omits to 
 state the necessary condition, namely that the sum of the two given numbers plus i must 
 be the sum of two squares.
 
 ao8 THE ARITHMETICA 
 
 Therefore, if we can make x* lie between these, we shall 
 
 solve the problem. 
 We must have x> || and < -ff. 
 Hence, in making 9^ a square, we must find 
 
 x>ft and < if. 
 Put 9 x z = (3 mxf, say, whence x = 6mf(m 2 + i ). 
 
 17 6m , 10 
 Therefore < - < . 
 
 12 W 2 +I 12 
 
 The first inequality gives Jim > iym z + 17 ; and 
 
 36 2 - 17. 17= 1007, 
 the square root of which 1 is not greater than 31 ; 
 
 therefore m $ 3I+36 , i.e. m^. 
 17 ^17 
 
 Similarly from the inequality igm*+ I9>?2m we find 2 
 
 ' 
 
 Let m = 3^. Therefore 9 ;r 2 = (3 3|^) 2 , and ^= 
 Therefore * 2 = fiHf, 
 
 and the segments of i are 
 
 ii. To divide unity into three parts such that, if we add the 
 same number to each of the parts, the results are all squares. 
 
 Necessary condition*. The given number must not be 2 or any 
 multiple of 8 increased by 2. 
 
 Given number 3. Thus 10 is to be divided into three 
 squares such that each > 3. 
 
 Take of 10, or 3^, and find x so that - 2 -f 3| may be a 
 
 square, or 3cur 2 + i =a square = ($x+ i) 2 , say. 
 Therefore ^=2,^ = 4, i jx* = , and 
 
 3V + 3i = W = a square. 
 Therefore we have to divide 10 into three squares each of 
 
 which is as near as possible to - 1 ^ 1 -. 
 
 Now 10 = 3 2 + i 2 = the sum of the three squares 9, ^f , -f^. 
 Comparing the sides 3, |, | with -y-, 
 
 or (multiplying by 30) 90, 24, 18 with 55, we must 
 
 make each side approach 55. 
 
 1 I.e. the integral part of the root is ~t> 31. The limits taken in each case are a fortiori 
 limits as explained above, pp. 61-3. 
 
 2 See p. 6 1, ante. 
 
 3 See pp. 108-9, ante -
 
 BOOK V 209 
 
 [Since then M = 3 ~M = f + M = f + fl> we put for the 
 sides of the required numbers 
 
 3-35*. 3i* + , 37*+f. 
 
 The sum of the squares = 3555^- n6>+ 10= 10 
 Therefore *=&&, 
 
 and this solves the problem. 
 
 12. To divide unity into three parts such that, if three different 
 given numbers be added to the parts respectively, the results are 
 all squares. 
 
 Given numbers 2, 3, 4. Then I have to divide 10 into 
 
 three squares such that the first > 2, the second > 3, 
 
 and the third > 4. 
 Let us add of unity to each, and we have to find three 
 
 squares such that their sum is 10, while the first lies 
 
 between 2, 2j, the second between 3, 3$, and the 
 
 third between 4, 4^. 
 It is necessary, first, to divide 10 (the sum of two 
 
 squares) into two squares one of which lies between 
 
 2, 2^; then, if we subtract 2 from the latter square, 
 
 we have one of the required parts of unity. 
 Next divide the other square into two squares, one of 
 
 which lies between 3, 3^; 
 subtracting 3 from the latter square, we have the second 
 
 of the required parts of unity. 
 Similarly we can find the third part 1 . 
 
 1 Diophantus only thus briefly indicates the course of the solution. Wertheim solves 
 the problem in detail after Diophantus' manner ; and, as this is by no means too easy, 
 I think it well to reproduce his solution. 
 
 I. It is first necessary to divide 10 into two squares one of which lies between i 
 and 3. We use the wapffb-nrrot dywyj. 
 
 The first square must be in the neighbourhood of a J ; and we seek a small fraction 
 
 -g which when added to t\ gives a square: in other words, we must make 4( 2 4+:is) 
 
 a square. This expression may be written 10 + ( - J , and, to make this a square, we put 
 
 107*+ i =(3^+1)*, say, 
 
 whence y = 6, ^=36, ^=144, so that *J+ ^= = (||j , which is an approximation 
 to the first of the two squares the sum of which is 10. 
 
 The second of these squares approximates to 7^, and we seek a small fraction -5 such 
 
 that 7i+^ is a square, or 30+ { - J =30 + ( -J , say =a square. 
 
 H. D. 14
 
 210 THE ARITHMETICA 
 
 13. To divide a given number into three parts such that the 
 sum of any two of the parts gives a square. 
 Given number 10. 
 
 Put 30jr + i = ($y + i ) 2 , say ; 
 
 therefore y 2, y- = 4, x 2 ~i6, so that 7^ + -^ = ' 2 * = / \ = ( ) . 
 
 Now, since io=i 2 + 3 2 , and i2=i+ , while ^ = 3-^-, 
 12 12 12 a 12' 
 
 we put (i + 7-*) 2 + (3 3^) a = 10, [Cf. v. 9] 
 
 so that x= , 
 
 Therefore the two squares into which 10 is divided are -s-^* and ^ , and the first of 
 
 041 041 
 
 these lies in fact between 2 and 2^. 
 
 II. We have next to divide the square ~ into two squares, one of which, which 
 we will call x' 2 , lies between 3 and 4. [The method of V. 10 is here applicable.] 
 Instead of 3, 4 take ^, , as the limits. 
 
 Therefore ^~ < x 3 < -\ , 
 
 10 16 
 
 4 4 
 
 , 6561 /8i , Y 
 And -g- *" must be a square =1 kx \ , say, 
 
 which gives X =^(^kr 
 
 k has now to be chosen such that 
 
 <'> id^r!' 
 
 from which it follows that ^<2'8..., 
 
 162/6 8 
 
 and (2) !<- 
 
 29 (i+^ 2 ) 4 
 
 whence ^>2'3.... 
 
 We may therefore put k = i-$. 
 
 Therefore x=^, jfl=^^ t 
 
 841 707281 
 
 6561 
 and g x- 
 
 The three required squares into which 10 is divided are therefore 
 1849 2624400 2893401 
 841 ' 707281 ' 707281 
 
 And if we subtract 2 from the first, 3 from the second and 4 from the third, we obtain 
 as the required parts of unity 
 
 14044? 50*557 64277 
 707281' 707281' 707281'
 
 BOOK V 211 
 
 Since the sum of each pair of parts is a square less than 
 10, while the sum of the three pairs is twice the 
 sum of the three parts or 20, 
 
 we have to divide 20 into three squares each of which 
 is < io. 
 
 But 20 is the sum of two squares, 16 and 4; 
 
 and, if we put 4 for one of the required squares, we 
 have to divide 16 into two squares, each of which is 
 < 10, or, in other words, into two squares, one of which 
 lies between 6 and 10. This we learnt how to do 1 
 [V. io]. 
 
 We have, when this is done, three squares such that 
 each is < io, while their sum is 20 ; 
 
 and by subtracting each of these squares from io we 
 obtain the parts of io required. 
 
 14. To divide a given number into four parts such that the 
 sum of any three gives a square. 
 Given number io. 
 
 Three times the sum of the parts = the sum of four squares. 
 Therefore 30 has to be divided into four squares, each of 
 
 which is < io. 
 (i) If we use the method of approximation 
 
 we have to make each square approximate to 
 
 1 Wertheim gives a solution in full, thus. 
 
 Let the squares be Jt 2 , 16-x 2 , of which one, jc 2 , lies between 6 and io. 
 
 Put instead of 6 and io the limits and 9, so that 
 
 To make 16 - x 2 a square, we put 
 
 i6-**=(4- 
 
 86 
 whence r= I+l 
 
 These conditions give, as limits for k, 2-84... and vti... . 
 We may therefore e.g. put k = i\. 
 
 Then *=*?, , = ??, i6-*=?3. 
 
 29 841 841 
 
 6400 7056 
 The required three squares making up 20 are 4, -g-y- , -gp . 
 
 Subtracting these respectively from io, we have the required parts of the given number 
 
 - 2010 1354 
 
 io, namely 6, -^, -^p 
 
 142
 
 212 THE ARITHMETICA 
 
 then, when the squares are found, we subtract each 
 
 from 10, and so find the required parts. 
 (2) Or, observing that 30= 16 + 9 + 4+ i,we take 4, 9 for 
 
 two of the squares, and then divide 17 into two squares, 
 
 each of which < 10. 
 If then we divide 17 into two squares, one of which lies 
 
 between 8| and 10, as we have learnt how to do 1 
 
 [cf. V. 10], the squares will satisfy the conditions. 
 We shall then have divided 30 into four squares, each of 
 
 which is less than 10, two of them being 4, 9 and the 
 
 other two the parts of 17 just found. 
 Subtracting each of the four squares from 10, we have the 
 
 required parts of 10, two of which are I and 6. 
 
 15. To find three numbers such that the cube of their sum 
 added to any one of them gives a cube. 
 
 Let the sum be x and the cubes JX*, 26**, 63^. 
 Therefore 96> 3 = ^-, or <)6x*= I. 
 
 But 96 is not a square; we must therefore replace it by a 
 square in order to solve the problem. 
 
 1 Wertheim gives a solution of this part of the problem. 
 
 I /2\ 2 
 
 As usual, we make 8 + - t , or 34 + ( - I , a square. 
 
 Putting - = - , we must make 34 + ( - ) a square. 
 
 Let 34J/ 2 + i = a square = (6> - i) 2 , 
 
 and we obtain }> = 6, y* = $6, ^=144. 
 
 Thus 8 i + ^ = ill5 = 
 
 144 i 44 
 
 and is an approximation to the side of each of the required squares. 
 
 Next, since 17= i 2 + 4 2 , and ^ = r + ^= 4 - ^ , 
 12 12 * 12 
 
 we put i7 = (i + 2 3 ;r) 2 + ( 4 -i3.r) 2 , 
 
 and we obtain x = -^- . 
 
 349 
 
 -^-^ 
 
 Subtracting each of these from 10, we have the third and fourth of the required parts 
 of 10, namely 
 
 185754 I79M 
 121801" 121801"
 
 BOOK V 2I3 
 
 Now 96 is the sum of three numbers, each of which is i less 
 
 than a cube; 
 therefore we have to find three numbers such that each 
 
 of them is a cube less i, and the sum of the three 
 
 is a square. 
 Let the sides of the cubes 1 be m + i, 2 - m, 2, whence the 
 
 numbers are m 3 + 3 m 2 + $m, 7 - 1 2m + 6m' - m 3 , 7 ; 
 
 their sum = gm 2 - gm + 14 = a square = (3^ - 4)', say ; 
 therefore m = -^, 
 and the numbers are 4M|, i&HJ 7. 
 
 o375 J 3375 * / 
 
 Reverting to the original problem, we put x for the sum of 
 the numbers, and for the numbers respectively 
 
 whence 
 
 that is (if we divide out by 15 and by x), 
 
 2916**= 225, and x=\\. 
 The numbers are therefore found. 
 
 1 6. To find three numbers such that the cube of their sum 
 minus any one of them gives a cube. 
 
 Let the sum be x, and the numbers \x*, ff x 3 , ff^ 3 . 
 
 Therefore fm^> = ^, 
 
 and, if ff|f were the ratio of a square to a square, the 
 
 problem would be solved. 
 But f f || = 3 - (the sum of three cubes). 
 Therefore we must find three cubes, each of which < i, 
 
 and such that (3 - their sum) = a square. 
 If, a fortiori, the sum of the three cubes is made < i, the 
 
 square will be > 2. Let 2 it be 2\. 
 
 1 If a 3 , 3, c 3 are the three cubes, so that aP + fi + c 3 - 3 has to be a square, Diophantus 
 chooses c 3 arbitrarily (8) and then makes such assumptions for the sides of a 3 , ft 3 , being 
 linear expressions in m, that, in the expression to be turned into a square, the coefficient 
 of m 3 vanishes, and that of w 2 is a square. If = 111, the condition is satisfied by 
 putting l> = 3# i -m, where k is any number. 
 
 3 Bachet, finding no way of hitting upon 2^ as tne particular square to be taken 
 in order that the difference between it and 3 may be separable into three cubes, and 
 observing that he could not solve the problem if he took another arbitrary square between 
 i and 3, e.g. 2^, instead of sj, concluded that Diophantus must have hit upon 2$, 
 which does enable the problem to be solved, by accident. 
 
 Fermat would not admit this and considered that the method used by Diophantus for 
 finding 2^ as the square to be taken should not be difficult to discover. Fermat accord- 
 ingly suggested a method as follows. 
 
 Let JT- i be the side of the required square lying between i and 3. Then 3 - (x - i) 2 
 = 2 + ix - jr 2 , and this has to be separated into three cubes. Fermat assumes for the sides
 
 THE ARITHMETICA 
 We have therefore to find three cubes the sum of which 
 
 =i or m\ 
 
 that is, we have to divide 162 into three cubes. 
 
 But 162= 125+64- 27 ; 
 
 and "we have it in the Porisms" that the difference of 
 
 two cubes can be transformed into the sum of two 
 
 cubes 1 . 
 Having thus found the three cubes 2 , we start again, and 
 
 x=2\x*, so that^=|. 
 The three numbers are thus determined. 
 
 of two of the required cubes two linear expressions in x such that, when the sum of their 
 cubes is subtracted from i + ix- x 2 , the result only contains terms in x* and x 3 or in x 
 and units. 
 
 The first alternative is secured if the sides of the first and second cubes are i x and 
 i + x respectively ; for 
 
 This latter expression has to be made a cube, for which purpose we put 
 
 j^_26^_ _W 3 * 3 
 
 ~ 4 * 27 ~^7~' Say> 
 
 which gives a value for x. We have only to see that this value makes -x less than i, 
 and we can easily choose m so as to fulfil this condition. 
 
 [E.g. suppose * = 5, and we find x= - , so that 
 
 i 13 i 20 72 
 
 -x = -2 , i x = , i+x=, 
 
 3 33 3 33 33 
 
 and the side of the third cube is - . 
 
 We then have three cubes which make up the excess of 3 over a certain square ; but, 
 while the first of these cubes is < i, the second is > i and the third is negative. Hence 
 we must, like Diophantus, proceed to transform the difference between the two latter 
 cubes into the sum of two other cubes. 
 
 It will, however, be seen by trial that even Fermat's method is not quite general, for 
 it will not, as a matter of fact, give the particular solution obtained by Diophantus in 
 which the square is 2^. 
 
 1 On the transformation of the difference of two cubes into the sum of two cubes, see 
 pp. 101-3, ante. 
 
 2 Vieta's rule gives 4 3 - 3 3 = (^ Y + (^Y- II follows that 
 
 3_i62_/s\ 3 /ioi\ 3 / 20 \ 3 
 
 o 
 
 and, since X s = , the required numbers are 
 
 9.8 
 
 210 27 
 
 8 4998267 _8_ 20338417
 
 BOOK V 215 
 
 17. To find three numbers such that each of them minus the 
 cube of their sum gives a cube. 
 
 Let the sum be x and the numbers 2jt*, 9**, 2&r*. 
 
 Therefore y^x- = i ; 
 
 and we must replace 39 by a square which is the sum 
 
 of three cubes + 3 ; 
 therefore we must find three cubes such that their sum 
 
 + 3 is a square. 
 
 Let their sides 1 be m, 3 m, and any number, say I. 
 Therefore gm* + 31 27^ = a square = (30* - 7)*, say, so 
 
 that m = , and the sides of the cubes are f, f , I. 
 Starting again, we put x for the sum, and for the numbers 
 
 whence 1445^=125, ^ = ^, and x=.fr. 
 The required numbers are thus found. 
 
 1 8. To find three numbers such that their sum is a square and 
 the cube of their sum added to any one of them gives a square. 
 Let the sum be X s and the numbers 3*-*, &r 8 , I $x*. 
 It follows that 26x* = i ; and, if 26 were a fourth power, the 
 
 problem would be solved. 
 To replace it by a fourth power, we have to find three 
 
 numbers such that each increased by I gives a square, 
 
 while the sum of the three gives a fourth power. 
 Let these numbers be /* 2tn*, m* + 2m, m*2m [the sum 
 
 being m*] ; these are indeterminate numbers satisfying 
 
 the conditions. 
 Putting any number, say 3, for m, we have as the required 
 
 auxiliary numbers 63, 15, 3. 
 Starting again, we put x* for the sum and 3**, 1 5**, 63** for 
 
 the required numbers, 
 and we have 8 !;** = .*, so that;r = . 
 
 The numbers are thus found ( , , J 
 
 19. To find three numbers such that their sum is a square and 
 the cube of their sum minus any one of them gives a square 
 
 [There is obviously a lacuna in the text after this enunciation ; 
 for the next words are " And we have again to divide 2 as before" 
 
 1 Cf. note on V. 15. In this case, if one of the cubes is chosen arbitrarily and m* 
 is another, we have only to put (3** - m) for the side of the third cube in order that, in the 
 expression to be made a square, the term in m* may vanish, and the term in m* may be a 
 
 square.
 
 216 THE ARITHMETICA 
 
 whereas there is nothing in our text to which they can refer, and 
 the lines which follow are clearly no part of the solution of V. 19. 
 
 Bachet first noticed the probability that three problems inter- 
 vened between v. 19 and V. 20, and he gave solutions of them. 
 But he seems to have failed to observe that the eight lines or so in 
 the text between the enunciation of V. 19 and the enunciation of 
 V. 20 belonged to the solution of the last of the three missing 
 problems. The first of the missing problems is connected with 
 V. 1 8 and 19, making a natural trio with them, while the second and 
 third similarly make with V. 20 a set of three. The enunciations 
 were doubtless somewhat as follows. 
 
 iga. To find three numbers such that their sum is a square 
 and any one of them minus the cube of their sum gives a square. 
 
 19 b. To find three numbers such that their sum is a given 
 number and the cube of their sum plus any one of them gives a 
 square. 
 
 19 c. To find three numbers such that their sum is a given 
 number and the cube of their sum minus any one of them gives a 
 square. 
 
 The words then in the text after the enunciation of V. 19 
 
 evidently belong to this last problem.] 
 The given sum is 2, the cube of which is 8. 
 We have to subtract each of the numbers from 8 and 
 
 thereby make a square. 
 Therefore we have to divide 22 into three squares, each 
 
 of which is greater than 6 ; 
 after which, by subtracting each of the squares from 8, we 
 
 find the required numbers. 
 
 But we have already shown [cf. V. n] how to divide 22 
 into three squares, each of which is greater than 6 
 and less than 8, Diophantus should have added. 
 [The above is explained by the fact that, by addition, 
 three times the cube of the sum minus the sum itself 
 is the sum of three squares, and three times the 
 cube of the sum minus the sum = 3.8 2 = 22.] l 
 
 1 Wertheim adds a solution in Diophantus' manner. We have to find what small 
 
 fraction of the form : , we have to add to or , and therefore to 66, in order to 
 x 2 - 39 
 
 make a square. In order that 66 + -j may be a square, we put 
 
 66* J + i = square = ( i + 8*) 2 , say, 
 which gives #= 8 and ^=64.
 
 BOOK V 217 
 
 20. To divide a given fraction into three parts such that any 
 one of them minus the cube of their sum gives a square, 
 
 Given fraction |. 
 
 Therefore each part = ^ -f a square. 
 
 Therefore the sum of the three = | = the sum of three 
 
 squares + ^. 
 Hence we have to divide f into three squares, "which is 
 
 easy 1 ." 
 
 21. To find three squares such that their continued product 
 added to any one of them gives a square. 
 
 Let the " solid content " = x"-. 
 
 We want now three squares, each of which increased by I 
 
 gives a square. 
 They can be got from right-angled triangles 2 by dividing 
 
 the square of one of the sides about the right angle 
 
 by the square of the other. 
 Let the squares then be 
 
 The continued product = s'iVffio-** = ^ by hypothesis. 
 Therefore J$$x* = I ; and, if jf$ were a square, the problem 
 would be solved. 
 
 We have therefore to increase 66 by ^- , and therefore 7^ by - , in order to make a 
 
 square. And in fact 7$ + 7^ = (^)~. 
 
 Next, since 22 = 3*4- 3 2 + 2 2 , and 65-48=17, while 72-65 = 7, we put 
 22 = (3 - 7jr) 2 + (3 - 7.r) s + (2 + 1 7*)*. 
 
 and *!. 
 
 387 
 
 Therefore the sides of the squares are ^ , 2 , -- , 
 
 1100401 1100401 1094116 
 
 the squares themselves - ^ , ^~t - f 
 
 149769 149769 149769 
 
 and the required parts of 2 are &1%-, !Z75 * 
 149769 149769 H9769 
 
 1 As Wertheim observes, J4 = -?- + + , and the required fractions into which 
 
 64 64 25 400 
 
 i . 250 89 6r 
 
 - is divided are % , ^ , ? . 
 4 loco i 600 1600 
 
 2 If a, b be the perpendiculars, c the hypotenuse in a right-angled triangle, 
 
 a 2 <-> 
 ^+i=^=asquare. 
 
 Diophantus uses the triangles (3, 4, 5), (5, 12, 13), (8, 15, 17).
 
 218 THE ARITHMETICA 
 
 As it is not, we must find three right-angled triangles such 
 
 that, if b's are their bases, and /'s are their perpen- 
 
 diculars, p\p<ipz bJ)J) 3 = a square ; 
 and, if we assume one triangle arbitrarily (3, 4, 5), we 
 
 have to make \2p l pj)J) z a square, or 3/A// 2 ^2 a square. 
 "This is easy 1 ," and the three triangles are (3, 4, 5), 
 
 (9, 40, 41), (8, 15, 17) or similar to them. 
 
 1 Diophantus does not give the work here, but only the result. Bachet obtains it 
 in this way. Suppose it required to find three rational right-angled triangles (h\, p, ^i), 
 (A 2 ,/2i ^2) and (/6 3 , / 3 , 3 ) such that /i/2/3/^i^3 is the ratio of a square to a square. 
 One triangle (A 1 , p l , b-^ being chosen arbitrarily, form two others by putting 
 
 and we have 
 
 If now A, = 5,^, = 4 , 1 = 3 , the triangles (/& 2 ,/2, 2 ) and (/5 3 ,/ 3 , *s) are ( 4 i, 9, 40) 
 and (34, 1 6, 30) respectively. Dividing the sides of the latter throughout by 2 (which 
 does not alter the ratio), we have Diophantus' second and third triangles (9, 40, 41) and 
 (8, 15, 17). 
 
 Fermat, in his note on the problem, gives the following general rule for finding two 
 right-angled triangles the areas of which are in the ratio m : n (;>). 
 
 Form ( i) the greater triangle from im + n, m - , and the lesser from m + in, m - n, 
 or (2) the greater from im -n, m + n, the lesser from in - m, m + n, 
 or (3) the greater from 6m, im - n, the lesser from 4/ + n, ^m - in, 
 or (4) the greater from m + \n, im 4, the lesser from 6n, m in. 
 The alternative (2) gives Diophantus' solution if we put m = $, n=i and substitute 
 m in for in m. 
 
 Fermat continues as follows : We can deduce a method of finding three right-angled 
 triangles the areas of which are in the ratio of three given numbers, provided that the 
 sum of two of these numbers is equal to four times the third. Suppose e.g. that m, n, q 
 are three numbers such that m + q = ^n (m>q). Now form the following triangles : 
 ( i ) from t + +n, im +n, 
 
 (2) from 6, m in, 
 
 (3) from 4 + q, 4 - iq. 
 [If A\ , AZ , AS be the areas, we have, as a matter of fact, 
 
 A-^m = Azfn = A^\q = - 6m 3 + $6m 2 n+ i ^mn 2 - 384 3 .] 
 
 We can derive, says Fermat, a method of finding three right-angled triangles the areas 
 of which themselves form a right-angled triangle. For we have only to find a triangle 
 such that the sum of the base and hypotenuse is four times the perpendicular. This is 
 easy, and the triangle will be similar to (17, 15, 8) ; the three triangles will then be formed 
 
 (1) from 17 + 4.8, 2 . 17-4. 8 or 49, 2, 
 
 (2) from 6.8, 17-2.8 or 48, i, 
 
 (3) from 4.8+15, 4.8-2. 15 or 47, 2. 
 
 [The areas of the three right-angled triangles are in fact 234906, i (0544 and 207270, 
 and these numbers form the sides of a right-angled triangle.] 
 
 Hence also we can derive a method of finding three right-angled triangles the areas 
 of which are in the ratio of three given squares such that the sum of two of them is equal to
 
 BOOK V 219 
 
 Starting again, we put for the squares 
 
 Equating the product of these to x*, we find x to be 
 rational \x = ifr , and the squares are , , 4 ]. 
 
 22. To find three squares such that their continued product 
 minus any one of them gives a square. 
 
 Let the solid content be x*, and let the numbers be 
 obtained from right-angled triangles, being 
 
 Therefore the continued product (^ 
 
 . 25600 m 
 
 1221025 
 If then - were a fourth power, i.e. if ~ - were 
 
 a square, the problem would be solved. 
 
 We have therefore to find three right-angled triangles 
 with hypotenuses /t ly 7i 2 , h 3 respectively, and with 
 A> A> A as one f *h e perpendiculars in each re- 
 spectively, such that 
 
 h\hJi*P\ P*Ps = a square. 
 
 Assuming one of the triangles to be (3, 4, 5), so that e.g. 
 Jt 3 p 3 = 5 . 4 = 20, we must have 
 
 S^iA^aA = a square. 
 
 This is satisfied if //iA = 5 ^2 A- 
 
 With a view to this we have first (cf. the last proposition) 
 to find two right-angled triangles such that, if x^, y^ 
 are the two perpendiculars in one and x^ y^ the two 
 perpendiculars in the other, x-^y-^ = ^x^y^. From such 
 a pair of triangles we can form two more right- 
 angled triangles such that the product of the 
 hypotenuse and one perpendicular in one is five times 
 the product of the hypotenuse and one perpendicular in 
 the other 1 . 
 
 four times the third, and we can also in the same way find three right-angled triangles of 
 the same area; we can also construct, in an infinite number of ways, two right-angled 
 triangles the areas of which are in a given ratio, by multiplying one of the terms of the 
 ratio or the two terms by given squares, etc. 
 
 1 Diophantus' procedure is only obscurely indicated in the Greek text. It was 
 explained by Schulz in his edition (cf. Tannery in Oeuvres de Fermat, I. p. 313, note).
 
 THE ARITHMETICA 
 
 Since the triangles found satisfying the relation ^ 1 y 1 = 
 
 are (5, 12, 13) and (3, 4, 5) respectively 1 , we have in 
 fact to find two new right-angled triangles from them, 
 namely the triangles (^i,/i,^)and (A 2 , A. ^a) sucn tnat 
 
 ^ A = 30 and h*p z = 6, 
 
 the numbers 30 and 6 being the areas of the two 
 triangles mentioned. 
 
 These triangles are (6|, f, [u^j) and (2 ^ j [ T 7_]) re- 
 spectively. 
 
 Starting again, we take for the numbers 
 
 [- 1 / divided by 2\ gives ff , and f divided by 6| gives |f$.] 
 
 The product =x^ : 
 
 therefore, taking the square root, we have 
 
 4.24.120 
 
 5-25.169 
 so that ^=||, and the required squares are found. 
 
 23. To find three squares such that each minus the product of 
 the three gives a square. 
 
 Having given a rational right-angled triangle (z, x, y), Diophantus knows how to find a 
 rational right-angled triangle (A, f, b) such that /&/ = - xy. We have in fact to put 
 
 * = !,,^2, whence * = *-^ = l(*Z^W^Y. 
 
 2 4\ S 2 / \ 22 / 
 
 Thus, having found two triangles (5, 11, 13) and (3, 4, 5) with areas in the ratio of 5 
 to i (see next paragraph of text with note thereon), Diophantus takes 
 
 ^! = ^.i3=6i,/i=^^ = ^; and similarly h z = * . 5 = 2^, / 2 = -^ = ~ . 
 
 Cossali (after Bachet) gives a formula for three right-angled triangles such that the 
 solid content of the three hypotenuses has to the solid content of three perpendiculars 
 (one in each triangle) the ratio of a square to a square ; his triangles are 
 
 (.),\*,/[,-=ipotenusa], 
 
 If /=5, ^ = 4, / = 3, we can get from this triangle the triangles (13, 5, 12) and 
 (65, 63, 16), and our equation is ' ' x 2 i. 
 
 1 These triangles can be obtained by putting ; = 5, n= i in Fermat's fourth formula 
 (note on last proposition). By that formula the triangles are formed from (9, 6) and 
 (6, 3) respectively ; and, dividing out by 3, we form the triangles from (3, i) and (2, i) 
 respectively.
 
 BOOK V 221 
 
 Let the "solid content" be x*, and let the squares be 
 formed from right-angled triangles, as before. 
 
 If we take the same triangles as those found in the last 
 problem and put for the three squares 
 
 each of these minus the continued product (**) gives a 
 
 square. 
 It remains that their product =x*\ 
 
 this gives * = f , and the problem is solved. 
 
 24. To find three squares such that the product of any two 
 increased by I gives a square. 
 
 Product of first and second + i = a square, and the third 
 is a square ; therefore " solid content " + each = a 
 square. 
 The problem therefore reduces to V. 21 above 1 . 
 
 1 De Billy in the Inventum Novum, Part II. paragraph 28 (Oeuvres de Fermai, III. 
 pp. 373-4), extends this problem, showing how to find four numbers, three of which (only) 
 are squares, having the given property, i.e. to solve the equations 
 .r 2 2.r 3 2 + 1 = r 2 , -rj 2 .v 4 + i = 2 , 
 
 First seek three square numbers satisfying the conditions of Diophantus' problem 
 v. 24, say ( -% , f -7T- ). the solution of V. i \ given in Sachet's edition. We have then 
 to find a fourth number (a-, say) such that 
 
 VW ( 
 
 are all squares. 
 
 Substitute y 1 + y for x, so as to make the first expression a square. We have 
 then to solve the double-equation 
 
 729 ' 729 
 which can be solved by the ordinary method. 
 
 De Billy does not give the solution, but it may be easily supplied thus 
 
 The difference = ( -f ^ V '_ O - ^ } (y* + V)
 
 222 THE ARITHMETICA 
 
 25. To find three squares such that the product of any two 
 minus I gives a square. 
 
 This reduces, similarly, to V. 22 above. 
 
 26. To find three squares such that, if we subtract the product 
 of any two of them from unity, the result is a square. 
 
 This again reduces to an earlier problem, V. 23. 
 
 27. Given a number, to find three squares such that the sum of 
 any two added to the given number makes a square. 
 
 Given number 15. 
 
 Let one of the required squares be 9 ; 
 
 I have then to find two other squares such that each 
 -f 24 = a square, and their sum + 15 = a square. 
 
 To find two squares, each of which + 24 = a square, take 
 two pairs of numbers which have 24 for their pro- 
 duct 1 . 
 
 Let one pair of factors be 4/ar, 6x, and let the side of one 
 
 square be half their difference or - - 3^. 
 
 X 
 
 Let the other pair of factors be $jx, Sx, and let the 
 side of the other square be half their difference or 
 
 Therefore each of the squares + 24 gives a square. 
 It remains that their sum + 15 =a square; 
 
 therefore ( - ^x J + ( - ^x\ + 1 5 = a square, 
 
 Equating the square of half the sum of the factors to the larger expression, we have 
 
 \ 3 2 7/ 9 9 
 
 whence y= ^ , and v 2 + i y= - -. ~^ . 
 
 ' 11520 ' f '* v2 
 
 Therefore A- = (y s +iy)=- , ,- , which satisfies the equations. In fact 
 9 T 74649600' 
 
 4 345/ 1 4o 
 
 But even here, as the value of Jt which we have found is negative, we ought, strictly 
 speaking, to deduce a further value by substituting y ^ , ; for x in the equations 
 
 and solving again, which would of course lead to very large numbers. 
 
 1 The text adds the words " and [let us take] sides about the right angle in a right- 
 angled triangle." I think these words must be a careless interpolation : they are not 
 wanted and give no sense; nor do they occur in the corresponding place in the next 
 problem.
 
 BOOK V 223 
 
 6i 
 or - + 2 ^x* - 9 = a square = 25*', say. 
 
 Therefore *" = f , and the problem is solved *. 
 
 28. Given a number, to find three squares such that the sum of 
 any two minus the given number makes a square. 
 Given number 13. 
 Let one of the squares be 25 ; 
 I have then to find two other squares such that each 
 
 + 12 = a square, and (sum of both) 13 = a square. 
 Divide 12 into factors in two ways, and let the factors be 
 
 (2*. 4/*) and (4*, 3 /;r). 
 Take as the sides of the squares half the differences of the 
 
 factors, i.e. let the squares be 
 
 Each of these + 12 gives a square. 
 
 It remains that the sum of the squares 13 = a square, 
 
 or -f + 6| -^ 25 = a square = -| , say. 
 Therefore x = 2, and the problem is solved 2 . 
 
 1 Diophantus has found values of f , 17, f satisfying the equations 
 
 Fermat shows how to find four numbers (not squares) satisfying the corresponding 
 conditions, namely that the sum of any two added to a shall give a square. Suppose 0=15. 
 
 Take three numbers satisfying the conditions of Diophantus' problem, say 9, , - . 
 
 Assume * 2 - 1 5 as the first of the four required numbers ; and let the second be 6jr + 9 
 (because 9 is one of the square numbers taken and 6 is twice its side) ; for the same 
 
 reason let the third number be - x H -- and the fourth or + - . 
 5 ioo 15 225 
 
 Three of the conditions are now fulfilled since each of the last three numbers added to 
 the first (x 2 - 15) plus 15 gives a square. The three remaining conditions give the triple- 
 equation 
 
 136, 522 I5= I|6 jr+ /77y = ^ 
 
 15 5 15 \i5/ 
 
 tf*; l .j-+&+.^.:a,4(&y*. 
 
 15 ioo 225 15 \6J 
 
 2 Fermat observes that four numbers (not squares) with the property indicated can 
 be found by the same procedure as that shown in the note to the preceding problem. 
 If a is the given number, put .r 2 + a for the first of the four required numbers.
 
 224 THE ARITHMETICA 
 
 29. To find three squares such that the sum of their squares is 
 a square. 
 
 Let the squares be ;r 2 , 4, 9 respectively 1 . 
 
 Therefore x* + 97 = a square = (x* - io) 2 , say ; 
 
 whence x* = ^. 
 
 If the ratio of 3 to 20 were the ratio of a square to a 
 
 square, the problem would be solved ; but it is not. 
 Therefore / have to find two squares (/ 2 , q z , say) and a 
 
 number (m, say) such t/iat m- p* q 4 Jias to 2m the 
 
 ratio of a square to a square. 
 Let J? = z*,f = 4. and m = z* + 4. 
 Therefore w 2 -/ 4 - q 4 - O 2 + 4) 2 - 2* - 1 6 = 8-z 2 . 
 Hence 8z 2 /(22* + 8), or 4^ 2 /(- s ' 2 + 4). mus t be the ratio of a 
 
 square to a square. 
 Put < sr 2 + 4 = (+i) 2 , say; 
 
 therefore z= \\, and the squares are p*=2\, q* = 4, while 
 
 m = 6\; 
 
 or, if we take 4 times each,/ 2 = 9, q*= 16, m = 2$. 
 Starting again, we put for the squares x*, 9, 16; 
 then the sum of the squares = ;r 4 + 337 = (-** 25) 2 , and 
 
 *=V- 
 
 The required squares are , g, 16. 
 
 30. [The enunciation of this problem is in the form of an 
 epigram, the meaning of which is as follows.] 
 
 A man buys a certain number of measures (%oe<?) of wine, some 
 at 8 drachmas, some at 5 drachmas each. He pays for them a 
 square number of drachmas ; and if we add 60 to this number, the 
 result is a square, the side of which is equal to the whole number 
 of measures. Find how many he bought at each price. 
 
 Let x= the whole number of measures ; therefore x* 60 
 was the price paid, which is a square (xmf, say. 
 
 If now 2 , fl, m 2 represent three numbers satisfying the conditions of the present 
 problem of Diophantus, put for the second of the required numbers ik* +/ 2 , for the third 
 2/JT + / 2 , and for the fourth 2//w+/ 2 . These satisfy three conditions, since each of the 
 last three numbers added to the first (x 2 + a) less the number a gives a square. The 
 remaining three conditions give a triple-equation. 
 
 i "Why," says Fermat, "does not Diophantus seek two fourth powers such that 
 their sum is a square ? This problem is in fact impossible, as by my method I am in 
 a position to prove with all rigour." It is probable that Diophantus knew the fact 
 without being able to prove it generally. That neither the sum nor the difference of 
 two fourth powers can be a square was proved by Euler (Commentatioms arithmeticae, j. 
 pp. 24sqq., and Algebra, Part II. c. xm.).
 
 BOOK V 225 
 
 Now of the price of the five-drachma measures + of 
 the price of the eight-drachma measures =x\ 
 
 so that x z 60, the total price, has to be divided into 
 two parts such that of one + of the other = x. 
 
 We cannot have a real solution of this unless 
 x > i (x"- - 60) and < \ (x* - 60). 
 
 Therefore $x < x' 2 60 < &r. 
 
 (1) Since x z > 5^ + 60, 
 x' i =t > x+ a number greater than 60, 
 whence x is 1 not less than 1 1. 
 
 (2) x'*<8x + 6o 
 
 or -r 2 = &r -t- some number less than 60, 
 
 whence x is 1 not greater than 12. 
 
 Therefore 11 <x< 12. 
 
 Now (from above) x (m 2 + 6o)/2w; 
 
 therefore 22 / < m* + 60 < 247/2. 
 
 Thus (i) 22m = m' 2 + (some number less than 60), 
 
 and therefore m is 2 not less than 19. 
 
 (2) 24# = w 2 + (some number greater than 60), 
 
 and therefore m is 2 less than 21. 
 
 Hence we put w = 20, and 
 
 x*-6o = (x- 2o) 2 , 
 
 so that*= \\%,x*= 132^, and * a - 60 = 72$. 
 Thus we have to divide 72^ into two parts such that 
 
 of one partptus | of the other = 1 1. 
 Let the first part be 5*. 
 Therefore (second part) = 1 1| - .#, 
 or second part = 92 82 ; 
 therefore 5* + 92 8^ = 72^, 
 
 Therefore the number of five-drachma %oe? 
 eight-drachma 
 
 1 For an explanation of these limits see p. 60, ante. 
 
 2 See p. 62, ante.
 
 226 THE ARITHMETICA 
 
 BOOK VI 
 
 1. To find a (rational) right-angled triangle such that the 
 hypotenuse minus each of the sides gives a cube 1 . 
 
 Let the required triangle be formed from x, 3. 
 
 Therefore hypotenuse = x- + 9, perpendicular = 6x, base 
 
 =**-9. 
 
 Thus x z + 9 (x z 9)= 1 8 should be a cube, but it is not. 
 Now 1 8 = 2 . 3* ; therefore we must replace 3 by m, where 
 
 2 . m- is a cube ; and m = 2. 
 We form, therefore, a right-angled triangle from x, 2, 
 
 namely (x* + 4, ^x, x* 4) ; and one condition is 
 
 satisfied. 
 
 The other gives x n - 4^ + 4 = a cube ; 
 therefore (x 2) 2 is a cube, or x 2 is a cube = 8, say. 
 Thus x = 10, 
 
 and the triangle is (40, 96, 104). 
 
 2. To find a right-angled triangle such that the hypotenuse 
 added to each side gives a cube. 
 
 Form a triangle, as before, from two numbers ; and, as 
 before, one of them must be such that twice its 
 square is a cube, i.e. must be 2. 
 
 We form a triangle from x> 2, namely .a-' 2 +-4, 4^-, 4 ;r 2 ; 
 therefore ^ + 4^ + 4 must be a cube, while x* must 
 be less than 4, or x < 2. 
 
 Thus x + 2 = a cube which must be < 4 and > 2 = %j-, say. 
 Therefore x=^-, 
 
 and the triangle is (-*$ , $% , & 7 \ , 
 or, if we multiply by the common denominator, (135, 
 352, 377) 
 
 3. To find a right-angled triangle such that its area added to 
 a given number makes a square. 
 
 Let 5 be the given number, (3^, 4^, $x} the required 
 triangle. 
 
 1 Diophantus' expressions are 6 tv rfj viroreivotiffrj, " the (number) in (or represent- 
 ing) the hypotenuse," 6 fv eKar^pg, ruv 6p6uiv, "the (number) in (or representing) each 
 of the perpendicular sides," 6 iv T inpaSy, "the (number) in (or representing) the area," 
 etc. It will be convenient to say "the hypotenuse," etc. simply. It will be observed 
 that, as between the numbers representing sides and area, all idea of dimension is ignored.
 
 BOOK VI 227 
 
 Therefore 6x z + 5 = a square = 9*', say, 
 
 or 3-^=5. 
 
 But 3 should have to 5 the ratio of a square to a square. 
 
 Therefore we must find a right-angled triangle and a 
 number such that the difference between the square 
 of the number and the area of the triangle has to 5 the 
 ratio of a square to a square, i.e. = \ of a square. 
 
 Form a right-angled triangle from ( m, \ ; 
 
 \ */ 
 
 thus the area is m z . 
 
 Let the number be ; + , so that we must have 
 
 101 
 
 4 . 5 H = of a square ; 
 
 therefore 4.25+ ^-~- = a square, 
 
 or ioow 2 + 505 = a square = (lorn + 5) 2 , say, 
 
 and m = ^. 
 
 The auxiliary triangle must therefore be formed from ^, 
 
 / T , and the auxiliary number sought is ^. 
 Put now for the original triangle (fix, px, bx\ where (//, /, b} 
 
 is the right-angled triangle formed from ^, ^ ; 
 this gives \pbx- + 5 = L l%ffix*; 
 and we have the solution. 
 [The perpendicular sides of the right-angled triangle are 
 
 5 2 
 whence 
 
 and the triangle is 
 
 4. To find a right-angled triangle such that its area minus a 
 given number makes a square. 
 
 Given number 6, triangle ($x, 4*, $x), say. 
 Therefore 6> 2 6 = square = 4-r 2 , say. 
 
 Thus, in this case, we must find a right-angled triangle 
 and a number such that 
 
 (area of triangle) - (number) 2 = of a square. 
 
 Form a triangle from /, . 
 m 
 
 15*
 
 228 THE ARITHMETICA 
 
 Its area is m 2 , and let the number be m . . 
 
 m* m 
 
 Therefore 6 = (a square), 
 
 or ^6m z 60 = a square = (6m 2) 2 , say. 
 
 Therefore m = f , and the auxiliary triangle is formed from 
 
 (f , |), the auxiliary number being |~|. 
 We start again, substituting for 3, 4, 5 in the original 
 
 hypothesis the sides of the auxiliary triangle just 
 
 found, and putting (ff) 2 .*" 2 in place of ^x-\ and the 
 
 solution is obvious. 
 
 [The auxiliary triangle is (*$$-, 2, tyffl), whence 
 0^2 _ 6 = (f|)^ 2 , and x = f , 
 
 so that the required triangle is (- 4 S ^, -V 6 - -Wf)-] 
 
 5. To find a right-angled triangle such that, if its area be 
 subtracted from a given number, the remainder is a square. 
 
 Given number 10, triangle (3^, 4*, $x), say. 
 Thus 10 6^ 2 = a square; and we have to find a right- 
 angled triangle and a number such that 
 
 (area of triangle) + (number) 2 = T ^ of a square. 
 
 Form a triangle from m, , the area being m* ; , 
 
 m m 
 
 and let the number be h 5w. 
 m 
 
 Therefore 26m 2 + 10 = ^ bf a square, 
 
 or 26cwz 2 + 100 = a square, 
 
 or again 6$m'* + 25 = a square = (Sm + 5) 2 , say, 
 
 whence m = 80. 
 
 The rest is obvious. 
 
 The required triangle is AOfff^, T ^, ^iHHHHM 
 
 6. To find a right-angled triangle such that the area added 
 to one of the perpendiculars makes a given number. 
 
 Given number 7, triangle (3^, 4^, 5^). 
 
 Therefore 6x z + yc = 7. 
 
 In order that this migJit be solved, it ivould be necessary that 
 
 (half coefficient of xj- + product of coefficient of x* and 
 
 absolute term should be a square ; 
 but (i^) 2 + 6.7 is not a square. 
 Hence we must find, to replace (3, 4, 5), a right-angled 
 
 triangle such that 
 
 (\ one perpendicular) 2 + 7 times area = a square. 
 Let one perpendicular be w, the other i.
 
 BOOK VI 229 
 
 Therefore ftm + = a square, or ijm + i = a square.) 
 
 Also," since the triangle is rational, m* + i = a square.) 
 
 The difference m* - i^m = m (m 14) ; 
 
 and putting, as usual, 7 2 = 147/2 + i, 
 
 we have m = ^-. 
 
 The auxiliary triangle is therefore (^, 1,2,?-) or (24, 7, 25). 
 
 Starting afresh, we take as the triangle (24^, 7.*-, 25*). 
 
 Therefore 84^ + ?x = 7, 
 
 and ^ = ^. 
 
 We have then ^6, ^ > j) as the solution 1 . 
 
 7. To find a right-angled triangle such that its area minus one 
 of the perpendiculars is a given number. 
 Given number 7. 
 As before, we have to find a right-angled triangle such that 
 
 ( one perpendicular) 2 + 7 times area= a square ; 
 this triangle is (7, 24, 25). 
 
 Let then the triangle of the problem be (7^, 24^-, 25^). 
 Therefore 843? - jx = 7, 
 
 *=l, 
 
 and the problem is solved 2 . 
 
 1 Fermat observes that this problem and the next can be solved by another method. 
 " Form in this case," he says, " a triangle from the given number and r, and divide 
 the sides by the sum of the given number and i ; the quotients will give the required 
 triangle." 
 
 In fact, if we take as the sides of the required triangle 
 (a 2 + i) .*, (a 2 - i ) x, iax, 
 where a is the given number, we have 
 
 (a 2 i) ax 2 + iax=a, 
 
 one root of which is x= --= - + -5 = ; , 
 
 a i fl* I a+ i 
 
 and the sides of the required triangle are therefore 
 
 a 2 + i a 2 - i va 
 a+i ' a+i ' a+ r ' 
 The solution is really the same as that of Diophantus. 
 
 2 Similarly in this case we may, with Fermat, form the triangle from the given number 
 and i, and divide the sides by the difference between the given number and i, and we 
 shall have the required triangle. 
 
 In VI. 6, 7, Diophantus has found triangles f, , r) (f being the hypotenuse), such that 
 
 (i) l - 
 
 and (2) \to 
 
 Fermat enunciates the third case 
 
 (3) *-
 
 230 THE ARITHMETICA 
 
 8. To find a right-angled triangle such that the area added to 
 the sum of the perpendiculars makes a given number. 
 Given number 6. 
 Again I have to find a right-angled triangle such that 
 
 (^ sum of perpendiculars) 2 -f 6 times area = a square. 
 Let m, i be the perpendicular sides of this triangle ; 
 therefore (m + i ) 2 + 3 m = {m* + ^\m + i = a square, 
 while m* + i must also be a square. 
 
 T-I r 
 
 Therefore 2 > are both squares. 
 
 The difference is 2m . 7, and we put 
 
 whence m = |f , 
 
 and the auxiliary triangle is (ff , i, ff), or (45> 2 ^> 53)- 
 
 Assume now for the triangle of the problem 
 
 (45*, 28*, 53*). 
 Therefore 630^ + 73* = 6 ; 
 
 * is rational [== T ' H ], and the solution follows. 
 
 9. To find a right-angled triangle such that the area minus the 
 sum of the perpendiculars is a given number. 
 
 Given number 6. 
 
 As before, we find a subsidiary right-angled triangle such 
 that (^ sum of perpendiculars)-+ 6 ti tries area = a square. 
 This is found to be (28, 45, 53) as before. 
 Taking (28*, 45*, 53*) for the required triangle, 
 
 630*-'- 73* = 6; 
 x = ws> an< ^ t ne problem is solved 1 . 
 
 10. To find a right-angled triangle such that the sum of its 
 area, the hypotenuse, and one of the perpendiculars is a given 
 number. 
 
 observing that Diophantus and Bachet appear not to have known the solution, but that 
 it can be solved "by our method." He does not actually give the solution ; but we may 
 compare his solutions of similar problems in the Inventum Novitm, e.g. those given in 
 the notes to VI. 1 1 and vi. 15 below and in the Supplement. The essence of the method 
 is that, if the first value of x found in the ordinary course is such as to give a negative 
 value for one of the sides, we can derive from it a fresh value which will make all the 
 sides positive. 
 
 1 Here likewise, Diophantus having solved the problem 
 
 Fermat enunciates, as to be solved by his method, the corresponding proble
 
 BOOK VI 23 T 
 
 Given number 4. 
 
 If we assumed as the triangle (hx.px, bx), we should have 
 
 and, in order that the solution may be rational, we must 
 find a right-angled triangle such that 
 
 \ (hyp. + one perp.) 2 + 4 times area = a square. 
 Form a right-angled triangle from i, m + i. 
 Then | (hyp. -f.jone perp.) 2 = J- (m* + 2m + 2 + m 2 + 2nif 
 
 = ;;z 4 + 4m 3 + 6m z + 4m+ i , 
 and 4 times area = 4 (m + i) (m 2 + 2m) 
 
 = 4m 3 + 1 2m 2 + 8m. 
 Therefore 
 
 m* + 8m 3 + i 8m 2 + 1 2m + i = a square = (6m + 1 - w 2 ) 2 , say, 
 whence ;// = |, and the auxiliary triangle is formed from 
 (i, f) or (5, 9). This triangle is (56, 90, 106) or 
 (28, 45, S3)- 
 
 We assume therefore 28^,45^, 53^ for the original triangle, 
 and we have 630^ + 8 \x = 4. 
 
 Therefore ^=i^, and the problem is solved. 
 
 II. To find a right-angled triangle such that its area minus 
 the sum of the hypotenuse and one of the perpendiculars is a given 
 number. 
 
 Given number 4. 
 
 We have then to find an auxiliary triangle with the same 
 
 property as in the last problem ; 
 therefore (28, 45, 53) will serve the purpose. 
 We put for the triangle of the problem (28^,45^, 53*), and 
 we have 63O,r 2 8 \x = 4 ; 
 
 x=-^, and the problem is solved 1 . 
 
