Fermat appears to have been satisfied with the actual solution', but later he points out that, although Frénicle and Wallis had given many particular solutions, they had not supplied a general proof" (ie. 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+7967 equal to a square. “I 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 2x+5=square "Bachet boasts, in his commentary on Diophantus', of having discovered 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.” 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 x2-Ay=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, 11. p. 402. "Relation des nouvelles découvertes en la science des nombres," Oeuvres, II. p. 433. See on Diophantus IV. 39, and above, pp. 80-81. 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 2x+5=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 18- Au2=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 12-3u2= -12, and the solution =6, x=4 is easily obtained. credited with the very same solution of the equation x3- Ay' = 1 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 règle générale pour résoudre 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 "règle" 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'. Euler, Lagrange, Gauss, Jacobi, Dirichlet, Kronecker are the great names associated with it. I will only add a few particulars with regard to Euler' as coming nearest to Fermat. 133 In a letter to Goldbach' of 10th August, 1730, Euler mentions that he requires the solution of the equation x' - Ay in order to make ax2 + bx + c a complete square. He goes on to observe that the problem of solving - Ay 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 1092 + 1 = x2 as being the most difficult case solved by them, whereas the most difficult examples actually solved were 433μ3 + 1 a and 313y2+ 1 = x2. = A paper of a year or two later contained the proof that the evolution of successive solutions of ax2 + bx + c=y2 when one is known requires that one solution of a§a + 1 = ŋa 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 = y3, the importance of which remark is emphasised by Lagrange". In the paper quoted in the last paragraph Euler finds any number of successive solutions of ax2 + bx+c=y3, and the law for forming them, when we are given one value n of x which will make ax + bx + c_a complete square and one value p of έ which will make a + 1 a complete square, or, in other words, when an3 + bn + c = m2 and ap2 + 1 = q2. He then takes the particular case ax2 + bx + d2 = y2 where (since x = 0, y = d satisfies the equation) we can substitute o for n and d for m in the expressions representing the successive solutions of ux2 + bx+c=y2. Then again, putting b=0 and d= 1, 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. • Cf. Konen, op. cit. pp. 47–58. › Correspondance mathématique et physique de quelques célèbres géomètres du XVIIIième siècle, publiée par P. H. Fuss, Pétersbourg, 1843, 1. p. 37. "De solutione problematum Diophanteorum per numeros integros" in Commentarii Acad. Petropol. 1732–3, VI. (1738), pp. 175 sqq.= Commentationes arithm. I. pp. 4–10. Algebra, Part 11. ch. VI. Additions to Euler's Algebra, ch. VIII. number of successive solutions of aga + 1 = n when one solution §=¿, 7=9 is known. The successive values of έ are the law of formation being in each case that, if A, B be consecutive values in either series, the next following is 29B-A. The question then arises how to find the first values p, 4 which will satisfy the equation. Euler first points out that, when a has one of many particular forms, values of p, q can at once be written down which satisfy the equation. The following are such cases with the obvious values of and y. p = 1, 9=6, - 1 ; a = a1eb + 2acb-1; p=e, q = ac1+1 ± 1 (where a may even be fractional provided ab-1 is an integer), But, if a cannot be put into such forms as the above, then the method explained by Wallis must be used. Euler illustrates by finding the least values p, which will satisfy the equation 31§o÷ 1 = y2, and then adds a table of the least solutions of the equation ag+ 1 = n2 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 Ja. For, since ap2 + 1 = q, we have Ja = √(ga − 1)/A and, if 9 (and therefore p also) is large, q/p is a close approximation to √a; the error is not greater than 1/(2a). Euler illustrates by taking 6. The first solution of 6a + 1 = n2 (after έ= o, n = 1) is p = 2, q = 5. Taking then the series of values above given for aέ2 + 1 = n2, namely $ = 0, p, 2pq, 4pq – p, ... A, B, 29B – A, and substituting = 2, 95, the successive corresponding values P, Q of, respectively become P=0, 2, 20, 198, 1960, 19402, 192060, 1901198, ... Q= 1, 5, 49, 485, 4801, 47525, 470449, 4656965, ... and the successive values Q/Pare closer and closer approximations to 6. It will be observed that the method of obtaining successive approximations H. D. · 19 by to a from successive solutions of ag2 + 1 = ŋ2 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 ag2 + 1 = n2 by developing a 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' returns to the problem of finding all the solutions of {{ax3 + bx + c = y2 or ax2+b=y2 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. and then "If x = x = a, y = b satisfies ax2+p=y3, x === c, y = d satisfies ax2 + q = y2, : bc ± ad, y = bd ± aac satisfies ax2 + pq = ya.” 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' in which Euler: obtained the solution of the equation x2 - Ay2 = 1 by the process of converting A 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 = p2 + 1, then q/p is an approximation to, and q/p is "such a fraction as expresses the value of √ 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 states the process generally thus. (13), √(61) and (67), after which he If z be the given surd and 7 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., 629 sq. 3 "De resolutione formularum quadraticarum indeterminatarum per numeros integros" in Novi Commentarii Acad. Petropol. 1762–3, IX. (1764), pp. 3 sqq. – Commentat. arithm. "I. pp. 297-315. * "De usu novi algorithmi in problemate Pelliano solvendo" in Novi Commentarii Acad. Petropol. 1765, x1. (1767), pp. 28-66 Commentat, arithm. 1. pp. 316-336. The paper seems to have been read as early as 15 Oct. 1759. (This is of course exactly the process given in text-books of Algebra, eg. Todhunter's.) Euler now remarks as follows. I. The numbers A, B, C, D... cannot exceed v; the first, A, is `equal to v; since a (v + A)/a, aa – A = B≤v, and so on. 2. Unless where one of the numbers a, ß, 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 27, the next quotients will be a, b, c, d... in the same order. 4. Similar periods occur with the letters a, ß, y, d... and the term of this series corresponding to a quotient av is always 1. The successive convergents to the continued fraction are then investigated and it is shown that, for successive convergents / beginning with v/1, 7a − zp2 = − a, +ß, −y, +8, - etc. in order. It follows that the problem is solved whenever one of the terms with a positive sign, ẞ, 8, etc., becomes 1. - Since unity for one of the terms a, B, y, 8 corresponds to the quotient 27, and each fresh period begins