 1 Diophantus has in vi. 10, n shown us how to find a rational right-angled triangle 
 f> > *? (f being the hypotenuse) such that 
 
 (1) Jfr+f+f=. 
 
 (2) l&l-(t+t) = a. 
 
 Fermat, in the Inventum Novum, Part III. paragraph 33 (Oeuvres de Fermat, in. 
 p. 389), propounds and solves the corresponding problem 
 
 (3) *+-;& = 
 
 In the particular case taken by Fermat = 4. He proceeds thus: 
 First find a rational right-angled triangle in which (since a = 4) 
 
 ^(f +)['- 4-^ = 
 
 a square.
 
 232 THE ARITHMETICA 
 
 Lemma I to the following problem. 
 
 To find a right-angled triangle such that the difference of the 
 perpendiculars is a square, the greater alone is a square, and further 
 the area added to the lesser perpendicular gives a square. 
 
 Let the triangle be formed from two numbers, the greater 
 
 perpendicular being twice their product. 
 Hence I must find two numbers such that (i) twice their 
 product is a square and (2) twice their product exceeds 
 the difference of their squares by a square. 
 This is true of any two numbers the greater of which 
 
 = twice the lesser. 
 Form then the triangle from x, 2.x, and two conditions are 
 
 satisfied. 
 
 The third gives 6x* + $x* = a square, or 6x- + 3 = a square. 
 I have therefore to find a number such that 6 times its 
 
 square + 3 = a square ; 
 one such number is i, and there are an infinite number of 
 
 others 1 . 
 If x = i, the triangle is formed from i, 2. 
 
 Suppose it formed from x+ i, x; the sides then are 
 
 f= IX* + 1 JC + I , %=2X+1, 1} 2JC^+2Ji. 
 
 Thus +)l. 
 
 = a square 
 
 = ( .r 2 - 2X + i ) 2 , say. 
 
 - 
 3 
 
 The triangle formed from - , - is ( ,,-) Thus we may take as the auxiliary 
 
 33 \9 9 9/ 
 triangle (17, 15, 8). 
 
 Take now ifx, 15.*, 8* for the sides of the triangle originally required to be found. 
 We have then 
 
 f +--)? = 32* -6cur- = 4; 
 
 whence x=-, and the required triangle is ( , , - ). 
 3 V 3 3 3/ 
 
 [The auxiliary right-angled triangle was of course necessary to be found in order to 
 make the final quadratic give a rational result.] 
 
 Bachet adds after vi. u a solution of the problem represented by 
 
 to which Fermat adds the enunciation of the corresponding problem 
 
 1 Though there are an infinite number of values of x for which 6.r 2 + 3 becomes a square, 
 the resulting triangles are all similar. For, if x be any one of the values, the triangle is
 
 BOOK VI 233 
 
 Lemma II to the following problem. 
 
 Given two numbers the sum of which is a square, an infinite 
 number of squares can be found such that, when the square is multi- 
 plied by one of the given numbers and the product is added to 
 the other, the result is a square. 
 Given numbers 3, 6. 
 
 Let x* + 2x + i be the required square which, say, when 
 multiplied by 3 and then increased by 6, gives a square. 
 We have 3_r 2 + 6.r + 9 = a square ; 
 
 and, since the absolute term is a square, an infinite number 
 
 of solutions can be found. 
 Suppose, e.g. yp + 6x + 9 = (3 - $x}\ 
 and x = 4. 
 
 The side of the required square is 5, and an infinite 
 
 number of other solutions can be found. 
 12. To find a right-angled triangle such that the area added 
 to either of the perpendiculars gives a square. 
 Let the triangle be ($x, \2x, 13^). 
 Therefore ( I ) 30^ + 1 2x a square = 36^, say, 
 and x = 2. 
 
 But (2) we must also have 
 
 3Q* 2 + 5-r = a square. 
 This is however not a square ivhen x = 2. 
 Therefore I must find a square ; 2 ^ 2 , to replace 36**, such 
 that I2/O 2 - 30), the value of x obtained from the 
 first equation, is real and satisfies the condition 
 
 30tr 2 + $x = a square. 
 This gives, by substitution, 
 
 (6om 2 + 25 2o)/(** 6om + 900) = a square, 
 or 6ow 2 + 2520 = a square. 
 
 This could be solved [by the preceding Lemma II] if 
 
 60+2520 were equal to a square. 
 Now 60 arises from 5 . 12, i.e. from the product of the 
 
 perpendicular sides of (5, 12, 13); 
 
 2520 is 30. 12. (12-5), i.e. the continued product of the 
 area, the greater perpendicular, and the difference 
 between the perpendiculars. 
 
 formed from .*, 2*, and its sides are therefore a* 2 , 4-r 2 , 5-* 2 ; that is, the triangles are all 
 similar to (3, 4, 5). Fermat shows in his note on the following problem, vi. 12, how to 
 find any number of triangles satisfying the conditions of this Lemma and not similar to 
 (3. 4- 5)- See p. 235, note.
 
 34 THE ARITHMETICA 
 
 Hence we must find an auxiliary triangle such that 
 (product of perps.) + (continued product of area, 
 greater perp. and difference of perps.) = a square. 
 
 Or, if we make the greater perpendicular a square and 
 divide out by it, we must have 
 (lesser perp.) + (product of area and diff. of perps.) 
 
 = a square. 
 
 Then, assuming that we have found two numbers, (i) the 
 product of the area and the difference of the perpen- 
 diculars and (2) the lesser perpendicular, satisfying 
 these conditions, we have to find a square (m 2 ) such 
 that the product of this square into the second of 
 the numbers, when added to the first number, gives 
 a square 1 . 
 
 1 The text of this sentence is unsatisfactory. Bachet altered the reading of the MSS. 
 So did Tannery, but more by way of filling out. The version above follows Tannery's text, 
 which is as follows : airdycrai eh ri> Mo &pt6/M>vs tvpbvras [for Suras of MSS. ] < rbv re vwb > 
 roO ^9a5oC Kot T?)J i>7repox?s rwv 6p0wv, </ca2 rbv tv Ty tXdffffovt rwv opffuv > , avOis [for 
 avrrjt of MSS.] farfiv QOV riva, 5s jro\Xair\a<na<T0ets eirl i-va rbv dod^vja, < Kai TT po<T\afiuv 
 rbvtrtpov>, troiel rerpdywvov. 
 
 The argument would then be this. If (A, />, l>) be the triangle (>/), we have to make 
 
 bp + - bp (i> -p) b a square, 
 
 or, if b is a square, / + - bp(b-p) must be a square. 
 
 The ultimate equation to be solved (corresponding to 6o/ 2 +252O = a square) is 
 bpm z + -bp(b -p) 6 = a. square, 
 
 or, if b is a square, pni 1 + - bp (b -/) = a square ; 
 
 and then fore, according to Tannery's text, "the problem is reduced to this: Having found 
 
 two numbers - bp (b -/) and / [satisfying the conditions, namely that their sum is a 
 
 square, while b is also a square], to find after that a square such that the product of it 
 and the latter number added to the former number gives a square." 
 
 The difficulty is that, with the above readings, there is nothing to correspond exactly to 
 
 the phraseology of the enunciation of Lemma I, which speaks, not of making / + - bp (b-p) 
 a square when b is a square, but of making b p, b and/-f - bp all simultaneously ^quares. 
 But \\ieparticular solution of the Lemma is really equivalent to making b and/> + - bp (b p) 
 simultaneously squares. For the triangle is formed from a, ia ; this method of ma 1 ing 
 b a square ( = 4<z 2 ) incidentally makes b p a square ( = a 2 ), and p + - bp becomes 3<z 2 + 6*, 
 
 while p + - bp (b -p) becomes 3a 2 + 6rt 6 . Since the solution actually used is ai, the 
 
 effect is the same whichever way the problem is stated. And in any case, whether the 
 expression to be made a square is 3 > a l i i + 6a^ or 3a?m? + 6a 6 , the problem equally reduces, 
 to that of making yn" 1 + 6 a square.
 
 BOOK VI 235 
 
 How to solve these problems is shown in the Lemmas. 
 The auxiliary triangle is (3, 4, 5). [Lemma I.] 
 Accordingly, putting for the original triangle (3*, 4*-, $x\ 
 
 we have 6**+ I b th squares. 
 
 Let .* = - be the solution of the first equation : 
 
 Ittr O 
 
 then x- = 
 
 m 4 \2m--\- 36 
 
 The second equation therefore gives 
 
 whence i2m* + 24 = a square, 
 
 and we have therefore to find a square (m*) such that 
 
 twelve times it + 24 is a square; this is possible, since 
 
 12 + 24 i s a square [Lemma II]. 
 A solution is m* =25, 
 whence x= -fa, 
 
 and f~. ', ) is the required triangle 1 . 
 \i9' 19' ip/ 
 
 13. To find a right-angled triangle such that its area minus 
 either perpendicular gives a square. 
 
 We have to find an auxiliary triangle exactly as in the 
 last problem ; 
 
 Bachet's reading is drdyertu efj ri> 5i/o apiOftwr dofftrrur TOV T( tuftadov, iced rijt 
 (\dffcrovot TUV irepl rrjv opff-^v, avrols fi/ret* rerpdyuvbi' TIVCL, 5j TO\XairXa<ria<rtftti IT'I 
 fva. run SoOtvruv, Kai T/xxrXa^Swi' rov erepov, Toifj rfrpdyuvoif. 
 
 1 Fermat observes that Diophantus gives only one species of triangle satisfying the 
 condition, namely triangles similar to (3, 4, 5), but that by his (Fermat's) method an infinite 
 number of triangles of different species can be found to satisfy the conditions, the first 
 being derived from Diophantus' triangle, the second from the new triangle, and so on. 
 
 Suppose that the triangle (3, 4, 5) has been found satisfying the condition that 
 
 where , 77 are the perpendicular sides and >if. 
 
 To derive'a second such triangle from the first (3, 4, 5), assume the greater of the two 
 perpendicular sides to be 4 and the lesser $+* 
 
 Then $, + $(_,,) . I n = 3 6- i2jr-8.r = a square. 
 
 Also f 2 = 2 + ij 2 =25 + dr + Jr s = a square. 
 
 We have therefore simply to solve the double-equation 
 36-i2jr-8* 2 = 2 V 
 25+ 6.r+ *2=^)' 
 
 which is a matter of no difficulty. As a matter of fact, the usual method gives 
 20667 . / 20667 23729i6s\ 
 
 ' +3= ^S9 - and the mangle is (593-89- < -SSr
 
 236 THE ARITHMETICA 
 
 this triangle is (3, 4, 5), and accordingly we assume for 
 the triangle of the problem (3^-, 4*-, $x). 
 
 One condition then gives 6r 2 4^= a square = m^x 2 , say 
 (*'<6), 
 
 and ;i: = ^ - 
 
 6 m 2 
 
 The second condition gives 6x 2 3^r= a square ; and, by 
 substitution, 
 
 96 12 
 
 > - ^ - - = a square, 
 m* 1 2 m* +36 6 w 2 
 
 or 24+1 2;/z 2 = a square. 
 
 This is satisfied by ;;/ = I, 
 
 whence ;r=f , and the required triangle is ( , , 4 J . 
 
 Or, if we do not wish to use the value I for m, 
 \etm = z+i, and (dividing by 4) we have 
 
 3w 2 + 6 = 3-sr 2 + 6z + 9 = a square ; 
 ^ must be found to be not greater than Jg 3 - (in order that 
 
 m* may be less than 6), and /// will not be greater than 
 
 % 2 -. The solution is then rational 1 . 
 
 14- To find a right-angled triangle such that its area minus the 
 hypotenuse or minus one of the perpendiculars gives a square. 
 Let the triangle be (p:, 4*, $x). 
 
 Therefore , 2 5 \ are both squares. 
 
 Making the latter a square (= m^\ we have 
 x =6^1tf (^ 2 <6). 
 
 1 Diophantus having solved the problem of finding a right-angled triangle f, TJ, 
 (f being the hypotenuse) such that 
 
 ire both squares, 
 
 Fermat enunciates, as susceptible of solution by his method, but otherwise very difficult, 
 the corresponding problem of making 
 
 both squares. 
 
 This problem was solved by Euler {Navi Commentarii Acad. Petropol. 1749, n. (1751), 
 pp. 49 sqq. Commentationes arithmeticae, I. pp. 62-72).
 
 BOOK VI 
 
 The first equation then gives 
 
 54 15 
 
 - 
 
 or 1 5; 2 36 = a square. 
 
 This equation we cannot solve because \ 5 is not tfie sum of 
 
 two squares*. Therefore we must change the assumed 
 
 triangle. 
 Now (with reference to the triangle 3, 4, 5) I5w 2 = the 
 
 continued product of a square less than the area, the 
 
 hypotenuse, and one perpendicular ; 
 while 36 = the continued product of the area, the perpen- 
 
 dicular, and the difference between the hypotenuse 
 
 and the perpendicular. 
 Therefore we have to find a right-angled triangle (h, p, b, 
 
 say) and a square (w 2 ) less than 6 such that 
 
 m^hp ^pb .p(]i p} is a square. 
 If we form the triangle from two numbers X lt X 3 and 
 
 suppose that p=2.X^X^ and if we then divide 
 
 throughout by (X l X^f which is equal to // -/, we 
 
 must find a square ,5 s [= m-l(X^ - XJf\ such that 
 
 z*/ip ^pb .p is a square. 
 Ttie problem can be solved if X^, X^ are "similar plane 
 
 numbers*? 
 Form the auxiliary triangle from similar plane numbers 
 
 accordingly, say 4, i. [The conditions are then 
 
 satisfied 3 .] 
 [The equation for m then becomes 
 
 8 . ijm- 4 . 1 5 . 8 . 9 = a square, 
 or 1 36; 2 4320 = a square.] 
 
 Let 4 m* 36. [This satisfies the equation, and 36 < area 
 
 of triangle.] 
 
 1 See p. 70 above. 
 
 4 Diophantus states this without proof. [A " plane number " being of the form a . b t 
 
 a plane number similar to it is of the form a. b or -^ ab.] 
 The fact stated may be verified thus. We have 
 
 * ( X? + Xf) vXiXz- Xi X 2 (X? - AV) 2X1 X 2 = a square. 
 
 The condition is satisfied if z*=X\X<i, for the expression then reduces to +X^ X.X. 
 In that case X\X<t is a square, or X^jX^ is a square. 
 
 3 Since ^=4, X 3 = i, we have A = i?,f=S, b= 15, 32=^^=4, and 
 
 z*hp-^pb.p=. 17. 8-4. 15. 8 = 2. 32=64, a square. 
 
 4 The reason for this assumption is that, by hypothesis, *=w/ 2 /(.V 1 - X) y , or 
 4 = w 2 /3 2 , and m z = $6.
 
 238 THE ARITHMETICA 
 
 The triangle formed from 4, i being (8, 15, 17), we assume 
 
 &tr, \$x, \jx for the original triangle. 
 We now put 6ox- - &r = 36^, 
 and x ~\- 
 
 The required triangle is therefore (-. c, ?2| 
 
 U 5 37 
 Lemma to the following problem. 
 
 Given two numbers, if, when some square is multiplied into 
 one of the numbers and the other number is subtracted from 
 the product, the result is a square, another square larger than 
 the aforesaid square can always be found which has the same 
 property. 
 
 Given numbers 3, 1 1, side of square 5, say, so that 
 
 3.25 11- = 64, a square. 
 Let the required square be (x + 5) 2 . 
 Therefore 
 
 3 (*+ 5) 2 - 1 1 = 3* 2 + 3<^+ 64 = a square 
 
 = (8 - 2x}\ say, 
 and x=62. 
 
 The side of the new square is 67, and the square itself 
 4489. 
 
 15. To find a right-angled triangle such that the area added 
 to either the hypotenuse or one of the perpendiculars gives a 
 square. 
 
 In order to guide us to a proper assumption for the 
 required triangle, we have, in this case, to seek a 
 triangle (//, /, b, say) and a square (m z ) such that 
 m* > \pb, the area, and 
 
 mthp \pb.p(}i p) is a square. 
 Let the triangle be formed from 4, i, the square (; 2 ) 
 
 being 36, as before ; 
 but, the triangle being (8, 15, 17), the square is not 
 
 greater than the area. 
 We must therefore, as in the preceding Lemma, replace 
 
 36 by a greater square. 
 
 Now hp = 136, and \pb .p (h p} = 60 . 8 . 9 = 4320, 
 so that 1 36m 2 4320 = a square, 
 
 which is satisfied by m*= 36 ; and we have to find a larger 
 square (^ 2 ) such that 
 
 4320 = a square.
 
 BOOK VI 39 
 
 Put s = in + 6, and we have 
 
 (m* -f 1 2nt + 36) 1 36 - 4320 = a square, 
 or 1 36; 2 -f 1 6$2m + 576 = a square = (km 24)*, say. 
 This equation has any number of solutions ; e.g., putting 
 k = 1 6, we have 
 
 m = 20, z = 26, and 2 s = 676. 
 
 We therefore put (&r, ir, \yx) for the original triangle, 
 and then assume 
 
 6ot- 2 + &r = 676*- 2 , 
 whence ^ = 7^, and the problem is solved 1 . 
 
 1 In vi. 14, 15 Diophantus has shown how to find a rational right-angled triangle 
 *7i I (where f is the hypotenuse) such that 
 
 are both squares, 
 
 are both squares. 
 
 In the Inventum Afovum, Part I. paragraphs 26, 40 (Oeuvres de Fermat, ill. pp. 341 
 -2, 349-50) is given Fermat 's solution of a third case in which 
 
 are both squares. 
 
 This depends on the Lemma : To find a rational right-angled triangle in which 
 f(f+^)-^i? = a square. 
 
 Form a right-angled triangle from JT+ i, r ; the sides are then 
 
 o^+2jr+a, jr*+2jr, 2JT+2. 
 We must therefore have 
 
 (.r 2 + wr + )(.* + 3jr + I)-(JT+I) (.i 3 +2-r) = a square, 
 or ** + 4* 3 + 6j.- a + 6^-+ 2 = a square 
 
 = (j^+2x+i) s , say. 
 
 Therefore r= --, and the triangle has one of its sides x- + ijc negative. Instead 
 
 therefore of forming the triangle from -, I or from i, 2, we form it from jr-f- 1, a aad 
 repeat the operation. The sides are then 
 
 j^+a-x + s, jt+2jr-3, 4^ + 4, 
 and we have 
 
 (.**+ 2.r+ 5) (jc 2 + 4jr- i) - (2JT+ 2) (jr 2 + 2J. - 3) = a square, 
 i =a square 
 = (i + iar-4r*)*, say,
 
 2 4 o THE ARITHMETICA 
 
 1 6. To find a right-angled triangle such that the number 
 representing the (portion intercepted within the triangle of the) 
 bisector of an acute angle is rational 1 . 
 
 A 
 
 C 
 
 Suppose the bisector AD = $x, and one segment of the 
 
 base (DB} = 3^- ; therefore the perpendicular = 4^-. 
 Let the whole base CB be some multiple of 3, say 3 ; then 
 
 0? = 3-3x 
 But, since AD bisects the angle CAB, 
 
 AC:CD = AB-.BD; 
 therefore the hypotenuse A C = f (3 3^) = 4 - 4*. 
 
 whence x = ~, and the required auxiliary triangle is formed from ^9 or from 29, 11, 
 the sides being accordingly 985, 697, 696. 
 
 (Fermat observes that the same result is obtained by putting y for x in the 
 expression x* + ^x 3 + ftx* + 6x + 2 ; for we must have 
 
 uare=^+5^-yy, say, 
 
 whence y = , so that #=^ = , and the triangle is formed from, i or from 
 
 29, 12, as before.) 
 
 We now return to the original problem of solving 
 
 We assume for the required triangle (985^, 697^, 696*) and we have - rj = 244556^, 
 
 so that 
 
 985^-242556^) 
 
 V must both be squares. 
 697^-242556^) 
 
 Assume that 697^ - 242556^= (697^)*, 
 
 and we have x - 348^ 2 =^97^ 2 , 
 
 whence x = , and the required triangle is ( , , ). 
 
 1045' \iQ45 io 45 ' io 45 ; 
 
 [The 985^-242556^ is a square by virtue of the sides 985, 697, 696 satisfying the 
 conditions of the Lemma; for 985^-242556^ = -^; - ^. , 9 ^' 9 2 - , which is a square 
 
 if 985. 1045 --.697.696 is a square, and 1045 ^697 + -.696.] 
 
 1 Why did not Diophantus propound the analogous problem " To find a right-angled 
 triangle such that the sides are rational and the bisector of the right angle is also rational"? 
 Evidently because he knew it to be impossible, as is clear when (a, c being the perpen- 
 
 diculars) the bisector is expressed as ^2. (Loria, op. cit. p. 148 .)
 
 BOOK VI 241 
 
 Therefore [by Eucl. I. 47] 
 
 16= 
 
 and * = . 
 
 If we multiply throughout by 32, the perpendicular = 28, 
 the base = 96, the hypotenuse = 100, and the bisector 
 = 35- 
 
 17. To find a right-angled triangle such that the area added 
 to the hypotenuse gives a square, while the perimeter is a cube. 
 
 Let the area be x and the hypotenuse some square 
 
 minus x, say 16 x. 
 The product of the perpendiculars = 2x ; 
 therefore, if one of them be 2, the other is x, and the 
 
 perimeter = 18, which is not a cube. 
 
 Therefore we must find some square which, when 2 is 
 added to it, becomes a cube 1 . 
 
 1 "Did Diophantus know that the equation w 2 +2 = w 3 only admits of one solution 
 =, ^=3? Probably not" (Loria, op. cit. p. 155). The fact was noted by Fermat 
 (on the present proposition) and proved by Euler. 
 
 Killer's proof (Algebra, Part II. Arts. 188, 193) is, I think, not too long to be given 
 here. Art. 188 shows how to find x, y such that cuP + cy* may be a cube. Separate 
 fl-r' + ry 2 into its factors x*Ja+y^]( -c), x<Ja -yJ( - f), and assume 
 
 the product (a/ 1 + cq^f being a cube and equal to cuP + ey*. 
 
 To find values for x and y, we write out the expansions of the cubes in full, and 
 
 -c)- yff Ja + cf J( - c), 
 whence a- = a/ 3 - yfty 3 , 
 
 For example, suppose it is required to make x*+y* a cube. Here a=i and e= 
 so that x-p^-^pq^, 
 
 y = 3?q-q*, 
 . If now/=2 and f=i, we find x=* and^= n, whence 
 
 Now (Art. 193) let it be required to find, if possible, in integral numbers, other squares 
 besides 25 which, when added to 2, give cubes. 
 
 Since 0^ + 2 has to be made a cube, and 2 is double of a square, let us first determine 
 the cases in which x lj riy L becomes a cube. Here a= i, c 2, so that 
 
 .r=/3_6> ? 2, y=&q*f; 
 
 therefore, since y= i i, we must have 
 
 ifq- if or q($p*-iq*)= I 5 
 consequently q must be a divisor of i. 
 
 Let, then, q=i, and we shall have 3/2 - 2 = i. 
 
 With the upper sign we have 3/^ = 3 and, taking /)= - i, we find x~ 5 ; with the lower 
 sign we get an irrational value of/ which is of no use. 
 
 H. D. l6
 
 24 2 THE ARITHMETIC A 
 
 Let the side of the square be m+i, and that of the cube 
 
 m i. 
 Therefore m z - yn 1 + yn i = m 2 + 2m + 3, 
 
 from which m is found 1 to be 4. 
 
 Hence the side of the square = 5, and that of the cube = 3. 
 Assuming now x for the area of the original triangle, 
 
 25 x for its hypotenuse, and 2, x for the perpen- 
 
 diculars, we find that the perimeter is a cube. 
 But (hypotenuse) 2 = sum of squares of perpendiculars ; 
 
 therefore x* 50*- + 62 5 = x* + 4 ; 
 x = &g^, and the problem is solved. 
 
 1 8. To find a right-angled triangle such that the area added 
 to the hypotenuse gives a cube, while the perimeter is a square. 
 
 Area x, hypotenuse some cube minus x, perpendiculars.*-, 2. 
 Therefore we have to find a cube which, when 2 is added 
 
 to it, becomes a square. 
 Let the side of the cube be m i. 
 
 Therefore m* yiP + yn + i = a square = ( \\m + i ) 2 , say. 
 Thus m = *, and the cube = (Jf) 3 = js. 
 Put now x for the area, x, 2 for the perpendiculars, and 
 
 -%\- x for the hypotenuse; 
 and x is found from the equation (-%%- xf = x* + 4. 
 
 nd the trianle is 2 1 
 
 19. To find a right-angled triangle such that its area added to 
 one of the perpendiculars gives a square, while the perimeter is 
 a cube. 
 
 Make a right-angled triangle from some indeterminate odd 
 
 number*, say 2x+ i ; 
 
 then the altitude = 2*+ i, the base = 2x"- + 2.x, and the 
 hypotenuse = 2x"- + 2x+ i. 
 
 It follows that there is no square except 25 which has the required property. 
 
 Fermat says ("Relation des nouvelles decouvertes en la science des nombres," 
 Oeuvres, II. pp. 433-4) that it was by a special application of his method of descente, 
 such as that by which he proved that a cube cannot be the sum of two cubes, that he proved 
 
 (1) that there is only one integral square which when increased by i gives a cube, and 
 
 (2) that there are only two squares in integers which, when added to 4, give cubes. The 
 latter squares are 4, 121 (as proved by Euler, Algebra, Part n. Art. 192). 
 
 1 See pp. 66, 67 above. 
 
 2 This is the method of formation of right-angled triangles attributed to Pythagoras. 
 If m is any odd number, the sides of the right-angled triangle formed therefrom are m, 
 i(w 2 -i), i(^ 2 -ri), for w2+ U( W 2- ,)1 2 = J1( W 2 + !)j. 2 . Cf . p roc i US) Comment. 
 on Rucl. i. (ed. Friedlein), p. 428, 7 sqq., etc. etc.
 
 BOOK VI 243 
 
 Since the perimeter = a cube, 
 
 and, if we divide all the sides by x + r, we have to make 
 2 a cube. 
 
 Again, the area + one perpendicular = a square 
 
 Before + . . square ; 
 
 But 4-r + 2 = a cube ; 
 
 therefore we must find a cube which is double of a 
 
 square ; this is of course 8. 
 Therefore 4^+2 = 8, and x=\\. 
 
 The required triangle is (- , 3 , %\ . 
 
 20. To find a right-angled triangle such that the sum of its 
 area and one perpendicular is a cube, while its perimeter is a 
 square. 
 
 Proceeding as in the last problem, we have to make 
 $x + 2 a square) 
 2x+ i a cube j " 
 We have therefore to seek a square which is double of a 
 
 cube; this is 16, which is double of 8. 
 Therefore 4-r+ 2= 16, and x=$\. 
 
 The triangle is (-, ^, %) . 
 \9 ' 9 ' 9J 
 
 21. To find a right-angled triangle such that its perimeter is 
 a square, while its perimeter added to its area gives a cube. 
 
 Form a right-angled triangle from x t i. 
 
 The perpendiculars are then 2x, x* i, and the hypotenuse 
 
 Hence 2x- + 2x should be a square, 
 
 and x 3 + 2x--\-x a cube. 
 It is easy to make 2x- + 2x a square ; let 2** + 2x 
 
 therefore x=2/(m 2 -2). 
 By the second condition, 
 
 o 82 
 
 _ 1- _ + _ must be a cube, 
 
 20Z 4 _ . 
 
 16 2
 
 2 44 THE ARITHMETICA 
 
 Therefore 2; 4 = a cube, or 2m a cube = 8, say. 
 
 Thus ;;z = 4, ^=T 2 4 = 7> x * = -$- 
 
 But one of the perpendiculars of the triangle is x-- I, and 
 
 we cannot subtract i from ^. 
 Therefore we must find another value for x greater than i ; 
 
 hence 2 < m 2 < 4. 
 
 And we have therefore to find a cube such that | of the 
 
 square of it is greater than 2, but less than 4. 
 If z* be this cube, 
 
 2<\&< 4, 
 
 or 8 < ^ < 16. 
 
 This is satisfied by ^ = -^-, or s 3 = ^-. 
 Therefore *=, ^ 2 = ||f, and *=fff, the square of 
 
 which is > i. 
 Thus the triangle is known 
 
 22. To find a right-angled triangle such that its perimeter is 
 a cube, while the perimeter added to the area gives a square. 
 
 (1) We must first see how, given two numbers, a triangle 
 
 may be formed such that its perimeter = one of 
 the numbers and its area = the other. 
 
 Let 12, 7 be the numbers, 12 being the perimeter, 7 the 
 area. 
 
 Therefore the product of the two perpendiculars 
 
 = 14 = ^. 14*. 
 If then -, 14-r are the perpendiculars, 
 
 hypotenuse = perimeter sum of perps. = 12 --- \^x, 
 
 % 
 
 Therefore [by Eucl. I. 47] 
 
 ~ + I96.r 2 + 172 - ^ - 336^= ^ + 196* ; 
 
 that is, 172 = 336^-+ , 
 
 3C 
 
 or 172^=336^ + 24. 
 
 This equation gives no rational solution, because 86 2 24 . 336 
 
 is not a square. 
 Now 172 = (perimeter) 2 + 4 times area, 
 
 24. 336= 8 times area multiplied by (perimeter) 2 . 
 
 (2) Let now the area = m, and the perimeter = any 
 
 number which is both a square and a cube, say 64.
 
 BOOK VI 245 
 
 Therefore { (642 + 4w)} 2 - 8 . 64" . m must be a square, 
 
 or 4^ 2 - 24576^ +4194304 = a square. 
 
 Therefore m 2 6 144772 + 1048576 = a square] 
 
 Also m + 64 = a square] ' 
 
 To solve this double-equation, multiply the second by 
 such a number as will make the absolute term the 
 same as the absolute term in the first. 
 
 Then, if we take the difference and the factors as usual, 
 the equations are solved. 
 
 [After the second equation is multiplied by 16384, the 
 double-equation becomes 
 
 m* -6i44m + 1048576 = a square) 
 16384^ + 1048576 = a square} ' 
 
 The difference is m z 22528;;?. 
 
 If m, m 22528 are taken as the factors, we find m = 7680, 
 which is an impossible value for the area of a right- 
 angled triangle of perimeter 64. 
 
 We therefore take as the factors 1 1 m, -fam 2048 ; then, 
 when the square of half the difference is equated to 
 the smaller of the two expressions to be made squares, 
 we have 
 
 (&m + 1024)* = 16384;^ + 1048576, 
 and m = 
 
 Returning now to the original problem, we put - , 2mx 
 
 for the perpendicular sides of the required triangle, 
 and we have 
 
 64 - - - 
 
 which leads, when the value of m is substituted, to 
 the equation 
 
 78848* - 8432^ + 225 = o. 
 
 The solution of this equation is rational, namely 
 _52723 = 25 _o_ 
 9856 448 176* 
 
 Diophantus would of course use the first value, which 
 would give (-^f-, ^, *$$) as the required right- 
 angled triangle. The second value of x clearly gives 
 the same triangle.]
 
 246 THE ARITHMETICA 
 
 23. To find a right-angled triangle such that the square of its 
 hypotenuse is also the sum of a different square and the side of 
 that square, while the quotient obtained by dividing the square 
 of the hypotenuse by one of the perpendiculars of the triangle is 
 the sum of a cube and the side of the cube. 
 
 Let one of the perpendiculars be x, the other x*. 
 
 Therefore (hypotenuse) 2 = the sum of a square and its 
 
 X* -4- X^ 
 
 side ; also = x 3 + x = the sum of a cube and its 
 
 x 
 
 side. 
 
 It remains that x* +** must be a square. 
 Therefore x 2 + I = a square = (x 2) 2 , say. 
 Therefore x= f , and the triangle is found [f , ^, {]. 
 
 24. To find a right-angled triangle such that one perpendicular 
 is a cube, the other is the difference between a cube and its side, 
 and the hypotenuse is the sum of a cube and its side. 
 
 Let the hypotenuse be x*+x, and one perpendicular 
 
 Therefore the other perpendicular = 2x z = a cube = X s , say. 
 Thus x= 2, and the triangle is (6, 8, 10). 
 
 It is on Bachet's note to vi. 2 2 that Fermat explains his method of solving 
 triple-equations^ as to which see the Supplement, Section v. 
 
 [No. 20 of the problems on right-angled triangles which Bachet 
 appended to Book vi. (" To find a right-angled triangle such that its area 
 is equal to a given number ") is the occasion of Fermat's remarkable note 
 upon the theorem discovered by him to the effect that The area of a right- 
 angled triangle the sides of which are rational numbers cannot be a square 
 number. 
 
 This note will be given in full, with other information on the same 
 subject, in the Supplement]
 
 ON POLYGONAL NUMBERS 
 
 All numbers from 3 upwards in order are polygonal, containing 
 as many angles as they have units, e.g. 3, 4, 5, etc. 
 
 " As with regard to squares it is obvious that they are such 
 because they arise from the multiplication of a number into 
 itself, so it was found that any polygonal multiplied into a 
 certain number depending on the number of its angles, with 
 the addition to the product of a certain square also depending 
 on the number of the angles, turned out to be a square. This 
 I shall prove, first showing how any assigned polygonal 
 number may be found from a given side, and the side from 
 a given polygonal number. I shall begin by proving the pre- 
 liminary propositions which are required for the purpose." 
 
 i. If there are three numbers with a common difference, then 
 8 times the product of the greatest and middle + the square of the 
 least = a square, the side of which is the sum of the greatest and 
 twice the middle number. 
 
 Let the numbers be AB, BC, BD in the figure, and we 
 have to prove SA . C + BD* = (AB + zBC}\ 
 
 E _ A O p _ B 
 
 By hypothesis A C= CD, AB= BC + CD, BD= BC- CD. 
 Now 8A B.BC=4A B.BC+(^BC* + 4BC.CD}. 
 Therefore ZAB.BC+BD* 
 
 . [Eucl. n. 8] 
 
 and we have to see how AB* + ^AB . BC+ ^BC* can 
 be made a square. 
 
 [Diophantus does this by producing BA to E, so that 
 AE = BC, and then proving that 
 
 It is indeed obvious that 
 
 2. If there are any numbers, as many as we please, in A.P., 
 the difference between the greatest and the least is equal to the 
 common difference multiplied by the number of terms less one.
 
 248 ON POLYGONAL NUMBERS 
 
 [That is, if in an A.P. the first term is a, the common 
 difference b and the greatest term /, n being the 
 number of terms, then 
 
 l- a = (n- i)&] 
 
 Let AB, BC, BD, BE have a common difference. 
 
 Now A C, CD, DE are all equal. 
 
 Therefore EA = ACx (number of terms AC, CD, DE} 
 = A C x (number of terms in series I ). 
 
 3. If there are as many numbers as we please in A.P., then 
 (greatest + least) x number of terms = double the sum of the 
 terms. 
 
 [That is, with the usual notation, 2s = n (/ + #).] 
 (i) Let the numbers be A, B, C, D, E, F, the number of 
 them being even. 
 
 Let GH contain as many units as there are numbers, 
 and let GH, being even, be bisected at K. Divide 
 GK into units at L, M. 
 
 Since F-D=C-A, 
 
 But F + A = (F+A).GL- 
 
 therefore C + D = (F+A) .LM. 
 
 Similarly E + B = (F + A ) . MK. 
 
 Therefore, by addition, 
 
 A + B + C + D + E + F = ( F + A ) . GK. 
 Therefore 2 (A ++ ...)= 2 (F + A). GK 
 
 = (F+A).GH. 
 (2) Let the number of terms be odd, the terms being 
 
 A, B, C, D, E. 
 
 F H L K G 
 
 Let there be as many units in FG as there are terms, 
 so that there is an odd number of units. 
 
 Let FH be one of them ; bisect HG at K, and divide HK 
 into units, at L.
 
 ON POLYGONAL NUMBERS 
 
 Since E-C=C-A, 
 
 E + A = 2C=2C.LK. 
 Similarly B + D=2C.LH. 
 
 Therefore A +E + B + D = 2C. HK 
 = C. HG. 
 
 Also C=C.HF- 
 
 therefore, by addition, 
 
 A+J3+C+ D + E=C.FG; 
 and, since 2C= A + E, 
 
 249 
 
 4. If there are as many numbers as we please beginning with 
 i and increasing by a common difference, then the sum of all 
 x 8 times the common difference + the square of (common 
 difference 2) = a square, the side of which diminished by 2 
 = the common difference multiplied by a number which when 
 increased by i is double of the number of terms. 
 
 [The A. p. being i, i + b, ... i +(n - \)b, and s the sum, 
 we have to prove that 
 
 s . &b + (b - 2f = {b (2n - i) + 2J 2 , 
 i.e. 8bs = 4#V/ 2 - 4 (b - 2) nb, 
 
 or 2s = bn z (b 2)n 
 
 = n {2 + ( i)b\. 
 
 The proof being cumbrous, I shall add the generalised 
 algebraic equivalent in a column parallel to the 
 text.] 
 
 Let AB, CD, EF be the terms in I i +&, i + 26, i + 36,.... 
 A. P. after i. 
 
 P A 
 
 K N 
 
 Let GH contain as many units 
 as there are terms including r. 
 
 Difference between EF and i 
 = (diff. between AB and i) x (GH- i). 
 [Prop. 2]
 
 2 5 
 
 ON POLYGONAL NUMBERS 
 
 Put AK, EL, GM each equal 
 to unity. 
 
 Therefore LF=KB.MH. 
 Make KN=2, and we have to 
 inquire whether 
 
 (sum of terms) x 8KB f NB- 
 
 = {2 + KB(GH+HM)}\ 
 Now sum of terms 
 
 .GH [Prop. 3] 
 
 = \ (KB . MH, GH + 2677), 
 since LF = KB .MH [above]. 
 
 Bisecting MH at O, we have 
 (sum of terms) 
 
 = KB.GH.HO+GH. 
 We have therefore to inquire 
 whether 
 
 (KB . GH. HO + GH} . 8KB + NB- 
 is a square. 
 
 Now KB. GH. HO. 8KB 
 .HO. KB* 
 
 Is then 
 
 4GH. HM . KB* + 8KB . GH + NB* 
 a square ? 
 
 Now 8GH.KB 
 = 4 GM. KB + 4 (GH+HM) KB. 
 Also 4GM. KB = 2NK . KB ; 
 and, adding NB*, the right-hand side 
 becomes KB* + KN*. [Eucl. II. 7] 
 
 Is then ^GH.HM.KB* 
 + 4 (GH + HM} KB + KB* + KN* 
 a square ? 
 
 Again, KB* + ^GH . HM . KB* 
 = GM*.KB* + ^GH. HM . KB* 
 = (GH+HM)* . KB*. [Eucl. n. 8] 
 Is then (GH+HM}* . KB* 
 
 + 4(Gtf+HM) KB + KN* 
 a square? 
 
 Make the number NO' equal to 
 (GH+HM).KB- 
 
 Call the expression on 
 the left-hand side X. 
 
 X=bn. H 
 
 = ( + - i}*b* 
 + 4 { + (
 
 ON POLYGONAL NUMBERS 
 
 thus (GH 
 
 will be proved later 1 . 
 
 Also ANO' = 2 . NO 1 . NK, since 
 
 Therefore 
 X = {( + n - i ) b + 2] 2 
 
 Is then NO* + NK* + 2NO' . NK 
 a square ? 
 
 Yes ; it is the square on KO. 
 
 And 
 
 aK - 2 = NO' = KB(GH + HM\ 
 while GH+HM+ i = (twice number 
 of terms). 
 
 Thus the proposition is proved. 
 
 " The above being premised, I say that, 
 
 [5] If there be as many terms as we please in A.P. beginning 
 from i, the sum of the terms is polygonal; for it has as many 
 angles as the common difference increased by 2 contains units, and 
 its side is the number of the terms set out including i." 
 
 The numbers being as set out in the figure of Prop. 4, we 
 have, by that proposition, 
 
 (sum of terms) . %KB + NB* = Ka\ 
 
 Taking another unit AP, we have KP ' = 2, while KN '= 2; 
 therefore PB, BK, BN are in A.P., so that 
 
 ZPB.BK + NB* = (PB + 2KB}* ; [Prop, i ] 
 
 while 3 + i = 2 . 2, or 3 is one less than the double of 2. 
 Now, since the sum of the terms of the progression 
 
 1 Deferred lemma. 
 
 To prove that (Gff+ffM) 2 . K& 
 
 = {(GH+HM). KB}*. 
 
 Place DE (equal to a) and EF (equal to ) in a 
 straight line. 
 
 Describe squares Dff, EL on DE, EF and com- 
 plete the figure. 
 
 Then DE:EF=Dff:HF, 
 
 and HE:EK=HF:EL. 
 
 Therefore ffFis a mean proportional between the two squares, 
 that is DH.FK=HF\ 
 
 or a./S=(a/3).
 
 252 ON POLYGONAL NUMBERS 
 
 including unity satisfies the same formula 1 [literally 
 " does the same problem "] as PB does, 
 
 while PB is any number and is also always a polygonal, 
 the first after unity (for AP is a unit and AB is the 
 term next after it), and has 2 for its side, 
 
 it follows that the sum of all the terms of the progression 
 is a polygonal with the same number of angles as PB, 
 the number of its angles being the same as the 
 number of units in the number which is greater by 2, 
 or PK, than the common difference KB, and that its 
 side is GH which is equal to the number of terms 
 including I. 
 
 And thus is demonstrated what is stated by Hypsicles in 
 his definition, namely, that, 
 
 "If there are as many numbers as we please beginning 
 from I and increasing by the same common difference, 
 then, when the common difference is I, the sum of all 
 the terms is a triangular number ; when 2, a square ; 
 when 3, a pentagonal number [and so on]. And the 
 number of the angles is called after the number 
 exceeding the common difference by 2, and the side 
 after the number of terms including I." 
 
 [In other words, if there be an arithmetical progression 
 
 i , i + b, i + 2b, . . . i + (n i ) b, 
 
 the sum of the n terms, or \n {2 + ( i) b} t is the 
 wth polygonal number which has {b + 2) angles.] 
 
 Hence, since we have triangles when the common dif- 
 ference is i, the sides of the triangles will be the 
 greatest term in each case, and the product of the 
 greatest term and the greatest term increased by i 
 is double the triangle. 
 
 - l Nesselmann (pp. 475-6), exhibits this result thus. 
 Take the A.P. i, i+, i + il>,... i+(n-i)l>. 
 
 If s is the sum, 8sl> + (/> - a) 2 = [b (in - i) + i} z . 
 
 If now we take the three terms b- 2, b, 6 + 1, also in A. P., 
 
 U(b + l) + (b-lY={(l> + l) + lb} Z 
 = ( 3 * +2)2. 
 
 Now 6+1 is the sum of the first two terms of the first series, and corresponds there- 
 fore to s when n = 2 ; and 3 2 . 2 - i , so that 3 corresponds to in- i. 
 
 Hence s and b + 2 are subject to the same law ; and therefore, as b + 1 is a polygonal 
 number with b + 1 angles, s is also a polygonal number (the wth) with b + 1 angles.
 
 ON POLYGONAL NUMBERS 253 
 
 And, since PB is a polygonal with as many angles as 
 
 there are units in it, 
 and 8P . (PB - 2) + (PB - 4)* = a square (from above, 
 
 BK being equal to PB - 2, and NB to PB - 4), 
 the definition of polygonal numbers will be as follows : 
 Every polygonal multiplied by 8 times (number of angles 
 
 2) + square of (number of angles 4) = a square 1 . 
 
 The Hypsiclean definition and the new one being thus 
 simultaneously proved, it remains to show how, when 
 the side is given, the prescribed polygonal is found. 
 
 For, having given the side GH and the number of angles, 
 we know KB. 
 
 Therefore (GH + HM} KB, which is equal to NO', is also 
 given ; therefore KO'(=NO'+NK or NO'+ 2) is given. 
 
 Therefore KO' Z is given ; 
 
 and, subtracting from it the given square on NB, we 
 obtain the remaining term which is equal to the 
 required polygonal multiplied by "&KB. Thus the 
 required polygonal can be found. 
 
 Similarly, given the polygonal number, we can find its 
 side GH. Q. E.D. 
 
 Rules for practical use. 
 
 (i) To find the number from the side. 
 
 Take the side, double it, subtract i, and multiply the 
 remainder by (number of angles 2). Add 2 to the 
 product ; and from the square of the sum subtract 
 the square of (number of angles - 4). Dividing the 
 remainder by 8 times (number of angles - 2), we 
 have the required number. 
 
 1 Hultsch points out (art. Diophantos in Pauly-Wissowa's Real-Encyclopadie der 
 dassischen Altertumswissenschafteri) that this formula 
 
 8P (a - -i ) + (a - 4 ) 2 = a square 
 
 shows that Diophantus intended it to be applied not only to cases where a is greater than 
 4 but also where = 4 or less. For 36, as Diophantus must have known, besides being 
 the second 36-gon, is also a triangle, a square, and a ij-gon, inasmuch as 
 
 ) 2 = 289= i? 2 , 
 -4) 2 = 576=242, 
 
 And indeed it is evident from Def. 9 of the Arithmetica that (3-4) 2 = '. while it is 
 equally obvious that (4-4)2 = 0.
 
 2 54 
 
 ON POLYGONAL NUMBERS 
 
 [If P be the wth -gonal number, 
 
 P . 8 (a - 2) + (a - 4) 2 = {2 + (2 - I ) ( - 2)j 
 
 
 
 (2) 
 
 To find the side from the number. 
 
 Multiply the number by 8 times (number of angles 2) ; 
 add to the product the square of (number of angles -4). 
 We thus get a square. Subtract 2 from the side of 
 this square and divide the remainder by (number of 
 angles - 2). Add I to the quotient, and half the 
 result gives the side required 1 . 
 
 = i rilP 
 L 2V 
 
 a-2 
 
 
 \ I 
 J'\ 
 
 Given a number, to find in how many ways it can be polygonal. 
 Let AB be the given number, BC [Algebraical equivalent.] 
 
 the number of angles, and in BC take 
 
 Number AB = P. 
 Number of angles BC=a. 
 
 Since the polygonal AB has BC 
 angles, 
 
 (i) S AB.BD + BE*= 
 say. 
 
 Cut off AH equal to i ; 
 therefore SAB.BD 
 
 = X\ say. 
 But SP(a-2) 
 
 Make DK equal to 4(AB+Btf\ 
 and for $AH . BD put 2BD . DE. 
 
 1 Fermat has the following note. " A very beautiful and wonderful proposition which 
 I have discovered shall be set down here without proof. If, in the series of natural 
 numbers beginning with i , any number n be multiplied into the next following, n + i , 
 the product is twice the nth triangular number; if n be multiplied into the (n+i)(h 
 triangular number, the product is three times the nth tetrahedral number ; if n be 
 multiplied into the (+ \}th tetrahedral number, the product is four times the nth triangulo- 
 triangular number {figured number of $th order} ; and so on, ad infinitum. I do not 
 think there can be, in the theory of numbers, any theorem more beautiful or more 
 general. The margin is too small, and I am not at liberty, to give the proof." (Cf. 
 Letter to Roberval of 4 November 1636, Oeuvres de Fermat, n. pp. 84, 85.) For a proof, 
 see Wertheim's Diophantus, pp. 318-20.
 
 ON POLYGONAL NUMBERS 
 
 255 
 
 Therefore 
 (2) FG Z =KD.DB+2BD.DE + BE\ 
 
 = KD.DB + BD* + DE\ 
 
 [Eucl. II. 7] 
 
 = KB.BD+DE*. [Eucl. n. i] 
 But, since 
 
 (3) 
 (4) 
 
 and DC = half 4 or 2 ; 
 
 therefore CK > CD. 
 
 Therefore, if DK be bisected at 
 L, L falls between C and K. 
 
 And, since Z^" is bisected at Z, 
 
 KB .BD + LD* = LB*, 
 whence KB . BD = LB* - LD\ 
 
 Therefore, by (4) above, 
 
 (5) F 
 or F 
 
 (6) or LD*-DE* = LB*-FG\ 
 
 Again, since ED = DC, and 
 is produced to , 
 
 therefore EL .LC= DL? - DC 2 
 
 = DL*-DE* 
 
 (7) =LB*-FG\ 
 
 Put FM=BL (for BL>FG, 
 since FG* + DL Z = BL* + ED\ 
 while > 2 > ^Z^ 2 ). 
 
 Therefore ^J/ 2 -FG* = EL.L C. 
 
 Now, Z?^ being bisected at Z and 
 being equal to 4 (^4 B + BH\ 
 
 And DC=2AH. 
 
 Therefore 
 or 
 
 But 
 
 therefore AB = \EL, 
 
 while BH=CL. 
 
 Therefore AB.BH= 
 or 
 
 LC, 
 
 EL .LC=i6AB.BH. 
 
 + 2 ( - 2) . 2 + (rt - 
 
 DL=2(2P-l) 
 
 -{2(2P-l)} 2 +2 2 
 {2(2P-l)) 2 -2 2 
 
 CL=2(2P-\}-2] 
 
 {2(2P- l)+ 
 
 FM=2(2P- 
 
 EL =
 
 256 ON POLYGONAL NUMBERS 
 
 (8) Therefore 
 
 1 6AB . BH = MF~ - FG* 
 
 Therefore GM is even. 
 
 Let GM be bisected at N . . 
 
 \6P(P- 
 
 \_={2(2P-2)-2(H-l}(a-2}Y 
 
 + 2 {2 +(2n- i)(-2)} 
 
 [Here the fragment ends, and the question of course arises whether 
 Diophantus ever actually solved the problem of finding in how many 
 different ways a given number can be a polygonal. Tannery went so far 
 as to call the whole of the fragment, from and including the enunciation 
 of the problem, the "vain attempt of a commentator" to solve it 1 . 
 Wertheim 2 has however shown grounds for thinking that Diophantus did 
 solve the problem and that the fragment is a genuine part of his argument 
 leading to that result. The equation 
 
 8P(a - 2) + (a - 4 ) 2 = {2 + (zn - i) (a - 2 )} 2 
 easily reduces (by algebra) to 
 
 8P(a - 2) = 4 (a - 2) {2 + ( - i) (a - 2)}, 
 or 2P-- n (2 + (n- i) (a - 2)}. 
 
 Wertheim has shown how this result can be obtained by a continuation 
 of the work, from the point where the fragment leaves off, in the same 
 geometrical form which is used up to that point 3 , and how, when the 
 
 1 Dioph. i. pp. 476-7, notes. 
 
 2 Zeitschrift fiir Math. u. Physik, hist. litt. Abtheilung, 1897, pp. 121-6. 
 
 3 The only thing, so far as I can see, tending to raise doubt as to the correctness of 
 this restoration is the fact that, supposing it to be required to prove geometrically, from 
 the geometrical equivalent of 
 
 that iP=n {2 + (- i) (a-i}\, 
 
 it can be done much more easily than it is in Diophantus' proposition as extended by 
 
 Wertheim. 
 
 For let FG=2 + (in- \)(a-i). Cut off FR equal to 2, and produce RFtv S so that 
 
 2n(o-2) 
 
 T 
 
 We have now 8P . SX=FG 2 - SF 2 
 
 Bisect SG at T, and divide out by 4 ; 
 therefore iP '. S = ST 2 -ST. SF 
 
 = Sr(ST-SF) 
 = ST.FT 
 
 Now ST=n .SR, and FJ? = i, while RT=(n- i). SK = (n- \)(a-i). 
 It follows that iP=n{i + (n-i) (a -2)}.
 
 ON POLYGONAL NUMBERS 
 
 257 
 
 formula is thus obtained, it can be used for the purpose of finding the 
 number of ways in which P can be a polygonal number. The portion of 
 the geometrical argument which has to be supplied is, it is true, somewhat 
 long, and its length and difficulty may, as Wertheim suggests, account for 
 the copyist having failed, as it were, to see his way through it and having 
 stopped through discouragement when he had lost his bearings. 
 
 I shall now reproduce Wertheim's suggested restoration of the rest of 
 the problem. The figure requires some extension, and I accordingly give 
 a new one after Wertheim. 
 
 A H 
 
 B- 
 
 S R Q N 
 
 The last step in the above fragment is 
 (9) 2FG . GM + GM 2 = i6AB.BH. 
 
 Bisect GM in JV, 
 so that GW= NM. 
 
 Therefore, if we divide by 4, 
 (10) FG. GN+ GN* = 4AB. BH, 
 
 Put now FR=2AB, and RS=GN, 
 so that GS = RN, and we have 
 
 FS=FR-RS = 2AB - RS, 
 
 FN= FR + RN= 2AB + Ri\, 
 GN=RS=2AB-FS. 
 Substituting in (n), we have 
 (12) (2AB+ RN}(2AB-FS)=$AB.BH, 
 
 { 2 ( 2 P- 2 )-2(n-i)(a-2)} 
 
 GN=NM 
 
 {2P+n(a- 2)} 
 
 ( 2 (/>-!)-(-,) ( -a)} 
 
 FS=2P 
 
 -{ 2 (P-i}-(n-i}(a-2)\ 
 
 = 2 +(n- t)(a-2) 
 FA 7 =2P+n(a-2), from above 
 GN=2(P-i)-(n-i}(a-2) 
 RN= AV - 2AB = n (a - 2) 
 \2P+n(a-2)} 
 
 '7
 
 2 5 8 
 
 ON POLYGONAL NUMBERS 
 
 (13) 4 AB?-2AB(FS-RN) 
 
 Therefore 
 (14) 2AB(FS-RN} 
 
 2P{2-(a-2)} 
 
 + (0-2) {(-i)(a -2) + 2} 
 
 2P(a-2} 
 
 Now RN= FN-FR = FM- NM - FR = FM-\ GM- FR 
 = BL-\ GM- 2AB = BD + \DK- 1 GM 
 = BD + 2AB + zBH- \ GM - 2AB 
 
 and 
 
 FS=FR-RS 
 
 Therefore 
 
 RN- 
 
 2BH- zAB 
 
 and RN- FS + 2 AH = BD. 
 
 Again, we have 
 
 RN= BD + 2BH- \ GM= BD + 2BH- 
 = BD+ 2BH- \BD -%DL + \FG 
 = \BD + 2BH-\DL + \FG 
 = \BD + 2BH- (AB + BH) + %FG 
 = \BD + BH- AB + ^FG 
 = \BD-AH+\FG 
 = | (BD + FG- 2 AH}. 
 
 But, from the rule just preceding this proposition, 
 
 therefore BD + FG = 2n . BD + 2, 
 
 or BD + FG - 2AH= 211 . BD ; 
 
 therefore RN = n . BD. 
 
 Accordingly the equation (15) above becomes 
 
 (16) 
 
 or 
 
 (17) 
 
 = n.BD.FS, 
 
 2P(a- 2 } 
 
 = n(a-2){(n-i)(a-2) 
 2P=n{(n- i)(a-2) + 
 
 Thus the double of any polygonal number must be divisible by its 
 side, and the quotient is the number arrived at by adding 2 to the product 
 of (side - i) and (number of angles - 2). 
 
 For a triangular number the quotient is n + i, and is therefore greater
 
 ON POLYGONAL NUMBERS 
 
 2 S9 
 
 than the side ; and, as the quotient increases by n - i for every increase 
 of i in the number of angles (a), it is always greater than the side. 
 
 We can therefore use the above formula (17) to find the number of 
 ways in which a given number P can be a polygonal number. Separate *P 
 into two factors in all possible ways, excluding i . zP. Take the smaller 
 factor as the side (). Then take the other factor, subtract 2 from it, 
 and divide the remainder by (n - i). If (n - i) divides it without a 
 remainder, the particular factors taken answer the purpose, and the quotient 
 increased by 2 gives the number of angles (a). If the second factor 
 diminished by 2 is not divisible by (n-i) without a remainder, the 
 particular division into factors is useless for the purpose. The number of 
 ways in which P can be a polygonal is the number of pairs of factors 
 which answer the purpose. There is always one pair of factors which will 
 serve, namely 2 and P itself. 
 
 The process of finding pairs of factors is shortened by the following 
 considerations. 
 
 2P=n\(n- i)(a- 2) + 2}; 
 therefore zPjn = 4 + an - a - 2n, 
 
 2(P-n) 
 
 and a = 2 + / : 
 
 n(n-i) 
 
 therefore not only 2 Pin but also 1 must be a whole number and, 
 
 n(n-i} 
 
 as a is not less than 3, 
 
 2(P-n) 
 n(n-i) 
 
 > or i, 
 
 and consequently 
 
 Thus in choosing values for the factor n we need not go beyond that 
 shown in the right-hand expression. 
 
 Example i. In what ways is 325 a polygonal number? 
 
 Here - i + V(i + &P) = - i + V(a6oi) = 50. Therefore n cannot be 
 greater than 25. Now 2 . 325 = 2 . 5 . 5 . 13, and the only possible values 
 for n are therefore 2, 5, 10, 13, 25. The corresponding values for a are 
 shown in the following table. 
 
 
 
 a 
 
 2 
 325 
 
 5 
 34 
 
 10 
 
 9 
 
 13 
 6 
 
 25 
 3 
 
 Example 2. P= 120. 
 
 - 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 8 
 6 
 
 10 
 
 12 
 
 '5 
 
 E 
 
 120 
 
 41 
 
 - 
 
 - 
 
 - 
 
 3 
 
 172
 
 CONSPECTUS OF THE ARITHMETICA 
 
 Equations of the first degree with one unknown. 
 
 i. 8. x + a = m (x + b). 
 
 i. 9. a-x = m(b~x). 
 
 i. 10. x + b = m (a x). 
 
 i. ii. x + b = m (x a). 
 
 i. 39. (a + x) b + (b + x) a = 2 (a + b} x, \ 
 
 or (a + b) x + (b + x) a = 2 (a + x} b, \ (a> b}. 
 
 or (a + b} x + (a + x) b = z (b + x) a, ) 
 
 Determinate systems of equations of the first degree. 
 
 i. i. x+y = a, x-y = b. 
 
 {i. 2. x +y = a, x = my. 
 
 i. 4. x-y = a, x = my. 
 
 i. 3. x+y a, x=my + b. 
 
 {i i , 
 i. 5- x+y = a. x + -y = 0. 
 m n 
 i. 6. x+y = a, x y = b. 
 li. 1 2. x 1 + x. 2 = y-L +y 2 = a, x l = my^ y l nx 2 , (x 1 > x$, y 1 >jy 2 ). 
 
 i. 15. x + a = m (y a), y + b = n (x b). 
 (i. 16. y + z = a, z + x = t>, x+y = c. 
 
 1 1. 18. y + z- x= a, 
 
 J L 19. y + z + 7v x = a, z + w + x y = b, w + x +y z c, 
 i x +y + z w d. 
 
 i. 20. x+y + z a, x + y = mz, y + z = nx. 
 
 i. 2 1. x =y -\ z . y = z + - x, z = a + - y, (x > y > z). 
 
 ***>*' aa ' A^ \ ^ / 
 
 /I \ / I \ /I 
 
 ii. 18*. x { x + a] + ( - z + f] = y { -y + b\ + [ x + a 
 \m ) \p ) ' \n j ) \ 
 
 ' " '-y + t>), 
 
 x +y -i z = a. 
 Determinate systems of equations reducible to the first degree 
 
 i. 29. x+y a, x 2 y^ = b. 
 
 * Probably spurious.
 
 CONSPECTUS OF ARITHMET1CA 261 
 
 i. 3 ! x = ni y> x* + y* n ( x + y)- 
 
 1.32. x = my, x 2 +y 2 = n (x-y). 
 
 i. 33. x = my, x* -f = n(x +y). 
 
 I. 34. x = my, x*-y 2 = n (x -y). 
 
 i. 34. Cor. i. x = my, xy = n(x+y). 
 Cor. 2. x = my, xy=n(x -y). 
 
 ( i. 35. x = my, y*=nx. 
 
 1 I. 36. x = my, j 2 = ny. 
 
 1.37. x=my,y* = n(x+y). 
 i. 38. x = my, y* = n(x -y). 
 
 1.38. Cor. x = my, x 2 =ny. 
 
 x = my, x* = nx. 
 x-my, x* = n(x+y). 
 x = my, xr = n(x y)- 
 n. 6*. x-y = a, x?-y* = xy + t>. 
 iv. 36. yz = m (y + z), zx = n(z + x), xy =p (x +-j). 
 
 Determinate systems reducible to equations of second degree. 
 1 1. 27. x+y = a, xy = b. 
 1 1. 3 . x -y = a, xy = b. 
 
 i. 28. x+y = < 
 Jiv. i. x s +y 3 = 
 \ iv. 2. x 3 -y 3 = a, x-y = b. 
 
 iv. 15. (y + z)x = a, (z + x)y = b, (x+y)z-c. 
 ( iv. 34. yz + (y + z) = a 2 - i , zx + (z + x) b* \ , xy + (x +y) = c 2 i . 
 
 iv. 37. yz = m(x+y + z), zx = n(x+y + z), xy=p(x+y+z). 
 Lemma to v. 8. yz a 2 , zx = IP, xy c*. 
 
 Systems of equations apparently indeterminate but really reduced, by 
 arbitrary assumptions, to determinate equations of the first degree, 
 i. 14. xy-m (x +y) [value of y arbitrarily assumed]. 
 
 I n. i*. (cf. I. 31.) x 2 +y 2 = m(x+y) 
 
 - n. 2*. (cf. i. 34.) x 2 -y 2 =m(x-y) j- [x assumed - 2y\. 
 n. 4*- (cf. i. 32-) x*+y*=m(x-y) 
 
 * n. 5*. (cf. i. 33.) x 9 -f = m (x +y) 
 
 11.7*. x t y z -m(x-y) + a [Diophantus assumes x -y = 2]. 
 
 i 22. x--x + -z=y--y + -x = z-^-z + -y [value of y assumed]. 
 m p n j m p /< 
 
 = z--z+-y = w--w + -z [value of v assumed]. 
 p n* q p L 
 
 * Probably spurious.
 
 262 CONSPECTUS OF ARITHMETICA 
 
 1.24. x+ - 
 
 i. 25. x + - (y + z + w) -y + -(z+w + x) 
 
 [value of y + z assumed]. 
 
 = z + - (w + x +y) = zv + -(x+y + z) 
 
 [value of y + z + w assumed]. 
 
 11.17*. (cf. i. 22.) x-( x + a ) + ( -z + c ] 
 \m ) \p ) 
 
 [ratio of x to y assumed]. 
 
 IV. 33. l l 
 
 
 Diophantus assumes l -y = i . 
 
 Indeterminate equations of the first degree. 
 
 Lemma to iv. 34. xy + (x +y) = a^ [Solutions iv dopurru. 
 
 iv. 35. xy - (x +y) = a\ y practically found in 
 
 iv. 36. xy = m (x +y) } terms of x.] 
 
 Indeterminate analysis of the second degree. 
 n. 8. 
 
 II. 9. 
 
 n. 10. x~ y 2 = a. 
 
 n. ii. x + a = u*, x + b = v*. 
 
 n. 12. a-x-u^, b-x = v*. 
 
 n. 13. x-a = u 2 , x-b = v*. 
 11.14 = 111.21. x+y a, x+ z z u 2 , y + z 2 = v 2 , 
 II. 15 = 111. 20. x+y = a, z 2 -x = u 2 , z*-y = v*. 
 
 n. 1 6. x = my, a 2 + x = u 2 , a 2 +y = v*. 
 
 11.19. x 2 -y 2 = m (y 2 - z 2 ). 
 
 ii. 20. x 2 + y = u 2 , y* + x = Z' 2 . 
 
 n. 21. x 2 -y = u*, y 2 -x = v 2 . 
 
 n. 22. x 2 + (x +y) = 2 , y 2 + (x +y) = v~. 
 
 n. 23. x* (x+y) = u 2 , y 2 - (x +y) = if: 
 
 ii. 24. (x +y) 2 + x = u 2 , (x +y) 2 +y = v 2 . 
 
 II. 25. (x+y) 2 -x=u 2 , (x+y) 2 -y = v>. 
 
 II. 26. xy + x = u 2 , xy+y = v 2 , u + v = a. 
 
 ii. 27. xy-x = u 2 , xy-y = v*, u + v = a. 
 
 II. 28. X 2 y 2 + x 2 = u 2 , X 2 y 2 +y 2 = v>. 
 
 n. 29. x 2 y 2 -x 2 = u 2 , x 2 y 2 -y 2 = tf. 
 
 n. 30. xy + (x +y) = u 2 , xy - (x +y) = v*. 
 * Probably spurious.
 
 CONSPECTUS OF ARITHMETICA 263 
 
 II. 31. xy + (x + y) = w 2 , xy~ (x +y) = v>, x+y = w l . 
 
 II. 32. y 2 + z = 2 , z- + x = v* t x^ +y = it?. 
 
 " 33- /-z^" 2 , z--x = v>, x?-y = w*. 
 
 ill. 34. x* + (x + y + z) = u\ y + (x +y + z) = tf, z* + (x +y + z) = w'. 
 
 II. 35. x 2 - (x +y + z) = u\ y - (x + y + z) = tf, z* -(x+y + z) = w>. 
 
 in. i* (x+y + z)-x t = u 2 , (x+y + z)-y* = v*, (x+y + z) -z* = w*. 
 
 I in. 2*. (x +y + zf + x = a 2 , (x +y + zf +y = i?, (x +y + z) 2 + z = -uf. 
 
 in. 3*. (x +y + zf -x = u\ (x +y + zf -y = v*, (x + y + z) 2 - z = W*. 
 
 in. 4* x - (x + y + zf = a 2 , y - (x +y + zf = v*, z- (x +y + z) 1 = w*. 
 
 in. 5. x+y + z=f*, y + z x-u 2 , z + xy = v t , x+y-z^w*. 
 
 in. 6. x +y + z = ?, y + z = w 2 , z + x - 
 
 in. 7. xy =y-z, y + z- if, z + x = 
 
 fin. 8. x +y + z + a = / 3 , y + z+ = *, 
 
 (in. 9. x +y + z -a = P, y+z a = u 3 , z + x-a = v*, x+y-a=w' 1 . 
 
 fin. 10. yz + a = 
 
 tin. ii. yz a = 
 
 fin. 12. yz + x = u 
 
 tin. 13. yz-x = u*, zxy = tf, xy-z-vf. 
 
 in. 14. yz + x 2 - u\ zx +y* = z? t xy + z 3 - to*. 
 
 fm. 15. yz + ( y + z) = u*, zx + (z-t-x) = v*, xy+(x+y) = w*. 
 
 tin. 16. yz-(y + z) = u*, zx-(z + x) = v t , xy-(x+y)=itr i . 
 
 fill. 17. xy + (x + y) = if*, 
 
 tin. 18. xy - (x +y) = w 2 , 
 
 in. 1 9. 
 
 (Xi + x z + x 3 + x 4 )* x 2 = , 
 
 (x 1 + Xs + x 3 + x t y x 3 = , 
 
 (^! + x 2 + x 3 + x 4 )* x 4 = 
 
 4. 
 5- 
 
 nv. 
 tiv. 
 
 iv. 13. Jf + i = f, y + i = *, a; H-j' + i = z/ 2 , ^ y + i = w*. 
 
 iv. i 4 . ^+y+s 2 =(^-jF 3 ) + (/- z 2 ) + (^-2 2 ) (^> 
 
 nv. 16. x+_y + z = / 2 > a^ +^ = a 2 , j>* + z = i? t & + x = ur. 
 
 tiv. 17. ^+^ + = ^, x a -y = ts i , y-z^tr 8 , 2 -o; = ^. 
 
 17. 
 
 iv. 19. ^0 + i = w 2 , z* + i = p 2 , ^y + i = a/ 2 . 
 
 iv. 20. #2*3 + i = r 2 , #3*1 + i = s 2 , x l x. 2 +i=f t , 
 
 x t x t + i - u*, x s x 4 + i=v*, x s x 4 + i = Z0 2 . 
 
 iv. 21. a:z=y, j:-j = 2 , ^-z = z^, y-z = ur (x>y>z). 
 
 {iv. 22. ^yz + ^^M 2 , xyz+y-z^, xyz + z = w*. 
 
 iv. 23. j -* = *, xyz-y = zt, xyz-z = ur. 
 
 * Probably spurious,
 
 264 CONSPECTUS OF ARITHMETICA 
 
 IV. 29- 
 
 iv. 3- x 2 +y 2 + z 2 + w 2 (x +y + z + w) = a. 
 
 iv. 31. x+y=i, (x + d)(y + b} = u 2 . 
 
 iv. 32. x +y + z = a, xy + z-u 2 , xy-z = tf. 
 
 iv. 39. x y = m (y - z), y + z = u 2 , z + x^v 2 , x+y = zv 2 . 
 
 iv. 40. x 2 -y 2 = m (y - z), y + z = u 2 , z + x=zr>, x+y = uf 
 
 v. 2. xz =y 2 , x + a = u 2 , y + a = v*, z + a = zv 2 . 
 
 yz + a tf 1 , zx + a v 2 , xy + a = w 2 . 
 v. 4. x - a = r"-, y - a = s 2 , z - a - f 2 . 
 
 yz a = u 2 , zx a = s , xy a = itr. 
 
 V. 6. X - 2 = r 2 , y - 2 = S 2 , Z - 2 = f 2 , 
 
 _yz _y - z u 2 , zx z x=zP, xy xy = 
 yz-x = u' 2 , zx y = v' 2 , xy-z = w' 2 . 
 Lemma i to v. 7. xy + x 2 +y* - u 2 . 
 
 v. 7. x 2 (x+y + z)= { J , / (x +y + z) = \ *\ , 
 
 ,, ,V 
 
 v. 8. yz(x +y + z) = l , 2 \ , zx (x +y + z) = 
 
 v. 9. (cf. II. ii.) x+y-i t 
 
 v. 1 1 . x +y + 2=1, x + a = u 2 , y + a^v 2 , z + a = w 2 . 
 
 v. 10. x+y= i, # + a = # 2 , y + b = v i . 
 
 v. 12. # +_y + z = i, x + a = u 2 , y + b-v 1 ^ z + c=w 2 . 
 
 v. 1 3. x +y + z = a, y + z = u 2 , z + x = v 2 , x +y = zv 2 . 
 
 v. 14. x+y + z + w = a, 
 
 x+y + z = s 2 ,y + z + w = t 2 , z + w + x-u 2 , 
 
 v. 21. X 2 y 2 z 2 + x 2 = u 2 , x 2 fz 2 +y 2 = v* t x 2 y 2 z 2 + z 2 = w 2 . 
 
 v. 22. X 2 y 2 z 2 -x 2 = u 2 , X 2 y 2 z 2 -y 2 ^zr', x 2 y 2 z 2 - z 2 = u?. 
 
 v. 23. x 2 - x 2 fz 2 = u\ y 2 - x 2 y 2 z 2 = z?, z 2 - X 2 y 2 z 2 = w 2 . 
 
 v. 24. y 2 z 2 + i = u\ z 2 x 2 +i=v*, x 2 }' 2 + i = or 8 , 
 
 v. 25. y 2 z 2 -i=u 2 , z 2 x 2 -i=v>, ^y- i=a<. 
 
 v. 26. i-y 2 z 2 = u 2 , i-z 2 x 2 = v i , i-x 2 y 2 = ur>. 
 
 v. 27. y 2 + z 2 + a = u 
 
 v. 28. y 2 + z 2 -a = u 
 
 v. 30. mx + ny=u 2 , 
 
 Lemma 2 to vi. 12. ax 2 + fr = u' 2 (where a + ^ = ^). 
 
 Lemma to vi. 1 5, ax 2 -t> = u 2 (where ad 2 - b = r is known)
 
 CONSPECTUS OF ARITHMETICA 265 
 
 [m. 15]. xy + x+y = u 2 , x+i=(y+i). 
 - = -~ z 
 
 [in. 16]. xy - (x +y) = u\ x-i=~ z (y- i). 
 
 1 2 / 
 
 [iv. 32]. *+i = -i(*-i). 
 
 [v. 2 1]. x 2 + i = u 2 , y 2 + i = v*, z 2 + i = w 2 . 
 
 Indeterminate analysis of the third degree, 
 iv. 3. x z y = u, xy = u 3 . 
 
 (iv. 6. * 3 
 I iv. 7. x 3 
 
 iv. 8. x +y s = u 3 , x+y = u 
 iv. 9. x +y 3 = u, x+y = u 3 
 iv. 10. x 3 +y 3 x+y 
 
 j iv. ii. -> = * --y^ the same m 
 
 UV. 12. 3?+y =}r + X) 
 
 Really reducible to second 
 degree. 
 
 iv. 1 8. 
 
 iv. 24. x+y = a, xy = tf u. 
 
 iv. 25. x+y + z = a, xyz = {(x-y) + (x-z) + (y-z)} 3 (x>y>z). 
 (iv. 26. 
 tiv. 27. 
 
 iv. 28. 
 
 iv. 38. (x+y + z)x=-%u(u+ i), 
 
 v. 15. (x +y + z) 3 + x = u 3 , (x+y + sf+y^v 3 , (x +y 4- z) 3 -f z = 
 
 v. 16. (x+y + z) 3 -x = u 3 , (x+y + zy-y = i?, (x +y + z) 3 - z = 
 
 v. 17. x-(x+y + z) 3 = u*, y-(x+y + z) 3 = v>, z-(x+y + z) 3 = 
 
 v. 18. x+y + z = t 2 , (x+y + z) 3 + x = u 2 , (x+y + z) 3 +y = v>, 
 
 (x +y + z) 
 
 v. 19. x+y + z = t\ (x+y + z) 3 -x = u 2 , (x + y + z) 3 -y = v>, 
 
 v.iga. x+y + z = f 2 , x-(x+y + z) 3 = u\ y - (x +y + z) 3 = v*, 
 v.igb. x+y + z = a, (x +y + z) 3 + x = 
 
 (x +y + z) 3 
 
 z = a, (x +y + z) 3 - x = u\ 
 
 v. 20. x + y + z = - , x - (x + y + z) 3 = u 2 , y - (x + y + z) 3 = tf, 
 
 [iv. 8]. x-y=i, 
 [iv. 9, 10]. x?+
 
 266 CONSPECTUS OF ARITHMETICA 
 
 [v. 15]. x 3 +/ + z 3 - 3 = u\ 
 [v. 1 6]. 3 - (x 3 +y + z 3 ) = u 2 . 
 [v. 1 7]. x 3 +y + z 3 + 3 = u 2 . 
 
 Indeterminate analysis of the fourth degree. 
 v. 29. x 4 +y* + z 4 = u 2 . 
 [v. 1 8]. x 2 +y 2 + z 2 3 = u 4 . 
 
 Problems of constructing right-angled triangles with sides in rational 
 numbers and satisfying various other conditions. 
 
 [N.B. I shall use x, y for the perpendicular sides and z for the 
 hypotenuse in all cases, so that the condition x 2 +y 2 = z 2 must be under- 
 stood to apply in every case in addition to the other conditions specified.] 
 
 Lemma to v. 7. xy x 1 y 1 = x a y a . 
 fvi. i. z x = u 3 , zy = v*. 
 Ivi. 2. z + x = tt 3 , z+y i^. 
 j vi. 3. \xy + a = u 2 . 
 JVI. 4. %xy-a = u 2 . 
 I VI. 5. a-^xy = u 2 . 
 fvi. 6. %xy + x = a. 
 tvi. 7. \xy-x-a. 
 fvi. 8. \xy + (x +y) - a. 
 tvi. 9. \xy-(x+y) = a. 
 fvi. 10. %xy + (x + z) = a. 
 tvi. ii. \xy - (x + z) = a. 
 
 Lemma i to vi. 12. x = u 2 , x-y = v 2 , *xy + y = w 2 . 
 
 (VI. 12. 
 
 tvi. 13. 
 fvi. 14. 
 tvi. 15. 
 
 vi. 1 6. + i) = 
 fvi. 17. \xy + z 
 tvi. 1 8. \xy-\-z 
 
 (vi. 1 9. \xy + x = u 2 , x +y + z = v*. 
 (vi. 20. ^xy + x = j^, x+y + z v 2 . 
 (vi. 21. x +y + 2 = u 2 , \xy + (x +y + z) - v 3 , 
 tvi. 22. x+y + z = u 3 , \xy + (x + y + z) = v 2 . 
 
 vi. 23. z 2 = u 2 + u, z 2 /x = 
 
 vi. 24, z = u 3 + u, x-n 3 
 [vi. 6, 7]. (\x} 2 + ^mxy = 
 [vi. 8, 9]. 
 
 [vi. 10, n], {| (z + x)} 2 + \rnxy = u 2 . 
 [vi. 12]. y + (x-y).%xy = u\ x = v 2 (x>y). 
 [vi. 14, 15]. u 2 zx - %xy . x (z - x) = ir ( 2 <or>
 
 SUPPLEMENT 
 
 ADDITIONAL NOTES, THEOREMS AND PROBLEMS BY FERMAT, 
 TO WHICH ARE ADDED SOME SOLUTIONS BY EULER 
 
 I HAVE generally referred to the notes of Ferrnat, and allied propositions 
 of his, on the particular problems of Diophantus which were the occasion 
 of such notes, illustrations or extensions ; but there are some cases where 
 the notes would have been of disproportionate length to give in the places 
 where they occur. Again, some further explanations and additional 
 theorems and problems given by Fermat are not in the notes to Diophantus 
 but elsewhere, namely in his correspondence or in the Doctrinae Analyticae 
 Inventum Novum of Jacques de Billy " based on various letters sent to 
 him from time to time by Pierre de Fermat " and originally included at the 
 beginning of the 2nd (1670) edition of Bachet's Diophantus (the Inrentum 
 Novum is also published, in a free French translation by Tannery, in 
 Oeurres de Fermat, Vol. in. pp. 323-398). Some of these theorems and 
 problems are not so closely connected with particular problems in Dio- 
 phantus as to suggest that they should be given as notes in one place 
 rather than another. In these circumstances it seemed best to collect the 
 additional matter at the end of the book by way of Supplement. 
 
 In the chapter on the Porisms and other assumptions in Diophantus 
 (pp. 106-110 above) I quoted some famous propositions of Fermat on the 
 subject of numbers which are the sums of two, three or four square numbers 
 respectively. The first section of this Supplement shall be devoted to 
 completing, so far as possible, the story of Fermat's connexion with these 
 theorems. 
 
 SECTION I. 
 
 ON NUMBERS SEPARABLE INTO INTEGRAL SQUARES. 
 
 As already noted, Fermat enunciated, on Diophantus iv. 29, a very 
 general theorem of which one part states that Every number is either a 
 square or the sum of /wo, three or four squares. We shall return to this 
 later, and shall begin with the case of numbers which are the sum of 
 two squares.
 
 268 SUPPLEMENT 
 
 i. On numbers which are the sum of two squares. 
 
 I may repeat the beginning of the note on in. 19 already quoted (p. 106). 
 
 " A prime number of the form 4^+1 is the hypotenuse of a right-angled 
 triangle in one way only, its square is so in two ways, its cube in three, its 
 biquadrate in four ways, and so on ad infinitum. 
 
 " The same prime number 472+1 and its square are the sum of two 
 squares in one way only, its cube and its biquadrate in two ways, its fifth 
 and sixth powers in three ways, and so on ad infinitum. 
 
 " If a prime number which is the sum of two squares be multiplied into 
 another prime number which is also the sum of two squares, the product 
 will be the sum of two squares in two ways ; if the first prime be multiplied 
 into the square of the second prime, the product will be the sum of two 
 squares in three ways ; if the first prime be multiplied into the cube of the 
 second, the product will be the sum of two squares in four ways, and so on 
 ad infinitum." 
 
 Before proceeding further with this remarkable note, it is natural to 
 ask how Fermat could possibly have proved the general proposition that 
 (a) Every prime number of the form 4^+1 is the sum of two square 
 numbers, which was actually proved by Euler 1 . Fortunately we have 
 in this case a clear statement by Fermat himself of the line which his 
 argument took. In his " Relation des nouvelles decouvertes en la science 
 des nombres" sent by Fermat to Carcavi and shortly after (14 August, 
 1659) communicated by the latter to Huygens, Fermat begins by a refer- 
 ence to his method of proof by indefinite diminution (descente infinie or 
 indefinie) and proceeds 2 thus: "I was a long time before I was able to 
 apply my method to affirmative questions because the way and manner 
 of getting at them is much more difficult than that which I employ with 
 negative theorems. So much so that, when I had to prove that every 
 prime number of the form 4^+1 is made up of two squares, I found myself 
 in a pretty fix. But at last a certain reflection many times repeated gave 
 me the necessary light, and affirmative questions yielded to my method, 
 with the aid of some new principles by which sheer necessity compelled me 
 to supplement it. This development of my argument in the case of these 
 affirmative questions takes the following line : if a prime number of the 
 form 4 + i selected at random is not made up of two squares, there will 
 exist another prime number of the same sort but less than the given 
 number, and again a third still smaller and so on, descending ad infinitum, 
 until you arrive at the number 5 which is the smallest of all numbers of 
 
 1 Novi Commentarii Academiae Petropolitanae 1752 and 1753, Vol. iv. (1758), 
 pp. 3-40, 1754 and 1755, Vol. v. (1760), pp. $-cfi=Coimentationes arithmetics 
 colledae, 1849, I. pp. 155-173 and pp. 210-233. 
 
 2 Oeuvres de Fermat, n. p. 432.
 
 THEOREMS AND PROBLEMS BY FERMAT 269 
 
 the kind in question and which the argument would require not to be made 
 up of two squares, although, in fact, it is so made up. From which we 
 are obliged to infer, by reductio ad absurdum, that all numbers of the kind 
 in question are in consequence made up of two squares." 
 
 The rest of the note to Diophantus in. 19 is as follows. 
 
 " From this consideration it is easy to deduce a solution of the problem 
 
 " To find in how many ways a given number can be the hypotenuse of 
 a right-angled triangle. 
 
 "Take all the primes of the form 4^ + i, e.g. 5, 13, 17, which measure 
 the given number. 
 
 "If powers of these primes measure the given number, set out the 
 exponents of the powers ; e.g. let the given number be measured by the 
 cube of 5, the square of 13, and by 17 itself but no other power of 17; 
 and set out the exponents in order, as 3, 2, i. 
 
 " Take now the product of the first of these and twice the second, and 
 add to the product the sum of the first and second : this gives 1 7. Multiply 
 this by twice the third exponent and add to the product the sum of 17 and 
 the third exponent : this gives 52, which is the number of the different right- 
 angled triangles which have the given number for hypotenuse. [If a, b, c be 
 the exponents, the number of the triangles is ^abc + 2 (be + ca + ab] + a + b + c.] 
 We proceed similarly whatever the number of divisors and exponents. 
 
 " Other prime factors which are not of the form 4 + i, and their 
 powers, do not increase or diminish the number of the right-angled triangles 
 which have the given hypotenuse. 
 
 " PROBLEM i. To find a number which is a hypotenuse in any assigned 
 number of ways. 
 
 "Let the given number of times be 7. We double 7 : this gives 14. 
 Add i, which makes 15. Then seek all the prime numbers which measure 
 it, i.e. 3 and 5. Next subtract i from each and bisect the remainders. 
 This gives i and 2. [In explanation of the process it is only necessary to 
 observe that, for example, 2 \\abc + 2 (bc + ca + ab) + a + b + c] + i is equal 
 to (2a + i)(2 + i)(2c+ i), and so on.] Now choose as many prime 
 numbers of the form 4^ + i as there are numbers in the result just arrived 
 at, i.e. in this case two. Give to these primes the exponents i, 2 re- 
 spectively and multiply the results, i.e. take one of the primes and multiply 
 it into the square of the other. 
 
 " It is clear from this that it is easy to find the smallest number which 
 is the hypotenuse of a right-angled triangle in a given number of ways." 
 
 [Fermat illustrates the above further in a letter of 25 December 1640 
 to Mersenne 1 . 
 
 To find a number which is the hypotenuse of 367 different right-angled 
 triangles and no more. 
 
 1 Oenvres de Fermat, II. pp. 214 sq.
 
 2 7 
 
 SUPPLEMENT 
 
 Double the number and add i ; this gives 735. Take all the divisors 
 which are prime numbers : these are 3, 5, 7, 7. Subtract i from each and 
 then divide by 2 ; this gives i, 2, 3, 3. We have then to take four prime 
 numbers of the form 4/2 + i and give them i, 2, 3, 3 respectively as ex- 
 ponents. The product of these powers is the number required. 
 
 To find the least such number, we must take the four least primes of the 
 form 4 + i, i.e. 5, 13, 17, 29, and we must give the smallest of them, 
 in order, the largest exponent ; i.e. we must take 5 3 , i3 3 , if and 29 in this 
 case, and the product of these four numbers is the least number which is 
 the hypotenuse of 367 right-angled triangles and no more. 
 
 If the double of the given number + i is a prime number, then there is 
 only one possible divisor. Suppose the given number is 20; the double 
 of \\.plus i is 41. Subtracting unity and bisecting, we have 20, so that the 
 number to be taken is some prime number of the form 4 + i to the power 
 of 20.] 
 
 " PROBLEM 2. To find a number which shall be the sum of two squares 
 in any assigned number of ways. 
 
 " Let the given number be 10. Its double is 20, which, when separated 
 into its prime factors, is 2.2.5. Subtract i from each, leaving i, i, 4. 
 Take three different prime numbers of the form 4^+1, say 5, 13, 17, and 
 multiply the biquadrate of one (the exponent being 4) by the product 
 of the other two. The result is the required number. 
 
 "By means of this it is easy to rind the smallest number which is the 
 sum of two squares in a given number of ways. 
 
 " In order to solve the converse problem, 
 
 " To find in how many ways a given number is the sum of two squares, 
 "let the given number be 325. The prime factors of the form 4 + i 
 contained in this number are 5, 13, the latter being so contained once only, 
 the former to the second power. Set out the exponents 2, i. Multiply 
 them and add to the product the sum of the two : this gives 5. Add i, 
 making 6, and take the half of this, namely 3. This is the number of ways 
 in which 325 is the sum of two squares. 
 
 "If there were three exponents, as 2, 2, i, we should proceed thus. 
 Take the product of the first two and add it to their sum : this gives 8. 
 Multiply 8 into the third and add the product to the sum of 8 and the 
 third: this gives 17. Add i, making 18, and take half of this or 9. This 
 is the number of ways in which the number taken in this second case is 
 the sum of two squares. [If a, b, c be the three exponents, the number 
 of ways is \ {abc + (be + ca + ab] + (a + b + c) + i } provided that the number 
 represented by this expression is an integer.] 
 
 "If the last number which has to be bisected should be odd, we 
 must subtract i and take half the remainder.
 
 THEOREMS AND PROBLEMS BY FERMAT 271 
 
 " But suppose we are next given the following problem to solve : 
 
 "To find a whole number which, when a given number is added to it, 
 becomes a square, and which is the hypotenuse of any assigned number of 
 right-angled triangles. 
 
 " This is difficult. Suppose e.g. that a number has to be found which 
 is a hypotenuse in two ways and which, when 2 is added to it, becomes 
 a square. 
 
 " The required number will be 2023, and there are an infinite number 
 of others with the same property, as 3362 etc." 
 
 2. On numbers which cannot be the sum of two squares. 
 
 In his note on Diophantus v. 9 Fermat took up a remark of Bachet's 
 to the effect that he believes it to be impossible to divide 21 into two 
 squares because " it is neither a square nor by its nature made up of two 
 squares." Fermat's note was: "The number 21 cannot be divided into 
 two squares (even) in fractions. That I can easily prove. And generally 
 a number divisible by 3 which is not also divisible by 9 cannot be divided 
 into two squares either integral or fractional" 
 
 He discusses the matter more generally in a letter of August 1640 
 to Roberval 1 . 
 
 "I have made a discovery a propos of the i2th [gth] proposition of 
 the fifth Book of Diophantus (that on which I have supplied what Bachet 
 confesses that he did not know and at the same time restored the corrupted 
 text, a story too long to develop here). I need only enunciate to you my 
 theorem, while reminding you that I proved some time ago that 
 
 "A number of the form qn i is neither a square nor the sum of two 
 squares, either in integers or fractions." 
 
 [This proposition was sent by Mersenne to Descartes, on 22 March 
 1638, as having been proved by Fermat] 
 
 " For the time I rested there, although there are many numbers of the 
 form 4 + i which are not squares or the sums of squares either, e.g. 21, 
 33 77> etc -> a f act which made Bachet say on the proposed division of 21 
 into two squares 'It is, I believe, impossible since 21 is neither a square 
 nor by its nature made up of two squares,' where the word rear (I think) 
 clearly shows that he was not aware of the proof of the impossibility. 
 This 1 have at last discovered and comprehended in the following general 
 proposition. 
 
 " If a gi^n number is divided by the greatest square which measures it, 
 and the quotient is measured by a prime number of the form 4*1-1, the given 
 number is neither a square nor the sum of two squares either integral or 
 fractional. 
 
 1 Oftevres de Fermat, II. pp. 203-4.
 
 272 SUPPLEMENT 
 
 " EXAMPLE. Let the given number be 84. The greatest square which 
 measures it is 4, and the quotient is 21 which is measured by 3 or by 
 7, both 3 and 7 being of the form 4^-1. I say that 84 is neither a square 
 nor the sum of two squares either integral or fractional. 
 
 "Let the given number be 77. The greatest square which measures it 
 is i, and the quotient is 77 which is here the same as the given number 
 and is measured by u or by 7, each of these numbers being of the form 
 4# i. I say that 77 is neither a square nor the sum of two squares, 
 either in integers or fractions. 
 
 " I confess to you frankly that I have found nothing in the theory of 
 numbers which has pleased me so much as the proof of this proposition, 
 and I shall be glad if you will try to discover it, if only for the purpose 
 of showing me whether I think more of my discovery than it deserves. 
 
 " Following on this I have proved the following proposition, which 
 is of assistance in the finding of prime numbers. 
 
 11 If a number is the sum of two squares prime to one another, I say 
 that it cannot be divided by any prime mtmber of the form ^n i. 
 
 "For example, add i, if you will, to an even square, say the square 
 10 ooo ooo ooo, making 10000000001. I say that 10000000001 cannot 
 be divided by any prime number of the form 4// i, and accordingly, 
 when you would try whether it is a prime number, you need not divide by 
 3, 7, ii etc." 
 
 (The theorem that Numbers which are the sum of two squares prime to 
 one another have no divisors except such as are likewise the sum of two squares 
 was proved by Euler 1 .) 
 
 3. Numbers (i) which are always, (2) which can never be, the sum 
 of three squares. 
 
 (i) The number which is double of any prime number of the form 
 8- i is the sum of three squares (Letter to Kenelm Digby of June i658) 2 . 
 
 E.g. the numbers 7, 23, 31, 47 etc. are primes of the form 8n i ; the 
 doubles are 14, 46, 62, 94 etc. ; and the latter numbers are the sums of 
 three squares. 
 
 Fermat adds " I assert that this proposition is true, though I do so in 
 the manner of Conon, an Archimedes not having yet arisen to assert it 
 or prove it." 
 
 Lagrange 3 remarks that he has not yet been able to prove the pro- 
 position completely. The form 8 i reduces to one or other of the three 
 
 1 Novi Commentarii Acad. Petropol. 1752 and 1753, Vol. IV. (1758), pp. 3-40 = 
 Commentationes arithmeticae, I. pp. 155-173. 
 
 2 Oeuvres de Fermat, II. pp. 402 sqq. 
 
 3 "Recherches d'Arithmetique" in Berlin Mtmoires 1773 and 1 775 = Oeuvres de 
 Lagrange, III. p. 795.
 
 THEOREMS AND PROBLEMS BY FERMAT 273 
 
 forms 24n - i, 24/2 + 7, 24;* + 15, of which the first two only are primes. 
 Lagrange had previously proved that every prime number of the form 
 24 + 7 is of the form x 2 + 6y>. The double of this is 2x 2 + 1 2 /, and 
 
 that is, 2x* + i2/ 2 is the sum of three squares. 
 
 The theorem was thus proved for prime numbers of the form Bn - i, 
 wherever n is not a multiple of 3, but not for prime numbers of the form 
 2472 i. 
 
 Legendre 1 , however, has the theorem that Every number which is the 
 double of an odd number is the sum of three squares. 
 
 (2) No number of the form 24/2+ 7 or \ m (z^n + 7) can be the sum 
 of three squares. 
 
 This theorem is substantially stated in Fermat's note on Dioph. v. n. 
 We may, as a matter of fact, substitute for the forms which he gives the 
 forms 8/2 + 7 and 4'" (8n + 7) respectively. 
 
 Legendre 2 proved that numbers of the form Sn + 7 are the only odd 
 numbers which are not the sum of three squares. 
 
 4. Every number is either a square or the sum of two, three or 
 four squares. 
 
 This theorem is also mentioned in the " Relation des nouvelles de- 
 couvertes en la science des nombres " already quoted, as a case to which 
 Fermat ultimately found himself able to apply the method of proof by 
 descente. He says 3 that there are some other problems which require new 
 principles in order to enable the method of descente to be applied, and the 
 discovery of such new principles is sometimes so difficult that they cannot 
 be arrived at except after very great trouble. 
 
 " Such is the following question which Bachet on Diophantus admits 
 that he could never prove, and as to which Descartes in one of his letters 
 makes the same statement, going so far as to admit that he regards it as 
 so difficult that he does not see any means of solving it. 
 
 "Every number is a square or the sum of two, three or four squares. 
 
 " I have at last brought this under my method, and I prove that, if 
 a given number were not of this nature, there would exist a number smaller 
 than it which would not be so either, and again a third number smaller 
 than the second, etc. ad infinitum ; whence we infer that all numbers are 
 of the nature indicated." 
 
 In another place (letter to Pascal of 25 September, i654) 4 , after quoting 
 the more general proposition, including the above, that every number is 
 
 1 Legendre, Zahlentheorie, tr. Maser, I. p. 387. 
 
 2 Ibid. p. 386. 
 
 3 Oeuvres de Fermat, n. p. 433. 
 
 4 Ibid. p. 313. 
 
 H. n. 18
 
 274 SUPPLEMENT 
 
 made up (i) of one, two, or three triangles, (2) of one, two, three or four 
 squares, (3) of one, two, three, four or five pentagons, and so on adinfinitum, 
 Fermat adds that " to arrive at this it is necessary 
 
 1 i ) To prove that every prime number of the form 4^+1 is the sum 
 of two squares, e.g. 5, 13, 17, 29, 37, etc. ; 
 
 (2) Given a prime number of the form 4^+1, as 53, to find, by a 
 general rule, the two squares of which it is the sum. 
 
 (3) Every prime number of the form 3^+1 is of the form x 2 + 3jr, 
 e.g. 7, 13, 19, 31, 37, etc. 
 
 (4) Every prime number of the form 8n + i or 8>n + 3 ts of the form 
 x*+2y 2 , e.g. n, 17, 19, 41, 43. etc - 
 
 (5) There is no rational right-angled triangle in whole numbers the 
 area of which is a square. 
 
 "This will lead to the discovery of many propositions which Bachet 
 admits to have been unknown to him and which are wanting in Diophantus. 
 
 " I am persuaded that, when you have become acquainted with my 
 method of proof in this kind of proposition, you will think it beautiful, and 
 it will enable you to make many new discoveries, for it is necessary, as you 
 know, that multi pertranseant ut augeatur scientia [Bacon]." 
 
 Propositions (3) and (4) will be mentioned again, and a full account 
 will be given in Section in. of this Supplement of Fermat's method, or 
 methods, of proving (5). 
 
 The main theorem now in question that every integral number is the 
 sum of four or fewer squares was attacked by Euler in the paper 1 (1754 
 1755) in which he finally proved the proposition (i) above about primes 
 of the form 4/2 + i ; but, though he obtained important results, he did not 
 then succeed in completing the proof. Lagrange followed up Euler's 
 results and finally established the proposition in i77o 2 . Euler returned 
 to the subject in 1772 ; he found Lagrange's proof long and difficult, and 
 set himself to simplify it 3 . 
 
 (The rest of the more general theorem of Fermat quoted above, the 
 portion of it, that is, which relates to numbers as the sum of n or fewer 
 n-gonal numbers, was proved by Cauchy 4 .) 
 
 1 Novi Commentarii Acad. Petropol. for 1754-5, Vol. V. (1760), pp. 3-58= Cow- 
 mentatioiies arithmeticae collcctae, 1849, * PP- 2IO ~ 2 33- 
 
 2 Nouveanx Memoires de V Acad. Koy. des Sciences de Berlin, annee 1770, Berlin 1772, 
 pp. 123-133= Oeuvres de Lagrange, in. pp. 187-201: cf. Wertheim's account in his 
 Diophantus, pp. 324-330. 
 
 3 " Novae demonstrationes circa resolutionem numerorum in quadrata," Acta Erudit. 
 Lips. 1773, p. 193; Acta Petrop. I. II. 1775, p. 48; Comment, arithm. I. pp. 538-548. 
 
 4 Cauchy, "Demonstration du theoreme general dc Fermat sur les nombres polygones," 
 Oeuvres, ii e Serie, Vol. vi. pp. 320-353. See also Legendre, Zahletitheorie, tr. Maser, 
 n. pp. 332-343-
 
 THEOREMS AND PROBLEMS BY FERMAT 275 
 
 Under this heading may be added the further proposition that 
 
 " Any number whatever of the form 8 - i can only be represented as the 
 
 sum of four squares^ not only in integers (as others may have seen) but .in 
 
 fractions also, as I promise that I will prove 1 ." 
 
 5. On numbers of the forms x?+2y*, x* + 3^, y* + 5^* respectively. 
 (i) Every prime number of the form 8 + i or Sn + 3 is of the form 
 
 X*+2f. 
 
 This is one of the theorems enunciated in the letter of 25 Sept., 1654, 
 to Pascal 2 and also in the letter of June, 1658, to Kenelm Digby 3 . 
 
 [In a paper of 1754 Euler says that he does not yet see his way to 
 prove either part of the theorem 4 . In 1759 he says 5 he can prove the 
 truth of the theorem for a prime number of the form Sn + i, but not for 
 a prime of the form Bn + 3. Later, however, he proved it for prime 
 numbers of both forms 6 . Lagrange 7 also proved it for primes of the form 
 
 (2) Every prime number of the form $n + i is of the form x? + 3^. 
 The theorem is stated in the same two letters to Pascal and Digby 
 
 respectively. 
 
 Lagrange naturally quotes it as "All prime numbers of the form 6+ i 
 are of the form x* + 3^," for of course yi + i is not a prime number unless 
 n is even. 
 
 The proposition was proved by Euler 8 . Lagrange proved 9 (a) that all 
 prime numbers of the form i2# 5 are of the form x* + 3^, (b) that all 
 prime numbers of the form \zn i are of the form 3^ }?, and (c) that 
 all prime numbers of the form i2+ i are of both the forms ^ + 3^ and 
 **-3/. 
 
 (3) No number of the form 3^-1 can be of the form c? + 3^*. 
 
 In the "Relation des nouvelles de'couvertes en la science des nombres 10 " 
 Fermat says that this was one of the negative propositions which he proved 
 by his method of descente, 
 
 1 Letter to Mersenne of Sept. or Oct. 1636, Oeuvres de Fermat, II. p. 66. 
 
 2 Oeuvres de Fermat, II. p. 313. 
 
 3 Ibid. II. p. 403. 
 
 4 " Specimen de usu observationum in mathesi pura (De numeris formae 2aa + bb)" in 
 Novi Commentarii Acad. Petrop. 1756-7, Vol. VI. (1761), pp. 185-2 30 ^Comment. 
 arithm. I. pp. 174-192. 
 
 5 Nmi Commentarii Acad. Petrop. 1760-1, Vol. vill. (1763). PP- "6-8= Comment. 
 arithm. I. p. 296. 
 
 6 Commentationes arithmetics, II. p. 607. 
 
 7 " Recherches d'Arithmetique " in Ofuvres de Lagrange, III. pp. 776, 784. 
 
 8 " Supplementum quorundam theorematum arithmeticorum, quae in nonnullis de- 
 monstrationibus supponuntur (De numeris formae aa + $bb)" in Novi Commetit. Acad. 
 P^trop. 1760-1, Vol. VIII. (1763), pp. 105-1 28 = Comment. arithm. I. pp. 287-296. 
 
 9 Op. fit., Oeuvres de Lagrange, III. pp. 784, 791. 
 
 10 Oeuvres de Fermat, n. p. 431. 
 
 18 2
 
 276 SUPPLEMENT 
 
 (4) If two prime numbers ending in either 3 o r 7 which are also of the 
 form 4 + 3 are multiplied together, the product is of the form x 2 + $y 2 . 
 
 This theorem also is enunciated in the letter of June, 1658, to Kenelm 
 Digby. Fermat instances 3, 7, 23, 43, 47, 67 etc. as numbers of the kind 
 indicated. Take, he says, two of these, e.g. 7 and 23. The product 161 
 will be the sum of a square and 5 times another square, namely 81 + 5.16. 
 
 He admits, however, that he has not yet proved the theorem generally : 
 " I assert that this theorem is true generally, and I am only waiting for 
 a proof of it. Moreover the square of each of the said numbers is the sum of 
 a square and 5 times another square : this, too, I should like to see proved." 
 
 Lagrange proved this theorem also 1 . He observes that the numbers 
 described are either of the form 20/2 + 3 or of the form 20/2+7, an d he 
 proves that all prime numbers of these forms are necessarily of the form 
 2x? 2xy + 3_y 2 . He has then only to prove that the product of two 
 numbers of the latter form is of the form x 2 + $y 2 . 
 
 This is easy, for 
 
 = (2xx + xy +yx' + $yy') 2 + 5 (xy' -yx') 2 . 
 
 6. Numbers of the forms x 2 - 2y 2 and 2x 2 -y 2 . 
 
 Fermat's way of expressing the fact that a number is of one of these 
 forms is to say that it is the sum of, or the difference between, the two smaller 
 sides, i.e. the perpendicular sides, of a right-angled triangle. Like Diophantus, 
 he " forms " a rational right-angled triangle from two numbers x, y, taking 
 as the three sides the numbers x 2 + y 2 , x 2 -y 2 , 2xy respectively. The sum 
 therefore of the perpendicular sides is x 2 + 2xy -y 2 or (x +y) 2 - 2y 2 , and 
 their difference is either x 2 - 2xy -y 2 or zxy - (x 2 -y 2 ), that is, either 
 (x-y) 2 - 2 y 2 or 2y 2 -(x-y} 2 . 
 
 The main theorem on the subject of numbers of these forms is, as 
 a matter of fact, contained, not in a letter of Fermat's, but in two letters 
 of Frenicle to Fermat dated 2nd August and 6th Sept., 1641, respectively 2 . 
 It is, however, clear (cf. the letter in which Fermat had on isth June, 1641, 
 propounded to Frenicle a problem on such numbers) that the theorem was 
 at any rate common property between the two. 
 
 Frenicle's two statements of the theorem are as follows : 
 
 " Every prime number of the form 8n i is the sum of the two smaller 
 sides of a (right-angled) triangle, and every number which is the sum of the 
 two smaller sides of a (right-angled) triangle with sides prime to one another 
 is of the form 8/2 i." 
 
 "Every prime number of the form 8n i, or which is the product 
 of such prime numbers exclusively, is the difference between the two 
 smaller sides of an infinite number of primitive right-angled triangles." 
 
 1 Op. dt., Oeuvres de Lagrange, II. pp. 784, 788-9. 
 
 2 Oeuvres de Fermat, \\. pp. 231, 235.
 
 THEOREMS AND PROBLEMS BY FERMAT 277 
 
 Lagrange 1 quotes the theorem in the form 
 
 All prime numbers of the form 8n i are of the form y 2 - 2/ 2 . 
 
 Lagrange himself proves 2 that all prime numbers of the form Sn - i are 
 of both the forms x 2 - 2y 2 and 2x 2 -y 2 , and observes 3 that this theorem is more 
 general than that of Fermat so far as prime numbers of the form Sn - i are 
 concerned. This, however, seems scarcely correct if the further explanations 
 given by Frenicle are taken into account. For Frenicle shows clearly, 
 in the second of the two letters referred to 4 , that he was fully alive to 
 the fact that numbers which are of the form x 2 - zy 2 are also of the form 
 2X 2 -y 2 ; and indeed it is obvious that he was aware that 
 x 2 -2y*=2 (x +yf -(x+ 2y) 2 . 
 
 Lagrange proved in addition 5 that 
 
 Every prime number of the form 8 + i is at the same time of the three 
 forms x 2 + 2y 2 , x 2 - 2y 2 , 2X 2 -y 2 . 
 
 This is, I think, really included in Frenicle's statements when combined 
 with Fermat's theorem (i) above to the effect that every prime number 
 of the form Sn + i is of the form x 2 + 2y 2 . 
 
 The problem propounded by Fermat to Frenicle in connexion with the 
 numbers now under consideration was: 
 
 Given a number, to find in how many ways it can be the sum of the two 
 smaller sides of a right-angled triangle. 
 
 Frenicle replied that this involved also the problem of finding a number 
 which will be the sum of the two smaller sides of a right-angled triangle in 
 an assigned number of ways and no more, and tried, but unsuccessfully 6 , 
 to bring these problems under a rule corresponding to that by which 
 Fermat found the number of ways in which a prime number of the form 
 4/z+ i can be the hypotenuse of a right-angled triangle (see p. 269 above), 
 but with a prime number of the form Sn i substituted for the prime 
 number of the form 4^+1. I cannot find that Fermat ever communicated 
 his own solution, at all events in the correspondence which we possess. 
 
 SECTION II. 
 
 EQUATION X 2 - Ay 2 - I. 
 
 History of the equation up to Fermat's time. 
 
 Fermat was not the first to propound, or even to discover a general 
 method of solving, the problem of finding any number of integral values of 
 x, y satisfying the above equation, wherein A is any integral number not 
 a square. But Fermat rediscovered the problem and was perhaps the first 
 
 1 Op. tit., Oeuvres de Lagrange, III. p. 775. 2 Ibid. p. 784. 3 Ibid. p. 788. 
 
 4 Oeuvres de Fermat, n. pp. -235-240. 
 
 5 Op. cit., Oeuvres de Lagrange, III. p. 790. 
 
 6 See Oeuvres de Fermat, n. pp. 231, 238 sqq.
 
 278 SUPPLEMENT 
 
 to assert that the general solution is always possible whatever be the 
 (non-square) value of A. The equation has a history of over 2000 years, 
 and that history, even in outline, requires, as it has now obtained, a book 
 to itself. This note will therefore be confined, practically, to recalling, in 
 the briefest possible way, the recorded stages anterior to Fermat, and then 
 to setting out somewhat fully the passages in Fermat's writings which throw 
 the most light on his connexion with the subject. 
 
 The Pythagoreans. 
 
 We have seen (p. 117 above) that the Pythagoreans had already 
 discovered a general solution of a particular equation of this type, namely 
 
 *-y=i, 
 
 by which all the successive values of x, y satisfying the equation were 
 ascertained. If x =/, y = q satisfies the equation 2X 2 y~ = + i, they proved 
 that the equation zx 2 -y* = + i is satisfied by 
 
 the equation 2X 2 y* = + i again by 
 
 and so on. As /= r, q= i satisfies 2x*-y* - + i, we have all the suc- 
 cessive solutions of 2x*y i = i by forming (p l , ^ x ), (p z , q^) etc. in accord- 
 ance with the law. 
 
 Archimedes. 
 
 The solution of the above equation by the Pythagoreans was evidently 
 used in order to obtain successive approximations to ^2. 
 
 Consequently, when we find Archimedes giving, without explanation, the 
 fractions fff anc ^ Vs 5 zr as being approximately equal to ^3, the hypothesis of 
 Zeuthen and Tannery that he arrived at these approximations by obtaining 
 successive solutions of equations of a similar form, but with 3 substituted 
 for 2, is one of the most natural that have been suggested 2 . The equations 
 are in this case 
 
 XT 3_)' 2 = - 2. 
 
 Tannery shows how the law for forming successive solutions of such 
 simple cases as these can easily be found when we have found by trial 
 (which is not difficult) the three simplest solutions. If we take the more 
 general equation 
 
 1 H. Konen, Geschichte der Gleichung ft - Z? 2 =i, Leipzig (S. Hirzel), 1901. 
 
 2 Zeuthen, " Nogle hypotheser om Arkhimedes kvadratrodsberegning," Tidsskrift for 
 Mathematik, vi. Raekke, 3. Aargang, pp. isosqq.; P. Tannery, "Sur la mesure du cercle 
 d'Archimede" in Memoires de la soc. des sciences phys. et nat. de Bordeaux, ii e Ser. iv., 
 1882, p. 303; see Giinther, "Die quadratischen Irrationalitaten der Alien und deren 
 Entwickelungsmethoden " in Abhandlungen zur Gesch. der Mathematik, Heft iv. 1882, 
 pp. 87-91; Konen, op. cit. p. 15.
 
 THEOREMS AND PROBLEMS BY FERMAT 279 
 
 of which x =p, y = #isai known solution, and put 
 
 it is sufficient to know the three simplest solutions in order to find a, ft, y, 
 8; for, substituting the values of (p, q\ (p lt 9l ) and (p 2 , ^ 2 ) where (/, ?2 ) 
 are formed from (p lt ^) by the same law as (A, ^) are formed from (p, q], 
 we have four simultaneous equations in four unknown quantities. Taking 
 the particular equation 
 
 * 2 -3/=i, 
 
 we easily find the first three solutions, namely (p=i,g = o), (A = 2 ft = 
 and (/ 2 = 7, ^ 2 = 4), whence 
 
 2=0, I - y, 
 
 "J 2a + f3, 4 = 2y + 8, 
 and a = 2, /3 = 3, y = i, 8 = 2, so that 
 
 But there is evidence that Archimedes dealt with much more difficult 
 equations of the type, for (as stated above, p. 123) the Cattle-Problem 
 attributed to him requires us to solve in positive integers the equation 
 x 2 - 4729494^=1. 
 
 There is this difference between this equation and the simpler ones 
 above that, while the first solutions of the latter can be found by trial, 
 the simplest solution of this equation cannot, so that some general method, 
 e.g. that of continued fractions, is necessary to find even the least solution 
 in integers. Whether Archimedes was actually able to solve this particular 
 equation is a question on which there is difference of opinion ; Tannery 
 thought it not impossible, but, as the smallest values of x, y satisfying the 
 equation have 46 and 41 digits respectively, we may, with Giinther, feel 
 doubt on the subject 1 . There is, however, nothing impossible in the 
 supposition that Archimedes was in possession of a general method of 
 solving such equations where the numbers involved were not too great for 
 manipulation in the Greek numeral notation. 
 
 Diophantus. 
 
 Tannery 2 was of opinion that Diophantus dealt with the equation 
 
 x 2 - Ay 1 - i 
 
 somewhere in the lost Books of the Arithmetica. Diophantus does indeed 
 say (Lemma to vi. 15) that, if a, b are any numbers and ax^-b is a square 
 when x is given a certain value p, then other values of x greater than p can 
 also be found which have the same property ; and Tannery points out that 
 
 1 Giinther, op. ctt., pp. 92-93 note. Cf. Konen, op. cit., p. 14. 
 
 2 Tannery, " L'Arithmetique des Grecs dans Pappus" in Memoires dt la see, des 
 sciences phys. et nat. de Bordeatix, u e Ser. III., x88o, pp. 370 sq.
 
 2 8o SUPPLEMENT 
 
 we can, by making suppositions of the same kind as Diophantus makes, 
 deduce a more general solution of the equation 
 
 when one solution (/, q] is known. 
 Put p 1 = mx- 
 
 and suppose 
 
 p? - Aq? = m*x* - 2mpx + p 2 - Ax 2 - 2Aqx - Aq 2 = i ; 
 
 therefore (since/ 2 - Aq 2 = i) 
 
 mp + Aq 
 m 2 -A ' 
 
 and, by substitution in the expressions for/!, y lt we have 
 
 _ (m 2 + A)p + 2Amq __ zmp + (m 2 + A] q 
 A= m*-A ' ^~ nf-A 
 
 and in fact/! 2 - Aq?= i. 
 
 If an integral solution is wanted, one way of obtaining it is to substitute 
 u/v for m where u z -Ai^-i, i.e. where u, v is another solution of the 
 original equation, and we then have 
 
 A = (^ + Av*)p + 2Auvq, q-i = 2puv + (u 2 + Av*) q, 
 
 But this is all that we can get out of Diophantus as we have him, and 
 it will be observed that here too we must have ascertained two solutions of 
 the one equation, or one solution of it and a solution of an auxiliary equation, 
 before we can apply the method 1 . 
 
 1 It may be observed that, in the particular case of the equation jr 2 -3_y 2 =i, the 
 assumption of u, v satisfying the equation will not enable us to obtain from the formula 
 
 p l = (u 2 + Av*)p + lAuvq, q\ ipuv + (u 2 + Av 2 ) q 
 above given the simpler formula otherwise obtained by Tannery (p. 279 above), namely 
 
 for, if (/i, ^i) is to be a different solution from (/, q), we cannot make =i, v = o, but 
 must take u = i, v=i, whence, putting A =3, we obtain 
 
 which is the same as/ 2 > ?2. the next solution to/i = 
 
 In order to get the latter we have to take u, v satisfying, not x 2 - $j/ 2 =i, but 
 
 **-3^=-* 
 
 The values = i, v= i satisfy x 2 - 3_y 2 = - 2, and 
 
 and of course p\ = +(2/ + 3?), q\= +(/ + ?) can be taken, since they equally satisfy
 
 THEOREMS AND PROBLEMS BY PERM AT 281 
 
 The Indian Solution. 
 
 If the Greeks did not accomplish the general solution of our equation, 
 it is all the more extraordinary that we should have such a, general solution 
 in practical use among the Indians as early as the time of Brahmagupta 
 (born 598 A.D.) under the name of the "cyclic method." Whether this 
 method was evolved by the Indians themselves, or was due to Greek 
 influence and inspiration, is disputed. Hankel held the former view 1 ; 
 Tannery held the latter and showed how, from the Greek manner of 
 deducing from one approximation to a surd a nearer approximation, it is 
 possible, by simple steps, to pass to the Indian method 2 . The question 
 presumably cannot be finally decided unless by the discovery of fresh 
 documents; but, so far as the other cases of solution of indeterminate 
 equations by the Indians help to suggest a presumption on the subject, 
 they are, I think, rather in favour of the hypothesis of ultimate Greek 
 origin. Thus the solution of the equation ax by = c given by Aryabhata 
 (born 476 A.D.) as well as by Brahmagupta and Bhaskara, though it 
 anticipated Bachet's solution which is really equivalent to our method of 
 solution by continued fractions, is an easy development from Euclid's 
 method of finding the greatest common measure or proving by that process 
 that two numbers have no common factor (Eucl. vn. i, 2, x. 2, 3) 3 , and 
 it would be strange if the Greeks had not taken this step. The Indian 
 solution of the equation xy = ax + by + c, by the geometrical form in which 
 it was clothed, suggests Greek origin 4 . 
 
 The "cyclic method" of solving the equation 
 
 is found in Brahmagupta and Bhaskara 5 (born 1114 A.D.) and is well 
 described by Hankel, Cantor and Konen 8 . 
 
 The method is given in the form of dogmatic rules, without any proof 
 of the assumptions made, but is equivalent to a preliminary lemma followed 
 by the solution proper. 
 
 1 Hankel, Zur Geschichte der Math, im Alterthum und Mittelalter, pp. 203-4. 
 
 2 Tannery, "Sur la mesure du cercle d'Archimede" in Mem. de la soc. des sciences 
 phys. et nat. de Bordeaux, il" Ser. iv., 1882, p. 325; cf. Konen, pp. 27-28; Zeuthen, 
 " L'Oeuvre de Paul Tannery comme historien des mathematiques " in Bibliotheca Mathe- 
 matica, VI 3 , 1905-6, pp. 271-273. 
 
 3 G. R. Kaye, "Notes on Indian mathematics, No. 2, Aryabhata" in Journal of the 
 Asiatic Society of Bengal, Vol. iv. No. 3, 1908, pp. 135-138. 
 
 4 Cf. the description of the solution in Hankel, p. 199; Cantor, Gesch. d. Math. 13, 
 p. 631. 
 
 5 The mathematical chapters in the works of these writers containing the solution in 
 question are contained in H. T. Colebrooke's Algebra with arithmetic and mensuration 
 
 from the Sanskrit of Brahmegupta and Bhaskara, London, 18(7. 
 
 6 , Hankel, pp. 200-203; Cantor, I 3 , pp. 632-633; Konen, op. cit., pp. 19-26.
 
 282 SUPPLEMENT 
 
 Lemma. 
 
 If x =p, y = q be a solution of the equation 
 
 Ay- + s = x 2 , 
 
 and x =p', y = q'& solution of the equation 
 Ay 2 + s' = x 2 , 
 then, say the Indians, x =pp' Aqq , y =pq p'q is a solution of the equation 
 
 Ay* + ss' = x 2 . 
 In other words, if 
 
 Aq 2 +s =/ a< 
 
 Aq' 2 + s' =p' 2 
 then A (pq' p'q} 2 + ss' = (// Aqq'} 2 . 
 
 This is easily verified 1 . 
 
 In particular, taking s = s', we find, from any solution x =/, y = q of 
 the equation 
 
 a solution x -p 2 + Aq 2 , y = 2pq of the equation 
 
 Ay 2 + s 2 = x 2 . 
 Again, particular use of the lemma can be made when ^ = 1 or .y = + 2. 
 
 (a) If ^ = + i, and x =p, y = q is a solution of 
 
 Ay 2 + i = x 2 , 
 
 then x =p 2 + Aq 2 , y = 2pq is another solution of the same equation. 
 If j = i, and x =/, y - q is a solution of 
 
 Ay z -i=x*, 
 then x =p z + Aq^ y = zpq is a solution of 
 
 Ay 2 + i - x 2 . 
 
 (b] If J - 2, and x = p, y = q is a solution of 
 
 then x=p' 1 + Ay 2 , y= zpq is a solution of 
 
 Ay 2 + 4 = x 2 . 
 
 In this case, since zpq is even, the whole result when the values of 
 x, y are substituted must be divisible by 4, and we have x = % (p 2 + A<f), 
 y -pq as a solution of the equation 
 
 Ay 2 +i=x 2 . 
 
 For, since s=f - Aq^, s'=/ 2 - Aq"*, 
 ss'=(p*-A q *)(p'*-Aq>*) 
 = (pp'Y+(Aqq')* - A (pq' 
 = { (pp'Y lApp'qq' + (Aqq'?} -
 
 THEOREMS AND PROBLEMS BY FERMAT 283 
 
 Solution proper of the equation x 2 -Ay*= i. 
 
 We take two numbers prime to one another, /, q, and a third number s 
 with no square factor, such that 
 
 the numbers being also chosen (in order to abbreviate the solution) such 
 that s is as small as possible, though this is not absolutely necessary. 
 (This is a purely empirical matter; we have only to take a rough ap- 
 proximation to ^A in the form of a fraction//^.) 
 
 [It follows that s, q can have no common factor ; for, if 8 were a 
 common factor of s, g, it would also be a factor of/ 2 , and/ 2 , q" 1 would have 
 a common factor. But /, q are prime to one another.] 
 
 Now find a number r such that 
 
 q l = + is a whole number. 
 
 [This would be done by the Indian method called cuttaca (" pulveriser "), 
 corresponding to our method by continued fractions.] 
 
 Of the possible values of r a. value is taken which will make r 2 - A 
 as small as possible. 
 
 Now, say the Indians, we shall have : 
 
 s l - is an integral number, 
 
 and Aq l * + s l = (2^ 
 
 (Again the proofs are not given; they are however supplied by Hankel 1 .) 
 
 1 Since q l = ^ + q> is an integral number, all the letters in q\s=p + qr represent 
 
 integers. 
 
 Further, s=p' i -Aq i ; 
 
 therefore, eliminating s, we have 
 
 or p(pq\- i)= 
 
 Since /, q have no common factor, q must divide pq\ i ; that is, 
 
 PQ\ ~ l 
 
 - =an integer. 
 
 We have next to prove that si = (r 2 - A)fs is an integer. 
 Now ^_ A = (ti*-tf-<<9> = 9i'*-Wi> + * t since 
 
 therefore - i is an integer) 
 
 and, since s, q have no common factor, it follows that 
 
 Also
 
 2 8 4 SUPPLEMENT 
 
 We have therefore satisfied a new equation of the same form as that 
 originally taken 1 . 
 
 We proceed in this way, obtaining fresh results of this kind, until we 
 arrive at one in which s = i or + 2 or +4, when, by means of the lemma, 
 we obtain a solution of 
 
 Ay* + i = x 2 . 
 
 Example. To solve the equation 6"jy 2 + i = x 2 . 
 
 Since 8 2 is the nearest square to 67, we take as our first auxiliary equation 
 67. i 2 -3 = 8 2 , so that/ = 8, q = i, * = - 3. 
 
 Thus q l = --- . We put r= 7, which makes q-^ an integer and at the 
 o 
 
 same time makes s 1 = - - --- = 6 as small as possible. 
 
 Thus ?i = -5> A = (/?i^ I )/^ = -4i, 
 
 and we have satisfied the new equation 
 
 Next we take q z = ' * 2 , and we put r 2 = 5, giving q z = 1 1 ; thus 
 
 r 2 6? 
 
 - 2 - = - 7, and / 2 = (p& - i)/^ = 90, and 
 
 Next q 3 = -- -, and we put r 3 - 9, giving q s = - 27 ; therefore 
 
 r? 67 90 . 27 i 
 
 s 3 = '- = -2, /, = = -221, and 
 
 6 7 .(2 7 ) 2 -2=(22l) 2 . 
 
 As we have now brought our s down to 2, we can use the lemma, and 
 
 67 (2 . 27 . 22l) 2 + 4 = (22I 2 + 67 . 27 2 ) 2 , 
 
 or 67(ii934) 2 + 4 = (9768 4 ) 2 ; 
 
 therefore, dividing by 4, we have 
 
 67(5 9 6 7 ) 2 -f-i=( 4 88 4 2) 2 . 
 
 Of this Indian method Hankel says, " It is above all praise ; it is 
 certainly the finest thing which was achieved in the theory of numbers 
 
 1 Hankel conjectures that the Indian method may have been evolved somewhat in 
 this way. 
 
 s=p' i is given, and if we put Aq'^ + s'=p' z , then 
 
 Now suppose^', q' to be determined as whole numbers from the equation^' - p'q= i, 
 and let the resulting integral value of pp' Aqq' be r. 
 
 Then A+ss' = r 2 , and accordingly r 2 - A must be divisible by s, or s'=(A-r 2 )js is 
 a whole number. 
 
 Eliminating p' from the two equations in p' ', q', we obtain 
 
 and, as stated in the rule, r has therefore to be so chosen that (/ + qr}ls is an integer.
 
 THEOREMS AND PROBLEMS BY FERMAT 285 
 
 before Lagrange " ; and, although this may seem an exaggeration when we 
 think of the extraordinary achievements of a Fermat, it is true that the 
 Indian method is, remarkably enough, the same as that which was redis- 
 covered and expounded by Lagrange in his memoir of 1768'. Nothing is 
 wanting to the cyclic method except the proof that it will in every case 
 lead to the desired result whenever A is a number which is not a square ; 
 and it was this proof which Lagrange first supplied. 
 
 Fermat. 
 
 As we have already said, Fermat rediscovered our problem and was 
 the first to assert that the equation 
 
 3?-Af=l, 
 
 where A is any integer not a square, always has an unlimited number 
 of solutions in integers. 
 
 His statement was made in a letter to Frenicle of February, i657 2 . 
 Fermat asks Frenicle for a general rule for finding, when any number not a 
 square is given, squares which, when they are respectively multiplied by tlie 
 given number and unity is added to the product, give squares. If, says 
 Fermat, Frenicle cannot give a general rule, will he give the smallest value 
 of y which will satisfy the equations 6ij^ + i = or 2 and 109^+ i =^? 3 
 
 At the same time Fermat issued a challenge to the same effect to 
 mathematicians in general, prefacing it by some remarks which are worth 
 quoting in full 4 . 
 
 " There is hardly any one who propounds purely arithmetical questions, 
 hardly any one who understands them. Is this due to the fact that up to 
 now arithmetic has been treated geometrically rather than arithmetically? 
 This has indeed generally been the case both in ancient and modern 
 works; even Diophantus is an instance. For, although he has freed 
 himself from geometry a little more than others have in that he confines 
 his analysis to the consideration of rational numbers, yet even there 
 geometry is not entirely absent, as is sufficiently proved by the Zetetica 
 of Vieta, where the method of Diophantus is extended to continuous 
 magnitude and therefore to geometry. 
 
 " Now arithmetic has, so to speak, a special domain of its own, the 
 theory of integral numbers. This was only lightly touched upon by Euclid 
 in his Elements, and was not sufficiently studied by those who followed 
 him (unless, perchance, it is contained in those Books of Diophantus of 
 
 1 "Sur la solution des problemes indetermines du second degre" in Mcmoires de 
 VAcad. Royale des Sciences et Belles-Lettres de Berlin, t. xxm. 1769 (=Oeuvres de 
 Lagrange, n. pp. 377 sqq.). The comparison between Lagrange's procedure and the 
 Indian is given by Konen, pp. 75-77. 
 
 2 Oeuvres de Fermat, II. pp. 333-4. 
 
 3 Fermat evidently chose these cases for their difficulty ; the smallest values satisfying 
 the first equation are ^=226153980, *= 1766319049, and the smallest values satisfying 
 the second are_y= 15140424455100, JT= 158070671986249. 
 
 4 Oeuvres de Fermat, II. pp. 334-5-
 
 286 SUPPLEMENT 
 
 which the ravages of time have robbed us); arithmeticians have therefore 
 now to develop it or restore it. 
 
 "To arithmeticians therefore, by way of lighting up the road to be 
 followed, I propose the following theorem to be proved or problem to 
 be solved. If they succeed in discovering the proof or solution, they will 
 admit that questions of this kind are not inferior to the more celebrated 
 questions in geometry in respect of beauty, difficulty or method of proof. 
 
 " Given any number whatever which is not a square, there are also given 
 an infinite number of squares such that, if the square is multiplied into the 
 given number and unity is added to the product, the result is a square. 
 
 "Example. Let 3, which is not a square, be the given number; when 
 it is multiplied into the square i, and i is added to the product, the result 
 is 4, being a square. 
 
 "The same 3 multiplied by the square 16 gives a product which, if 
 increased by i, becomes 49, a square. 
 
 "And an infinite number of squares besides i and 16 can be found 
 which have the same property. 
 
 " But I ask for a general rule of solution when any number not a square 
 is given. 
 
 " E.g. let it be required to find a square such that, if the product of the 
 square and the number 149, or 109, or 433 etc. be increased by i, the 
 result is a square." 
 
 The challenge was taken up in England by William, Viscount Brouncker, 
 first President of the Royal Society, and Wallis 1 . At first, owing apparently 
 to some misunderstanding, they thought that only rational, and not neces- 
 sarily integral, solutions were wanted, and found of course no difficulty in 
 solving this easy problem. Fermat was, naturally, not satisfied with this 
 solution, and Brouncker, attacking the problem again, finally succeeded in 
 solving it. The method is set out in letters of Wallis 2 of zyth December, 
 1657, and 3oth January, 1658, and in Chapter xcvin. of Wallis' Algebra; 
 Euler also explains it fully in his Algebra*, wrongly attributing it to Pell 4 . 
 
 1 An excellent summary of the whole story is given in Wertheim's paper "Pierre 
 Fermat's Streit mit John Wallis" in Abhandlungen zur Gesch. der Math. ix. Heft 
 (Cantor-Festschrift), 1899, pp. 557-576. See also Konen, pp. 29-43. 
 
 2 Oeuvres de Fermat, ill. pp. 457-480, 490-503. Wallis gives the solution of each 
 of the three difficult cases last mentioned. 
 
 3 Euler, Algebra, Part II. chap. vn. 
 
 4 This was the origin of the erroneous description of our equation as the " Pellian " 
 equation. Hankel (p. 203) supposed that the equation was so called because the solution 
 was reproduced by Pell in an English translation (1668) by Thomas Brancker of Rahn's 
 Algebra; but this is a misapprehension, as the so-called "Pellian" equation is not so 
 much as mentioned in Pell's additions (Wertheim in Bibliotheca Mathematica, 1113, 
 1902, pp. 124-6; Konen, pp. 33-4 note). The attribution of the solution to Pell was a 
 pure mistake of Euler's, probably due to a cursory reading by him of the second volume 
 of Wallis' Opera where the solution of the equation ax 2 + i =j 2 is given as well as informa- 
 tion as to Pell's work in indeterminate analysis. But Pell is not mentioned in connexion 
 with the equation at all (Enestrom in Bibliotheca Mathematica, II1 3 , 1902, p. 206).
 
 THEOREMS AND PROBLEMS BY FERMAT 287 
 
 Fermat appears to have been satisfied with the actual solution*, but 
 later he points out that, although Frenicle and Wallis had given many 
 particular solutions, they had not supplied a general proof 2 (i.e. presumably 
 that the solution is always possible and that the method will always lead 
 to the solution sought for). He says, " I prove it by the method of 
 descente applied in a quite special manner.... The general demonstration 
 will be found by means of the descente duly and appropriately applied." 
 
 Further on, Fermat says he has discovered "general rules for solving 
 the simple and double equations of Diophantus." 
 
 " Suppose, for example, that we have to make 
 2X 2 +7967 equal to a square. 
 
 " / have a general rule for solving this equation, if it is possible, or 
 discovering its impossibility, and similarly in all cases and for all values 
 of the coefficient of x? and of the absolute term. 
 
 " Suppose we have to solve the double-equation 
 2X + 3 square j 
 2X + 5 = square J ' 
 
 " Bachet boasts, in his commentary on Diophantus 3 , of having dis- 
 covered a rule for solving in two particular cases ; I make it general for 
 all kinds of cases and can determine, by rule, whether it is possible or not 4 ." 
 
 Thus Fermat asserts that he can solve, when it is possible to solve 
 it, and can determine, by a general method, whether it is possible or 
 impossible to solve, for any particular values of the constants, the more 
 general equation 
 
 &-Ap=B. 
 
 This more general equation was of course solved by Lagrange. How 
 Fermat solved it we do not know. It is true that he has sometimes been 
 
 1 Letter of June, 1658, to Kenelm Digby, Oeuvres de Fermat, n. p. 402. 
 
 2 " Relation des nouvelles decouvertes en la science des nombres," Oeuvres, n. p. 433. 
 
 3 See on Diophantus IV. 39, and above, pp. 80-82. 
 
 4 With this should be compared Fermat's note on Dioph. iv. 39, where he says, 
 similarly : 
 
 " Suppose, if you will, that the double-equation to be solved is 
 i x + 5 = square I 
 6x + 3 = square | ' 
 
 " The first square must be made equal to 16 and the second to 36; and others will be 
 found ad infinitum satisfying the question. Nor is it difficult to propound a general rule 
 for the solution of this kind of question." 
 
 No doubt the double-equation in this case, as in the others referred to in the "Relation," 
 would be transformed into the single equation 
 
 t*-Au*=B 
 
 by eliminating x. I think this shows how Fermat was led to investigate our equation : 
 a question which seems to have puzzled Konen (p. 29), in view of the fact that the actual 
 equation is not mentioned in the notes to Diophantus. The comparison of the two places 
 seems to make the matter clear. For example, the two equations mentioned above in 
 this note lead to the equation / 2 -3 2 = -12, and the solution t = 6, = 4 is easily 
 obtained.
 
 288 SUPPLEMENT 
 
 credited with the very same solution of the equation x 2 Ay* = i as that 
 given by Brouncker and Wallis ; but this idea seems to be based on a 
 misapprehension of a sentence in Ozanam's Algebra (1702). Ozanam 
 gives the Brouncker-Wallis solution as " une regie generale pour resoudre 
 cette question, qui est de M. de Fermat " ; and possibly the ambiguity 
 of the reference of " qui " may have misled Lagrange and others into 
 supposing that the "regie" was due to Fermat. 
 
 For the history of the equation after Fermat's time I must refer to 
 other works and particularly that of Konen 1 . Euler, Lagrange, Gauss, 
 Jacobi, Dirichlet, Kronecker are the great names associated with it. I 
 will only add a few particulars with regard to Euler 2 as coming nearest 
 to Fermat. 
 
 In a letter to Goldbach 3 of loth August, 1730, Euler mentions that he 
 requires the solution of the equation x 2 Ay* i in order to make 
 as? + bx + c a complete square. He goes on to observe that the problem 
 of solving x? Ay 2 = i in integers was discussed between Wallis and 
 Fermat and that the solution (which he already attributes to Pell) was 
 set out in Wallis' Opera. There is an indication in this very passage that 
 Euler had then only read the Brouncker-Wallis correspondence cursorily, 
 for he speaks of the equation iogy 2 + i = x 2 as being the most difficult 
 case solved by them, whereas the most difficult examples actually solved 
 were 433_y 2 + i = x 2 and 3 1 ^y~ + i = x 2 . 
 
 A paper of a year or two later 4 contained the proof that the evolution 
 of successive solutions of ay? + bx + c =y- when one is known requires that 
 one solution of at? + i = tf must also be known. Similarly, in his Algebra*, 
 he shows that the solution of the latter equation is necessary for finding all 
 the possible solutions of the equation ax~ + b =y 2 , the importance of which 
 remark is emphasised by Lagrange 6 . 
 
 In the paper quoted in the last paragraph Euler finds any number 
 of successive solutions of ax 2 + bx + c =_>' 2 , and the law for forming them, 
 when we are given one value n of x which will make ax 2 + bx + c a 
 complete square and one value p of which will make a 2 + i a complete 
 square, or, in other words, when an 2 + bn + c = m 2 and a/ 2 + i = q 1 . He 
 then takes the particular case ax 2 + bx + d 2 =y 2 where (since x = o, y-d 
 satisfies the equation) we can substitute o for n and d for m in the 
 expressions representing the successive solutions of ax 2 + bx + c=y 2 . Then 
 again, putting b - o and d=i, he is in a position to write down any 
 
 1 Konen, op. cit.; cf. Cantor's Geschichte der Mathematik, iv. Abschnitt xx., as 
 regards Euler and Lagrange. 
 
 2 Cf. Konen, op. cit. pp. 47-58. 
 
 3 Correspondence mathematique et physique de quelques celebres geometres du xvi I !*?/ 
 siecle, publiee par P. H. Fuss, Petersbourg, 1843, I. p. 37. 
 
 4 " De solutione problematum Diophanteorum per numeros integros" in Commentarii 
 Acad. Petropol. 1732-3, VI. (1738), pp. 175 sqq. = Coinnientationes arithm. I. pp. 4-10. 
 
 6 Algebra, Part II. ch. VI. 
 
 6 Additions to Euler's Algebra, ch. vm.
 
 THEOREMS AND PROBLEMS BY FERMAT 289 
 
 number of successive solutions of a? + i = rf when one solution =/, 
 t\-q is known. The successive values of | are 
 
 and the corresponding values of rj are 
 
 i, 4, 2f-i, 44* -34, ... 
 
 the law of formation being in each case that, if A, B be consecutive values 
 in either series, the next following is 2qB - A. 
 
 The question then arises how to find the first values /, q which will 
 satisfy the equation. Euler first points out that, 'when a has one of many 
 particular forms, values of/, q can at once be written down which satisfy 
 the equation. The following are such cases with the obvious values of 
 / and q. 
 
 a = & + i ; p = ze, q = 
 
 a = aV* + 2CU? 6 - 1 ; p = e, q = a<? 6+1 i 
 
 (where a may even be fractional provided o^ 6 " 1 is an integer), 
 a = (ae b + fte^)' 2 + 2a^~ l + 2 fie*- 1 p = e, 4 = a b + l + fie* 
 a = 
 
 But, if a cannot be put into such forms as the above, then the method 
 explained by Wain's must be used. Euler illustrates by finding the least 
 values /, q which will satisfy the equation 3i 2 + i -rf, and then adds a 
 table of the least solutions of the equation a^ + i = rf for all values of 
 a (which are not squares) from 2 to 68. 
 
 The important remark follows ( 18) that the above procedure at once 
 gives a very easy way of finding closer and closer approximations to the 
 value of any surd *]a. For, since a/ 2 + i = <f, we have >Ja - J(f - i)//, 
 and, if q (and therefore p also) is large, qjp is a close approximation to ,Ja ; 
 the error is not greater than ij(2/ 2 Ja). Euler illustrates by taking ^6. 
 The first solution of 6^ + i = rf (after = o, -q- i) is / = 2, q - 5. Taking 
 then the series of values above given for a 2 + i = rf, namely 
 = o, /, 2pq, Atpf -p, ... A, B, 2qB - A, 
 il=i, q, 2f-i, 44* -&,..., F, 2qF-E, 
 
 and substituting p - 2, q = 5, the successive corresponding values P, Q 
 of |, t] respectively become 
 
 P=o, 2, 20, 198, 1960, 19402, 192060, 1901198,... 
 
 <2=i, 5, 491-485, 4801, 47525. 47449. 4656965. 
 
 and the successive values QjP are closer and closer approximations to J6. 
 
 It will be observed that the method of obtaining successive approximations 
 
 H. D. 19
 
 290 SUPPLEMENT 
 
 to Ja from successive solutions of a- + i - rf is the same as that which, 
 according to the hypothesis of Zeuthen and Tannery, Archimedes used in 
 order to find his approximations to ^3. 
 
 The converse process of finding successive solutions of a$? + i -rf by 
 developing ,Ja as a continued fraction did not apparently occur to Euler 
 till later. In two letters' to Goldbach of 4th Sept. 1753 and 23rd Sept. 
 1755 he speaks of a "certain" method and of improvements which he had 
 made in the " Pellian " method but gives no details. His next paper on 
 the same subject 2 returns to the problem of finding all the solutions of 
 ax 1 + bx + c = y* or ax' 2 + l>=y 2 when one is known, and in the course of 
 his discussion of the latter he arrives at " the following remarkable theorem 
 which contains within it the foundation of higher solutions. 
 
 "If x = a, y = l> satisfies cur* +p y 2 , 
 
 and x = c, y = d satisfies cur 2 + q =/-, 
 
 then x be + ad, y bd + aac satisfies ax' 2 + pq =y 2 ." 
 
 That is to say, Euler rediscovers and recognises the importance of the 
 lemma to the Indian solution, as Lagrange did later. 
 
 More important is the paper of about three years later 3 in which Euler 
 obtained the solution of the equation x 2 - Ay- = i by the process of con- 
 verting ,JA into a continued fraction, this course being the reverse of that 
 which was, according to the hypothesis of Tannery and Zeuthen, followed 
 by Archimedes, and to the feasibility of which Euler had called attention 
 in 1732-3. He begins by stating, without proof, that, if q- - Ip 2 + i, then 
 q\p is an approximation to ^//, and qlp is " such a fraction as expresses 
 the value of Jl so nearly or exceeds it so little that a closer approximation 
 cannot be made except by bringing in greater numbers." Next he develops 
 certain particular surds, namely N /( I 3)> V(^ 1 ) an ^ V(^7)> a ^ ter which he 
 states the process generally thus. If ^z be the given surd and v the root 
 of the greatest integral square which is less than z, the process will give 
 
 the successive quotients a, b, c, d, being found by means of the process 
 shown in the following table : 
 
 1 Correspondance etc., ed. Fuss, pp. 614 sq., 62959. 
 
 2 "De resolutione formularum quadraticarum indeterminatarum per numeros integros " 
 in Novi Conimentarii Acad. Petropol. 1762-3, IX. (1764), pp. 3 i,<\<\. = Coinmentat. arithm. 
 I. pp. 297-315. 
 
 3 "De usu novi algorithm! in problemate Pelliano solvendo" in Novi Conimentarii 
 Acad. Petropol. 1765, xi. (1767), pp. 28-66= Continental, arithm. \. pp. 316-336. The 
 paper seems to have been read as early as 15 Oct. 1759.
 
 THEOREMS AND PROBLEMS BY FERMAT 
 
 291 
 
 Take 
 
 II. B=aa-A 
 
 III. C=B/>-. 
 
 yc-C 
 
 '. = M-D 
 
 and 
 
 z-Fp 
 
 It follows that 
 v + A 
 
 <* 
 
 7 
 
 v + D 
 8 
 
 8 
 
 etc. etc. 
 
 (This is of course exactly the process given in text-books of Algebra, 
 e.g. Todhunter's.) 
 
 Euler now remarks as follows. 
 
 1. The numbers A, B, C, D ... cannot exceed v ; the first, A, is equal 
 to v; since a ^ (v + A)/a, aa - A - B ^ z/, and so on. 
 
 2. Unless where one of the numbers a, ft, y, 8... is equal to unity, 
 none of the corresponding quotients a, b, c, d ... can exceed v. 
 
 3. When we arrive at a quotient equal to 2v, the next quotients will be 
 a, l>, c, d . . . in the same order. 
 
 4. Similar periods occur with the letters a, ft, y, 8... and the term 
 of this series corresponding to a quotient 2V is always i. 
 
 The successive convergents to the continued fraction are then investi- 
 gated and it is shown that, for successive convergents q[p beginning 
 with v/i, 
 
 q- - zp* = - a, + ft, -y, +8, - e etc. in order. 
 
 It follows that the problem is solved whenever one of the terms with a 
 positive sign, ft, 8, etc., becomes i. 
 
 Since unity for one of the terms a, ft, y, 8 corresponds to the quotient 
 2v, and each fresh period begins with 2V, the first period will produce 
 a convergent qjp such that q* zp- = i ; and the negative sign will apply 
 if the number of quotients constituting the period is odd, while the positive 
 sign will apply if the number of quotients is even. In the latter case we 
 have a solution of our equation at once; if, however, q--zp'* = -i, we 
 must go on to the end of the second period in order to get an even number 
 of quotients and so satisfy the equation ' q' 1 zp- + i. Or, says Euler, 
 instead of going on and completing the second period, we can satisfy 
 the latter equation more easily thus. 
 
 Suppose q- - zp* = - i, and assume 
 
 Then 
 
 q'- - zp'- = 4^ + 4^ + i - 
 
 192
 
 292 SUPPLEMENT 
 
 [This last derivation of a solution of y 2 zx 2 i from a known solution 
 of y 2 - zx 2 = - i is of course the same as the Indian method of doing the 
 same thing, for they assumed/' = zpq, q' = q 2 + zp 2 , and q 2 + zp 2 - q 2 + (q 2 + i).] 
 
 We thus see that in Euler's method there is everything necessary to the 
 complete solution of our equation except the proof that it must always lead 
 to the desired result. Unless it is proved that the quotient 2V will actually 
 occur in the development of the continued fraction in every case, we cannot 
 be sure that the equation has any solution except x - o, y - i. 
 
 I cannot, I think, do better than conclude by a quotation from 
 H. J. S. Smith 1 , the first part of which is well known 2 . " Euler observed 
 that [if T 2 - DU- = i] TIU is itself necessarily a convergent to the value 
 of JD, so that to obtain the numbers T and U it suffices to develop JD 
 as a continued fraction. It is singular, however, that it never seems to 
 have occurred to him that, to complete the theory of the problem, it was 
 necessary to demonstrate that the equation is always resoluble and that 
 all its solutions are given by the development of >JD. His memoir 
 contains all the elements necessary to the demonstration, but here, as 
 in some other instances, Euler is satisfied with an induction which does 
 not amount to a rigorous proof. The first admissible proof of the re- 
 solubility of the equation was given by Lagrange in the Melanges de la 
 Societe de Turin, Vol. iv. p. 4i 3 . He there shows that in the development 
 of ,JD we shall obtain an infinite number of solutions of some equation of 
 the form T 2 -DU 2 = A and that, by multiplying together a sufficient 
 number of these equations, we can deduce solutions of the equation 
 T 2 -DU 2 -i. But the simpler demonstration of its solubility which 
 is now to be found in most books on algebra, and which depends on 
 the completion of the theory (left unfinished by Euler) of the development 
 of a quadratic surd as a continued fraction, was first given by Lagrange 
 in the Hist, de F Academic de Berlin for 1767 and 1768, Vol. xxm. p. 272, 
 and Vol. xxiv. p. 236"*, and, in a simpler form, in the Additions to Euler's 
 Algebra 6 , Art. 37." 
 
 1 "Report on the Theory of Numbers, Part ill.," British Association Reports for 1861, 
 London, 1862, p. $1$= Collected Works, Vol. i., Oxford, 1894, p. 192. 
 
 2 It is given in Cantor, Gesch. d. Math. iv. 1908, p. 159, and referred to by Konen, 
 op. tit. p. 51. 
 
 3 "Solution d'un probleme d'Arithmetique," finished at Berlin on 2oth Sept. 1768 
 and published in Miscellanea Taurinensia, iv. i'j66-i'j6<)=0euvres de Lagrange, I. 
 pp. 671-731. 
 
 4 The references are: "Sur la solution des problemes indetermines du second degre," 
 read 24th Nov. 1768 and published in the Memoires de VAcadeniie Royale des Sciences 
 et Belles-lettres de Berlin, Vol. xxm., 1769, pp. 165-310 =#>; de Lagrange, II. 
 PP- 377-535 > "Nouvelle methode pour resoudre les problemes indetermines en nombres 
 entiers," read 2ist June, 1/70, and published in Memoires de rAcademie Royale des 
 Sciences et Belles-lettres de Berlin, Vol. xxiv., 1770, pp. i%i-i$() = Oeuvres de Lagrange, 
 II. pp. 655-726. 
 
 5 The Additions of Lagrange were first printed as an appendix to lei>iens d' ' Algcbre 
 par M. L. Euler traduits de Vallemand, Vol. II., Lyons, 1774; second edition, Paris, 
 
 1798; they were thence incorporated in Oeuvres de Lagrange, vn. pp. 158 sqq.
 
 293 
 
 SECTION III. 
 
 THEOREMS AND PROBLEMS ON RATIONAL RIGHT-ANGLED TRIANGLES. 
 
 i. On No. 20 of the problems about right-angled triangles added 
 by Bachet to Book vi. ("To find a right-angled triangle such that its 
 area is equal to a given number") Fermat has a note which shall be 
 quoted in full, not only for the sake of the famous theorem enunciated 
 in it, but because, exceptionally, it indicates the lines on which his proof 
 of the theorem proceeded. 
 
 "The area of a right-angled triangle the sides of which 
 are rational numbers cannot be a square number. 
 
 "This proposition, which is my own discovery, I have at length 
 succeeded in proving, though not without much labour and hard thinking. 
 I give the proof here, as this method will enable extraordinary develop- 
 ments to be made in the theory of numbers. 
 
 " If the area of a right-angled triangle were a square, there would exist 
 two biquadrates the difference of which would be a square number. Con- 
 sequently there would exist two square numbers the sum and difference of 
 which would both be squares. Therefore we should have a square number 
 which would be equal to the sum of a square and the double of another 
 square, while the squares of which this sum is made up would themselves 
 [i.e. taken once each] have a square number for their sum. But if a square 
 is made up of a square and the double of another square, its side, as I can 
 very easily prove, is also similarly made up of a square and the double of 
 another square. From this we conclude that the said side is the sum of the 
 sides about the right angle in a right-angled triangle, and that the simple 
 square contained in the sum is the base and the double of the other square 
 the perpendicular. 
 
 "This right-angled triangle will thus be formed from two squares, 
 the sum and the difference of which will be squares. But both these 
 squares can be shown to be smaller than the squares originally assumed 
 to be such that both their sum and their difference are squares. Thus, 
 if there exist two squares such that their sum and difference are both 
 squares, there will also exist two other integer squares which have the same 
 property but have a smaller sum. By the same reasoning we find a sum 
 still smaller than that last found, and we can go on ad infinitum finding 
 integer square numbers smaller and smaller which have the same property. 
 This is, however, impossible because there cannot be an infinite series 
 of numbers smaller than any given integer we please. The margin is too 
 small to enable me to give the proof completely and with all detail. 
 
 " By means of these considerations I have also discovered and proved 
 that no triangular number except i can be a biquadrate."
 
 294 SUPPLEMENT 
 
 As Wertheim says, it may have been by following out the indications 
 thus given by Fermat that Euler succeeded in proving the propositions 
 that x 4 -y* and x 4 +y* cannot be squares, as well as a number of other 
 theorems connected therewith (Commentationes arithmeticae collectae, i. 
 pp. 24 sqq. ; Algebra, Part n. Chapter xin.). 
 
 Zeuthen 1 suggests a method of filling out Fermat's argument, thus. 
 The sides of a rational right-angled triangle can be expressed as 
 
 x*+y 2 , x 2 -y 2 , 2xy. 
 
 As a common factor in the sides would appear as a square in the 
 number representing the area, we can neglect such a factor, and assume 
 that x 2 -y" and therefore also x+y and x-y are odd numbers and thnt 
 x, y are prime to one another, so that x, y, x +y, x -y are all prime to 
 one another. 
 
 We have now to test the assumption that the area of the triangle 
 
 xy (x -y) (x +y) 
 is a square. If so, the separate factors must be squares, or 
 
 x = u\ y = v 2 , 
 u 2 + v 2 =p 2 , u 2 -z? = g 2 . 
 
 (" There would exist two biquadrates the difference of which [# 4 z/*] would 
 be a square, and consequently there would exist two squares the sum and differ- 
 ence of which [u 2 + v z , u- - v~] would both be squares" Fermat.) 
 
 From the last two equations we obtain 
 
 (" We should have a square number which would be equal to the sum 
 of a square and the double of another square [p 1 = 21? + q^\" Fermat.) 
 
 Now p + q and p q are both even numbers because, on the above 
 assumptions, p* and q" are both odd; but they cannot have any other 
 common factor except 2, since u 2 and v 2 are prime to one another. It 
 follows therefore from the last equation that 
 
 (2m 2 (n 2 
 
 where ;/ is an even number. 
 We obtain, therefore, 
 
 The whole numbers m 2 and are therefore sides of a new right-angled 
 
 triangle with the square area ---- . 
 4 
 
 1 Zeuthen, Geschichte der MatJiematik im XVI. and XVII. Jahrhundert, 1903, p. 163.
 
 THEOREMS AND PROBLEMS BY FERMAT 295 
 
 (" If a square is made up of a square and the double of another square 
 [/ 2 = 21? + g*], its side is, as I can very easily prove, also made up of a 
 
 square and the double of another square p = tri*+ 2 {-] From this we 
 
 conclude that the said side is the sum of the sides about the right angle 
 in a right-angled triangle, the square [w 2 ] being the base and the double of 
 
 the other square 2J the perpendicular," Fermat.) 
 
 That the sides of the new triangle are less than those of the original 
 triangle is clear from the fact that the square on its hypotenuse 2 or 
 x is a factor of one of the perpendicular sides of the original triangle 1 . 
 
 As now an infinite series of diminishing positive whole numbers is 
 impossible, the original assumption from which we started is also impossible. 
 
 It will be observed, as Zeuthen says, that the proof includes also the 
 proof of the fact that u 4 - v* cannot be a square and therefore cannot be 
 a fourth power, from which it follows that the equation rf = z/ 4 + w 4 cannot 
 be solved in whole numbers, and consequently cannot be solved in rational 
 numbers either. 
 
 The history of this theorem would not be complete without an account 
 of a "proof originating with Fermat" which Wertheim has reproduced 2 . 
 In the small paper of Fermat's entitled " Relation des nouvelles decouvertes 
 en la science des nombres 3 " containing a statement of his method 
 of "diminution without limit" (descents infinie or indefinie) and of a 
 number of theorems which he proved by means of it, there is a remark 
 that he had sent to Carcavi and Frenicle some proofs based on this 
 method. And, sure enough, Frenicle gives a proof by this method of 
 the theorem now in question in his "Traite des triangles rectangles en 
 nombres 4 ." Wertheim accordingly concludes that we have here a proof 
 of Fermat's. A short explanation is necessary before we come to Frenicle's 
 proof. 
 
 We obtain a right-angled triangle 2, x, y in rational numbers (z 2 = x 2 +/) 
 if, a, b being any integers and a>t>, we put 
 
 z = a 2 + b\ x = a 2 - P, y = 2ab. 
 
 If a is prime to b and one of these numbers is even, the other odd, then 
 it is easily shown that the greatest common measure of x,y, z is i. 
 
 In the right-angled triangle a 2 - 1? and zab are the perpendicular sides, 
 
 1 Zeuthen's inference at this point diverges slightly in form from what we actually find 
 in Fermat's own statement of his argument. Fermat does not actually say that the new 
 right-angled triangle is a triangle in smaller numbers than the original triangle and with 
 the same assumed property, but that its formation gives us two new square numbers the 
 sum and difference of which are squares, and which are smaller than the two squares 
 originally assumed to have this property. 
 
 2 Zeitschrift filr Math. u. Physik, hist. litt. Abtheilung XLIV. 1899, pp. 4-6. 
 
 3 Ontvres de Fermat, Vol. n. pp. 431-6. 
 
 4 Memoires de VAcademie Royale des Sciences, v., Paris, 1729, pp. 83-166.
 
 296 SUPPLEMENT 
 
 and (a 2 - P) ab is the area, a, b are called the generating numbers (the 
 numbers from which the triangle is formed) and if a is prime to b, and one 
 of them is odd and the other even, so that x, y, z have no common factor 
 except i, the triangle is called & primitive triangle. 
 
 If (cP b^ab is the area of a primitive right-angled triangle and it 
 is enough to prove the proposition for such each of the three numbers 
 o? - P, a, b is prime to the other two. If, then, the product is a square 
 number, each of the three factors must be square, and in that case a 2 - b 2 
 will be the difference between two fourth powers. The theorems 
 
 (1) the area of a right-angled triangle in rational numbers cannot be 
 a square number, and 
 
 (2) the difference of two fourth powers cannot be a square, 
 accordingly state essentially the same fact. 
 
 The proof which Frenicle gives of the first of these propositions depends 
 on the following Lemmas. 
 
 Lemma I. If the odd perpendicular of a primitive right-angled triangle 
 is a square number, there exists a second primitive right-angled 
 triangle with smaller sides which has for its odd perpendicular 
 the root of the said square number. 
 
 If a 2 - 1? = r 3 , it follows that a? = I? + <*, so that a, b, c are the sides 
 of a right-angled triangle. The odd perpendicular of this second triangle 
 is c, for by hypothesis c 3 is odd; consequently the even perpendicular is 
 b, while a is the hypotenuse. The triangle is "primitive" because a 
 common divisor of any two of the three numbers a, b, c would divide 
 the third, while by hypothesis a, b have no common factor except i. 
 Next, the second triangle has smaller sides than the first, since c<c 2 , 
 a<a 2 + t> 2 , b<zab. 
 
 By this lemma we can from the triangle 9, 40, 41 derive the triangle 
 3, 4, 5, and from the triangle 225, 25312, 25313 the triangle 15, 112, 113. 
 Lemma II. If in a primitive right-angled triangle the hypotenuse as 
 well as the even perpendicular were square, there would exist a 
 second primitive right-angled triangle with smaller sides which 
 would have for hypotenuse the root of the hypotenuse of the first, 
 for odd perpendicular a square number, and for even perpendicular 
 the double of a square number. 
 
 Let the sides of the first triangle be a? + b z , a- - b-, zab. If zab were 
 a square, ab would be double of a square ; therefore, since a, b are prime 
 to one another, one of these two numbers, namely the odd one, would 
 be a square, and the other, the even one, would be double of a square. 
 Let a be the odd one of the two, b the even. If now the hypotenuse 
 a 2 + b- were a square number c*, we should have a second right-angled 
 triangle a, b, c which would necessarily be primitive and in which the sides 
 would be smaller than those of the first triangle; for c<c, b<2ab and 
 a < a" b~ since a + b > a, a b^ i .
 
 THEOREMS AND PROBLEMS BY FERMAT 297 
 
 By means of the above two lemmas combined we can now prove that 
 the area of a primitive right-angled triangle cannot be a square number. 
 
 Let the sides of the triangle be c? + lr, a 2 - F, zab. If now the area 
 were square, the product of the perpendicular sides would be double of 
 a square. But the perpendicular sides are prime to one another. There- 
 fore the odd perpendicular a- - P would be a square, and the even 
 perpendicular zab the double of a square. But, if cr-lr were equal to 
 ^, we could (by the first Lemma) find a second primitive triangle with 
 smaller sides in which the odd perpendicular would be t, the even per- 
 pendicular fi, and the hypotenuse a. Again, since zab would be double 
 of a square, ab would be a square, and, since a is prime to b, both a and 
 b would be squares. The second triangle would accordingly have a square 
 number both for its hypotenuse (a) and for its even perpendicular (b). 
 That is, the second primitive triangle would satisfy the conditions of the 
 second Lemma, and we could accordingly derive from the second primitive 
 triangle a third primitive triangle with still smaller sides which would, 
 exactly like the first triangle, have a square number for its odd perpendicular, 
 and for its even perpendicular the double of a square number. 
 
 From this third triangle we could obtain a fourth, and by means of the 
 fourth we could obtain a fifth with the same property as the first, and so 
 we should have an unending series of primitive right-angled triangles, each 
 successive triangle having smaller sides than the one before, and all being 
 such that the odd perpendicular would be a square number, the even 
 perpendicular the double of a square number, and consequently the area 
 a square number. This, however, is impossible since there cannot be an 
 unending series of integral numbers less than any given integral number. 
 
 Frenicle proves, by similar considerations, that neither can the area of a 
 right-angled triangle in rational numbers be Hie double of a square number. 
 
 In enunciating Fermat's problems on right-angled triangles I shall in 
 future for brevity and uniformity use 17, to denote the three sides, while 
 will always represent the hypotenuse and & t\ the two perpendicular sides. 
 
 2. To find a right-angled triangle (, 17) such that 
 
 t + V = 
 
 [Since "' = * + if, this problem is equivalent to that of finding .r, y such 
 that 
 
 which is Question 17 in Chapter xiv. of Euler's Algebra, Part 11.] 
 
 first method. 
 
 Form a right-angled triangle from the numbers x + i, x; the sides will 
 then be
 
 298 SUPPLEMENT 
 
 We have then the double-equation 
 
 2X* + 2X+ I = U- | 
 2X 2 + 4X + I = V 2 j 
 
 The ordinary method of Diophantus gives the solution x = - ^- > trie 
 triangle will therefore be formed from -f- and -- 1 /- or, if we take the 
 numerators only, -5 and - 12, and the triangle is (169, - 119, 120) which 
 is equally the result of forming a triangle from + 5 and + 12. 
 
 But, as one of the perpendiculars is negative, we must find another 
 value of x which will make all three sides positive. 
 
 We accordingly form a triangle from x + $ and 12, instead of from 
 5 and 12, and repeat the operation. This gives for the sides 
 
 t,- x 2 +10^+169, g = x- + IQX- 119, 77-- 243;+ 120, 
 and we have to solve the double-equation 
 
 x 2 + iox+ 169 = u\ 
 x 2 + 343; + \-v i . 
 Making the absolute term the same in each, we have to solve 
 
 x 2 + lox + 169 = u 2 , 
 1693? + 57463; + 169 = v' 2 . 
 
 The difference is i68x- + 57363:, which we may separate into the factors 
 143:, i2x + -i- 8 T 6 -. (the sum of the terms in x being 26x or 2 . 133:). 
 
 Equating the square of half the sum of these factors to the larger 
 expression, or the square of half their difference to the smaller, we find 
 in the usual way 
 
 The triangle is therefore formed from -|iyf-, I2 > or ^ rom 2I 595> 
 246792, and the triangle itself is 
 
 4687298610289, 4565486027761, 1061652293520, 
 the hypotenuse and the sum of the other two sides being severally squares. 
 
 Second method. 
 
 This is the same as the first method up to the forming of the triangle 
 from x + 5 and 12 and the arrival at the double-equation 
 3? 2 + iox+ 169 = u 2 , 
 x 2 + 343; + i = v 2 . 
 
 Multiply the two expressions together, and we must have 
 x* + 44X 3 + 5io3: 2 -f 57563;+ 169 = a square 
 
 this gives, as a matter of fact, the same value of x, namely
 
 THEOREMS AND PROBLEMS BY FERMAT 299 
 
 and the triangle is the same as before 1 . 
 
 In his note on Diophantus vi. 22 Fermat says that he confidently 
 asserts that the above right-angled triangle is the smallest right-angled 
 triangle in rational numbers which satisfies the conditions. 
 
 [The truth of this latter assertion was proved by Lagrange 2 . Lagrange 
 observes that, since + rj =y-, g + rf = #*, say, we have, if we put z for - >/, 
 
 Z 2 +/ = 2*, 
 
 or 2X* -y* = z 2 , 
 
 and, if x, y is any solution of the latter equation, 
 
 1 For comparison we may give Euler's solution (Algebra, Part n., Art. 240; Commen- 
 tationes arithmeticae, n. p. 398). 
 We have to solve the equations 
 
 First make x 2 +y 2 a square by putting x=a? - IP, y = iab, so that 
 
 To make the last expression a fourth power put a=p^-q i , b = ipq, so that 
 
 a> + *=(/* + ?*)', 
 
 and accordingly ** + y* = (p* + q*f. 
 
 We have now only to make x +y a square. 
 
 Now x = a?-lP=p*-6p z q' i + q*, y = iab=4p*q-4pq 3 ; 
 
 therefore / 4 + \p*q - 6>V 2 - 4A/ 3 + ?* = a square. 
 
 In solving this we have to note that /, q should be positive, p must be >q (for other- 
 wise^ would be negative), and a>b in order that x may be positive. 
 
 Put 
 
 and we obtain $fq - 6ffi 2 = - $fq + 6p V, whence p\q = | . 
 
 But, if we put/ = 3, q = i, we find x= - 1 19, a negative value. 
 
 To find fresh values, we can substitute for / the expression %y + r and solve for the 
 ratio q\r; then, by taking for q the numerator and for r the denominator of the fraction 
 so found, we find a value for / and thence for x, y. This is Euler's method in the 
 Algebra. But we avoid the necessity for clearing of fractions if (as in the Comment. 
 ariihm.) we leave 2 as the value of q and substitute 3 + z for 3 as the value of/. 
 
 We then have / 4 = 81 + io8z/+54 2 + 
 
 6/ 2 ^ 2 = 216 i 44"' 24^ 2 , 
 -4/^3 = _ 96-32^, 
 
 + <?*= 16, 
 
 whence x+y= i -t- 148^+ iO2Z> 2 +2OZ' 3 + z/ 4 = a square=(i + 74W-z^) 2 , say; 
 and we obtain 
 
 1343 = 42^, or -=, and / = 3 + z'=-^, while q = ^. 
 Taking integral values, we put/=i469, ^ = 84. 
 Therefore a= 1385 . 1553 = 2150905, =168.1469 = 246792, 
 and ^ = 4565486027761, y= 1061652293520, 
 
 which is the same as Fermat's solution. 
 
 2 N. Mhnoires de fAcad. Royale des Sciences et Belles-lettres de Berlin, annee 1777, 
 Berlin, 1779= Oeuvres de Lagrange, IV. pp. 377-398.
 
 300 SUPPLEMENT 
 
 He sets himself therefore to find a general solution of the equation 
 2X 4 _y 4 = z 2 and effects it by a method which is a variation of Fermat's 
 descente, one of the most fruitful methods, as Lagrange observes, in the 
 whole theory of numbers. The modified method consists of two parts, 
 (i) a proof that, assuming that there exist integral values of x, y greater 
 than r which satisfy the condition 2X* J>* = z 2 , there are still smaller 
 integral values which will also satisfy it, (2) the discovery of a general 
 method of deducing the latter from the former. This being done, and 
 it being known that x- i, y = i are the minimum values, the successive 
 higher values are found by reversing the process. Lagrange found that 
 the four lowest values for x, y give the following pairs of values for , 77, 
 namely 
 
 (1) =I , 77 = 0, 
 
 (2) =120 , 77 = -ii9, 
 
 (3) -2276953 , 77 = -473 3 o4, 
 
 (4) =1061652293520, 77 = 4565486027761, 
 
 so that the last pair (4) are in truth, as Fermat asserts, the smallest possible 
 values in positive integers.] 
 
 3. To find a right-angled triangle , , 77 such that 
 
 [This is of course equivalent to solving 
 x-y = / 2 | 
 x* + / = it? \ 
 Form a triangle from the numbers x + i, i ; the sides will then be 
 
 = x'*+2X + 2, = X 2 + 2X, f)=2X+2. 
 
 We have then to solve the double-equation 
 
 Solved in the ordinary way, this gives x = - 1|- ; consequently the 
 triangle is formed from - T 5 ^, i, or from -5, 12. 
 
 We could proceed, as in the last problem, to deduce a new value for x, 
 but we observe that the triangle formed from 5, 12, i.e. the triangle 169, 
 119, 1 20, satisfies the conditions. 
 
 4. To find a right-angled triangle , , 77 such that 
 
 + mr) 
 where m is any number. 
 
 Fermat takes the case where m = 2. 
 
 Form a triangle from x, i; the sides are then = x'' 1 + i, f = # 2 - i, 
 7 = 2*.
 
 THEOREMS AND PROBLEMS BY FERMAT 301 
 
 Therefore 
 
 I must both be squares. 
 
 The difference = 24.*, and by the usual method we find x - T \. 
 But = x 2 - i is negative unless x> i. We therefore begin afresh and 
 form a triangle from x + 5, 12. 
 The sides of this triangle are 
 
 x*+iox+i6(), x?+iox 119, 24^+120. 
 We have therefore to solve the double-equation 
 x?+ iox+ 169 = &t 2 | 
 x*+ $Sx+ 121 = z> 2 ) 
 
 Fermat multiplies the two expressions together and puts 
 x 4 + 6&X 3 + &JOX 2 + noi2x + 20449 = a square 
 
 = (143 + -Yfar* - ffff|&f ^f, say ; 
 
 and the triangle is formed from 103447257961, 17749110120. 
 
 The double-equation could also have been solved by the usual 
 Diophantine method, as in the next problem to be* given. 
 
 5. To find a right-angled triangle , , rj such that 
 
 _.!*} 
 
 where m is any number. 
 Suppose that m - 2. 
 Form a triangle from x + i, i, so that the sides are 
 
 = X?+ 2X + 2, = X 2 + 2X, 1} 2X + 2. 
 
 Therefore we have to solve 
 
 X 2 + 2X + 2 = U* ' 
 
 Solving in the usual manner, we obtain x = - $i, so that the triangle is 
 formed from -^, i, or from - 5, 12, and is therefore (169, 119, 120). 
 
 We have to replace the value of x by a value which will avoid the 
 negative sign. Form a triangle, then, from # 5, 12. 
 
 The sides are or 2 - lox + 169, x 2 - 10^-119, 24^-120. 
 The double-equation now becomes 
 
 x 2 - iox+ 169 = 
 x 3 58.*+ 121 
 Multiply the second equation by iff, and we nave to solve 
 
 X s - iox+ 169 = u* ) 
 T -If *-" ~ ^nrr x + l ^9 = v ) 
 
 = w 2 ) 
 =1? }
 
 302 SUPPLEMENT 
 
 The difference = ^x* - - 8 ^ 2 -x = T \ x (ftx - f^). 
 Equating the square of half the difference of the factors to the smaller 
 expression (or the square of half the sum to the larger), we have 
 
 -^ttm 1 . 
 
 and the required triangle is formed from 4363225, 552552, the sides being 
 
 19343046113329, 18732418687921, 4821817400400. 
 Or again in this case we can multiply the expressions or 10^+169 
 and x 2 - 580; +121 and put their product 
 
 x 4 6&X 3 + S"jox 2 1 10120; + 20449 a square 
 
 / T A t 5 5 6 A* _i_ .v2\2 our 
 
 = ( J 43 - -\-^- x + x ) > sa y> 
 
 and the result will be the same as before, 
 
 X 
 
 6. To find a right-angled triangle , , 77 J^ /$/ 
 
 = 2 ) 
 
 ztf/jm- m is any given number. 
 
 Let m = 3. Form a triangle from # + i, i ; its sides will be 
 
 X 2 + 2X + 2, = X~ + 2X, Tf] 2X + 2. 
 
 We have therefore to solve the double-equation 
 
 Solving this in the ordinary manner, we shall find x = T V. 
 
 Hence the triangle is formed from j!*, i, or (in whole numbers) from 
 13, 12 ; the sides are therefore 313, 25, 312. 
 
 Fermat also finds the solution by multiplying the two expressions and 
 making the product a square; 
 
 x 4 + lox 3 + 22# 2 + i2x a square 
 
 = (x 2 + $x |^) 2 , say. 
 
 This gives the same value of x as before, x = ^; and the triangle 
 is 3i3, 25, 312. 
 
 7. Jb find a right-angled triangle , ^, 17 such that 
 
 where m is a given number. 
 Fermat takes the case m 3. 
 
 Remembering that in the corresponding problem with a plus sign we 
 found the triangle 313, 25, 312 which is formed from 13, 12, we form the 
 triangle in this case from x 13, 12 ; its sides are 
 
 = # 2 26# + 313, = # 2 - 26.*; + 25, 17=24^-312. 
 We have then x' 2 - 26x + 25 = u- 
 
 x-- gSx + 96 !=".
 
 THEOREMS AND PROBLEMS BY FERMAT 303 
 
 Multiplying the first expression by ^U, we have to solve the double- 
 equation 
 
 * 2 - 98* + 961 = 
 The difference = Y/-* 3 - m^-x = ^x (^-x - i-W-)- 
 Proceeding as usual, we find * = .*fj8l ; the triangle is formed 
 
 from x- 13, 12, or (in whole numbers) from 23542921, 3820440, and the 
 
 sides are 
 
 568864891005841, 539673367418641, 179888634210480. 
 The same result is obtained by multiplying the expressions x~ - 26* + 25 
 and x~ qSx + 961 and making the product a square; we put 
 x 4 - 124^ + 3534# 2 - 27436^+ 24025 -a, square 
 
 and the result is x = l^ffi|i, as before. 
 
 8. To find a right-angled triangle , , *} such that 
 
 + mt] - v 2 
 where m is any given number. 
 
 Suppose m - 2. Form a triangle from x + i, i ; the sides are 
 
 = X? + 2X + 2, = ^ 3 + 20-', 
 
 We have then to solve the double-equation 
 
 X 2 + 2X - U- 
 
 The usual method gives x= \, and the triangle is formed from f , i, or 
 (in whole numbers) from 5, 4, being the triangle (41, 9, 40). 
 
 Since + r; = :x; 2 +4:r + 4 = a square, we have actually solved the problem 
 of finding a right-angled triangle , f, rj such that 
 
 = 
 + ri = 
 
 9- To find a right-angled triangle , , 17 such that 
 
 i= a 
 
 #/?7 = V ) 
 
 where m is a given number. 
 
 Suppose m - 2. 
 
 Since the corresponding problem with a plus sign just preceding has 
 the solution (41, 9, 40) formed from the numbers 5, 4, we form a triangle 
 in this case from re -5, 4; the sides are 
 
 = x' 2 - IQ.V + 4 r, = re 2 IQX + 9, i/ - 8* - 40.
 
 3 o 4 SUPPLEMENT 
 
 We have then to solve the double-equation 
 
 9 = 
 
 X 2 26x + 121 = 
 
 De Billy (or Fermat) observes that this double-equation "seems to admit 
 of solution in several ways, but it will be found that it is hardly possible to 
 find a practical solution except by the new method " (expounded earlier in 
 the In-ventum Nbvum) of making the absolute terms equal (instead of using 
 the equal terms in x 2 , which method gives, in fact, the value x - o). That 
 is to say, we make the absolute terms in the two expressions equal by 
 multiplying the first by if 1 ' and the double-equation becomes 
 
 1 2 1 ~2 1 210 v , T _ T , >: 
 
 p XT TJ X + I 2 I = U 
 
 X 2 - 26X+ 121 = V' 
 
 The difference = ^x 2 - ^x = x (2 x - ^). 
 
 Equating the square of half the difference of the factors to v 1 , or the 
 square of half their sum to u" 2 , we find x = --. 
 
 Therefore the triangle is formed from 4^ 3 -, 4 or (in whole numbers) 
 from 493, 132, and the sides are 260473, 225625, 130152. 
 
 Since -77 = ^- i&c + 8i =a square, the above actually amounts to the 
 solution of the problem of finding a right- angled triangle , , T/ such that the 
 three conditions 
 
 are simultaneously satisfied. 
 
 De Billy (or Fermat) observes however that, while the above one solution 
 satisfies the conditions of both problems, it is not so with all solutions of 
 the problem involving the two conditions only; but v\\\y primitive triangles 
 satisfying the conditions of that problem satisfy the additional condition. 
 Thus the triangle (624, 576, 240) is such that one of the perpendicular 
 sides is a square and the difference between the hypotenuse and twice the 
 other perpendicular is also a square, but the hypotenuse minus the latter 
 perpendicular is not a square. 
 
 10. To find a right-angled triangle , , r/ such that 
 
 Assume x, i - x for the sides , 17 about the right angle respectively. 
 This supposition satisfies the second condition. 
 
 Again, since gr] = x-x 2 , the third and fourth conditions are satisfied, 
 for tf = x\ w*=i
 
 THEOREMS AND PROBLEMS BY FERMAT 305 
 
 It remains to satisfy the conditions 
 
 n=i -x = t* ) 
 
 t. 
 
 and 4 -e +T= ! 2* + 2# 2 = a square) 
 
 The difference = 2X 2 x = \x (4^-2), and we find, in the usual way, 
 The triangle is (^, |-, ^). 
 
 n. To find a right-angled triangle , , r/ such that 
 
 = a cube j 
 f>7 = a square] 
 
 Fermat assumes =i, f\ = x, so that the first condition is satisfied, 
 i being a cube. 
 
 We must now have i-x = u*) 
 
 and also " 2 = i 2 + tf = i + y? = v 1 ] 
 
 The difference = x 2 + %x = x (4^ + 2), and we find x = - f |f . 
 
 In order to derive a positive value for # we substitute y 1-|4 for x in 
 the equations, which gives 
 
 Make the absolute term in the first equation equal to that in the latter 
 by multiplying by , and we have to solve 
 
 The difference =/ - 
 We find accordingly 
 
 and 
 
 The triangle is then 
 
 12. To find a right-angled triangle , ^, 77 jw// that 
 
 Form a triangle from the numbers x+ i, x ; the sides are 
 
 =2* 2 +2.r+I, =2X+I, r)= 
 
 Thus ^ + ^ = 2X* + $x* + $x + i must be a square. 
 Suppose 2 x 3 + 5** + 3* + i = (|* + i ) s ,
 
 306 SUPPLEMENT 
 
 Now substitute y - -- for x in the expression to be made a square, and 
 we have 
 
 2/ - ^/ + Hy + |f | = a square 
 
 whence 7 = *&&, and * = 
 
 The triangle is accordingly found. 
 
 13. To find a right-angled triangle , , 77 such that 
 
 Form a triangle as before from x+ i, x, and in this case we shall have 
 2X* + 3# 2 + 3# + i = a square 
 
 = (f*+i) 2 , say, 
 whence x = f . 
 
 Substitute ^ - f for ^ in the expression to be made a square ; thus 
 2^ = a square 
 
 whence .7 = Wf . and 
 
 the triangle being therefore 
 
 ttttffl, 
 
 14. To find a right-angled triangle ^ , t\ such that 
 
 Let = A;, >j = i ; then 2 = a^ + i = a square] 
 
 Also, by the condition of the problem, iv 2 + f # + i = a square) 
 The usual method of solution gives # = ! 
 
 Substitute therefore j - 1-| for x in the two expressions, and we have 
 the double- equation 
 
 Or, if we make the absolute term in the first expression the same as in 
 the second by multiplying by jf||, 
 
 The difference = f f ^/ - Ullj = Hy (Wy ~ f Hf ). 
 and we find j = ffiff , so that x=y- |f = ^AV- 
 
 Therefore the two perpendicular sides of the triangle, in whole numbers, 
 are 39655, 129648, and the hypotenuse is 135577.
 
 THEOREMS AND PROBLEMS BY FERMAT 307 
 
 15. To find a right-angled triangle , , ij such that 
 
 ( + -t]f + b&l = a square. 
 
 This problem is mentioned in Fermat's letters to St Martin of 3ist May 
 and to Mersenne of ist September i643 x . The result only is given (in 
 the letter to Mersenne), and not the solution ; but it can easily be worked 
 out on the lines of the solution of the preceding problem. 
 
 Let = x, i) = i ; we must therefore have 
 
 L 
 
 both squares. 
 
 Solving in the usual way by splitting the difference f x into the factors 
 zx we find x = f$. 
 Substitute _y f$ for x in the two expressions, and we have to solve 
 
 Multiply the last by (f f ) 2 so as to make the absolute terms the same 
 and we have to solve 
 
 The difference = {(f|) 2 - 
 
 We therefore put (y - ^^f =/ - fty + 
 whence y (%& - ft) = < 
 
 and^-^^^, so that *=^-ft = ^ 
 
 The required triangle is therefore (V&Vo 8 * V/sWi or > in 
 numbers, (205769, 190281, 78320). 
 
 1 6. To find a right-angled triangle , 
 
 ^ + m . %rj = a square. 
 Fermat takes the case where m = 2. 
 Form a triangle from the numbers x, i ; the sides are then 
 
 =*+!, (=3?- I, 1J=2X. 
 
 Thus we must have (* + i) 2 + 2^: (- i) a square, that is, 
 x* + 2^ + 2^* - 2X + i = a square 
 
 = (* + * + |) 2 , say, 
 whence x ^. 
 
 1 Ofuvrfs de Fermat, II. pp. 260, 263.
 
 3 o8 SUPPLEMENT 
 
 But this value makes x 2 - i negative ; so we must seek another by 
 putting y + \ for x in the expression to be made a square. 
 We have f + 3/ + Qy 2 - T \jv + \\ \ = a square 
 
 This gives y = HI HI, and x =y + } = |||. 
 
 Therefore the triangle is generated (in whole numbers) from 571663 
 and 436440. 
 
 De Billy adds that there is one case in which the problem is impossible. 
 Tannery observes in a note that this remark seems to refer to the case in 
 which m = 8. 
 
 17. To find a right-angled triangle , , i] such that 
 
 -\&\ = a square. 
 Form a triangle from xi, 4 ; the sides will then be 
 
 = * 2 -2* + i7, = x*-2x-is, r l = 8x-&. 
 
 Thus (x z -2x- \$f-(4x- 4) (x 2 - 2^-15) must be a square, that is, 
 x* Sx 3 1 4# 2 + ii2x + i65 = a square 
 
 - (x 2 - 4* -is) 2 , say. 
 
 This gives x = *-/-, and accordingly, to find another value, we substitute 
 y - Y for x in the expression to be made a square. 
 We must therefore have 
 
 y - 3 8/ + i<y>-I/ - *8iy + &3^A = a square 
 
 This gives y = VAS and * = j - Y = VTIT- 
 
 The triangle is therefore formed from - 6 ^$p, 4, or (in whole numbers) 
 from 6001, 2280. 
 
 The sides are therefore 41210401, 30813601, 27364560. 
 
 1 8. To find a right-angled triangle , , r\ such that (if > ij) 
 ( - r/f - 2tf = a square. 
 
 This problem is enunciated in Fermat's note on vi. 22. He merely 
 adds that the triangle (1525, 1517, 156) formed from 39, 2 satisfies the 
 conditions, but does not give the solution. 
 
 The solution is however easy to obtain by his usual method, thus. 
 
 Form a triangle from x, i, so that 
 
 = *"+!, = # 2 -I, t]=2X. 
 
 Then (t-r ) Y-2rf = (x z -2X-iY-%x i 
 
 = x*- 4X 3 -
 
 THEOREMS AND PROBLEMS BY FERMAT 309 
 
 This has to be a square ; let it be equal to (x 2 - 2x - $y, say ; this gives 
 = 2ox + 25, 
 
 x - - 
 
 The triangle formed from -f, i, or from -3, 2, will have one side 
 negative. To avoid this, we proceed as usual to form a triangle from 
 -. 2. 
 
 Thus =y-6.r+i3, t=y*- 
 
 and (-i?)-2if = (/-iqy+ 17)'- 2(47-12)' 
 
 =y 2qy s + I02J 2 - 1487 + i. 
 
 In order that this may be a square, suppose it equal to (jr- loy - i) 2 or 
 y* - 2oy* + gSy + 2oy + i. 
 
 It follows that 1023? 1487 = gSy 2 + 2oy, 
 
 and y = 42. 
 
 The triangle required is formed from y~3, 2, that is, from 39, 2, 
 and is accordingly 1525, 1517, 156. 
 
 Fermat does not tell us in the note on vi. 22 what use he made of this 
 problem, but the omission is made good in a letter to Carcavi 1 , where he 
 says that it was propounded to him by Frenicle (who admitted frankly that 
 he had not been able to solve it), and that it served to solve another 
 problem which had occupied Frenicle. The latter problem is the 
 following. 
 
 19. To find a right-angled triangle , , t] such that 
 
 1 
 
 - >/ j- are all squares. 
 
 *-lJ 
 
 Fermat does not actually give the solution, but presumably it was 
 somewhat as follows. 
 
 Form a triangle from two numbers x, y ; the sides are then 
 
 Now - T/ - x? +y - 2xy and is ipso facto a square. 
 
 The other conditions give 
 
 x 3 +y* = a square, 
 
 and or ! -y ! -2^>' = (^-j>') 2 -2/ = asquare. 
 
 These conditions are satisfied by the two perpendicular sides of the 
 triangle of the last problem, that is, by x= 1517, y= 156. 
 i Oeuvres de Fermat, n. p. 265.
 
 310 SUPPLEMENT 
 
 The triangle required is therefore formed from 1517, 156 and is 
 (2325625, 2276953, 4733 4)- 
 
 The present seems to be the appropriate place for a problem contained 
 in a letter from Fermat to Frenicle the date of which was probably 
 15 June 1641'. 
 
 20. To find all the right-angled triangles in integral numbers such that 
 the perpendicular sides differ by i. 
 
 If a right-angled triangle is formed from x, y, the difference between 
 the two perpendiculars is either x*y i 2xy or 2xy-(x 2 -y*), that is 
 to say, either (x-yf2y i or zy i -(xyf. As this difference is to be i, 
 we have to find all the integral solutions of the equation 
 
 2y t -(x-y)* = l. 
 
 Those who are familiar with the history of Greek mathematics will here 
 recognise an old friend. The equation is in fact the indeterminate 
 equation 
 
 *-**! i, 
 
 which the Pythagoreans had already solved by evolving the series of 
 "side-" and "diagonal-" numbers described by Theon of Smyrna, the 
 property of which they proved by means of the geometrical theorems 
 of Eucl. n. 9, to. 
 
 If x, y are two numbers such that 
 
 2X 2 y^ + i, 
 then the numbers x +y, 2X +y will satisfy the equation 
 
 2*-,f = -i; 
 
 fresh numbers formed from x +y, 2x +y by the same law will satisfy the 
 equation 
 
 2^-7, 2 = +I, 
 
 and so on. 
 
 Take now the equation 
 
 2)> 2 - (* -j) 2 = I, 
 
 where x, y are two numbers from which a right-angled triangle has been 
 formed. We can deduce a right-angled triangle formed from x', y' where 
 
 2/ 2 -(*'-/) 2 =+i; 
 
 for by the above law of formation we have only to take 
 
 y=y + (x-y) = x, 
 x' -y' = 2 y+(x -y) = x +y, 
 whence also x' = 2X +y. 
 
 1 Oeuvres de Fermat ', n. pp. 321 sqq.
 
 THEOREMS AND PROBLEMS BY PERM AT 311 
 
 Fermat gave two rules for the formation of this second triangle. The 
 first rule is in the letter above quoted. 
 
 First Rule. If h, /, b be any right-angled triangle satisfying the con- 
 dition (h being the hypotenuse, / > b and / - b = i), then, if a triangle be 
 taken in which 
 
 the least side =2h+p + 26, 
 the middle side =2/i+j>+ 26+ i, 
 the greatest side = 3^ + 2 (/ + 6), 
 this triangle also will be a right-angled triangle satisfying the condition. 
 
 To verify this from the above considerations we have to consider two 
 cases, according as 2xy is greater or less than x 2 -y 2 . 
 
 Take the case in which 2xy > x 2 -y 2 ; then 
 
 2/-(*-jf=+i, 
 and accordingly 
 
 2/ 2 -(*'-/)* = -i, 
 or x' 2 -y' 2 > 2x'y'. 
 
 The least side, therefore, of the second triangle 
 
 2x'y = 2X (2X +y) = 2 (x 2 +/) + (axy) + 2 (x 2 -/) ; 
 the middle side 
 
 x' 2 -y' 2 = 2x'y + i ; 
 and the hypotenuse 
 
 X 1 * +/ 2 = (ax +yf + x 2 = 3 (* 2 +/) + 2 (x 2 -/ + axy). 
 The expressions on the right hand are those given by Fermat's rule. 
 Second Rule. 
 
 This rule is given in a letter of 31 May 1643 probably addressed 
 to St Martin 1 . 
 
 Fermat says : Given any triangle having the desired property, then, to 
 find another such triangle from it, "subtract from double the sum of 
 all three sides each of the perpendiculars separately [this gives two of the 
 sides of the new triangle], and add to the same sum the greatest side [this 
 gives the third side]." 
 
 That is to say, the sides of the new triangle are respectively 
 
 2Xy)- 2Xy, 
 
 1 Oeuvres de Fermat, II. p. 259.
 
 3 i2 SUPPLEMENT 
 
 In fact the three expressions reduce as follows : 
 
 2 (2X 2 + 2xy) (x 2 y 2 ) = 33? + ^xy +y 2 = (2X + y} 2 x 2 , 
 
 2(2X 2 +2Xy)-2Xy=2x(2X +y), 
 
 2 (2X* + 2xy) + x 2 +y z = (2x +y) 2 + x 2 ; 
 and the result agrees with the formation of the triangle from x', y' above. 
 
 From the triangle (3, 4, 5) we get (20, 21, 29); from the latter the 
 triangle (119, 120, 169), and soon. The sixth such triangle is (23660, 
 23661, 33461). 
 
 21. To find all the rational right-angled triangles in ivhole numbers 
 which are such that the two perpendiculars differ by any given number. 
 
 To his explanation of the First Rule above, applicable to the case 
 where the given number is i, Fermat adds in his curt way : "same method 
 for finding a triangle such that the difference of the two smaller sides is 
 a given number. I omit the rules, and the limitations, for finding all the 
 possible triangles of the kind required, for the rule is easy, when the 
 principles are once admitted." 
 
 He adds, however, to his Second Rule 1 its application to the case 
 where the given number is 7. 
 
 There are, he says, two fundamental triangles with the desired property, 
 namely 5, 12, 13 and 8, 15, 17. [In the case of the former 2xy > x 2 -y*, 
 and in the case of the second x 2 -y 2 > 2xy.] 
 
 From the first triangle (5, 12, 13) we deduce, by the Rule, a triangle 
 with the sides 2 . 30- 12, 2 . 30 -5, 2 . 30 + 13 or (48, 55, 73) ; from the 
 second a triangle with the sides 2.40-15, 2.40-8, 2.40+17, or 
 (65, 72, 97)- 
 
 And so on, ad infinitum. 
 
 Next to the explanation of the first of the above Rules Fermat 
 mentions, in the same letter, the problem 
 
 22. To find right-angled triangles in integral numbers , , 77 (>/) 
 
 such that 
 
 v \ 
 
 are both squares. 
 
 He observes that alternate triangles of the series in which the two 
 smaller sides differ by i satisfy the conditions, those namely in which the 
 smallest side 77 is 2xy and not x 2 -y 2 ; for x*+y 2 -2xy is a square, and 
 -7, being equal to i, is also a square. 
 
 1 Oeuvres de Ffrmat, n. p. 250.
 
 THEOREMS AND PROBLEMS BY FERMAT 313 
 
 Thus, while 3, 4, 5 does not satisfy the conditions, (20, 21, 29) does, 
 and, while the next (119, 120, 169) does not satisfy the conditions, the 
 triangle after that, namely (696, 697, 985), does. 
 
 Frenicle naturally objected, in his reply, that the triangles should not 
 be limited to those in which the smaller square representing the difference 
 between the perpendicular sides is r, and proposed the problem in the form 
 
 To find all the triangles (, , 77) such that 
 
 are both squares, 
 
 and one square does not measure the other. 
 
 Fermat seems to have, in the first instance, formed the triangle 
 from two numbers x, y where 
 
 x = r*+ i, y= 2r-2, 
 
 and then to have given the more general rule of forming a triangle from 
 x = r s + s 2 , y=2(r-s)s, 
 
 where r, s are prime to one another (Letter from Frenicle of 6 Sept. 
 1641)'. 
 
 It appears from a letter of Fermat's to Mersenne of 27th January 
 i643 2 that St Martin propounded to Fermat the problem, apparently 
 suggested by Frenicle 3 , 
 
 Given a number, to find how many times it is the difference between the 
 [perpendicular ?] sides of a triangle which has a square number for the 
 difference between its least side and each of the two others respectively. 
 
 The number given was 1803601800, and Fermat replies that there are 
 243 triangles, and no more, which satisfy the conditions. He adds " The 
 universal method, which I will communicate to him if he asks for it, is 
 beautiful and noteworthy, although I doubt not that Frenicle has already 
 given him everything on the subject of these questions." 
 
 23. To find two triangles, , , rj and ', ', 77' ( > 77, % > 77') such that 
 
 Suppose the two triangles formed from (x, y) and (x', /) respectively, 
 the sides being 
 
 = *+/, t=2xy, 17 = -/. 
 
 1 Oeuvres de Fermat, n. p. 233. 
 
 2 Ibid., p. 250. 
 
 3 Ibid., p. 247.
 
 3M SUPPLEMENT 
 
 Then we must have 
 
 (*-J>) 2 =2/ 2 -(*'- 
 
 and 2f-(x-y}* = (x'-yJ 
 
 which equations show that y =y', and that 
 
 are three squares in arithmetical progression. 
 
 Suppose that these squares are i, 25, 49 respectively ; thus y = 5 ; 
 x-y=i, so that #=1 + 5; x'-y^y, so that #' = 5 + 7. 
 
 Fermat accordingly gives the rule : Find three squares in arithmetical 
 progression ; then form the first triangle from (i) the sum of the sides of 
 the first and second squares and (2) the side of the second, and the 
 second triangle from (i) the sum of the sides of the second and third 
 squares and (2) the side of the second 1 . 
 
 In the particular case, the triangles are formed from (6, 5) and from 
 (12, 5) respectively; the triangles are therefore (61, 60, n) and (169, 120, 
 119) respectively. 
 
 For solving the problem of finding three square numbers in arithmetical 
 progression Fermat seems first to have given a rule which was not general, 
 and then in a later document to have formed the sides of the three squares 
 as follows : 
 
 r^-zs 2 , r* + zrs + 25*, r* + \rs + 2S 2 . 
 
 Frenicle formed them thus 2 : 
 
 the latter form agreeing with Fermat's if p = r + s, and q = s. 
 
 Fre'nicle expresses his formula neatly by saying that we take for the 
 side of the middle square the hypotenuse of any primitive triangle formed 
 from p, q, i.e. p^ + q"*, for the side of the smallest square the difference 
 between the perpendicular sides of the same triangle, i.e. p^-q^zpq, and 
 for the side of the largest square the sum of the perpendicular sides of the 
 same triangle. 
 
 Suppose the primitive triangle is (28, 45, 53) formed from (7, 2). 
 Then the sides of the three squares in arithmetical progression are 17, 53 
 and 73, the squares themselves being 289, 2809, 5329. The triangles 
 derived from these squares and having the above property are formed from 
 (7> 53) an d from (126, 53) respectively, and are therefore (7709, 7420, 
 2091) and (18685, i335 6 13067)- 
 
 1 Oeuvres de Fermat, n. p. 225. 
 
 2 Ibid., II. pp. 234-5.
 
 THEOREMS AND PROBLEMS BY FERMAT 315 
 
 24. To find two right-angled triangles (, & t\) and (', ', 17') such that 
 
 Form the triangle , , r; from the numbers x, i ; then 
 
 =* 2 +i, = # 2 --i, 17 =2*. 
 Thus '= x z + i ; and, since ' - 17' = - 17 = # 2 - 2* - i, it follows that 
 
 T\ = 2X + 2. 
 
 It remains to secure that ' 2 + >/ a = (a; 2 + i) 2 + (2X + 2) 2 shall be a square, 
 that is, 
 
 x* + 6x* + 8x + 5 = a square 
 
 = (* 2 +3 ) 2 ,say; 
 therefore x = ^. 
 
 Hence the triangle , ^, 17 is formed from , i or from 1,2; but this 
 solution will not do, as it gives a negative value for . Accordingly 
 we must find a fresh value for x, which we obtain by forming the triangle 
 from x + i, 2. 
 
 The sides are then 
 
 (;= + 2* -i- 5, = * 2 + 2* - 3, 17 = 4^ + 4; 
 thus ' = .x: 2 + 2^ + 5, T ? / = ^'-(jc 2 -2^-7) = 4^+ 12. 
 
 Therefore (x* + 2X -t- 5) 2 + (4^ + i2) 2 must be a square, or 
 # 4 + 4X 3 + 30^ + i i6x + 169 = a square 
 
 ^Oa + yf*-* 2 ) 2 ' sa y 
 
 from which we obtain # = --V$n and the triangle is formed from -, 2, 
 or (in whole numbers) from - 979, 1092. 
 
 " We can use these numbers as if both were real and form the triangle 
 from 1092, 979. We thus obtain the two triangles 
 
 2150905, 2138136, 234023, 
 
 2165017, 2150905, 246792, 
 which satisfy the conditions of the question." 
 
 25. To find two right-angled triangles (, , 17) and (C, , V) such that 
 
 Form the triangle , & t\ from the numbers x + i, i ; then 
 
 = # 2 + 2# + 2, = ^+2*, 17 = 2* + 2. 
 Thus ? = x*+2x+ 2, and 77' = + *7-'= 2a; -
 
 316 SUPPLEMENT 
 
 We must now have ' 2 + rj' 2 - (x 2 + 2X + 2) 2 + (2#) 2 a square ; that is, 
 x 4 + 4X 3 + 1 2X t + Sx + 4 - a square 
 
 - (x* + 2X + 4) 2 , say ; 
 whence x= |. 
 
 Accordingly we substitute y - f for x, and we must have 
 
 y4 _ 2 yS + IByZ _ 20 y + 1 69 _ & S q Uar6 
 
 This gives j = f , and * = f - 1 = ^ 
 
 The triangle , , y is therefore formed from f, i, or from 29, 26, and 
 is therefore 1517, 165, 1508 ; the triangle ', ', 77' is 1525, 1517, 156. 
 
 Or again we may proceed thus from the point where we found x = - f . 
 The triangle , , >; may be formed from - J, i or from - i, 2. 
 
 We therefore form a triangle from x i, 2 and start afresh. 
 
 The sides are 
 
 = * 2 -2*+5, = x*-2x-3, 77 = 4^-4. 
 
 Thus ' = x" - 2X + 5, and ?/ = + 77 - ^' = $x - 12. 
 
 Hence (x 2 - 2X + s) 2 + (4^ - i2) 2 must be a square ; that is, 
 x* - 43? + 30# 2 - n6x+ 169 = a square 
 
 = (i3-ff* + :x;2 ) 2 > sav - 
 
 This gives ^^yf, and the triangle , ^, 77 is therefore formed from 
 y|^, 2, or from 29, 26, as before. 
 
 The remaining problems on rational right-angled triangles in the 
 Inventum Novum are cases given in Part n. of that collection to illustrate 
 the method of the Triple-Equation due to Fermat and explained by him on 
 Diophantus vi. 22 as well as, at greater length, in the Inventum Novum. 
 An account of the method will be found in a later section of this Supple- 
 ment ; but the problems applying the method to right-angled triangles 
 will be enunciated here. 
 
 26. To find a right-angled triangle , , 77 such that 
 
 By Problem 2 above find a right-angled triangle h, /, b (h being the 
 hypotenuse) in which h, p + b are both squares ; the first condition is thus 
 satisfied.
 
 THEOREMS AND PROBLEMS BY FERMAT 317 
 
 To find , , 77, put = hx, =px, ri = bx. 
 
 The three remaining conditions thus give a "triple-equation" in x. 
 [The numbers would of course be enormous.] 
 
 27. To find a right-angled triangle. , , 77 such that 
 
 where m is any given number. 
 
 Fermat supposes m = 2. 
 
 Assume for the required triangle (3*, 4*, 5*); we have then the 
 triple-equation 
 
 144^+ 3* = M 
 
 144** + 4x=v* U 
 
 the solution of which gives x = ^^TF> an d the triangle is 
 
 507 .676 _4A| 
 
 8016* 8lTTS"> iSOl6' 
 
 28. To find a right-angled triangle , ^, t\ such that 
 
 Suppose w = 2. 
 
 Find a triangle (Problem 3 above) in which , - 17 are both squares, 
 say the triangle (119, 120, 169). Put 119*, 1200:, 169* for the sides of 
 the required triangle, and we have the " triple-equation " 
 166464;*?+ 
 
 166464^+ 
 
 1 66464^ + 338^ = 
 29. To find a right-angled triangle , ^, t\ such that 
 
 (f + i ? + ) a + C = 
 
 where m is any given number. 
 
 * The enunciation has ( - ^T;) instead of ^ - 17 ; but ( - Jfi;) is inconsistent with 
 the solution given, and I have therefore altered it so as to correspond to the solution.
 
 318 SUPPLEMENT 
 
 Take a right-angled triangle in which i,/ are the sides about the right 
 angle and are such that i - \p is a square (Problem 1 1 above). 
 
 Let q be the hypotenuse of the triangle so taken, so that q - V(/ 2 + l )> 
 and take as the sides of the required triangle x, px, qx ; we thus have 
 the triple-equation 
 
 (i +/ + ?)***+/* = 
 (i +/ + qf y? + mqx = 
 
 30. To find a right-angled triangle , , t] such that 
 
 First find a triangle in which one of the perpendiculars is a square, and 
 the sum of the perpendiculars is also a square, say 40, 9, 41. 
 
 Take 403:, yx, 41 x as the sides of the required triangle; and we there- 
 fore have the triple-equation 
 
 Sioox 2 + ^ox u z 
 8 1 oox 2 + X = 
 
 SECTION IV. 
 
 OTHER PROBLEMS BY FERMAT. 
 
 31. To find two numbers , t\ such that 
 
 (1) (-(?-*)} 
 
 (2) r)-(g*- rf) > are all squares. 
 (3), ( 4 ) tr,H?-W 
 
 Let ^ + i] = i 2x, t) 2x, so that $ = , i\ = ^ 2x, and 
 
 e-rf=2X-4X*. 
 Thus (3), (4) are both satisfied. 
 The other conditions (i) and (2) give 
 
 4X 2 - 2X + = 
 
 The difference - 2X = $x . \ ; and, putting (2X + ^) 2 = 4^ 2 - 2X + \ , we
 
 THEOREMS AND PROBLEMS BY FERMAT 319 
 
 The required numbers are therefore J, ^ 5 T . 
 
 Another pair of numbers satisfying the conditions will be obtained by 
 substituting^ + T 7 F for x in the expressions to be made squares, and so on. 
 
 32. To find two numbers , 17 such that 
 
 Let the numbers be + x, % -x; therefore -TJ, as well as ?-if, is 
 equal to 2x, The sum f + 17 = i. 
 
 Therefore i 2X must be a square, or we have the double-equation 
 IH- zx 
 
 Replace x by |y + j so as to make i + zx a square ; therefore 
 i - 2y y* = a square 
 
 = (i-37) 2 , say, 
 whence y = f , and # = ^y 2 + _y = |. 
 
 The required numbers are therefore f, ^. 
 
 33. To find two numbers , 17 jr that 
 
 (+ 9 )(*+f)?=Mfe 
 
 Assume = #, 17 = 2 - # ; therefore 
 
 ( + ?) ( 2 + V 8 ) = 2 (z* 2 -4* + 4) = a cube 
 
 = (2 -**), say. 
 
 This gives # = - f ; and to get a " real " value of x we must substitute 
 y \ for x in the expression to be made a cube. 
 Thus 4^-44^ + 125 = a cube 
 
 and j = ^, so that *=>>-! = 
 
 The required numbers are therefore f|!inr, 
 COR. We observe : 
 
 (1) that the numerators 26793, 15799 satisfy the conditions ; 
 
 (2) that we have in fact solved the problem To divide 2 into two parts 
 such that twice the sum of their squares is a cube ; 
 
 (3) that we can solve in the same manner the problem To find two 
 numbers such that any multiple of the sum of their squares is a cube. Thus 
 suppose that the multiple is 5 ; we then assume x and 5 - x for the 
 numbers and proceed as above;
 
 320 SUPPLEMENT 
 
 (4) that we can also deduce the solution of the following " very fine 
 problem " : 
 
 To find two numbers such that their difference is equal to the difference of 
 their biquadrates or fourth powers. 
 
 In other words, we can solve the indeterminate equation 
 *-**.- ^r 
 
 For we have only to take the two numbers found above, namely 26793 
 and 15799, and divide by (as a common denominator) the root of the cube 
 formed by multiplying their sum by the sum of their squares. 
 
 This common denominator is 34540, and the two required numbers are 
 
 This latter problem is alluded to in Fermat's note to Diophantus iv. i r 
 in these terms : " But whether it is possible to find two biquadrates the 
 difference between which is equal to the difference between their sides is a 
 question to be investigated by trying the device furnished by our method, 
 which will doubtless succeed. For let two biquadrates be sought such 
 that the difference of their sides is i, while the difference between the 
 biquadrates themselves is a cube. The sides will, in the first instance, be 
 -^ and |f. But, as one is negative, let the operation be repeated, in 
 accordance with my method, and let the first side be x--j^\ the second 
 side will be # + |f, and the new operation will give real numbers satisfying 
 the condition of the problem V 
 
 34. To find two numbers , tj such that 
 
 4 + S 7 ? 4 = a square, 
 
 Fermat (or De Billy) observes that it must be required that the first 
 biquadrate ( 4 ) shall not be unity, for in that case the problem would be 
 too easy, since 1 + 3.1 = 4 and i + 3 . 16 = 49. 
 Assume = x, ti = x-i ; therefore 
 
 4^ -12:^ + 18^ - i zx + 3 = a square 
 
 '= (zx z - 3* + f ) 2 , say. 
 
 This gives ^=\ 1 -, #-i = f; and a solution in whole numbers is 
 i = n, T/ = 3. In fact n 4 + 3 . 3 4 = 14641 + 243= 14884 or i22 2 . 
 
 We can also take any equimultiples of (n, 3), as (22, 6) and (33, 9) ; 
 and the latter pairs of numbers severally satisfy the condition of the 
 problem. 
 
 1 It gives in fact ^ , * as a solution of the subsidiary problem, and from this 
 we can obtain the same solution of the main problem as that given above 
 /26T93 i5799\ 
 \3454' 3454/ '
 
 THEOREMS AND PROBLEMS BY FERMAT 321 
 
 SECTION V. 
 
 FERMAT'S TRIPLE-EQUATIONS. 
 
 Fermat's own description of his method of " triple-equations," which is 
 contained in his note on vi. 22, is as follows : 
 
 "Where double-equations do not suffice, we must have recourse to 
 triple-equations, which are my discovery and lead to the solution of a 
 multitude of elegant problems. 
 
 If, for example, the three expressions 
 
 .v + 4, 2.r + 4, 5* + 4 
 
 have to be made squares, we have a triple-equation the solution of which 
 can be effected by means of a double-equation. If for x we substitute a 
 number which when increased by 4 gives a square, e-g.jr + 4y [Fermat says 
 ^ + 4-r], the expressions to be made squares become 
 
 f + w + 4, 2jr + &y + 4, 5^ + 2 qy + 4. 
 
 The first is already a square ; we have therefore only to make 
 
 2y*+ Sj + 4 
 ^y 2 + 201- + 4 J 
 
 severally squares. 
 
 That is to say, the problem is reduced to a double-equation. 
 
 This double-equation gives, it is true, only one solution : but from 
 this solution we can deduce another, from the second a third, and so on. 
 In fact, when we have obtained one value for y [say _> = ], we substitute 
 for y in the equations the binomial expression consisting of y plus the value 
 found [i.e. y + a\. In this way we can find any number of successive 
 solutions each derived from the preceding one.' ? 
 
 The subject is developed in the Doctrinae Analyticae Inventum Novum 
 of De Billy already mentioned so often. 
 
 It will be observed that the absolute term in all the three expressions 
 to be made squares is a square. It need not be the same square in the 
 original expressions ; if the absolute terms are different squares, the three 
 expressions can, so far as necessary, be multiplied by squares which will 
 make the absolute terms the same, when the method will apply. 
 
 We may put the solution generally thus. Suppose that 
 
 have to be made squares (a, b, c or some of them may be negative as well 
 as positive). 
 
 Put ax =f + 2py, 
 
 which makes the first expression a square (or of course we could put 
 ax = a*y* + zay). 
 
 H. D. 21
 
 322 SUPPLEMENT 
 
 Substitute (y 2 + 2py)\a for x in the second and third expressions. 
 Therefore ~ 
 
 must both be squares ; or, if we multiply the first expression by r 2 and the 
 second by q* (so as to make the absolute terms the same), we have to solve 
 the double-equation 
 
 ' 
 
 - a q- (f + 2py} + fr* - v*. 
 
 The difference = - . j 2 + 2p . - - .y. 
 
 This has to be separated into two factors of the form \y, py + v, where 
 v must be equal to 2qr (in order that, when ^ {(A + /*)j + v} is squared and 
 equated to the first, or when i{(A-//,)j- v} 2 is equated to the second, of 
 the two expressions, the absolute terms ^V 2 may cancel each other). 
 
 A different separation into factors is possible if b[a and c\a are both 
 squares ; but otherwise, as Fermat says, the method gives only one 
 solution in the first instance ; the above difference must necessarily be 
 split into the factors 
 
 p(b^-cf\ , qr 
 
 *J-y and y -y + 2ar. 
 aqr p j 
 
 Half the sum of these factors 
 
 -" q 
 
 aqr p 
 
 f_-cfq^ r 
 apqr ) 
 
 Squaring this and equating it to (y~ + 2py) 4- /r 2 , we have 
 
 fi (aq-r i + brp--cp 1 q- 
 
 \\y( 
 
 t 2 - 7 \ apqr 
 
 therefore 
 
 ap
 
 THEOREMS AND PROBLEMS BY FERMAT 323 
 
 that is, 
 
 - zcafq'i* - zabfft*) 
 
 = wpft* (br*f + cff - aft*), 
 
 or _ + cff - a?**) 
 
 J 
 
 whence *!=?-. ^-M is found. 
 \ a J 
 
 Exx. from the Inventum Novitm. 
 2* + 4 -j 
 yc + 4 - to be made squares. 
 
 (i) 2* + 4 -j 
 
 - to be 
 
 Here a = 2, ^ = 6, ; r= 3, / = q = r = 2 ; therefore 
 
 4.2.32(6. 16 + 3. 16- 2. 16) 
 
 and 
 
 i6 2 (4 + 36 + 9 - 2 . 6 . 3 - 2 . 3 . 2 - 2 . 2 . 6) 
 
 16.7 
 23 ' 
 
 I / ., x I (112^ 4-II2) 56, 1120 
 
 -( r + 4,r) = - -I ^=^-,(112-4.2^) = - . 
 
 2 V - 2\2 3 2 23. j 23^ 529 
 
 i-| 
 
 4 j- to be 
 
 J 
 
 3* + 4 - to be made squares. 
 
 2X + Q 
 
 Here a= i, ^ = 3, ^= 2, p = i, q 2, r=3; therefore 
 
 36 2 + 9. 81 +4 . 16 12 . 36 16 . 36 6 . 36. 9 
 
 - 4 3 6 = 144 
 
 - 36 . 46 + 9 . 81 + 4 . 16 1 ~ 863 ' 
 
 and * = / + 2j = (i**) 2 + 2 (|4|) = ff|4f. 
 
 The disadvantage of the method is that it leads so soon to such very 
 large numbers. 
 
 Other examples from the Inventum Novum are the following, which, 
 like those above given, can be readily solved ab initio without using the 
 above general formula. 
 
 (3) To solve 
 
 i + 50: = "or) 
 ic se 
 
 -} 
 
 Put x =y* + 2>-, and substituting in the second and third expressions we 
 have only to solve the double-equation
 
 324 SUPPLEMENT 
 
 The difference = 3 (f + ay) = y (y + 2). 
 
 Equate the square of half the difference of the factors to the smaller 
 expression ; thus 
 
 (y i Y = zyr + 4y + i , 
 
 whence y = - 6, and x =y + zy = 24. 
 
 (4) Equations x + 9 = u*' 
 
 3* + 9 = 
 
 5* + 9 = < 
 In this case we put x -y 1 + 6y, and we have to solve 
 
 5 (y* + 6y) + 9 = w ' 2 ) ' 
 The difference = 2 (y* + 6y) = zy (y + 6) ; we then have 
 
 and y = - f, so that x =y* + 6y = -^Vr- 
 (5) Equations i+ x-u 2 
 
 If we assume x.=y* + 2y, we find y T 2 T and x = T 4 ^. 
 There are two other problems of the same sort which are curiously 
 enunciated. 
 
 (6) " To find three cubes such that, if we add their sum to numbers 
 proportional to the cubes respectively, we may have three squares." 
 
 What Fermat really does is to take three cubes (a 3 , ft\ c' A ) such that their 
 sum is a square (this is necessary in order to make the term independent 
 of x in each of the three expressions a square) and then to assume 
 a a x, Px, c>x for the numbers proportional to the cubes. He takes as the 
 cubes i, 8, 27, the sum of which is 36. Thus we have the triple-equation 
 36+ x = u* | 
 36 + Sx = z? \- . 
 
 T > 6 + 2'lX = W i } 
 
 We put x =y + 1 2y in order to make the first expression a square. 
 Then, solving the double-equation 
 
 36+8 (y- + izy) = v 2 
 
 we obtain y = ^3* and x =y- + i2y = -f|-. 
 
 (7) "To find three different square numbers such that, if we add 
 to them respectively three numbers in harmonic progression, the three 
 resulting numbers will be squares." 
 
 Fermat first assumes three square numbers i, 4, 16 and then takes 
 2X, 3#, 6# as the required numbers in harmonic progression. (He observes
 
 THEOREMS AND PROBLEMS BY FERMAT 325 
 
 that, of the three numbers in harmonic progression, the greatest must be 
 greater than the sum of the other two.) We thus have the triple-equation 
 
 I + 2X = li l 1 
 
 4 + 3*=^ \, 
 
 or, if we make the absolute terms the same square, 
 16 + 32.* = u"~ 
 
 l6+ I2X = V"* 
 
 Making the last expression a square by putting / + ^y for x, we solve 
 as usual and obtain y = ^- and x = \ ( y* + By) = ^^ 
 
 Fermat observes that triple-equations of the form 
 
 x 2 + $x = it? 
 that is to say, of the form 
 
 f-y? + ax = u> 
 fy? + bx - 1? 
 jZx* 4- ex - v? 
 
 can be similarly solved, because they can be reduced to the above linear 
 form by putting x=\\y and multiplying up by/. 
 Examples. 
 (i) To solve the triple-equation 
 
 43? + 6x = v 1 
 ^x 2 + gx ix? 
 If x = i/y, this is equivalent to 
 
 Putting y = l?z 2 + 22 and solving as usual, we find 
 
 -*TO7HP+8S=fftf and x = % 
 
 (2) Equations * + x = 
 
 90:- + 20; 
 This is equivalent to 
 
 v + i 
 
 * + x = i? } 
 4^r + 3^- = i? | 
 90:- + 20; = a* 2 J 
 
 ay 4- 9 = r</' 2 
 We put y = z- + 2s and, solving the double-equation 
 
 e find s - ^i, .r - f ^^, so that * - |UHS-
 
 326 SUPPLEMENT 
 
 (3) " To find three square numbers such that, if we add their sum to 
 each of their roots respectively, we obtain a square." 
 
 Choose, says Fermat, three squares such that their sum is a square and 
 such that the root of the greatest is greater than the sum, of the roots of the 
 other two (the reason for this last condition will shortly appear) ; e.g. let 
 the squares be 4, 36, 81, the sum of which is 121. 
 
 Let 4# 2 , 36^, Si.* 2 be the three square numbers required; therefore 
 
 I2IX 2 + bx^V* 
 
 = it? 
 
 The solution, arrived at as above, is x = 
 
 Fermat actually used his triple-equations for the purpose, mainly, of 
 extending problems in Diophantus where three numbers are found 
 satisfying certain conditions so as to find four numbers satisfying like 
 conditions. The cases which occur are in his notes to the problems 
 in. 15, iv. 19, 20, v. 3, 27, 28; they, are referred to in my notes on 
 those problems. 
 
 De Billy observes (what he says Fermat admitted he had not noticed) 
 that the method fails when, the absolute terms being the same square, the 
 coefficient of x in one of the linear expressions to be made squares is equal 
 to the sum of the coefficients of x in the other two. Thus suppose that 
 
 i + 2x, i + 3*, i + 5jc 
 
 have to be made squares. To make the first expression a square put 
 x = 2^ + 2j. The other expressions then become 
 
 i + 6y + 6y, i +' iqy + ioy-. 
 The difference is 4^ + %y = zy (2y + 2), and the usual method gives 
 
 (27 + i) 2 = io/ 2 + ioy + i, 
 or 6y 2 + 6y = o, 
 
 so that y = i , and consequently x = 2y* + 2y = o. 
 
 It does not however follow, says De Billy, that a set of expressions so 
 related cannot be made squares by one value of x. Thus i + $x, i + i6x 
 and i + 2ix are all squares if # = 3, the squares being 16, 49, 64. He 
 adds ( n) that "we must observe with Fermat" that the triple-equation 
 
 2 
 
 not only cannot be solved by the above method, but cannot be solved at 
 all, because " there cannot be four squares in arithmetical progression" which 
 however would be the case if the above equations had a solution and we 
 took i for the first of the four squares.
 
 THEOREMS AND PROBLEMS BY FERMAT 327 
 
 The subject of triple-equations has been taken up afresh in a recent 
 paper by P. v. Schaewen 1 . The following are the main points made. 
 (i) The equations ax+JP = 
 
 = it? 
 can be reduced to the form 
 
 i + ax = u" 
 
 i + b'x' - v' 
 
 i + c'x' = w 
 
 by substituting MX' for x, where m is the least common multiple of/' 2 , ? 2 , r 2 . 
 (2) The method of Fermat has the disadvantage that, with one 
 operation, it only gives one value for x and not by any means always the 
 smallest solution. From this point of view there is a better method, namely 
 that of finding the general solution of the first two equations, substituting the 
 general value of x so found in the third equation and solving the resulting 
 equation in a new unknown. Consider the equations 
 
 i + ax - it 1 \ 
 i + bx = i? > . 
 
 i + ex = -up j 
 
 Suppose i + ax = / 2 , some square. Therefore 
 i+6x=i +-(/ 3 -i), 
 
 and, multiplying by a 3 , we have to make 
 
 abp- + a 2 ab a square. 
 
 This is a square if/=i ; and we therefore substitute q + i for/. Thus 
 abq- + zabq + a 2 = a square 
 
 say. 
 
 Therefore (ab - ^ q = 2 ( a - 
 
 zan (m nti) 
 and 9 = ~>-> ' 
 
 whence p=g+i- ^j^TT^i 
 
 ((2amn abn 2 tn?Y 
 
 and 
 
 - 4 ;// w 
 
 Substituting this value of x in the expression I'+ex, we have a biquad- 
 ratic expression in m which has to be made a square, namely 
 M 4 - ^ctrfn + {4 (a + b) c - 2ab\ m*ri 2 - ^abcmi? 
 
 1 Bibliotheca Mathematica, IX 3 , 1909, pp. 289-300.
 
 328 SUPPLEMENT 
 
 Example. Find x such that 
 
 i x, i + 4#, i + "jx are all squares. 
 
 First find the general value of x which will make the first two ex- 
 pressions squares ; this is 
 
 or, if we substitute k for 2n/m, 
 
 We have now to make i + TX a square ; that is, 
 
 k* + 14/fc 3 + 23/ 2 i4/fc + i - a square. 
 
 The first solution of this is k = i, and by means of these values we get 
 the further values k = f and k - \^ (cf. Euler's solution of the problem of 
 making x 2 + i and x + i simultaneously squares quoted in my note on 
 pp. 84, 85). The corresponding values of a- are respectively 
 
 3 120 . 120120 
 ~ , -- o and --- j i- . 
 
 4 "2Q 2 421- 
 
 Fermat's method gives, as the next solution after f , the value 
 
 (3) v. Schaewen observes that the problem of finding x such that three 
 different expressions of the form mx + n are all squares can always be solved 
 provided that we know one solution ; in this case the absolute terms need 
 not be squares. I doubt however if he is right in supposing that the 
 possibility of solution in this case was not known to Fermat or De Billy. 
 I think it probable that Fermat at least was aware of the fact ; for this case 
 of the triple-equation is precisely parallel to that of the double-equation 
 
 2x + 5 = t? 
 
 6x + 3 = w- 
 
 given as a possible case by Fermat in his note on Sachet's conditions for 
 the possibility of solving double-equations (cf. note on p. 287 above). 
 Fermat says that the square to which 2x + $ should be made equal is 16 
 and that to which 6x + 3 should be made equal is 36 (corresponding to 
 x = 5^), adding that an infinite number of other solutions can be found. 
 
 (4) Lastly, v. Schaewen investigates the conditions under which the 
 equations 
 
 i + ax = 2 , i + bx = v 2 , i + (a + b} x = w~, 
 
 which cannot be solved by Fermat's method, are nevertheless capable of 
 solution, and shows how to solve them when they have a solution other 
 than x = o.
 
 SOLUTIONS BY EULER. PROBLEM i 329 
 
 SECTION VI. 
 
 SOME SOLUTIONS BY EULER. 
 
 PROBLEM i. To solve getierally the indeterminate equation* 
 
 Vieta solved this equation on the assumption that two of the four 
 numbers are taken as known. 
 
 [I noted on p. 102 Euler's remark that, if 3 3 + 4 3 is turned into the 
 difference between two cubes by the direct use of Vieta's second formula, 
 the formula gives 3 3 + 4 3 = (A^) 3 - (f-f but not 3 s + 4 3 - 6 3 - 5'. I ought 
 however to have observed- that the latter can be obtained from Vieta's first 
 formula if we multiply throughout by a 3 + ^. The formula then becomes 
 
 a 3 (a 3 + 6*f = 6* (<? + P) 3 + c? (a 3 - zPf + P (20 3 - *) 3 . 
 Putting a=2, 6=1, we have i8 3 =9 3 + 12 3 + 15 3 , which gives (after division 
 by 3 3 ) 6 3 = 3 3 + 4 3 + 5 3 . The next solution, obtained by putting a = 3, b= i, 
 is 84* = 28 3 + 53 3 + 75 3 ; if a - 3, b - 2, we have ios 3 = 33* + yo 3 + Q2 3 ; and 
 so on. Similarly Vieta's second formula gives 
 
 a 3 (a 3 + 2 ^) 3 - a 3 (a 3 - ^) 3 + ^ (a 3 - ^) 3 + P (20* + Pf, 
 and we obtain other integral solutions ; thus 
 
 if a = 2, b = i, we have 2o 3 = 7 3 + i4 3 + 1 7 :! , 
 if a = 3, b=\, we have &f = 26' + $$ 3 + 78*; 
 and so on.] 
 
 (i) A more general solution can be obtained by treating only one of 
 the three numbers x, y, z as known. 
 
 To solve cf + x 3 +y = v 1 , 
 
 put x =pu + r, y = qu r; 
 
 therefore 
 
 say; 
 and we obtain, after dividing out by (/ + q) u*, 
 
 1 N. Comment. Acad. Petrof. 1756-57, Vol. vr. (1761), pp. IJ5 sqq- = Commtnt. 
 arithm. I. pp. 193-206. Cf. pp. 101-2 above. 
 
 - See Nesselmanivs "Anmerkungen zu Diophant" in the Zcitsfhrift fiir Math. u. 
 Physik, XXXVII. (1892), Hist. litt. Abt. p. 123.
 
 330 SUPPLEMENT 
 
 - ?") - ^ (/ + ?)'"' 
 
 _ 30 V 4 (/ + ?) - 
 
 
 where c, r and the ratio / : q may be given any values we please. 
 
 (2) A more general solution still is obtained if we regard none of the 
 first three cubes as known. 
 
 Suppose that, in the equation 
 
 x* + y* 4- z* = v"; 
 
 x = mt + pu, y=nt + qu, z = - nt + ru. 
 Therefore 
 
 o? +y 3 + 2 3 = 
 
 Put now Z 
 
 and we have, after division by u", 
 
 3 / {w/ + (f -r>)} + u (f 
 
 n* (<? + r)} 3 , 
 
 m 6 
 
 whence, neglecting a common factor which may be chosen arbitrarily, we 
 have 
 
 / - m 6 (f + g 3 + r 3 )- \m>p + j? (q + r}f, 
 
 u = yn* {nfp + n 2 (y + r}Y - 3^' {mp^ + n (f - r 1 )}, 
 or, if we divide by the factor q + r, 
 
 t = ; 6 (q- -qr + r*}- yn*ri i p' i ^tr^n^p (q + r) - n* (q + r) 3 , 
 u = - 3m e n (q r) + 6m 5 n 2 p + yri?n* (q + /-), 
 so that x, y, z and v can be written down. 
 The solution is, however, still not general. 
 
 (3) General solution. 
 
 To find generally all the sets of three cubes the sum of which is a cube. 
 
 Suppose A s +& + C 3 = D\ or A 3 + 3 = D l - C'\ 
 and assume A-p +y, B=p - q, C=r~s, D = r + s. 
 
 Then A :s + 3 = 2 p A + 6pq\ & - C s = 2 j 8 + 6r-s, 
 
 so that ' + '
 
 SOLUTIONS BY EULER. PROBLEM i 331 
 
 This equation cannot subsist unless f + $#*, s 2 + ^ have a common 
 divisor. Now it is known that numbers of this form have no divisors 
 except such as are of the same form. 
 
 To find them, we introduce six new letters to take the place of 
 /, </, r, s, thus : let 
 
 p = qx + $by, s = yy ~ dx t 
 
 whence f- + tf = ( rf 2 + 3 2 ) (* + 3^), s 2 + 3^ = (d- + 
 and our equation, divided by x 2 + 3^-, becomes 
 
 (ax + 3 ty) ( a + 3^) = (yy- dx) (d* + y*) ; 
 so that * - - 3* <* + 
 
 y. a +3 
 
 and we may put *=- yib (a- + yV) + yic (d- + y*), 
 
 y - na (a* + 3^) + nd (d- + 3^). 
 Hence the values of/, (/, r, s are found to be 
 
 g = (Zbc - ad] (<i* + y*) - n (<? + 3 ^) 2 , 
 r-n (d* + 3 ^) 2 - ( 3 6f - ad) (a 2 + 3^), 
 s = 3 (or + ^) (a 2 + 3^), 
 and ^ = n (^ac + ybc -ad+ 3 A/) (^ 2 + 3^) - n (a 2 + 
 
 -ad+ 
 
 D= n (d* + 3^) 2 + n (ytc - 
 These values satisfy the equation 
 
 A s + & + C* = Z^, 
 
 and, since no restriction has been introduced, the solution is capable of 
 giving all the sets of three cubes which have a cube for their sum. 
 
 More special forms for A, , C, D can of course be obtained by putting 
 zero for one of the letters a, b, r, d, and still more special forms by co'm- 
 bining with the assumption a = o or l> = o the assumption d=c, or com- 
 bining with the assumption c= o or d - o the assumption b = a. 
 Two cases are worth noting. 
 First, suppose b= o, d = c, and we have 
 A = Svaf-na*, B = i6nat* + na 4 , C= i6nc* - 2>ia*c, D = i 
 If further we write 2a for a and #/i6 for , we have 
 A = na(S-a 3 ), 8 = na ( 2 <? + a 3 ), C=nc(<?-c?\ D = nc 
 which is equivalent to Vieta's solution of his second problem. 
 Secondly, suppose d = o, b = a, and we have - 
 
 A -i Snat? 1 6//a 4 , B = 1 6a 4 , C = 9^ - _24o t V, D *= 
 or, if we write ia for a, 
 
 A = gnac 3 - a 4 , j9 = a 4 , C" = gnc* yid*c> D = gt
 
 332 SUPPLEMENT 
 
 which, if n - a = c i, gives the simplest solution of all 
 
 4 = 8, JB=i, C = 6, Z>= 9 , and i a + 6 s + 8 8 = 9 8 . 
 
 In proceeding to other solutions we have to remember that, while 
 
 A, , C, D must be integral, they should all be prime to one another; for 
 
 those solutions in which A, Jt, C, D have a common factor are not new 
 
 solutions in addition to that from which the common factor is eliminated. 
 
 Thus, while giving any values, positive or negative, to the numbers 
 a, , c, d in the formulae 
 
 x = yic (d* + 30 ~ yd (a* + 3^), 
 y = nd(d- + $r) + na (a- + 3^'), 
 
 we have to choose for n such a fraction as will make x, y prime to one 
 another. We then form 
 
 p = ax + $by, q=bx-ay> r - dy + ex, s = yy - dx ; 
 and, after again eliminating any common factor, we put 
 
 and we shall have A^ + ^+C 3 ^ I?. 
 
 (The cases in which one of the three cubes A"', J?\ C 3 is negative will 
 give the solutions of the equation X s +y* = z z + v\) 
 
 While any values of a, l>, c, d may be taken, it is necessary, if we want 
 a solution in which A, B, C, D will be small numbers, to choose , l>, c, d 
 so that cr + 3^ 2 , d* + 3^ may have a common factor. Euler accordingly 
 sets out a table of all numbers of the form m z + yr less than 1000 (giving 
 m values from i to 31 and // values from i to 18), and then chooses out 
 cases in which a 2 + 3<5 2 , d 2 + y~ have a tolerably large common factor. 
 
 Now, assuming that a 2 + 3^ 2 = mk, 
 
 we have (supposing further that ac+bd=f, $bc - ad ' = g) 
 
 In these formulae/, g may be either positive or negative, the signs of 
 a, l>, c, d being ambiguous ; and we may put 
 
 either A(" + A 1 or /=(^-^) } 
 
 ' er ^=(3^-^)l. g=(& + ad}}- 
 
 But, if/ changes sign while g remains unaltered, we get numbers of the 
 same form, only in different order ; therefore we may confine ourselves to 
 the positive sign in/
 
 SOLUTIONS BY EULER. PROBLEM i 
 Example i. Let 
 
 333 
 
 19, so that a = 4, b=^, 
 eP + 3^ = 76. so that </= i j or </= 7 1 or //= 8 1 
 
 Then ///= i, = 4, k= 19. 
 
 The following values for f, g result, viz. 
 
 I. /=2F, II. /=I 9 , III. /=I 9 , 
 
 ^=11, ^=19, ^=19, 
 
 IV. /=5, V. /=i6, VL /=o, 
 
 ^=37, ^-26, ^=38, 
 
 while, since m i, = 4, >= 19, 
 
 The values (VI) /= o, ^= 38 are excluded because, if /= o, A = - B 
 and C=D. 
 
 The values (I) give 
 
 that is, 
 
 = 315 
 
 C= 241 
 
 C*=230 
 
 The values (II) and (III) give, after division by 19, 
 
 A = 11 4, that is, i /* = 1501-^ = 5 A=^ 
 ^=13 + 4 B= 9 = 3 =17 
 
 (7=13 + 1 C=i2 C = 4 
 
 The values (IV) give 
 A= 411148, that is, 
 = 79+148 
 
 ^ = 319+ 37 
 Lastly, the values (V) give 
 A = 173 104, that is, 
 2? = 21 1 + 104 
 (7=256+ 26 
 
 26 
 
 D=2Q 
 
 A=i&qoTA= 63 i A = -ioj 
 B=-6<) #=-23 B= 227 
 
 c= 252 c= 84 I ^^326 
 
 Z?= 282 Z>= 94 | D= 356 
 
 =230 
 
 ^4 = 69 or A = 23 
 ,#=315 #=105 
 ^=282 C= 94 
 Z>=*78 Z>=i26
 
 334 SUPPLEMENT 
 
 Thus from the one assumption for a- + 3^, d- + ^c we have the 
 following solutions : 
 
 22-j s + 23o 3 + 277 3 = 356 3 1 io7 3 + 356 3 = 227 3 + 326 3 
 
 io7 3 + 230 + 277 3 = 326 3 I 23 3 + 94 s = 63*+ 84* 
 
 ., o , ,., ! 
 
 23 + 94+105126 
 
 3 3 + 4 3 + 5 S = 6 s 
 Example 2. Assuming 
 
 a 2 + 3^ 2 = 28, so that a - 
 
 d 2 + y 2 = %4, so that d=$\ or ^=6) or d=q\ 
 we have k= 28, m=- 1, n = 3, and the following solutions will be obtained : 
 
 I 3 + I2 3 = Q 3 + I0 3 
 
 PROBLEM 2. To find three numbers x, y, z such that 
 
 x +y, x + z, y + z, 
 
 xy, x- s, y-z, 
 are all squares, 
 First solution J . 
 
 Assume that x y =f, x-z = <f, y-z = r* ; 
 
 therefore y- x-p z , z = x-$ 2 , and <7 2 =/ 2 + r ! . 
 The first three formulae now become 
 
 x +y = 2X p^, x + z = 2X q*, y + z= 2X p"* q*. 
 
 Suppose that 2x -p*-<f = t 2 , so that 2x = f 1 + p* + q* ; therefore we 
 have to make / 2 + ^ and / 2 +/ 2 squares, while in addition q* =# 2 + r-. 
 
 Let q = a z + l>\ p^a*-P, r = tab ; 
 
 then t- + (rt 2 + <^) 2 - / 2 + a 4 + b* + zdW | 
 
 and t* + (rt 2 - ^) 2 = f 2 + a 4 + b* - 2aW } 
 
 must be made squares. 
 
 Comparing now / a + a 4 + b* with r + d- and 2 2 // 2 with 2cd, let us 
 suppose cd=c?P=pg*&ff, c=fY, d=tf&, a*=fW, P^g*}? (or a=/fi, 
 b = gfc) ; then the assumption / 2 + ^ 4 + fr = c z + </ 2 will assume the form 
 
 Algebra, Part II. Art. 235.
 
 SOLUTIONS BY EULER. PROBLEM 2 
 
 335 
 
 Hence the problem is reduced to finding the differences of two pairs of 
 fourth powers, namely / 4 -/ 4 and g* -h* t the product of which is a square. 
 
 For this purpose Euler sets out a table of values of m 4 - n* corre- 
 sponding to different values of m, n, with a view of selecting pairs of 
 values of m* - 4 the products of which are squares. 
 
 m* 
 
 t* 
 
 m l - n> 
 
 **+* 
 
 w 4 - < 
 
 4 
 
 I 
 
 3 
 
 5 
 
 3- 5 
 
 9 
 
 I 
 
 8 
 
 10 
 
 16. 5 
 
 9 
 
 4 5 
 
 13 
 
 5- i3 
 
 16 
 
 1 1 5 
 
 17 
 
 3- 5- 1 7 
 
 16 
 
 9 7 
 
 25 
 
 25- 7 
 
 25 
 
 i 24 
 
 26 
 
 16. 3.13 
 
 2 5 
 
 9 
 
 16 
 
 34 
 
 16 . 2.17 
 
 49 
 
 i 
 
 48 
 
 5 
 
 25. 16. 2. 3 
 
 49 
 
 16 
 
 33 
 
 65 
 
 3- 5-H-I3 
 
 64 
 
 i 
 
 63 
 
 65 
 
 9- 5- 7-i3 
 
 81 
 
 49 
 
 32 
 
 130 
 
 64- 5-3 
 
 121 
 
 4 
 
 117 
 
 I2 5 
 
 25- 9- 5-13 
 
 121 
 
 9 
 
 112 
 
 130 
 
 16. 2. 5.7. 13 
 
 121 
 
 49 
 
 72 
 
 170 
 
 144. 5.17 
 
 144 
 
 25 
 
 II 9 
 
 169 
 
 169. 7.17 
 
 I6 9 
 
 i 
 
 168 
 
 170 
 
 16. 3. 5. 7- 17 
 
 I6 9 
 
 81 
 
 88 
 
 250 
 
 25. 16. 5. n 
 
 22 5 
 
 64 
 
 161 
 
 289 
 
 289. 7.23 
 
 
 
 
 
 
 One solution is obtained from/ 2 = 9, t? = 4, g 2 = 81, h* = 49, whence 
 
 / = (/ 4 - 4 ) (^ - // 4 ) = 5 - i3 64 5 i3 = (52o) 2 = 270400. 
 Therefore 
 
 a =fh= 21, bgk\^^p = (f b' i = 117, $ = a 2 + fi* = 765, r = 2fl/ = 756 : 
 therefore 2.r = / 2 -h/ 2 + </' = 8693i4, or jc = 434657] 
 y = x-f?= 420968 V. 
 s = x -f = -i 5 o 5 6S\ 
 
 The last number z may be taken positively; the difference then becomes 
 the sum and the sum becomes the difference ; therefore 
 
 x= 434657, *+y= 855625 = (925) 2 , x-y= 13689 = (i 1 7) 2 , 
 
 y = 420968, x + z = 585225 = (765)2, x - z = 284089 = (533) 2 , 
 2=150568, J + z=57 1536 = (75 6 ) 2 > y- 2 = 270400 = (520)'-. 
 We might also have taken/ 2 = 9, #* = 4, ^ 2 = 121, - 2 = 4, which would 
 equally have given a solution.
 
 336 SUPPLEMENT 
 
 Second solution^. 
 
 This later solution (1780) of Euler's is worth giving on account of the 
 variety of the artifices used. 
 
 We can make x +y and x -y squares by putting x =/ 2 + g*, y = 2/y. 
 Similarly x + z, x-z will be squares if x = r- + s 2 , z = 2rs. 
 
 Therefore four conditions will be satisfied if only p l + f = r* + s' 2 . 
 Now [cf. Diophantus in. 19 and pp. 105-6 above] if we put 
 
 x = (a 2 + P) (c* + d*\ 
 
 x can be made the sum of two squares in two ways ; in fact 
 / = ac + bd, r = ad + be, 
 q ad be, s ac bd, 
 and 
 
 y 2pq = 2 (a?cd + abd^ abc* b*cd), z = zrs = 2 (cPcd + abc* - abd* Ircd], 
 so that y + z = ^cd (a- - //), y-z* $ab (d? - <*). 
 
 These latter expressions have to be made squares. 
 First make their product y 1 - z 2 a square ; this means that 
 ab(a*-l?) . cd(d* -c*) must be made a square. 
 
 To effect this, let us assume that cd(d' 2 c-} = ri*ab (a* - b*} ; we may 
 further, since the question depends on the relations between the pairs of 
 letters a, b and c, d, suppose that d^a. 
 
 We have then c (a 2 - r 2 ) = r?b (a* - //'), 
 
 whence a 2 = ?-, - , which fraction has accordingly to be made a square. 
 ri*b-c 
 
 2 A3 _ .-3 
 
 Suppose that a~b-c y so that ^Y_ - =1? 2bc + 1 3 , and we have 
 
 o = -(2 2 +i)-V+( 2 + 2 )<V; 
 
 b n*+2 
 therefore - = ^^ - 
 
 Put b = ri l + 2 and c - zri- + i ; therefore a = \n 1 -d. 
 
 As we have now made the product of the expressions ab (d~ - c 1 ) and 
 cd (a 1 - &*) a square, it only remains to make either of them singly a square, 
 say ab (d* - <?}. 
 
 But ab (d- - c*) = ab (d - c) (d + c) = 3/z 2 (tr - i ) (n 9 - + 2)% 
 
 We have therefore only to make 3 (# 2 - i) a square, which is easy, since 
 n 3 - i has factors ; for we have only to put 
 
 2 - 2 
 
 which gives 3 ( - i ) -g ( + i ), or n = 7-5^ . 
 
 o O6 J 
 
 1 Mtmoires de F Acadtmie Imptriale des Sciences de St Peter sbourg, 1813-14, vi. (181 8), 
 pp. 54 sqq.= Commentationes arithmeticae, II. pp. 392-5.
 
 SOLUTIONS BY EULER. PROBLEM 2 337 
 
 [Euler had previously tried the supposition a = b + c, which would 
 require 3(# 2 +i) to be made a square, which is impossible.] 
 
 All the conditions are now satisfied, and we have to find a, b, c, d etc. 
 
 n terms 
 
 As the whole solution depends on the ratios of a, b, c, d, we can 
 multiply throughout by the common denominator, divide by 3, and put 
 
 whence p = - 8/V (/< + 9^), r=/ 
 
 [Euler took a to be 2 i instead of i n- and consequently obtained 
 positive signs for the values of/ and s ; he also has f = (f 4 - 9g*)~, which 
 appears to be a slip.] 
 
 Assuming therefore any values for/ g in the first instance, we first find 
 values for a, b, c, d, then values for p t q, r, s, and lastly values for x, y, z. 
 It is to be observed that it is a matter of indifference whether we get negative 
 values or not ; for positive values can be substituted without danger. 
 
 Euler gives four examples. 
 
 If/= i, g= i, we find that x, y have equal values; this solution there- 
 fore does not serve our purpose. 
 
 The same is the case if/= 3, g- i. 
 
 Suppose then that/= 2, g- i; therefore a = //=- 16, b- 17, c= 33; and 
 (taking positive signs) we have 
 
 / = 8oo, ^ = 305, r=8i7, ^ = 256, 
 and # = 733025, ^ = 488000, .2 = 418304, 
 
 If/= i, g- 2, we have a = d= 16, b- 137, c- 153, and 
 
 ^ = 4640, ^=20705, r= 21217, ^=256, 
 leading to large numbers for x, y t z. 
 
 Euler adds that, if x, y, z satisfy the conditions of the problem, another 
 solution is furnished by X, Y, Z where 
 
 H. D.
 
 338 SUPPLEMENT 
 
 PROBLEM 3. To find three squares such that the difference of any pair 
 is a square, or to find x, y, z such that 
 
 x 2 JF 2 , x 2 z 2 , y 2 z 2 are all squares. 
 
 Any solution of the preceding problem will satisfy this, hut the numbers 
 would be large and we can get smaller solutions 1 . 
 
 x 2 y 2 
 Dividing by z~, we have to find three squares, 5-, 3 and i, such that 
 
 are all squares. 
 
 The last two conditions are satisfied if we put 
 x l> 2 +i , y <? 2 +i 
 ~z f^\ and * j^I J 
 
 and we have only to make p ~ "p = /\L 4 "" rt 4 a S( l ua re. 
 
 Now 
 
 Therefore (/V J - i) (/-/ 2 ) or (/ s ^ a -i) fC - i") has to be made 
 square. 
 
 (i) The latter expression is a square if 
 
 And /^ . ^ = ^ 2 , a square ; therefore 
 
 . ~ orfg (f* + g*) . hk (A 2 + &} must be a square. 
 
 , g = a-b, h = c+d, k = c-d, the expression becomes 
 4 (<z 4 - b 4 } (<? - d*), which must be a square. 
 
 From the Table to the last problem we may take the values 
 
 a 2 = 9 , F = 4, r = 8t, d* = 49, 
 which make the expression a square. 
 
 Then/= 5, g= i, /&= 16, k=2,pq = ^-, q\p = ^ = |f, so that ^ = ^, 
 ^ = - 1 /-, and therefore / = 4- 
 
 y 
 
 - 9 - - -- , - , --- 
 z / 2 -i 9 2 ^-i 153 
 
 1 Algebra, Part II., Arts. 236, 237. 
 
 . . .. 
 is the solution.
 
 SOLUTIONS BY EULER. PROBLEM 3 339 
 
 To obtain whole numbers, we put 2=153 an <3 then ^ = -697 and 
 7=185. 
 
 Thus * = 4858091 and s-y = 45 1584 = (6 72)', 
 
 y= 342251 f-z 
 
 z-= 23409] .r ! -z 
 
 (2) Without using the Table, we may make (/V 2 - J ) ( nj ~ J ) a square 
 in another way. 
 
 Put qlp = m or q = w/>, and (#z 2 / 4 - i) (nf - i) has to be made a square. 
 
 This is a square when / = i ; substitute therefore i + s for p and we 
 have 
 
 (w 2 - i) (m- - i + 4n?s + 6wV + 4?n z s 3 + m 2 * 4 ). 
 
 Dividing out by (w 2 -i) 2 and, for brevity, putting a for / 2 /(w 2 i), 
 we have 
 
 i + 4as + 6as 2 + 4i 3 + as*, 
 
 which has to be made a square. 
 
 Equating this to (i +fs+gs*Y, let us determine/,^ so that the first 
 three terms disappear ; 
 
 therefore 2/= 40, or /= 20, 
 
 and 6a = zg +f 2 or g = i (6a -T 2 ) = 30 - 2<z 2 . 
 
 Lastly, the equation gives 40 + a.r= ^fg + g^s, so that 
 
 4 a ~ 2 fg _ 40-120*+ So 3 4 - 1 20. + Sa" _ 4 (za- i) 
 
 4 a 3 - 
 
 Now m in the expression for a may have any value. 
 Ex. i. Let m = 2, so that a = | ; 
 therefore s = 4 . = ~~^T / = ~^7' ^ = ~2~3 
 
 * 949 y _ 6005 
 whence z = ^' -z~w' 
 
 Ex.2. Let w = |, so that a = f; 
 
 13-5 26 ^ 2 49 747 
 
 therefore s = 4 - , ^= - 77 ' ^--^1' 
 
 whence */a, j/z are determined. 
 
 Euler considers also the particular case in which a = m*/(m*-i) is a 
 square, P say. 
 
 The expression i + 4^-r + 60V + 4<*V + ^J 4 is then equated to 
 (i-f 2^ + Ar 2 ) 2 , 
 
 I-2b-2P , . 1-2? 
 
 and we obtain s= ^ - and /= ^ . 
 
 222
 
 34 o 
 
 SUPPLEMENT 
 
 Ex. a is a square if m = |, and in that case b = f. Therefore p=- ^ 
 q - nip = ^, and accordingly 
 
 z 145 
 
 PROBLEM 4. To find three square numbers such that the sum of each 
 pair is also a square^ i.e. to find numbers x,y, z such that 
 
 X? +/, X* + Z 2 , / + Z 2 
 
 are all squares*. 
 
 Dividing by z 2 , we have to make 
 
 x* y* x 2 y- 
 
 * + ;?' ^ + I 'i* +I 
 
 all squares. 
 
 The second and third are made squares by putting 
 
 z 2p z zq 
 
 and it only remains to make 
 
 This can hardly be solved generally, and accordingly we resort to 
 particular artifices. 
 
 i. Let us make the expression divisible by (/ + i) 2 , which is easily 
 done by supposing/ + i =^ i, or g=p+ 2, so that </+ i becomes/ + 3. 
 
 Thus (/ + 2) 2 (p - 1) 2 +/ 2 (p + 3) 2 , or 2/ 4 + S/ 3 + 6f - A + 4, must be a 
 square. 
 
 Suppose 2/ 4 + 8/ + 6/ 2 - ^p + 4 = (^/ 2 +^ + 2) 2 , 
 
 and let us choose/, g such that the terms in p, / 2 vanish ; therefore/- - i , 
 and 4-+i=6, or g=%. 
 
 We now have 2p + 8 = g*p + 2^- 
 
 _ 25 J. 5 
 
 "'Tvf-y* 
 
 so that / = - 24, and ^ = - 2 2, whence 
 
 *=^1 I = _S75 j = r~ I = 483 
 
 z 2/ 48 ' z 2^ 44 ' 
 
 Making 0=16.3. 1 1, the least common multiple of 48 and 44, we have 
 the solution 
 
 x= ii. 23. 25 = 6325, y=i2.2i .23 = 5796, 2 = 3. ii . 16 = 528, 
 and 
 
 /+z 2 =i2 2 ( 4 8 3 2 + 44 2 )=i2 2 . 
 1 Algebra, Part II., Art. 238.
 
 SOLUTIONS BY EULER. PROBLEMS 3, 4 341 
 
 2, 3. Euler obtains fresh solutions by assuming, first, that 
 
 ?-i = 2 (/+ i), 
 and, secondly, that q - i = (p - i). 
 
 4. Lastly, he makes our expression divisible by both (/ f i) 2 and 
 (p-if at the same time. 
 For this purpose he takes 
 
 whence ? + 1 
 
 Substituting in the formula ? (f - if + f (<? - if the value of q in 
 terms of/, / and then dividing by (/ 2 - i) 2 , we have the expression 
 
 and we have to make (// + i) 2 (/ + /) 2 +/> (/ 2 - i) 2 a square, 
 
 or /y + 2t (/ 2 + i)/ 3 + {2/ 2 + (/ 2 + i) 2 + (/ 2 - i) 2 }/ 2 + 2/(/ 2 + i)p + /* 
 
 must be a square. 
 
 We now equate this to {tjr + (t- + i)p - /} 2 , 
 whence we have 
 
 { 2 / 2 + (/ 2 + i) 2 + ((*- i) 2 }/ + 2/(/ 2 + i) = {(/ 2 + i) 2 - 2/ 2 }/ - at (t- + i), 
 which gives {4** + (t- - i) 2 }/ + 4/ (/ 2 + i ) - o, 
 
 and P = -> 
 
 therefore //+ i = - , / + /= 
 
 and 
 
 where t can be chosen arbitrarily. 
 
 Ex. Let /= 2 ; then/ = -f, ^ = - -^ and 
 
 x = f~*= 39 ^ = g'" I = "_ 7 . 
 z 2/ 80 ' z zq 44 * 
 
 Putting 2 = 4.4.5.11, the least common multiple of 80 and 44, 
 we have 
 
 # = -3. ii . 13 =- 4 2 9 
 ^ = -4. 5. 9. 13 = -2340, 
 z = 4.4.5.11= 880,
 
 342 SUPPLEMENT 
 
 and * 2 +/=3 2 . i3 2 (121 + 3600) = 3 2 . if. 6i 2 , 
 
 ^ + z 2 = ii 2 (1521 +6400) = n 2 .89 2 , 
 
 J 2 + Z 2 = 20 2 (13689+ 1936) =20 2 .I25 2 . 
 
 PROBLEM 5. (Extension of Dioph. iv. 20 to five numbers.) 
 To find five numbers such that the product of every pair increased by unity 
 becomes a square^. 
 
 Euler had already shown (see pp. 181, 182 above) that, if mn+ i = / 2 , 
 then the following four numbers which we will call a, b, c, d have the 
 property, viz. 
 
 a = m, b = n, c-m + n + 2l, d = $l(l + m) (l+n). 
 
 If now z is the fifth required number, the four expressions 
 
 i + az, i + bz, i +cz, i + dz 
 must all be squares. 
 
 If, says Euler, we had to satisfy these conditions singly, the difficulties 
 would be insuperable. But here too it happens, as in the former case, 
 that, if we make the product of the four expressions a square, the 
 expressions are all severally squares. 
 
 Let the product be i +pz + qz* + rz* + sz*, 
 where accordingly 
 
 p = a + b + c + d, = ab + ac + ad + be + bd + cd, 
 
 r abc + abd + acd + bed, s = abed. 
 Suppose now that 
 
 i +pz + ?z* + rz s + sz* = {i + \pz + (\g- i/ 2 ) z"} 2 ; 
 therefore, since the absolute term and the terms in z, z 1 vanish, we have 
 
 whence 
 
 Now it will be found (see the proof lower down) that 
 
 If.-V i(+*)'i 
 
 the denominator of the fraction will therefore be ^ (s i) 2 ; that is, the said 
 denominator fortunately turns out to be a square ; if it were not so, the 
 single expressions i + az, i + bz, i + cz, \+dz could not have been made 
 squares. 
 
 As it is however, substituting for \q - \p* its value in the numerator 
 and denominator of the fraction for z, we have 
 
 1 Commentationes arithnteticae, n. pp. 50-52.
 
 SOLUTIONS BY EULER. PROBLEMS 4, 5 343 
 
 and all the conditions will be fulfilled, that is, all the expressions 
 ab + i, af+i, ad+i, h+i, bd + i, 
 cd+i, az+i, bz+i, cz + i, dz + i 
 
 will be squares. 
 
 Lemma. To prove the fact (assumed above) that 
 
 k-J^=-H*+i). 
 
 For brevity, put m + n + /=/, / (/ + /) (/ + ) = , so that k =fl* + Imn ; 
 and, since a = m, b = n, c=f+ /, d=^k, we have a + b + c- 2/, and therefore 
 
 /=2/+ 4 
 
 Again, since q = (a + b + c) d + (a + H) c + at>, 
 
 q = 8/ + (m + nY+ 2/ (m + n) + mn ; 
 and, since mn + i = /, the latter expression becomes 
 
 ^ = 8/^+/ 2 -i. 
 
 Moreover, s - abed - ^mnk (f+ /) ; 
 
 therefore i + q + s = %fk +/ 2 + $mtik (/+ /), 
 
 and we have to see whether the right-hand expression is equal to \p~* 
 Now i/'=/ 2 +4//& + 4 /& 2 . 
 
 Assume then, as a hypothesis, that 
 *fk +f 2 
 
 or, if we divide throughout by 4^, 
 
 f + mn (f+l)=k f/ 2 + Imn, from above ; 
 that is, f+fmn =ff\ 
 
 which is of course true, since mn + t =-- A 
 Consequently it is proved that 
 
 , or -) = - J +0 
 
 Ex. i. Assume i = i, n = 3, so that 1=2; therefore 
 
 =i, ^ = 3, f = 8, ^=120, 
 whence / = i3 2 > ?= J 47S ^=4224, ^ = 2880, 
 
 and we deduce that 
 
 _ 4. 42 24 + 264^881 __ 777480 
 " 2 " ' 
 
 8288641 
 
 The conditions are satisfied, for 
 
 ab + i = 2 2 , <:+ i = 3 2 ,
 
 344 SUPPLEMENT 
 
 Ex. 2. To get smaller numbers (since we must put up with fractions) 
 let us put m = ^, n =f, so that /=f ; therefore 
 
 = .*, = f, <r=6, //=48, 
 
 whence / = S7 ? = 45 I T ^=93 I I> J = 3 6o > 
 
 _ 4.934 + 114.361 _ 44880 
 359 2 ~ 128881' 
 
 PROBLEM 6. Euler has a general solution of the problem of Dioph. in. 
 15, viz. 
 
 To find three numbers x, y, z such that 
 
 xy + x +y, xz + x -f 0, yz +y + z 
 are all squares*. 
 
 (i) Put x + i = A, y + i = B, z + i = C, so that AB - i, AC-\ and 
 BC- i have to be made squares. 
 
 Let AB=p*+i, AC=f+i, BC = r*+i; 
 
 therefore ABC= J{(p + i) (f + i)(r* + i)}. 
 
 To make this expression rational, let us regard /, q as given and put 
 (/ 2 + i) (q* + i) = m* + n?, so that m-pgi y n=p~+q\ therefore 
 ABC= J{(m* + a ) (r* + i)} = J{(mr + nf + (nr - mf\. 
 Put the latter root equal to mr + n + t (nr m) ; therefore 
 nr-m- zmrt + znt + nrf mP 
 
 (m* + 2 ) (f 2 + i ) 2 
 Therefore r 2 
 
 ,--- - TO> ~r ~ - - 
 
 \n (t 2 - i ) + 2tnt\* n (/ 2 - i ) + 2 // ' 
 
 thus, since C=t 3 + i, we have 
 
 (/ 2 - i) + zw/ 
 ^* t*+i ' 
 
 and, since w 2 + n- = (p* + i) (^ 2 + i), 
 
 (/ 2 +i)(/ 2 +i) 
 ' 
 
 n(f* 
 where w =/^ + i, =/ + ^. 
 
 This solution is very general, inasmuch as we may choose /, q as we 
 please, thus equating AB i, AC-i to any given squares; and, as / 
 can be chosen arbitrarily, we have an infinite number of square values for 
 SC-i. 
 
 (2) Euler adds two methods of obtaining solutions in integers, the 
 second of which is interesting. 
 
 1 " Considerationes circa analysin Diophanteam," Commentationes arithmeticae, n. p. 577.
 
 SOLUTIONS BY EULER. PROBLEMS 5, 6 345 
 
 Take two fractions-; and -so related that ad-bc = +i; and form a 
 o a 
 
 third fraction ^^> which is similarly related to either of the former 
 fractions. 
 
 Then the following three numbers will satisfy the conditions : 
 
 For, since ad-bc = i, 
 
 (Cf. Dioph. in. 19.) 
 
 Simple solutions are seen thus : 
 
 a 
 3 
 
 c 
 d 
 
 a+c 
 J+d 
 
 A B C 
 
 p 
 
 i 
 
 i 
 
 
 
 
 
 
 i 
 
 f 
 
 f+* 
 
 
 i 
 
 f~ l 
 
 f 
 
 ft f ft f 
 
 r 
 
 f" 
 
 f+i 
 
 2 if -2/+I 2/ z +2/+I 
 
 i 
 
 f 
 
 /+ 
 
 
 *- 
 
 if -i 
 
 2 ^-i j o a/ i/ nl o/'yT 
 
 and so on. 
 
 (3) If two of the numbers A, B are given such that AB-\ =/*, we 
 can find an infinite number of values for a third, C, which with A t B 
 will satisfy the conditions. 
 
 For, since AC i and BC - i have to be squares, take their product 
 ABC* - (A + B) C+ i and equate it to (mC+ i) 2 ; we have then 
 
 A + B + 2m (A + mf (B + m)* 
 
 C = AB-m" ' 
 
 Therefore we have only to make AB-m* a square; that is, 
 jp + i - m* - a square = ri* say, so that m 2 + n* =/ 2 + i. 
 
 Take now two fractions a and a such that 2 + a 2 = i, and let m = aj> + o, 
 
 // = op - a ; then 
 
 -< 
 where a, a are determined by giving any values whatever to / g in the 
 
 expressions
 
 346 SUPPLEMENT 
 
 PROBLEM 7. To find four numbers such that the product of any pair 
 plus the sum of that pair gives a square; or, in other words, to find four 
 numbers A, B, C, D such that the product of any pair diminished by i is a 
 square, that is, such that 
 
 AB-i, AC- i, AD- i, BC-i, BD-i, CD - i 
 are all squares 1 . (Cf. Diophantus iv. 20.) 
 
 Let us regard two of the numbers A, B as given, being such that 
 AB - i =f, or AB=p* + i. 
 
 Let a, a be such fractions that a 2 + a-= i, and put 
 
 Similarly let P + ft? = i, and put for the fourth number 
 A+B + 2(ty+ft) 
 
 ($p-b? 
 Thus five conditions are satisfied, namely, that 
 
 AB-i, AC -i, BC-i, AD -i, BD - i are all squares. 
 The sixth condition, that CD - i shall be a square, gives 
 (A + B) 2 + 2 (A + B} {(a + b)p + a + ft} + 4 (ap + a) (bp + ft) 
 
 - (ap - a) 2 (ftp - b)* = a square, 
 
 where AB has at the same time to be equal to/ 2 + i. 
 Regarding a, a, b, ft and / as given, we have 
 
 p*+ i A*+p*+ i 
 A + B = A +^ = -- ^ - ' 
 
 A A 
 
 and the expression to be made a square becomes the following expression 
 in powers of A, 
 
 + 4 A? (ap + a) (bp + ft) + 2 A (/ 2 + i) (a + ft) 
 
 Equate this to the square of 
 
 A* + A (a + b)p- (f + i) 
 
 and we have 
 
 A* {(a + ) 2 / 2 + 2 (a + ti) (a + ft)/> +( a + ftY- 4 (/> + i) 
 
 whence A is found. 
 
 Euler goes on to some particular cases, of which the following may 
 be given. 
 
 1 Commentationes arithmeticae, n. pp. 579582.
 
 SOLUTIONS BY EULER. PROBLEMS 7, 8 347 
 
 Suppose b = - a and /3 = - a ; we then have 
 _ A + B + 2 (at + a) 
 
 -of 
 
 and the expression above in A which had to be made a square becomes 
 
 ^ 4 +2^ 2 (/ 2 +l) + (/ 2 +l) 2 
 
 - 4A 2 (ap + a) 2 
 
 - A n - 
 This can be put in the form 
 
 by virtue of the relation a 2 + a 2 = i. 
 
 Our expression is clearly a square if 4 (ap - of = o, or ap-a = 2, 
 that is, / = (2 + a)/a, and 
 
 (2 + a) a 2a+'i 
 
 ap + a = - - + a = - , 
 a a 
 
 and in that case 
 
 A+J3+2(2a+i)/a a(A + ) 
 - = - 
 
 4 4 a 
 
 where A can be chosen quite arbitrarily. 
 
 Putting =(/ 2 - 2 )/(/ 2 + /)> a = zfgKf* + g*), we obtain the following 
 as a solution, where m, n can be any integers whatever. 
 
 . 
 
 Zmnfg Smng 
 
 Ex. Suppose /= i, g=2, m = 5, n = 6; 
 therefore ^ = ||, = ft, C=|||, />-/ 
 
 and AB-i=r' 
 
 PROBLEM 8. To find four nutnbers such that the product of any pair 
 added to a given number n gives a square 1 . 
 
 (i) A particular solution is found in this way. Let A, B, C, D be 
 the required numbers, and, since AB + n has to be a square, put 
 
 A = na 2 -P, B=nc z -d*, 
 
 so that AB=(nac-bdf-n(ad-bcf. [Cf. the Indian formula above, 
 p. 282.] 
 
 1 Commentationes arithmetical, u. pp. 581-3.
 
 348 SUPPLEMENT 
 
 The condition that AB + n is a square is therefore fulfilled, provided 
 
 that ad-bc = \: therefore we have to take fractions T , - such that 
 
 o a 
 
 ad- bc= i ; and, when this is done, the fractions -. , and T 7 will have 
 
 o + a bd 
 
 the same property in relation to either of the former fractions. 
 We accordingly put 
 
 C=n(a + cf -(b + d) 2 , D=n (a-cf - (b-d) 2 . 
 
 Thus five conditions are satisfied, and it only remains to make CD + n 
 a square ; that is, 
 
 2n (ad - bcf \ = a square. 
 
 or, since (ad- bcf = i, 
 
 n 2 (a 2 - c 2 } 2 - n { 2 (ab cd} 2 + i } + (b 2 - d 2 } 2 = a square. 
 
 (2) We obtain a general solution by the same method as that applied 
 above (p. 345) in the problem of making AB i, BC ' i, etc. squares. 
 
 Put AB=p 2 -n; then, to make AC+n, BC+n both squares, take the 
 product of these expressions and equate it to (n + Cx) 2 ; therefore 
 n 2 + n (A + B} C + ABC 2 = n 2 + 2nCx + C 2 x 2 , 
 
 whence C= 2 -j-=- - , and AC+ n = ^ . * , 
 
 so that (x 2 - AB]\n must be a square. 
 
 Let then y? - AB = x 2 -p 2 + n = ny 2 , or x 2 - ny 2 =p 2 - n. 
 Similarly let us put v 2 - nz 2 =/ 2 - n, so as to get 
 
 A + B-2V 
 
 and it remains to make CD + n a square, 
 
 that is, (A + B} 2 - 2 (* + v) (A + B) + ny 2 z 2 + $xv 
 
 must be a square. 
 
 But, since B = * and A + B- ^ , the expression becomes 
 
 (after multiplication by A 2 ) 
 
 A 4 - 2 A* (x + v) + 2 A 2 (p 2 - n} - 2 A (p 2 -n)(x + v) + (p 2 - n) 2
 
 SOLUTIONS BY EULER. PROBLEMS 8, 9 349 
 
 which must be a square = {A 2 - A (x + v) - (p z - n)} 2 say ; therefore 
 
 A 2 {(x + vf - 4 (f - n} - ny-z* - 4x0} + ^A (x + v) (/ 2 - ) = o, 
 so that 
 
 A=- 
 
 = 
 nfz 2 - 2n (y 2 + z 2 ) + (v + xf ' 
 
 (3) A particular solution is obtained by assuming that v = -x, so that 
 y, and 
 
 while AB=f-n = x*- nf. 
 
 For then we have to make 
 
 A 4 + A*{2(? z -n) + ny t - 4^} + (/> _ ) a square ; 
 that is, (A 2 -f + rif + nA*f (y* - 4) = a square. 
 
 This is satisfied if we put_y = 2, so that / 2 = .r 2 - 3. 
 
 Suppose p = x-t, and we have 
 
 3 - t* \ni? - / a -inu- + t- 
 
 - and p = , or ;> = - and x = * - 
 
 2t 2t 2tU 2tu 
 
 . (nu 2 - 1 2 ) (onu* - 1 2 ) 
 
 and hence AB = ^- -- ^E - ^ 
 
 4/ 2 w 2 
 
 We may therefore put 
 
 * - - 
 
 2gtu 2/tU 
 
 ,n_ 
 
 Bfetu S/gtu 
 
 It will be seen that in this solution C+ D = | (A + B}. 
 
 PROBLEM 9. To find four numbers such that the product of any pair 
 added to the sum of all gives a square 1 . 
 
 First find four numbers A, B, C, D such that the product of any pair 
 increased by a number n gives a square (Problem 8). 
 
 Take as the numbers sought mA, mB, mC, mD, and, since m*(AB + //) 
 is a square or nPAB + nfn is a square, we have only to make m 2 u equal to 
 the sum of the four numbers or m(A + B + C + Z)), whence 
 
 A + B+C+D 
 
 m = . 
 
 n 
 
 1 Commentationes artthweticae, II. pp. 583-5.
 
 350 SUPPLEMENT 
 
 But, since in the other problem C + Z>-^(A + ^\ this gives 
 
 2H 
 
 where n as well as/, g, t and u can be chosen as we please. 
 
 Since n may be chosen arbitrarily, take / 2 = # 2 - 3;;, as in the last 
 problem, so that n-\(af-f) t and AB =p* - n = ^ (^f - x?). 
 
 Accordingly we may put 
 
 A _f( 2 p + X) R _g(2j-x). 
 
 S3. -- - y JJ ~ - ) 
 
 therefore j+s.'W 
 
 and hence c ,'C^ 
 
 and 
 
 therefore m = (A + X+C+ D}\n = 
 
 Now two of the numbers, A, , can be chosen arbitrarily, and 
 
 herefore 
 sothat 
 
 -Bj* A+B 
 
 anu o j - *' = - ~~ 
 
 while 
 
 If, in order to get rid of fractions, we put A = $afg, B = $bfg, we have 
 
 Ex. Let/= 2, g= i ; therefore 
 
 A = 8a, ^-8^, C=6b-a, D = $a-2 
 
 1 2 (a + b~) 1 2 (a + b) 
 
 - a)'
 
 SOLUTIONS BY EULER. PROBLEMS 9, 9 A, 10 351 
 The following are simple cases : 
 
 (1) a= 5, t>=t, whence^ = 40, = &, C= i, Z>=23, M=||. 
 
 (2) a -ii, b=2, whence ^-88, B = 16, <7= i, Z>=5i, = |. 
 If/= S*^ !> w e can obtain integral solutions, thus. 
 
 A = 2oa, B = zob, C = 30^ + 2a, D = 8a - 20^, 
 .. _ 3 ( a + 6) 
 
 -a)' 
 Assuming then a = 19, = 7, we have 
 
 ,4 = 380, ,5=140, =248, Z>=i2, ; = , 
 so that the required numbers are 
 
 475. 175. 31, 15, 
 the sum of which is 975. 
 
 We can also solve the corresponding problem : 
 
 9 A. To find four numbers such that the product of any pair minus the 
 sum of all gives a square. 
 
 For we have only to give m a negative value. 
 
 PROBLEM 10. To find three numbers x, y, z such that 
 
 x+y + z \ 
 
 yz + zx + xy L are all squares 1 . 
 
 xyz j 
 
 (This may be expressed as the problem of finding />, q, r such that the 
 equation ^-p^ + q^-r-o has all its roots rational while/, q t r are 
 all squares.) 
 
 Take nx, ny, nz for the numbers required, so that 
 n(x+y + z) \ 
 
 n*(xy + xz +yz) \ must all be squares. 
 tfxyz ) 
 
 If the first and third conditions are satisfied, we must have, by 
 multiplication, 
 
 xyz (x + y + 2) = a square. 
 
 Put therefore xyz (x +y + z) = v 1 (x +y + z}\ 
 whence xyz = tf (x +y + z\ and z = 
 
 Since xyz = ' ^ we must nave> m or d er that nxyz may be 
 
 a square, 
 
 n = ni-xy (x +y) (xy - #*). 
 
 1 ffffi'i Commentarii Acad. Petropol. 1760-61, Vol. VIII. (1763), P- 6 4 sqq. = 
 lationes arithineticae, I. pp. 239-244.
 
 35 2 SUPPLEMENT 
 
 If the values of z, n thus found be taken, the first and third conditions 
 are satisfied, and the three numbers will be 
 
 nx - t/iVy (x +y) (xy - v*), 
 ny = trfxy* (x +y) (xy v*), 
 nz = nt&xy (x +y) 2 . 
 The second condition requires that 
 
 z/ 2 (x + yf 
 xy + z(x +y) xy-\ ! p- = a square. 
 
 Suppose for this purpose that xy-tf^u 2 (this introduces a restric- 
 tion because there are doubtless plenty of solutions where xy - tf is not 
 a square); therefore 
 
 _ v 2 + u z ^ _ z> 2 (x +y) 
 *~ x ' u 2 ' 
 
 and xy = v* + a 2 , x +y = , so that we must make 
 
 X 
 
 , 
 v* + u 2 + 5-^ a square. 
 
 22 
 
 Put x = tv, so that y = , and 
 
 tv 
 
 * + * + VL\LL = a square, 
 
 i) + 2 } 2 
 /V 
 
 or / 2 wV + /V + v* (f 2 + i) 2 + 2V (t 2 + i) + u* = a square, 
 
 i.e. if (/ 2 + i) 2 + wV 2 (3/ 2 + 2) + 4 (/ 2 + i) = a square 
 
 = {v 2 (/ 2 + i) + .rw 2 } 2 , say. 
 
 Therefore w 8 (3^ + 2) + w 2 (/ 2 + i) = 2^ (/* + i) + J 2 ^ 2 , 
 
 ^2 /2 + : _ ,y2 
 
 and j = -j-^ r 5 = a square. 
 
 Further, let s = t r, and we shall have 
 v 2 2rtr* + i 
 
 Multiply the numerator and denominator by 2r/ - r 2 + i, and we have 
 
 The problem is accordingly reduced to making the denominator of this 
 fraction a square. If we suppose this done, and Q to lie the square root, 
 while / and r are determined as the result of equating the denominator to 
 Q 2 , we shall have 
 
 v 2rt-r*+ i , v 2 + u 2 
 
 -= __-,ail*=^ ; y = - sr - > 
 
 whence we can derive the numbers required.
 
 SOLUTIONS BY EULER. PROBLEM 10 
 
 353 
 
 Now the denominator to be made a square is easily made such if the 
 coefficient of t*, or the absolute term, is a square; and the absolute term 
 is a square if 2 (r - i) is a square. 
 
 Case I. Suppose r=i; the coefficient of / 4 is then a square and the 
 absolute term vanishes. 
 
 We have 4/ 4 - lo/ 3 + 4^ - 8/ = Q*, while v/u = 2tjQ. 
 Suppose Q = 2t- - 1/, and we have 
 
 4 / 2 -8/ = W, and / = -V - = = ; 
 4/-5 '73 
 
 we therefore put # = -36, #=173, t = -$, and #:=/z;= 128; further 
 
 _ * + 3 _ 31225 _ 25 . 1 249 
 y ~ tv 128 1^8 ' 
 
 173*. 128 ' 
 and, since xy-v* = u*, the required numbers will be 
 
 nx = i z8 2 . 25. 1 249- 47609- 1 73 3 ^ 
 
 128. 25 2 . 1 24 9 2 . 47609^ 73* 
 
 ~"i28.178~~ ' 
 
 36 2 . 128.25. 1249.47609'- , 
 
 128. I28 2 
 
 In order to get rid of fractions, put m = J-f^, and we have 
 
 nx= i28 2 . i73 2 . 1249. 47609 = 490356736. 59463641, 
 _y = 5 2 .i73 2 . I249 2 . 47609 = 934533 02 5 59463641, 
 nz = 36" . 1249 . 476o9 2 = 61701264 . 59463641. 
 
 The product of the three numbers is obviously a square; their sum is 
 found to be 25 . 5946364I 2 ; and the sum of the products of pairs 
 = 1 73 2 - 59463641" 18248924559376 
 = (i73 59463641 
 
 Case II. Put r = f ; then V - = E2 ~ 5 , and 
 
 whence 6/ 1 - 
 
 Accordingly Q = 
 
 Therefore * = 
 u. D. 
 
 ^ - ^, and / = f$. 
 if. and we put = 19, = 14-
 
 354 SUPPLEMENT 
 
 and the required numbers are 
 
 That is, ## = 705600.2315449=1633780814400, 
 
 ny= 109172.2315449= 252782198228, 
 nz = 1500677 . 2315449 = 3474741058973. 
 
 The product of the three numbers is clearly a square; their sum will be 
 found to be 23I5449 2 ; and the sum of the products of pairs 
 
 = i4 2 - 23I5449 2 6631333489 
 = (14. 2315449. 8i433) 2 . 
 These numbers are much smaller than those first obtained. 
 
 If fractional numbers are admitted, we may divide those found in the 
 last solution by 23 15449% and the solution will be 
 
 PROBLEM n. To find four numbers x, y, 2, u such that 
 
 xy+yz + ... 
 
 > are all squares 1 . 
 xyz + yzu + zux + uxy I 
 
 xyzu J 
 
 A general solution being apparently impossible, some particular as- 
 sumption simplifying the problem had to be made. Euler therefore 
 assumed as the four numbers 
 
 Mob, Mbc, Mcd, Mda, 
 
 which assumption, although five letters are used, involves the restriction 
 that the product of the first and third numbers is equal to the product of 
 the second and fourth. 
 
 We must therefore have 
 
 M(ab + bc + cd + da) \ 
 
 Af 2 (ab*c + b?d + cd*a + da*b + 2abcd) I dl res 
 M 3 (a&fd + ab<*d* + tfbcd* + a*Pcd) | 
 
 J 
 
 1 Novi Commentarii Acad. Petrop., 1772, xvil. (1773), pp. 24 sqq. = Commentationes 
 arithmeticae, I. pp. 450-5.
 
 SOLUTIONS BY EULER. PROBLEMS 10, n 355 
 
 The above assumption therefore has the advantage of making the 
 product of the four numbers automatically a square and also of making the 
 third formula take the form 
 
 APaibcd (ab + bc + cd + da). 
 
 Since the first formula M \ab + be + cd + da) must be a square, it follows 
 that abed must be made a square. 
 
 In order to make the first and third formulae squares, take 
 
 or, if the latter expression has a square factor, say/ 5 , put 
 M= (ab + bc + cd + da)\f. 
 
 We now have only two conditions remaining to be satisfied, namely 
 abcd=3i square .............................. (i), 
 
 aPc + bc*d+ cd*a + dd*b + 2abcd= a square ............ (2). 
 
 The expression in (2) reduces to 
 
 (a* + c a )bd+ac(P + d*) + zabcd, 
 or bd(a^ + c 3 ) + (b + df ac = a square. 
 
 We have therefore only to find numbers a, b, c, d satisfying these two 
 conditions. It is further to be noted that a, c are connected by a relation 
 similar to that between b, d, and the whole question depends on the ratios 
 a : c and b : d. We may therefore assume a, c prime to one another and 
 likewise b, d prime to one another, for, if either pair had a common divisor, 
 it could be omitted and the relation would still be satisfied. 
 
 Consider now the second condition as being the more difficult Although 
 two ratios a : c, b : d are involved, neither can be arbitrarily assumed. -For 
 suppose e.g. that b : d= 2 : i ; then 2 2 + 2t* + <)ae would have to be made 
 a square ; this however is seen to be impossible, for, if we put a=p + q, 
 c=p q> we obtain the expression 13^- 5^, which cannot be made a 
 square. The same impossibility results if we put b : d=$ : i. Therefore 
 the ratios a : c and b : d can only be certain particular ratios. 
 
 Obviously the first class of ratios adapted for our purpose are square 
 ratios. Assume then that b : d =fr : q 3 , and put 
 
 *) + <u(f -f fY = pqa + ,, say; 
 
 therefore a (/* + ^) 8 a + tfffc = zmnpqa + nfc, 
 
 _= n ?- n *f? 
 
 c~ i? (f + f)* - zmnpq* 
 
 or, if m = kpq,
 
 35.6 SUPPLEMENT 
 
 Now, if values could be found for k, n, /, q such as would make ac or 
 
 (& - n-) {n (p* + g*Y 2kff\ a square, 
 
 we should have a solution of the problem, since, bd being already a square, 
 abed would then be a square. Euler however abandons the investigation 
 of this general problem as too troublesome and as certain, in any case, to 
 lead to very large numbers; and, instead, he proceeds to seek solutions 
 by trial of particular assumptions. 
 
 Particular values of ajc in terms of p, q are the following, which are 
 obtained by putting > = 2, n=i ; k = 3, = i ; etc. 
 
 in. ' , 5g .... iv. a - ^f- 
 
 c 4(y+<rrr<r c 
 
 VI f= 2 4/V 
 
 ^ (^ 2 +7 2 ) 2 
 
 a 
 c 
 
 Taking now the simplest values of bjd-p^lq^, let us write down the 
 simplest corresponding values for a\c : 
 
 if ^ = T. * becomes f, -|, |, / F , -- 1 /-, ^, ^, ^; 
 
 if _ 4 ^ liprnmpc 4 12 a2 3 2 5 5 60 20. 
 
 " T Becomes 3, TT , - T -, T5 -, T7 , yy, - T -, T , 
 
 if ^ hprniTIPC 108 45 28 64 fi4 
 
 11 ^ T> ~ uccujiies -2-g-> FT' TF> T' ^s' 
 
 The last assumption gives, "praeter exspectationem," two cases in which 
 ajc becomes a square ; and these give two solutions of our problem. 
 
 1. Putting a = 64, b = 9, c - 49, ^=4, we have 
 
 M=ab + bc + cd + da = $'](> + 441 + 196 + 256 
 
 = 1469* 
 and the four numbers are 
 
 1469.196, 1469.256, 1469.441, 1469.576. 
 
 2. Putting a = 64, /> = 9, ^=289, d - 4, we obtain 
 
 J/=576+ 2601 + 1156 + 256 = 4589, 
 and the four numbers are 
 
 4589.256, 4589.576, 4589.1156, 4589.2601.
 
 SOLUTIONS BY EULER. PROBLEM n 357 
 
 Again, the form of the expressions abed, bd (c? + e*~) + ac (b + df to be 
 made squares shows that any values for aje obtained by the above process 
 may be taken as values for bid. Also a and c may be interchanged. Euler 
 accordingly sets down as values of b\d\.Q be tried the following: 
 
 f , f , I, , , f*. , , etc. 
 
 He obtains no solution from the assumption bjd~^, but he is more 
 successful with the assumption bjd=\. 
 
 Putting bld = ^ t we have to make 
 
 20 (a 2 + r 2 ) + 8io<r=a square. 
 
 This is satisfied by a\c=\; let therefore aje = i + x, and we have to 
 make 20 {(i + x) z + i} + 8i (i + x) a square; that is, 
 
 121 + I2ix + 20x^-3. square = (n + xyf, say; 
 
 121 - 22V ,a V 2 22V+IOI ttt* + 
 
 therefore x = = - , and - =< - ^- 
 * 
 
 22mtl + 
 
 - , 
 c y-2o m*-2on- 
 
 and, by putting m 5, n i we obtain a/r= ^. 
 
 This solution serves our purpose, since it makes abed a square. 
 Putting a - 16, =5, ^=5, <^=4, we have 
 
 80 + 25 + 20 + 64 189 
 
 M= ~7^- 7 r> 
 
 and, if /^ = 3, M= 2 1 ; the four numbers are therefore 
 
 21. 20, 21.25, 21.64, 21.80. 
 This is a solution in much smaller numbers ; and 
 
 the sum of the numbers =9.21-, 
 
 the sum of products of pairs = 1 1 o 2 . 2 1 2 , 
 
 the sum of products of sets of three = i2o 2 . 2i 4 , 
 and the product of all four = i6oo 2 . 21*. 
 
 When one solution is known, others can be found. Take, for example, 
 the last solution in which, for bfd-^, we found that 
 a y* 22y+ IQI 
 
 C ~ jP20 
 
 In order that abed may be a square, we must have 
 
 5(y s -2o)(y-22j'+ 101)= a square. 
 This is satisfied by y = 5. Substitute 2 + 5 for y, and we have 
 
 5 (z 2 + 102 + 5) (z 2 - 1 20 + 1 6) = a square, 
 or 400 + 5000-4950"- 102^ + 52* = a square.
 
 358 SUPPLEMENT 
 
 Equate this expression to (20 + ^/-z ^-z*) 2 , and we have 
 
 \ 3 2 2 / 32 
 
 whence 
 
 3- "55. = 3- '5 = 33g 
 
 . 
 
 266321 242II 2201 2201 
 
 therefore j = 2 + 5 = ^/^- ; and the resulting values of a, c are large 
 numbers which Euler does not trouble to develop. Asa matter of fact, 
 
 a _ 55696 _ 4. n8 2 
 c "109465205 "5. 4679* ' 
 It follows that 
 
 f 2 M= 5 . 55696 + 5 . 109465205 + 4 . 109465205 + 4 . 55696 
 = 278480+ 547326025+ 437860820+ 222784 
 = 985688109; 
 and, putting /= 9, we have 
 
 M= 12168989. 
 
 The four numbers are therefore 
 12168989 . 278480, 12168989 . 547326025, 
 
 12168989 . 437860820, 12168989. 222784. 
 PROBLEM 12. To find three numbers x, y, z such that the expressions 
 
 X* +y > + z 2 , x 2 f + x 2 z 2 +y 2 z 2 
 are both squares^, 
 
 In order to satisfy the first condition, we have only to put 
 
 x =p 2 + q 2 r 2 , y = 2pr, z = 2qr, 
 
 for then x 2 +/ + z 2 = (/ + q 2 + r 2 ) 2 = P 2 , say. 
 
 The second condition requires that 
 
 therefore, since y* + z 2 - ^r 2 (/ 2 + ^), 
 
 <2 2 = ^ (p* + y 2 ) (p* + f- r 2 ) 2 
 or <2V 4 ^ - (/ 2 + f) (f + f- r 2 ) 2 
 
 In order to get rid of the sixth power of p and so make p* the highest 
 power of/, suppose that r-p-nq (which introduces no restriction); 
 therefore 
 
 2 2 / 4 (/ - nq? = (p 2 + q 2 } \znpq + (i - n 2 } q 2 } 2 + 4 /V (/ - nqf, 
 or <2 2 / 4 ? 2 (/ - nqf * (/ 2 + ? 2 ) { 2 / + (i - n 2 ) q} 2 + 4/ 2 (/ - ?) 2 - 
 
 1 Ada Acad. Scient. Imp. Petropol., 1779, Vol. III. (1782), pp. 30 sqq. = Commenta- 
 tiones arithmetic ae, n. pp. 457 sqq.
 
 SOLUTIONS BY EULER. PROBLEMS n, 12 359 
 
 Let the latter expression be denoted by fi 2 , so that Q= 2q(p-nq) R; 
 and 
 
 ^ = 4(1+ 2 )/ 4 - 4* (i + fP)fq + (i + 6rr* + ft 1 )/?? 
 
 This may be made a square in two ways, either (i) by taking advantage 
 of the fact that the last term is a square, or (2) by making the first term, 
 i.e. making i + 2 , a square. 
 
 (i) Put .ff s ={( I - 2 )r + 2 /V + a ; > y> and make the term in q* 
 disappear by choosing a so that i + 6* + 4 = 4 2 + 20(1 -*), or 
 
 I + 2# 2 + 4 
 
 a = - - - ; we then have 
 
 whence (i5~35 2 + i3 4 - 6 )/ = 8(i -*) (3 - 
 
 this divides throughout by 3 - a 2 , and 
 
 ^ 5 
 
 Let/=-8(i -w 2 ), ^=5 io 2 + 4 ; then r=p-nq = n (3 + 2*- 4 ), 
 while 
 
 and a:, j, s can be expressed in terms of . 
 Ex. i. Suppose = 2; then 
 
 / = -48, ^ = -19, r = -io, ^ = 7035, <2=3 8 
 x = 2565, y = 960, s = 380, or (dividing throughout by 5) x = 5130'= 1 9 2 
 2= 76 (in which case =106932, P=>^^). 
 
 Ex. 2. Suppose = 3; then p = - 192, ^ = - 4, r = - 180, or (dividing 
 by -4) 
 
 ^ = 48, ^=i, '=45, ^=14120, =1270800; 
 
 x - 280, _y = 90 . 48, z- 90. 
 Dividing the values of x,y, z by 10, we have 
 
 * =28, ^ = 432, = 9, and Q= 12708, /*=433- 
 (2) To make the first term in IP a square, suppose i + 2 = w 2 , which 
 is the case whenever n = (a 2 - tr)/2al>. 
 We have then 
 
 IP = 4m*p 4 - 4/ 2 /V + (* + 4** a )/y + 4 (i - 8 )// ^ ( r ~ "T / 
 Euler solves this in three ways.
 
 360 SUPPLEMENT 
 
 First, he puts R=2mp 2 nmpq + (i-n 2 }q 2 ; and from this, by taking 
 a = 2, b = i, so that n - f , m = f , he obtains the particular solution 
 
 * = -39 2 > 7=i3 86 , * = -ios6, 
 or x= 196, 7= 693, z = 528. 
 
 Secondly, he puts R = im f + 2npq + (i - n 2 ) ? 2 , and deduces, by the same 
 particular assumptions, 
 
 * = 936, 7=74, 2 = 355 2 , 
 or * = 4 68, 7 = 37, 2=1776. 
 
 Thirdly, he supposes 
 
 w 4 + 3 2 
 
 K = 2mp mnpq H a 2 . 
 
 ^m 
 
 where however the last term should apparently be q~. 
 
 Euler's son, J. A Euler, gave, in a Supplement to his father's paper, 
 another solution as follows. 
 We know that 
 
 (/ - i) 2 + 4/> 2 = (/ + i) 2 and (g n - - i) 2 + 4^ = (f + i) 2 . 
 Multiply the first equation by 4^ 2 and the second by (p- + i)-; this gives 
 
 or _ I/+ 
 
 Therefore the three numbers 
 
 satisfy the first of the conditions. 
 
 The sum of the squares of the products of pairs of these numbers must 
 now be a square; after dividing out by 4^ 2 , this gives 
 
 (? 2 - 2 (/ 4 - i) 2 + 4/(? 2 - i) 2 (/+ i) 2 + i6/V (/- i) 2 = a square. 
 
 But (/ 4 -i) 2 + 4/ 2 (/ 2 +i) 2 = (/ 2 + i) 4 ; 
 
 therefore (/ 2 + i) 4 (^ 2 - i) 2 + i6/V 2 (/ 2 - i) 2 must be a square. 
 
 For brevity, let A 2 = (/ 2 + i) 4 , * = io> 2 (/ 2 - i) 2 , and 
 
 A 2 (f 2 - i) 2 + ^ 2 , or AY + (B 1 - 2 A*} q* + A 2 , 
 must be a square. 
 
 Put A 2 q i +(JP 
 
 A1 _
 
 SOLUTIONS BY EULER. PROBLEMS 12, 13 361 
 
 Now both the numerator and denominator of this fraction are squares 
 if v* = A- - ?, for the numerator becomes B* and the denominator 
 
 which is the square of A + ,J(A* - &}. 
 
 But, putting for A, B their values in terms of p as above, we have 
 
 <P-&=S 
 therefore 
 
 ' 
 
 and the numbers required will be 
 
 8/ 
 
 or, if we multiply by (/ 2 i) 2 , 
 
 (6/ -/-!)(/ + 1), 
 The sum of the squares of these numbers turns out to be (/* + 1) 6 , 
 which is not only a square but a sixth power, while the sum of the squares 
 of the products of pairs is found to be 
 
 or 
 
 Ex. Put p- 2, and we have 
 
 -16.4.9. 337 2 = 8o88 2 , 
 the solution being in smaller numbers than Euler's own. 
 
 PROBLEM 1 3. To find 1 three positive integral numbers x, y, z such that 
 
 x +y + z = 
 
 To make ar'+y + z 2 a square, put x = a 2 + $*-c i ) }> = 2af, z = 2h, and we 
 have x 2 +f + z* = (a 2 + P + r 8 ) 2 . 
 
 We have now to make a 2 + ^ + r 8 a square, and we put similarly 
 
 af> ij rq i r L ^ b = 2pr, c2qr\ 
 we have then x 2 +y> + 2 = (f + <f + r 2 ) 4 . 
 
 Now let us express x, y, z in terms of/, ^, r ; this gives 
 x =p* + f + r 4 + 2ff + 2/V - 
 
 1 Commentationes arithmeticae, II. pp. 399-400.
 
 362 SUPPLEMENT 
 
 therefore 
 x +y + z =p* + q* + ? A + 2/V + 2/V - 
 
 (1) Arrange this according to powers of/, and 
 x+y + z =/ 4 + 2 (^ + r) 2 / 2 + 8/tyr 2 
 
 In making this a square, we have to see that />, g, r are all positive, and 
 also / + ^ > r'. Also o 2 + 2 must be > <?. 
 
 Equate the expression to {/ 2 + (q + r) 2 } 2 , and we have 
 
 8/^r 2 + / + 4?V- 6/r 2 - ^r 3 + r* - (? + r) 4 , 
 whence 8/tyr 2 = 1 2^ 2 ^ + S^r 3 , or / = \q + r. 
 
 Therefore a = \ 3 -^ 2 + ^r, b = $qr + 2^, c = zqr, 
 where both the letters ^, r may be given any positive values. 
 
 Ex. i. Suppose 4=2, r= i ; therefore 
 
 / = 4, a -19, l> = 8, c = 4; 
 accordingly the numbers are 
 
 # = 409, .7=152, 2 = 64, 
 
 and ^+^ + 2 = 625 = 25 2 , x i +y i + z i = 194481 = 44i 2 = 2 1 4 . 
 
 Ex. 2. Let q = 2, r = 2 ; therefore 
 
 / = '5, = 25, ^=20, r=8, 
 and #=961, _y = 4oo, 2 = 320; 
 
 therefore #+jy + 2= 1681 =4i 2 , ^ 2 +_y 2 + 2 2 = 1185921 = 33 4 . 
 
 (2) Arrange the expression for x+y + z according to powers of g ; 
 this gives 
 
 x+y + z = # t + 44? r + 2^ (p* - y"} + ^qr (f- + zpr - r*) + (p- + ^) 2 . 
 
 In order that the terms in q* and q* and the absolute term may vanish, 
 equate the expression to 
 
 te 2 +2^-(/ 2 + ^)} 2 , 
 
 2pr (p + r) 
 
 whence we obtain q - ^- ^ . 
 
 2r--p* 
 
 Ex. Suppose/ = i, r = i ; therefore 
 
 q = 4, a= 16, 6=2, c = S, 
 or (if we divide by 2) a = 8, b - i, c = 4 ; 
 therefore x = 49, y = 64, 2 = 8; 
 
 and #+_>' + = ii 2 , # 2 +y j + 2 2 =656i =8i 2 = 9 4 . 
 
 These numbers are no doubt the smallest which satisfy the conditions. 
 
 The case of three numbers is thus easier than that of two (see p. 299, 
 note). Euler solves the same problem for four and five numbers, and shows 
 how the method may be extended to six numbers, and so on indefinitely.
 
 SOLUTIONS BY EULER. PROBLEMS 13, 14 363 
 
 PROBLEM 14. To find three numbers x, y, z, positive and prime to one 
 another, such that both x+y + z and x?+y 2 + z li are fourth powers 1 . 
 
 As in the above problem, put 
 
 x = a z + b* c 2 , y = 2ac, z 2bc, 
 and further a=fl* + qr i -r*, b- 2pr, c = zqr, 
 
 and make the expression in /, q, r for x + y + z a square by equating it 
 to {f + (q + rff as before. This gives / = \q + r; but we have now, in 
 addition, to make / 2 + (q + rf a square. 
 
 therefore g* (? + r) = zfgp +f> (q + r). 
 
 Substitute f q + r for/, and this becomes 
 
 whence 
 
 The problem may therefore be solved in this way. 
 
 Take q =/ 2 + 2 fg-g* and r = g* - 3 fg-/*, 
 
 so that / = ^(/ 2 -" 2 ), then find a, t>, c, and then again x, y, z, in terms 
 of/.?. 
 
 Ex. Let /= 1,^=3; therefore 
 
 # = -2, r = -i, p = -4, 
 or 4=2, r=i, p = 4. 
 
 Thus aig, ^ = 8, ^=4, 
 
 and ^ = 409, ^=152, 2 = 64; 
 
 so that x+y + z = 62$ = $*, x 2 +y 2 + z*= 194481 - 2 1 4 . 
 
 To find limits for the values off, g, change the signs of q, r, putting 
 V=g*-2fg-f\ r=f^ Z fg-g\ / = iQr 2 -/ 2 )- 
 
 In order that q may be positive, 
 
 glf> i +V2>2.4i4..., 
 and, in order that r may be positive, 
 
 Suppose e.g. that /= 2, g = 5 ; then 
 
 y=i, r=<), p = Q, 
 
 or in integers f-2, r=iB, p = 2i; 
 
 hence a =121, =756, ^=72, 
 
 ^=580993, JF= 17424, 2=108864, 
 = 707281 = 29 4 , ^+j 2 + 2 2 
 
 1 Comtnentationes arithmeticae, n. p. 402.
 
 364 SUPPLEMENT 
 
 PROBLEM 15. (Problem in Fermat's note on vi. 13.) 
 
 To find a right-angled triangle (in rational numbers) such that either 
 of the sides about the right angle less the area gives a square^. 
 
 Let the perpendiculars be , - , so that the area is -^ ; and 
 
 2X XV 
 
 --- ^ or 2xz xy 
 
 2 
 
 - - -^- or yz - xy 
 
 y 
 as well as -- or 4^ +y 
 
 have to be made squares. 
 
 Since the first two expressions are to be squares, their product must be 
 so also ; therefore 
 
 2xyz* tot? yz xy*z + x*y* = a square 
 
 and, after dividing byyz, we have 
 
 2p 
 
 2x2 - 2oc - xy = - xy + 
 
 * L * 
 Whence 
 
 Thus = 
 
 p*xy + fxy - 2pqxy = xy (p - qf 
 2<?x-fy 2<fx-J*y> 
 
 x (2qx -pyf x 2 (2qx - pyf 
 and 2xz -xy= -$2L j > 
 
 Therefore the two expressions are squares if 24* x 2 p* xy is a square. 
 
 therefore (2^ - r*} x = p*y, or x/y=f-/(2? z -r*). 
 
 It is sufficient for our solution to know the ratio x/y, since a common 
 denominator z has already been introduced. 
 Therefore we may put 
 
 1 Novi Commentarii Acad. Petrop., 1749, Vol. II. (1751), pp. 49 sqq. = Commentationes 
 arithmeticae, I. pp. 62-72.
 
 SOLUTIONS BY EULER. PROBLEM 15 365 
 
 whence ,-, = * 
 
 and ^ 
 
 . 
 
 It only remains to make 4*? +y* a square ; that is, 
 
 4p' + 42 4 ~44 2 r 2 + ? A must be a square. 
 
 A general solution of this equation giving all possible values of /, q, r 
 is impossible. We must therefore be satisfied with particular solutions. 
 
 Particular solutions (i) and (2). 
 Put 4/ + 4?* 
 
 therefore 4 
 
 and / ! = + (? 2 -^), 
 
 that is, either 
 
 (') /^Vfe'-O, or (2) / = 
 
 (i) Now / = ^ V^ 2 " **) is satisfied by q = + <t\ r = 2cd, whence 
 
 or we may put 
 
 / = (t* + <P) (r - 
 and we thus find values for 
 
 Ex. i. Suppose c= 2, d- i ; then 
 
 / = 5-3 =I 5> ^ = 4-5 = 20 r= 4.4=16, 
 
 544.25 25.89 2225 
 - -- =-- 
 
 and the triangle is 
 
 2 44 J 435 2 
 
 = = 
 
 2 ~ 89 ' 2 25.89' * 25.89 
 
 Ex. 2. If r = 3, d= i, we get the triangle
 
 3 66 SUPPLEMENT 
 
 (2) In this case p - - J(r* - q z \ and we have to put 
 
 2Cd(<*-< 
 
 \ g= 2cd, whence p = 
 
 or p = 2cd (<? - d z ), q= 
 
 while 
 
 Here, since 2<f must be > r 2 , 
 
 &<?d* > (c z + d*f and 2cd j 
 therefore d*>(c-d >J2) Z , and 
 
 either d>c-d^2, so that -> 
 
 c i + V 2 ' 
 or d>d^2-c, so that -< 
 
 C *J2 - I 
 
 If therefore d=i, either c<^2 + i or 0^2-1. The second 
 alternative is satisfied by c> i. 
 
 Ex. Let c - 2, d= i, and we have 
 
 ^ = 4.3=12, ^ = 4.5 = 20, ^=5.5 = 25; 
 therefore 
 
 The triangle is therefore 
 
 2X = 288.25 = 45_o . J = 25-I75 = 4375 . V(4^ 2 +y) = 25.337 = 8425 
 z 4048 253' z 4048 4048' * 4048 4048' 
 
 Particular solution (3). 
 Put 4 / 4 + 4^ 
 
 therefore ^ - 4? V = 8/V 2 , and / 2 = 
 
 therefore either / = ^(zr 3 - 8^), or p = 
 
 The first value is however useless, since 2q- - r 2 > o, or 
 
 We have therefore / = J(Sa 2 - 2^), 
 , .. 4f 
 
 while 
 
 Since 8^ 2 2^ must be a square, put
 
 SOLUTIONS BY EULER. PROBLEM 15 367 
 
 therefore 4^ - 2r = ^ (zq + r\ or ^d' 2 q - 2^/V = 2<?q + <?r, 
 whence 
 
 2r*) = Scd, and therefore * = 
 Multiplying by zd' 1 + , we have in integers 
 
 while x, y, z have the values above stated. 
 Ex. i. Put c= i, d= i ; therefore 
 
 / = 4 ? = 9> r =6; x=i6, y=i 2 6, ^(4^ +/)= 130, 
 
 j . 126 . 25 207 
 
 and 2=16 + -- 7-^= ^-. 
 
 36 2 
 
 The triangle is therefore 
 
 6 4 22 * 2 + 260 
 
 207' z ~ 207 ' z ~~ 207' 
 
 This is the triangle in the smallest numbers satisfying the conditions, as 
 Euler proves later. 
 
 Ex. 2. Since 2^ a >r 2 , it follows that ' c\d > 2 - ^2 ; but it does not 
 matter whether 2d' 2 > <? or not, since /, y, r may be negative as well as 
 positive. 
 
 Put then d= 2, c=$; therefore 2d' 2 - <? = - i, and 2d' 2 + <? = 17. 
 We then have p - - 24, ^ = 289, r - - 34 ; 
 
 The triangle is therefore 
 
 ?f = 2 34 J = 28.41. if J(4X*+f) = 4-5-53-3 I 3 
 z 28118255' z 28118255' -z ' 28118255 
 
 It is to be observed that in all the above examples it matters not 
 whether c, d are negative ; it will only result in the values of / or q or r 
 becoming negative, but the values of x and y will not be thereby changed. 
 Only z will vary, since z may be either 
 
 y(p- ?) 2 y (P + v? 
 
 x + , or x + J ^ . 
 
 After remarking that the problem of making 
 
 4/ 4 + 4^ - 4? 2 ^ + ^ or 4/ 4 + (24* - r 2 ) 2 a square 
 may be solved generally by equating it to 
 
 Euler passes to his general solution.
 
 368 SUPPLEMENT 
 
 General solution. 
 If , - are the perpendicular sides of the triangle, let x - ab, 
 
 y = a*-lr*i the triangle is then 
 
 2ab rf-P a* + t? 
 
 and the area is 
 
 Now we found above, at the beginning of the investigation, that 
 _ = 2q i x* + q^xy - 2pqxy = ^ [ xy (p - q? 
 2q l x-fy zfx-fy* 
 
 or, since q can be taken positively as well as negatively, 
 
 where x = ab, y = a* IP. 
 
 And we took 2?*x 2 -p z xy = r* 
 
 y(pgf (^ 
 
 whence z = x + I = a b+^ - 
 
 We have therefore only to satisfy the equation 
 
 and, since xy = ab(d i -b y ), we have to find such numbers for a, b that 
 ab(<?-F) may be of the form 2/ 2 -^ 2 or (2/*-g n -}h\ 
 
 Suppose now that such numbers a, b have been found that 
 ab(o?-b^ = (2 
 
 Then, since x = ab, 
 
 and a natural inference is suggested, namely that 
 aba abr . 
 
 Let now p ab, and accordingly 
 
 the triangle is then 
 
 2ab 
 
 + (a 2 - P) (ab fKf ' 
 
 abg'W + (a 2 - P) (ab fhf ' 
 
 abgW + (a 2 - P) (ab fKf '
 
 SOLUTIONS BY EULER. PROBLEM 15 369 
 
 Also from any particular values of a, b any number of triangles can be 
 derived satisfying the conditions. 
 
 For, if p ab, and 
 
 a& (<*-&) = (*/*-?), 
 
 we have (a/ 3 - g*) h* = 2<?- r>, 
 
 or 
 
 Put 2(/a + ?) = (tf* + r) f 9*&fk- 9 = 
 
 therefore q = *mngh -##) jk ^ 
 
 -)gh - qmnfh 
 
 21? -n? 
 
 or, in integers, p-(2f^ m-) ab, 
 
 q = 2mngh - (20* + m 2 )fh, 
 r = (2# 2 + * 2 ) gh 
 
 while z 
 
 rr, , - . 
 
 Thus the triangle I , - , - I is known. 
 \ z z a / 
 
 Lastly, to find suitable values for a, b, Euler writes down all the 
 numbers from i to 200 which are of the form 2/ 2 - w 2 , including all the 
 squares arising from the supposition that u = t, and all the doubles of 
 squares corresponding to u - o. Inspection shows that the table contains 
 (within the limits) all the prime numbers of the form 8/ + i, and no other 
 primes, the doubles of the primes, the products of the primes into all 
 squares and into one another, and the doubles of those products. 
 
 Now, since the product a . b (a + b} (a - b} is to be of the form z/ 2 - ir 1 , 
 and the factors a, b, a + b, a b are either prime to one another or at the 
 most have 2 as a common divisor, while 2 is itself contained in the form 
 2/ 2 - 2 , the several factors must all be of that form, in which case the 
 product will be of that form. 
 
 We have therefore first to take some value of b in the table and then see 
 whether there are in the table three other numbers a&, a, a + b differing 
 by b. Euler gives a second table showing values of a corresponding to 
 values i, 7, 8, 9, 16, 17 etc. of b. 
 
 The values of a in the table corresponding to b=i are 8, 17, 63, 
 72, 127. 
 
 Ex. i. Take b= i, a = 8 ; therefore 
 
 ab = S, <?-P = 63, a(a a - 2 ) = 8.9.7 = 4.9. 14, 
 
 and 4.9. 14 = A 2 (z/ 2 -^ 2 ), so that h = 6, 2/*-g*=i4, and accordingly 
 /=3> ^= 2 - 
 
 H. D. 24
 
 370 SUPPLEMENT 
 
 We have therefore in this case 
 
 / = 8(2 2 -0* 2 ), ^ = 24/;m-i8(2 2 + ; 2 ), r= 12 (zn? + m*) - >j 2tnn ; 
 or, dividing by 2, 
 
 Thus there are any number of values of z from which triangles may be 
 obtained satisfying the conditions. 
 
 The simplest value is found by putting m=i, n = o, whence 
 
 and either z = 8 + . 25 = !, 
 
 or z = 8 + f.i6 9 = lY JL - 
 
 The first value gives the triangle in smallest numbers above found 
 
 (P- 367), 
 
 2ab_ _ _64_ a 2 -^ 2 _ 252 cP + P = 260 
 
 z ~ 207' z 207' z 207' 
 Substituting 1215 for 207, we have the sides of the triangle corre- 
 
 sponding to the second value of z. 
 
 The particular triangles are also directly obtained from the values of 
 
 a, t>, f, g, h without bringing in m, n ; for 
 
 thatis , 
 
 Ex. 2. Take ^ = 41, a 112; therefore 
 
 0^=7.16.41, a 2 -^ 2 = 71 . 9. 17, 
 and ab(c? - P) = 16 . 9 . 7 . 17 . 41 . 71 = ( 2 / 2 
 
 whence h - 12, and 7 . 17 . 41 . 71 = 2/ 2 g*. 
 
 The simplest solution is/ =41 7, = 37. 
 
 is easily found, and consequently the triangle 
 zab a*-P a* + 
 
 [Euler finds values for/ g by using the formulae 
 (2a 2 - / 8 2 )( 2 y 2 -8 2 ) = (2ay^8) 2 -2( ) 8 y 
 and x* - 2f = 2 (# j>/) 2 - (a: 2j>;) 2 .
 
 SOLUTIONS BY EULER. PROBLEMS 15, 1 6 371 
 
 He does not actually give the steps leading up to the particular 
 solution 7=41 7, -=37, but it can be obtained thus. 
 Since 7 = 2. 2 2 - i and i7 = 2.3 2 -i, we have 
 
 7. l7=(2.2.3+I.l) 2 - 2 (3. I + 2.I) 2 
 
 = i3 2 - 2 .5 2 = 2 (i3-5) 2 -(i3-2. 5 ) i = 2 .8 2 -3 2 . 
 Again, since 41 = 2 . 5 2 - 3 2 and 71 = 2 . 6 2 - i 2 , it follows that 
 4I . 71^(2.5. 6-3. i) 2 - 2 ( 3 . 6-5. i) 2 
 
 = 57 2 -2.i3 a =2(57-i3) a -(57-2.i3) 2 = 2 .44 2 -3i 8 . 
 Therefore, by multiplication, 
 
 - 7 97 2 - 2 . 3 8o 2 = 2 (797 -38o) 2 - (797 - 2 tfY= * 4i7 2 -37 J -] 
 
 PROBLEMS 16. "De problematibus indeterminatis, quae videntur plus 
 quam determinata 1 ." 
 
 We have seen that by means of certain " Porisms " stated without 
 proof Diophantus is able to obtain relations between three numbers 
 x, y, z which have the effect that, when they are satisfied, a quite 
 appreciable number of symmetrical expressions in x, y, z are auto- 
 matically (as it were) made squares. 
 
 It is clear therefore, says Euler, that, if a general method of finding 
 "porisms" of this kind can be discovered, the whole subject of Diophantine 
 analysis will be appreciably advanced. Accordingly he proceeds to discuss 
 such a method. 
 
 The method depends on a Lemma the truth of which is evident. 
 
 Lemma. If values have been found for the letters z, y, x etc. which 
 satisfy the equation W-Q, where W is any function of those letters, 
 and P, Q, R etc. are other functions of the letters such that P+W, 
 Q + W, RW etc. are squares, then, if the values of z,y,x etc. are 
 taken which satisfy W=o, the resulting values of P, Q, R etc. will 
 also be made squares. 
 
 Cor. P, Q, R etc. will similarly be made squares if P+ a W, Q + ft W, 
 yW&c. or, more generally, if 
 
 etc. 
 are squares. 
 
 Conversely, If such values for z, y, x etc. have been assigned as will 
 satisfy W= o, all formulae such as P 2 + a W, Q? + p W, R* + yW etc. will 
 at the same time be made squares. 
 
 1 Novi Commentarii Acad. Petropol., 1756-57, vi. (1761), pp. 85 sqq. = Commenta- 
 tiones arithmeticae, I. pp. 145-259.
 
 372 SUPPLEMENT 
 
 And, as the number of such formulae is subject to no limit, it is clear 
 that an unlimited number of conditions can be prescribed which are all 
 satisfied provided that the one condition W= o has been satisfied. 
 
 The same Lemma can be extended to the case of cubes or any higher 
 powers ; for, if W= o has been satisfied by certain values, all expressions of 
 the form P 3 + a W will thereby be made cubes, all expressions of the form 
 P* + a W will be made fourth powers, and so on. 
 
 While it is plain that, if values for z, y, x etc. are found which satisfy 
 the condition W= o, all the expressions P 2 + a W, Q 2 + ft W, R 2 + y W etc. 
 will be made squares by the same values of x, y, z, the difficulty will be, 
 when a number of expressions P 2 + a W, Q 2 + ft W etc. are given which are 
 capable of being made squares in this way, to identify and separate the 
 expression W the equating of which to zero will make the rest of the 
 several expressions automatically squares. It would indeed be easy so to 
 hide away the composition of the expressions as to make this separation 
 itself a most arduous problem. On the other hand it is easy and in- 
 teresting to begin with W= o, and then investigate the simpler formulae 
 which can by this means be made squares. Before proceeding to the 
 particular cases, Euler observes further that it is convenient to take for W 
 an expression in which z, y, x etc. enter symmetrically and can be inter- 
 changed ; for then, if P 2 is such a square that P 2 + a W is a square, and 
 z, y, x etc. are interchanged so as to turn P 2 into Q 2 , R 2 etc., Q 2 + a.W, 
 R" 1 + a W etc. will also be squares. Also, since solutions in rational 
 numbers are required, z, y, x etc. should not enter in any higher power 
 than the second into the expression W. Euler begins with expressions 
 containing two unknowns 2, y only. 
 
 Problem (i). Given W=y + z - a = o, to find the more simple formulae 
 which by means of this equation can be made squares. 
 
 When the equation y + z-a = o is satisfied, it is clear that the general 
 formula P 2 + M(- a +y + z) will become a square whatever quantities are 
 put for P and M. Accordingly Euler, by giving /*, M various values, 
 obtains without difficulty 44 different expressions which become squares 
 when y + z a o. 
 
 He supposes M- 2, - 2, 2, -y, - z -y, y + z + a, n (y + z + a), 
 (y + z + a)(y-z + a)(z-y + a), and 3 and n 2 - i times the last expression 
 respectively, and with each of these assumptions he combines one or more 
 forms for P. I need only quote a few expressions which are thus made 
 squares, e.g. 
 
 (y - i) 2 + 2 (- a +y + z} =y 2 + 22 
 (2 - i) 2 + 2 (- a +y + 2) = 2 2 + 2y + 
 
 + i - 2aJ '
 
 SOLUTIONS BY EULER. PROBLEMS 16 373 
 
 (y - z + i ) 2 - 2 (- a + y + z) = (y - zf - 40 + i + 2a\ 
 
 y(-a+y + z) =y z + znz + n" 2na, 
 
 -(y + z) (- a+y + z) = ay + az 
 
 (yz -nY + n(y + z + a) (- a + y + z) =y i z* + tiy 2 + nz* + 2 - 
 (y*+ z 2 +a?) 2 + (y+z + a) (y-z + a) (z-y+ a) (-a+y + z) 
 
 and so on. Wherever a new expression can be got by interchanging z 
 and y, this may be done. 
 
 Taking the more particular case of W=y + 0-1=0, Euler obtains the 
 following expressions which are thereby made squares, 
 
 y + 42, f-y + z, y + z, y-yz, 
 z* + ty, z*-z +y, z -yz, 
 
 /Z 2 +/ + Z 2 , 2/+2Z 2 -!, 
 
 which indeed are easily seen to reduce to squares if we put y + z=i or 
 y=i-z. 
 
 The fact that j 2 2 2 +y* + z 2 is a square if y = i z or, more generally, if 
 y = iz, is included in the Porism in Dioph. v. 5. Similarly 
 
 is made a square if we put y = a z. 
 
 The last expression but one in the first of the above lists, namely 
 y*z z + ny* + nz* + ri*- na\ becomes a square whatever value n has. If a = t, 
 it becomes 
 
 y>z* + ny* + nz* + n 2 - n 
 
 or (f+n)(z' i + n)-n. 
 
 That this is a square when z -y i is part of Diophantus' assumption 
 in v. 4 (see p. 104 above). 
 
 Euler's Problems (2) and (3) similarly show how to find a number of 
 formulae which are all made squares by values of y, z satisfying the 
 equations W-yz -a(y + z) + t> = o and W=y* + z 2 2nyz -a-o. 
 
 He then passes to the cases where there are three unknowns. 
 
 Problem (4). Given W- x+y + z-a = o, to find the more noteivorthy 
 formulae which can be made squares by satisfying this equation. 
 
 In this case the general expression P* + M(x +y + z-a) becomes a 
 square.
 
 374 SUPPLEMENT 
 
 Put M- 2n and P= one of the expressions x - n, y n, z-n, or one 
 of the expressions y + z n, z + x n, x +y n. 
 
 These assumptions make the following expressions squares : 
 z 2 + 2n (x +y) + n?-2na, and the two other similar expressions, 
 (y + zf + 2nx + ri* - 2na, 
 
 If M 2nyz, P=yz ny nz, and so on, 
 
 jV + 2nxyz + 2 y + V + 2n (n - a)yz 
 and the two other similar expressions are squares. 
 
 If M= - (a + x +y + z), /> =j> + 2 # and corresponding expressions, 
 
 2 4yz - 4yx are all squares. 
 a?-4zx-4zy ] 
 
 In particular, if n = 2a, a- \, the following six expressions are made 
 squares by putting x +y + z = ^ : 
 
 y* + z + x, 
 
 z* + x +y, (x +y) 2 + z. 
 If a = 2, we make the expressions 
 
 i xy xz 
 i-yz-yx 
 i zx xy 
 all squares by putting x +y + z = 2. 
 
 Problem (5) finds expressions which are made squares if 
 
 W=yz + zx + xy a (x +y + z) + b = o 
 is satisfied. 
 
 Problem (6). Given W= x 2 +y z + z 2 - zyz 2zx 2xy -a = o, to find 
 the more simple formulae which can be made squares by means of solving that 
 equation. 
 
 Here the general formula will be 
 
 If J/=-i, P=x+y + z, 
 
 4yz + 4zx + ^xy + a = a square. 
 If M= - i, P= y + z-x, etc., 
 
 "I 
 
 4zx + a h are squares. 
 
 \xy + a 
 I(Af=-i, P=y-z, etc., 
 
 a + 2 (y + z) x - 
 
 a + 2 (z + x)y y* V are squares. 
 
 a + 2 (x +y) z 
 
 - x 2 "i 
 
 y* V 
 
 z 2 J
 
 SOLUTIONS BY EULER. PROBLEMS 16 375 
 
 In the particular case where a - 4*1, so that 
 
 * = 2yz + 2zx + 2xy + 4*, 
 
 zx + n 
 - xy + n ar e all squares; 
 
 yz f zx + xy + n J 
 
 or our formula gives the means of solving the elegant Diophantine 
 problem : 
 
 Given any number n, to find three numbers such that the product of any 
 pair added to n gives a square, and also the sum of the products of the pairs 
 added to n gives a square. 
 
 By solving the equation 
 
 IV 3? +y* + z 3 2yz zzx 2xy 4 = o, 
 we obtain z = x + y 2 J(xy + n). 
 
 We assume, therefore, such numbers for x, y as will make xy + n a 
 square; suppose xy + n = u*, and we then have two values for z, namely 
 z = x +y 2u, each of which along with x, y will satisfy the conditions. 
 
 In fact, if z = x + y 2u, while u - J(xy + n), 
 
 J(xz + n) = %(x + z -y) = xu, 
 
 
 (Cf. Euler's solution of Dioph. in. 10, p. 160 above.) 
 
 Problem (7). Given 
 
 W- y? +y* + z* - 2yz 2zx - 2xy 2a(x+y + z)-6 = o, 
 to find the more noteworthy expressions which can be made squares by 
 satisfying this equation. 
 
 The general expression here is 
 
 P* + M{x* +y* + z*- 2yz - zzx - 2xy - 20 (x +y + 2) - b}. 
 If M- - i, P= x+y + z + a, we have 
 
 (a) 4yz + 4zx + 4*y + 4a(x+y + z) + a* + 6 = 3i square. 
 =-i, P=x+y+z-a, 
 
 ^) 472 + 4zx +4xy + a l + 6 = a. square. 
 =-i, P=y + z-x + a, etc., 
 
 4yz + 40 (y + z) + a* + b \ 
 (c) 4zx + 40 (z + x) + s + b [ are all squares. 
 4xy + 40 (x +y) + a* + b
 
 376 SUPPLEMENT 
 
 If M= i, P=y + z - x a, etc., 
 
 4yz + 4ax + a? + b \ 
 (d) ^zx + 4ay \-cP-\-b L are all squares. 
 
 4xy + 4<zz + d* + & ) 
 
 Cor. i. In order to solve the problem represented by (c), equate the 
 expression 4xy + 40 (x + y) + a? + b to a square # 2 , whence 
 
 4 (x + a) (y + a) = u* - b + 30?, 
 or (x + a) (y + a) = % (u* -b + 3a 2 ) ; 
 
 x and y are then determined by splitting \(u*-b + 3<z 2 ) into two factors 
 and equating x + a, y + a to these factors respectively. Next, solving, 
 for z, the equation 
 
 x 2 +jv 2 + z 2 - 2yz - zzx - 2xy - za (x +y + z)-t> = o, 
 we find, since 4xy + 40 (x +y) + cP + b = u*, that 
 
 z x +y + a + u. 
 Cor. 2. If b = a 2 , then, by solving the equation 
 
 x 2 +y 2 + z 2 = 2yz + 2zx + 2xy + 2a (x + y -f z) a 2 , 
 
 we make all the following formulae severally squares, 
 
 yz + a (y + z), yz + ax, 
 
 zx + a(z + x), zx + ay, 
 
 xy + a (x +y}, xy + az, 
 
 yz + zx + xy, 
 
 yz + zx + xy + a (x +y + z), 
 by assuming 
 
 z = x+y + a+ 2,J{xy + a (x +y)} = x +y + a+ 2u, 
 
 where (x + a) (y + a) is put equal to 2 + a 2 . 
 
 An interesting case of this last problem is that in which a = i ; and 
 from this case we can deduce a solution of a new problem in which the 
 corresponding expressions with # 2 , y 2 , z 2 in place of x, y, z are all squares. 
 The problem is 
 
 To find three square numbers such that (i) the product of any two added 
 to the sum of those two, (2) the product of any two added to the third, (3) the 
 sum of the products of pairs, (4) the sum of the products of pairs added to 
 the sum of the numbers themselves, all give squares. 
 
 We have to find values of x z , y z , z"* which will make all the following 
 expressions squares, 
 
 /Z 2 +/ + Z 2 , /Z 2 + X 2 ,
 
 SOLUTIONS BY EULER. PROBLEMS 16, 17 377 
 
 As we have seen, all these will be squares if 
 
 z* = x* +f + i + 2 y(*y + of + f). 
 
 We have also seen (Problem (i) above) that x?f + x>+f becomes a 
 square if only y = x + i. Put then y = x + i, when we have 
 
 2 2 =2^ 2 + 2X+ 2 2 ,J(x*+2 X 3 + ^X 2 + 2X + i); 
 
 that is, = 4 (* + * + i). 
 
 It only remains to make x 2 + x + i a square. Equate this to (- x + f)\ 
 and we have 
 
 whence 
 
 Therefore the roots of the required squares are 
 
 Or, putting t=(r q)\zq, the values become 
 
 4qr $qr zqr 
 
 Let ^ = i, r = 2, and we have a; = f , y - 1, a = J ; or, if we put / = 2 in 
 the values expressed in terms of t, the values are x = f , y = f , z = Y- 
 
 PROBLEM 17. To find two fourth powers A 4 , 4 such that their sum 
 is equal to the sum of two other fourth powers^. 
 
 In other words, to solve the equation A* + '= * + >', or (what is 
 the same thing) A* - >* = C* - \ 
 
 It is proved, says Euler, that the sum of two fourth powers cannot be 
 a fourth power, and it is confidently affirmed that the sum of three fourth 
 powers cannot be a fourth power. But the equation A* + * C* = 2)* is 
 not impossible. 
 
 First solution. 
 
 Suppose A =p + <?, D=p-q, C-r + s, B = r- s; 
 thus the equation A 4 - Z> 4 = C 4 - * 
 
 becomes pq (p- + q*) = rs(r* + s?). 
 
 Put p = ax, q = by, r = kx and s =y, and we obtain 
 ab (aV + bY) 
 
 1 Navi Commentarii Acad. Petropol., 1772, Vol. XVII. (1773), pp. 64 &\<\.=.Commen- 
 tationes arithmeticae, I. pp. 473-6; Mt 'moires de FAcad. Imp. de St Pitersbourg, XI. (1830), 
 pp. 49 sq.= Comment, arithm., II. pp. 450-6. 
 
 24~5
 
 378 SUPPLEMENT 
 
 therefore ^= ^ -r , which fraction has therefore to be made a square. 
 One obvious case is obtained by putting k = ab, for then 
 
 whence _y = a, # = i, so that p = a, q = ab, r-ab, s = a, and the result is 
 only the obvious case where # = s, q = r. 
 
 Following up this case, however, let us put k = ab(i + z). 
 We then have 
 
 x?~ ab (P-i- z) Ir - i - z ' 
 
 therefore, multiplying numerator and denominator by P - i - z and ex- 
 tracting the square root, we obtain 
 
 To make the expression under the radical a square, equate it to 
 
 and assume /, g such that the terms in z, z 2 vanish. 
 
 In order that the term in z may vanish, /= | (3/r- i), and, in order 
 that the term in z 2 may disappear, 
 
 3^ 2 - 2) = 2 (P- i) g+f>= 2 (^- 
 
 t_*- 
 
 whence 
 
 __. 
 
 The equation to be solved is then reduced to 
 
 Now b can be chosen arbitrarily; and, when we have chosen it and 
 thence determined z, we can put 
 
 x b 2 i 2, y 
 and accordingly 
 
 -\ +fz + gz*), s = a(F-i +fz -f gz"), 
 where we may also divide out by a. 
 
 If x, y have a common factor, we may suppose this eliminated before 
 J>, q, r, s are determined.
 
 SOLUTIONS BY EULER. PROBLEM 17 379 
 
 Ex. i. Let b = 2 (for b cannot be i, since then g would be oo ). 
 
 TViorAfnrA f ii <T zs -_6OO 
 
 Ineretore /-->r> ^-~TT 2 -T?27- 
 
 As a does not enter into the calculation, we may write i for it : 
 therefore 
 
 6600 2187 ii 6600 25 /66oo\ s 
 
 =* --- = - - = ---- "- 
 
 / 
 
 2929 2929 2 2929 24 
 
 = - + 55407 -"QQ = 3 2889494 
 
 2929* 2929* 
 
 But the ratio x :y is what we want, and 
 
 y = 3 . 28894941 = 28894941 = 3210549 = 1070183 
 
 .Y 2187.2929 2929.729 2929.81 27.2929' 
 
 so that we may put 
 
 x = 79083, .7=1070183. 
 
 Therefore 
 
 / = 79083, r= 27. 19058 =514566, 
 
 ^=2.1070183 = 2140366, 5=1070183. 
 
 Consequently 
 
 A=p + q= 2219449, C=r + s= 1584749, 
 ^ = r-5 = -5556i 7 , D = p-q = -2061283, 
 and A 4 + *=C 4 + D*. 
 
 Ex. 2. Let b = 3 ; therefore /= 13, g = f , * = fr J 
 ^^ 3-369 no7_27-4i. 
 
 ***< s +^"i6r"is9~ i6 9 ' 
 
 200 8.144 
 *= -T6^ = ~T69~' 
 
 5 200\ - 200 2447 _ 8.89736 
 
 * + 4'-^) = + ^'~^~ 16? 
 
 x 8.144-169 6.169 
 Therefore - = - 8T g^ '^3-9' 
 
 and we may put -v = 1014, y = 3739- 
 
 Accordingly / = i o 1 4, r= ~* . i o 1 4 - 6642, 
 
 ^=11217, ^ = 3739. 
 
 and therefore ^ = 12231, (7=10381, 
 
 .5=2903, Z> = - 10203, 
 and again A 4 + 4 = C<
 
 380 SUPPLEMENT 
 
 Another solution in smaller numbers. 
 
 In the second of the papers quoted Euler says that, while investigating 
 quite different matters, he accidentally came across four much smaller 
 numbers satisfying the conditions, namely, 
 
 ,4 = 542, .5=103, ^=359. ^> = 5 I 4, 
 which are such that A* + E* = C* + D 4 . 
 
 He then develops two methods of analysis leading to this particular 
 solution; but, while they illustrate the extraordinary ingenuity which he 
 brought to bear on such problems, they are perhaps of less general interest 
 than the above.
 
 INDEX. 
 
 [The references are to pages.] 
 I. GREEK. 
 
 "impossible," 53 
 
 01X0705 ( = " undescribed " apparently), 
 Egyptian name for certain powers, 41 
 
 dipto-ros, indeterminate: xX^tfos /j.ovdSuv 
 dopiffrov, an undetermined number of 
 units = the unknown, dpi0/i6j, i.e. x, 32, 
 115, 130 ; iv T(J> dopiffTif, indeterminately, 
 or in terms of an unknown, 177 
 
 apifffjujriK^ distinguished from XoytorwnJ, 4 
 
 dptOfibs, number, used by Dioph. as techni- 
 cal term for unknown quantity (=x), 
 32, 115, 130; symbol for, 32-37, 130 
 
 dptOfMffTbr (=!/.*) and sign for, 47, 130 
 
 S.TOTTOS, " absurd," 53 
 
 StirXij Iff6n)t or SirXoib-inj j, double-equation, 
 
 square," used for square of un- 
 known ( = x i ): distinguished from rrrpd- 
 yuros, 37-38 ; sign for, 38, 129 ; Terpav\ij 
 Svva/jus, "quadruple-square," Egyptian 
 name for eighth power, 41 
 5vva./ju>8vvanis, fourth power of unknown 
 (=jr t ), sign for, 38, 129 
 
 f, submultiple of 8\n>a/4odv- 
 (= i/* 4 ) and sign, 47, 130 
 i>/3os, "square-cube" (=x 5 ), sign 
 for, 38, 129 
 SwapoKvpoffTor, submultiple of 5wa/t6/ci>/3oj 
 
 (=ilx 5 ) and sign, 47, 130 
 5wanoffT&v, submultiple of SuVa^us ( = i/jr 2 ) 
 and sign, 47, 130 
 
 elSos, "species," used for the different terms 
 in an algebraic equation, 7, 130, 131 
 
 AXei^-tJ, "deficiency": *r tXXei'feaf riva, 
 etSii, "any terms in deficiency," i.e. 
 "any negative terms," 7, 131 
 
 , "existent," used for positive 
 terms, 7, 130 
 
 ttrdvffijfJM ("flower" or "bloom") of 
 Thymaridas, 114-116 
 
 &roy, equal, abbreviation for, 47-48 
 
 Kv/36jrv/3o?, "cube-cube," or sixth power 
 of unknown (=x 9 ) t and sign for, 38, 
 129 
 
 KvfioKvf)o<rr6v (= i fx*) and sign, 47, 130 
 iru/Joj, cube, and symbol for cube of un- 
 known, 38, 1 29 ; <n'</3oi e|Xi/tT6f , Egyp- 
 tian term for ninth power (x 9 ), 41 
 ( = !/-**) and sign, 47, 130 
 
 Xefare ir, to be wanting: parts of ve 
 
 to express subtraction, 44 ; Xehrorra eldy, 
 negative terms, 130 
 
 X^, "wanting," term for subtraction 
 or negation, 130; Xety (dat.) = minus, 
 sign common to this and parts of verb 
 \fi-reif, 41-44 
 
 X0yrTuti), the science of calculation, in; 
 distinguished from dfH0/MjruoJ, 4 
 
 *- I 44- 
 
 /i/x)5, "part," =an aliquot part or sub- 
 multiple ; fi^prj, " parts," used to describe 
 any other proper fraction, 191 
 
 /ttijXfTijj dpidfjidf (from /x^Xor, an apple), 4, 
 
 "3 
 
 pord*, "unit," abbreviation for, 39, 130 
 Mopto<mrd, supposed work by Diophantus, 
 
 3-4 
 fwpiov, or iv i^pitf, expressing division or 
 
 a fraction, 46, 47 
 /xt/ptAi TPUTTJ, dtvrtpa, 47-48
 
 382 
 
 INDEX 
 
 6/ju>w\i)9r} (etSi}), (powers of unknown) 
 "with the same coefficient," 7 
 
 *, value of, calculated by Archimedes 
 and Apollonius, 122 
 
 trapiff(>Tris, TrapiffbnjTos dyuy/i, approxima- 
 tion to limits, 95 sq., 207, 208, 209 ., 
 
 211 
 
 ir\a<Tna.TiKt>v, "formativum" (Tannery), 
 
 meaning of, 140 ft. 
 irXfvpd, side, = square root, 65 . 
 w\rj0os, "number" or "multitude," used 
 
 for "coefficient," 64 . 
 wp6raffis, proposition, 9 
 
 ffTfpe6s, solid, used of a number with 
 three factors, 183 . 
 
 ftovASuv, "heap" or collection of 
 units, 37 
 aapds, "heap," 37 
 
 rpaTrX^ Svva/jus, "quadruple-square," 
 Egyptian term for eighth power, 41 
 
 , "existence," denoting a positive 
 term (contrasts with \^ts), 41 
 inrdpxovra (et8ri), "existent" or positive 
 (terms), i^on. 
 
 <f>ta\lTrfi apiff(i,6s (from 0iX??, a bowl), 4, 
 "3 
 
 of Apollonius, 122 
 , determinate, 115 
 
 II. ENGLISH. 
 
 Abu'l-Faraj, i 
 
 Abu'1-Wafa al-Buzjanl, 6, 19 
 
 Achmim Papyrus, 45 
 
 Addition, expressed in Dioph. by juxta- 
 position, 42; Bombelli's sign for, 22; 
 first appearance of + , 49 . 
 
 Ahmes, 112 
 
 Alfraganus, 20 
 
 Algebra: three stages of development, 
 49-51 
 
 Algebraical notation : Diophantus, 32-39, 
 41-44; Bombelli, 22, 38; earlier Italian 
 algebraists, 38 ; Xylander, 38, 48 ; Bachet 
 and Fermat, 38 ; Vieta, 38, 39, 50 n. ; 
 beginnings of modern signs, 49-50 n. 
 
 Aljabr, 64 
 
 al-Karkhi, 5, 41 . 
 
 al-Khuwarazmi, Muhammad b. Musa, 
 34. 5 
 
 Almukabala, 64 
 
 Amthor, 122 
 
 Anatolius, 2, 18 
 
 Andreas Dudicius Sbardellatus, 17, 25 
 
 Angelus Vergetius, 16 
 
 Anthology, arithmetical epigrams in, 113- 
 114; on Diophantus, 3; indeterminate 
 equations in, 114 
 
 Apollonius of Perga, 5, 6, 12, 18, 122 
 
 Approximations: Diophantus, 95-98; Py- 
 thagoreans, 117-118, 278; Archimedes, 
 278-279 
 
 Arabian scale of powers of unknown 
 compared with that of Diophantus, 
 40, 41 
 
 Arabic versions and commentaries, 19 
 
 Archimedes, n, 12, 35, 278, 279, 290; 
 Codex Paris, of, 48 ; Cattle-Problem, 
 121-124, 2 795 Arenarius, 35, 122 
 
 Arenarius of Archimedes, 35, 122 
 
 Arithmetica of Diophantus : different titles 
 by which known, 4-5 ; lost Books, 5-12; 
 division into Books, 5, 17-18; notation 
 in, 32-53 ; conspectus of problems in, 
 260-266 
 
 Arithmetical progression, summation of, 
 248-249 
 
 Ars rei et census, 20 
 
 Aryabhata, 281 
 
 Auria, Joseph, 15, 18 
 
 Bachet, 12, 16, 17, 21, 22., 25, 26-29, 35, 
 38, 45, 48, 80-82, 87 n., 101, 107 ., 
 109, no, 140 ., 173 ., 196-197 ., 
 
 2I3., 22O ., 23O.,232.. 234~235., 
 
 246, 271, 273, 287, 293 
 "Back-reckoning," 56, 89, 93 
 Baillet, 45 . 
 
 Bessarion, Cardinal, 17, 20 
 Bhaskara, 281 
 Bianchini, 10 
 Billy, Jacques de, 28, 165 ., 166 ., 184 ., 
 
 221 ., 267, 304, 308, 320, 321, 326
 
 INDEX 
 
 383 
 
 Bodleian MSS. of Dioph., 15, 34, 35 ; 
 
 MS. of Euclid, 35 
 Bom belli, Rafael, n, 27; Algebra of, 
 
 21, 22; symbols used by, 22, 38 
 Brahmagupta, 281 
 Brancker, Thomas, 286 n. 
 Brouncker, William, Viscount, 286, 288 
 
 Camerarius, Joachim, 21 
 
 Cantor, Moritz, 3 ., 6, 6$n., 112, n8., 
 
 120 ., 125 ., 281 
 Cardano, 21, 23, 40 
 Cattle- Problem of Archimedes, n, 12, 
 
 121-124, 279 
 Cauchy, i88., 274 
 Censo, or Zensus, = square, 40, 41 
 Charmides, scholiast to, in, 113, 121- 
 
 122 
 
 Chasles, n 
 
 Cleonides, i6. 
 
 Coefficient, expressed by w\TJ0ot, multitude, 
 39, 64 n. 
 
 Colebrooke, 6, 281 . 
 
 Com, =the unknown, 22, 40 
 
 "Coss," 23 
 
 Cossali, i, 21 ., 40, 41, 140 ., 
 220 n. 
 
 Cracow MS. of Dioph., 5 ., 14, 18 
 
 Cube : Vieta's formulae for transforming 
 the sum of two cubes into a difference 
 of two cubes and vice versd, 101-103; 
 Fermat's extensions, ibid. ; a cube cannot 
 be the sum of two cubes, i44. ; Euler's 
 solution of problem of finding all sets of 
 three cubes having a cube for their sum, 
 329-334; sign for cube of unknown or 
 x 3 , 38, 129 
 
 " Cube-cube " ( = sixth power of unknown, 
 or .z 6 ), sign for, 38, 129 
 
 Cubic equation, simple case of, 66-67, 
 242 
 
 Cuttaca ("pulveriser"), Indian method of, 
 283 
 
 Definitions of Diophantus, 32, 38, 39, 
 129-131 
 
 "Denominator," 137 
 
 Descartes, 271, 273; notation, sow. 
 
 Determinate equations : of first and second 
 degrees, 58 ; pure, 58-59 ; mixed quad- 
 ratics, 59-65; simultaneous equations 
 leading to quadratics, 66 ; a particular 
 cubic, 66-67 
 
 "Diagonal-" numbers, 117, 118, 310 
 
 Dionysius, 2 ., 9, 119 
 
 Diophantus : spelling of name, i ; date, 
 1-2; epigram on, 3; works, 3-13; in 
 Arabia, 5-6, 19; "Pseudepigraphus," 
 12, 31 ; MSS. of, 14-18; commentators 
 and editors, 18-31 ; notation of, 3253 ; 
 methods of solution, 54-98 ; porisms 
 f| 3> 8-10, 99-101 ; other assumptions, 
 103 sqq.; theorems in theory of numbers, 
 105-110; on numbers which are the 
 sum of two squares, 105-106; numbers 
 which are not the sum of two squares, 
 107-108; numbers not sum of three 
 squares, 108-109; numbers as sums of 
 four squares, no; Dioph. not inventor 
 of algebra, in-n6; nor of indeter- 
 minate analysis, 115-124; his work 
 a collection in best sense, 124; his ex- 
 tensions of theory of polygonal numbers, 
 127 
 
 Division, how represented by Dioph., 
 
 44-47 
 
 Doppelmayer, ion. 
 
 Double-equations (for making two ex- 
 pressions in x simultaneously squares), 
 ii, 73-87, 91-92; two expressions of 
 first degree, 73-80, 80-82 n. ; two ex- 
 pressions of second degree or one of first 
 and one of second, 81-87 ; general rule 
 for solving, 73, 146 ; double equations 
 for making one expression a square and 
 another a cube, 91-92 
 
 Dudicius Sbardellatus, Andreas, 17, 45 
 
 Egyptians: hau, sign for, 37; names for 
 successive powers, 41 ; beginnings of alge- 
 bra, ^^-calculations, 111-112; method 
 of writing fractions, 1 1 2 
 
 Eisenlohr, 1 1 2 n. 
 
 Enestrom, 63 ., 286 n. 
 
 Epanthema of Thymaridas, 114-116 
 
 Epigrams, arithmetical, in Anthology, 113- 
 114; on Diophantus, 3; one in Dio- 
 phantus (v. 30), 124 
 
 Equality: abbreviation for, 47-48; sign in 
 Xylander, 48 ; the sign = due to Recorde, 
 50 . 
 
 Equations, see Determinate, Indeterminate, 
 Double, Triple, etc. 
 
 Eratosthenes, 121 
 
 Euclid, 8, n, 12, 19, 63, 117, 124, 132 ., 
 144 ., 191 
 
 Eudoxus, 114 
 
 Euler, 56, 71-72 ., 83-85 ., 86 ., 9O.,
 
 384 
 
 INDEX 
 
 ioo., 102 ., 107, no, 145 w., 151 ., 
 160 ., 162 w., 178 ., 181-182 ., 
 188 ., 224 ., -236 ., 241 ., 242 ., 
 268, 272, 274, 275, 286, 288-292, 
 294, 297, 299 ., Supplement, 329-379 
 passim 
 
 Euler, J. A., 360 
 
 Eutocius, 5, 6 
 
 Exponents, modern way of writing due 
 to Descartes, 50 . 
 
 Fakh-ri, 5, 41 n. 
 
 "False supposition," use of, in Egypt, 
 112-113 
 
 Fermat, 28, 29, 30, 38, 78, 90, 101, 
 102, 103, 106, 107, 108, 109, no, 
 144-145 ., 163 ., 173 ., 179-180 w., 
 182 ., 183 ., 184 ., 188 ., 190- 
 191 ., 197 ., 202 ., 204 ., 205 ., 
 213-214 ., 218 ., 220 ft., 223 ., 
 ^29 n., 230 ., 231 ., 232 ., 233 ., 
 235 , 236 w., 239 ., 240 ., 241 ., 
 742 ., 246, 254 n., Supplement, 267- 
 328 passim, 364 ; " great theorem of 
 Fermat," 144-145 .; Fermat on num- 
 bers which are, or are not, the sums of 
 two, three, or four squares respectively, 
 106110, 267-275 ; on numbers of form 
 
 x* iyi or ix 2 y 2 , 276-277, of form 
 j^-f 3j/ 2 , 275, and of form ^ 2 4-&y 2 , 276, 
 277; on equation x*-Ay z i, 285-287 ; 
 x*-y* = z 2 cannot be solved in integers, 
 224, 293-297; problems on right-angled 
 triangles, 204-205 ., 2i8-2i9., 220 w., 
 229 n., 230 ., 23/-233M., 235 w., 236 ., 
 239-240 ., 297-318; Fermat's "triple- 
 equations," 321-328 
 
 Fractions : representation of, in Diophantus, 
 44-47; sign for i, 45; for |, 45; 
 sign for submultiple, 45-47 
 
 Frenicle, 102 ., 276, 277, 285, 287, 
 295-297, 39, 3io, 313, 314 
 
 Gardthausen, 35, 36 
 
 Geminus, 4 
 
 Georg v. Peurbach, 20 
 
 Georgius Pachymeres, 18, 19, 31, 37 
 
 Girard, Albert, 30, 106 . 
 
 Gnomons, 125 
 
 Gollob, 14, 1 8 
 
 Grammateus (Schreiber), Henricus, 49 n. , 
 
 50 . 
 
 Greater and less, signs for, 50 n. 
 Gunther, 6, 278., 279^. 
 
 Hankel, 6, 54-55, 108 ., 281, 283, 284, 
 
 286 w. 
 
 Harriot, 50 n. 
 Hau ( = "heap"), the Egyptian unknown 
 
 quantity, 37, 112 
 Heiberg, 35, 48 ., 118, 205 n, 
 Henry, C., 13* 28 . 
 Herigone, 50 n. 
 Heron, 12, 13, 35, 36, 43, 44, 45, 63, 
 
 129 n. 
 
 Hippocrates of Chios, 63, 124 
 Holzmann, Wilhelm, see Xylander 
 Hultsch, 2., 3, 4, 9, 10, ir, 12, ign, t 
 
 35,36, 37,47M 63., ii8., 122, 253 . 
 Hydruntinus, loannes, 16 
 Hypatia, 5, 6, 14, 18 
 Hypsicles, 2 ; on polygonal numbers, 
 
 125-126, 252, 253 
 
 lamblichus, 2, 3,37 n., 49, 50, 115-116, 126 
 
 Ibn abi Usaibi'a, 19 
 
 Ibn al-Haitham, 19 
 
 Identical formulae in Diophantus, 104, 105 
 
 Indeterminate equations : single, of second 
 degree, 67-73 ; of higher degrees, 87- 
 91 ; how to find fresh solutions when one 
 is known, 68-70 ; double-equations for 
 making two expressions simultaneously 
 squares, 11, (i) two expressions of first 
 degree, 73-80, 80-82 ., (2) two of second 
 degree, or one of second and one of first, 
 81-87 ; double-equations for making one 
 expression a square and another a cube, 
 91-92 ; rule for solving double-equations 
 in which two expressions are to be made 
 squares, 73, 146 ; indeterminate equations 
 in Anthology, 114; other Greek ex- 
 amples, 118121; ix^ _y 2 = A i solved 
 by Pythagoreans, 117-118, 278, 310 
 
 "Indian method," 12-13, 21 . 
 
 Indian solution of x*-Ay z =i, 281-285, 
 290, 292 
 
 Inventum Novum of J. de Billy, 28, 
 165 ., 184 ., 198 n., 204 ., 205 ., 
 221 ., 230 n., 231 ., 239 ., Supple- 
 ment, 267-328 passim 
 
 loannes Hydruntinus, 16 
 
 Ishaq b. Yunis, 19 
 
 Italian scale of powers, 40, 41 
 
 Jacobi, 108 ., 288 
 
 Kab, Arabic term for cube of unknown, 
 41 n.
 
 INDEX 
 
 385 
 
 al-Karkhl, 5, 41 . 
 
 Kausler, 31 
 
 Kaye, G. R., 281 
 
 Kenyon, 45 
 
 Konen, 278 ., 279 n., 281 ., 285 n. t 
 
 286 n., 288, 292*. 
 Kronecker, 288 
 Krumbiegel, 122 
 Kummer, 145 n. 
 
 Lagrange, 72 ., no, 188 ., 272, 273, 
 274, 275, 276, 277, 285, 287, 288, 
 290, 292, 299, 300 
 
 Lato, " side," 40 n. 
 
 Legendre, 107 ., 109 ., i88., 273 
 
 Lehmann, 35 
 
 Lejeune-Dirichlet, 145 ., 288 
 
 Leon, 124 
 
 Leonardo of Pisa, n, 41 ., 120 
 
 Less and greater, signs for, 50 . 
 
 Limits : method of, 57, 94, 95 ; approxi- 
 mation to, 95-98 
 
 Logistica speciosa and Logistica numerosa 
 distinguished by Vieta, 49 
 
 Loria, 62 ., 157 ., 168 ., 175 n., 
 i-j6n., 195 ., 197 ., 207 ., 240 ., 
 241 n. 
 
 Lousada, Abigail, 31 
 
 Luca Paciuolo, 21, 40 
 
 Madrid MS., 14, 15, 16 
 Mdl, Arabic term for square, 41 . 
 Manuscripts of Diophantus, 14-18 
 Maximus Planudes, 13, 14, 19, 21, 31, 
 
 43. 44. 45. 4 6 > 4 8 
 Measurement of a circle (Archimedes), 
 
 122 
 
 Mendoza, 17 
 Melrica of Heron, 43, 44, 45, 63, 129 .; 
 
 MS. of, 118 
 Metrodorus, 5, 113 
 Minus, Diophantus' sign for, 41-44, 
 
 130 ; same sign in Heron's Metrica, 
 
 43, 44 ; Bombelli's abbreviation, 22 ; 
 
 modern sign for, 49 .; Vieta's 
 
 sign for difference between ( = for ~), 
 
 50*. 
 
 Montchall, Carl v., 18 
 Montucla, 28 
 
 Moriastica of Diophantus, 3-4 
 Muhammad b. Musa al-Khuwarazmi, 34, 
 
 5 
 
 Multiplication, signs for, 50 . 
 Murr, Ch. Th, v., ion. 
 
 Negative quantities not recognised by 
 Diophantus as real, 52-53 
 
 Nesselmann, 6-10, i ., 25, 26 ., 29, 
 33. 34. 49-5L 55-58. 67, 87, 89, 93, 
 108 ., 140 ., 173 ., 204 ., 207 . , 
 252 n,, 329 n. 
 
 Nicomachus, 2, 126, 127 
 
 Notation, algebraic: three stages, 49-51; 
 Diophantus' notation, 32-49, 51-52 
 
 Numbers which are the sum of two squares, 
 105-107, 268-271 ; numbers which are 
 not, 107-108, 271-272; numbers which 
 are the sum of three squares, 272-273 ; 
 numbers which are not, 108-109, 273 ; 
 numbers not square are the sum of two, 
 three or four squares, no, 273, 274; 
 corresponding theorem for triangles, 
 pentagons, etc., 188, 273 
 
 Numtrus, numero, term for unknown quan- 
 tity, 38, 40 
 
 Nunez, 23 
 
 Oughtred, 50 n. 
 Ozanam, 288 
 
 Pachymeres, Georgius, 18, 19, 31, 37 
 
 Paciuolo, Luca, 21, 40 
 
 Pappus, 11, 13 
 
 Papyrus Rhind, 112; Berlin papyrus 
 6619, 112 
 
 Paris MSS. of Diophantus, 15, 16, 18 
 
 Pazzi, A. M., 21 
 
 Pell, John, 31, 286 n., 288 
 
 "Pellian" equation, origin of this er- 
 roneous term, 286 
 
 Peurbach, G. von, 20 
 
 Philippus of Opus on polygonal numbers, 
 125 
 
 Planudes, Maximus, 13, 14, 19, 21, 31, 
 43. 44. 45. 46, 48 
 
 Plato, 4, 38*., in, 113, 116, 125 
 
 Plus, signs for, 22, 49*.; expressed in 
 Diophantus by juxtaposition, 39 
 
 Plutarch, 127 
 
 Polygonal Numbers, treatise on, 3, n-n, 
 247-259; sketch of history of subject, 
 124-127 ; began with Pythagoreans, 124- 
 125; figured by arrangement of dots, 
 125; Hypsicles on, 125-126, 252, 253; 
 Diophantus' extensions, 127 
 
 Porisms of Diophantus, 3, 8-10, 99-101, 
 201, 202, 214 
 
 Poselger, 30, 98 
 
 Powers of unknown quantity and signs
 
 386 
 
 INDEX 
 
 for,, 37-39* 139; Italian and Arabian 
 scale (multiplicative) contrasted with Dio- 
 phantine (additive), 40-41 ; Egyptian 
 scale, 41 
 
 Proclus, 4., 113* 116, ii7., 118, 242 w. 
 
 Psellus, 2, 14, 18, 41, m 
 
 Ptolemy, 18, 44 
 
 Pythagoreans : 3 ; on rational right-angled 
 triangles, 116, 242 .; on polygonal 
 numbers, 124-125; on indeterminate 
 equation ix 2 -y 2 =i, 117-118, 278, 
 310 
 
 Quadratic equations in Diophantus, 7-8, 
 59-66 ; in Hippocrates of Chios, 63 ; 
 in Euclid, 63; in Heron, 63, 64 
 
 Quadratic inequalities in Diophantus, 
 60-63 ! limits to roots, 60-63, 65, 95 
 
 Qusta b. Luqa, 19 
 
 Radice ( = x), 40 
 
 Radix (=x), 38 
 
 Rahn, 50 ., 286 n. / , 
 
 Ramus, i6n. 
 
 Rationality, Diophantus' view of, 52-53 
 
 Recorde, Robert, 50 . 
 
 Regiomontanus, 5, 17, 20, 23, 49 
 
 " Regula falsi" in Egypt, 112-113 
 
 Relato, Italian term for certain powers 
 of unknown, 41 
 
 Res, alternative for radix, in sense of 
 unknown quantity, 38 
 
 Rhind Papyrus, 112 
 
 Right-angled triangles in rational numbers 
 in Diophantus, 93-94, 105-106 ; method 
 of "forming," 93-94; other methods of 
 forming attributed to Pythagoras, 116- 
 117, and to Plato, i 16-1 1 7 ; Euclid's for- 
 mula for, 117, 120; Pythagorean formula 
 once used by Diophantus, 242 ; Greek 
 indeterminate problems on, other than 
 those of Dioph., 119-121 ; Fermat's 
 theorems and problems on, 204-205 ., 
 218-219 ., 220 ., 229 ., 230 ., 231- 
 232 ., 235 ., 236 ., 239-240 ., 293- 
 318, 364-371 
 Rodet, 34, 35 
 Rosen, 50 
 
 Jtudioj 63 n. 
 
 Rudolff, Christoff, 23, 50 . 
 
 Salmasius, Claudius, 17 
 Sand-reckoner of Archimedes, ia 
 Saunderson, N., 27 . 
 
 Schaewen, P. v., 327, 328 
 
 Schmeisser, 31 
 
 Schone, 43, 45, 118 
 
 Schreiber, H., see Grammateus 
 
 Schuler, Wolfgang, 24 
 
 Schulz, 9, n, 18, 30, 31, 108 ., 140 ., 
 219 n. 
 
 Sebastian Theodoric, 24 
 
 Serenus, 12 
 
 " Side " = square root, 65 . 
 
 " Side- " and " diagonal-" numbers, Py- 
 thagorean solution of ix 2 -y i =i by 
 means of, 117-118, 278, 310 
 
 Simon Simonius Lucensis, 25 
 
 Simplicius, 63 n. 
 
 Sirmondus, J., 27 
 
 Smith, H. J. S., 292 
 
 " Species" (ef5r?) of algebraical quantities, 
 
 7 '3, 131 
 
 Speusippus on polygonal numbers, 125 
 "Square-cube" ( = x 6 ), sign for, 38, 
 
 129 
 Square root, sign V f r 5<>.; =v\evpd 
 
 (side), 65 . 
 "Square-square" (=x*), sign for, 38, 
 
 129 
 Squares : numbers as sum of two, three, 
 
 or four, no, 273, 274; of two, 105-107, 
 
 268-271; not of two, 107-108, 27F-272; 
 
 of three, 272-273 ; not of three, 108, 
 
 109, 273 
 
 Stevin, Simon, 29, 30 n. 
 Stifel, M.., 23, 49 M., 50 . 
 Submultiples, sign for, 454? ; decom- 
 position of fractions into, 46, 112; 
 
 submultiples of unknown and powers, 
 
 47 
 
 Subtraction, symbol for, 41-44 
 Suidas, i, 1 8, 22 
 
 Surdesolides, sursolida or super solida, 41 
 Surds, 23-24 
 Suter, H., 19 . 
 
 Tannery, P., i ., 3, 5, 6, 8, 10-12, 14-19, 
 25, 28 n., 31, 32-37, 43-44, 45, io8w., 
 in, 118 ., 125 n., 135 ., 138 ., 
 144 ., 148 ., 150 ., 156 ., 160 ., 
 198 ., 219 ., 234 ., 256, 278, 279, 
 280, 281, 290, 308 
 
 Tanto, unknown quantity, in Bombelli, 22 
 
 Tartaglia, 21, 40 
 
 Theaetetus, 124 
 
 Theon of Alexandria, 2, 18 
 
 Theon of Smyrna, 2, 36, 117, 126, 310
 
 INDEX 
 
 387 
 
 Theudius, 124 
 
 Thompson, D'Arcy W., 37 
 
 Thymaridas, Epanthema of, 114-116 
 
 " Triple-equations " of Fermat, 163 ., 
 
 179 ., 182/1., 102 n., 223 ., 224 ., 
 
 246, 321-328 
 
 " Units " (/xo'd5es) = absolute term, 39-40; 
 abbreviation for, 39, 130 
 
 Unknown quantity ( = x), called in Dio- 
 phantus dpi.9ij.6s, "number": definition 
 of, 32, 115, 130; symbol for, 32-37, 
 130; signs for powers of, 38, 129; 
 signs for submultiples of unknown and 
 powers, 47, 130 ; Italian- Arabian and 
 Diophantine scales of powers, 40, 41 ; 
 Egyptian scale, 41; x .first used by 
 Descartes, 50 . ; other signs for, i, used 
 by Bombelli, 22, 38, N (for Numerus) 
 by Xylander, Bachet, Fermat and others, 
 38, R (Radix or Res), 38, Radice, Lato, 
 Cosa, 40 . 
 
 Vacca, G., io6. 
 
 Valla, Georgius, 48 
 
 Vatican MSS. of Diophantus, 5 ., 15, 16, 
 
 17 
 Vergetius, Angelus, 16 
 
 Vieta, 27, 38-39, 49, sow., 101, 102, 
 
 214 ., 285, 329, 331 
 Vossius, 31 
 
 Wallis, 40., 286, 287, 288, 289 
 
 Weber, Heinrich, 3 n. 
 
 Weber and Wellstein, 107 n., 145 . 
 
 Wertheim, 30, no ., 137 ., 138 ., 
 145 ., 151 ., 161 n., 209 ., 2ii ., 
 2I2W., 216 n., 217 ., 254 ., 256, 257, 
 286 ., 294, 295 
 
 Westermann, 125 . 
 
 Widman, 49 w. 
 
 Wieferich, 145 . 
 
 Woepcke, 5 . 
 
 "Wurm's problem," 123 
 
 JT for unknown quantity, originated with 
 
 Descartes, 50 . 
 Xylander, 17, 22-26, 27, 28, 29, 35, 38, 
 
 107-108 ., 140 .; Xylander's MS. of 
 
 Diophantus, 17, 25, 36 
 
 Zensus (=Ctnso), term for square of un- 
 known quantity, 38 
 Zetetica of Vieta, 27, 101, 285 
 Zeuthen, 118-121, 205 ., 278, 281 ., 
 290, 294-295 
 
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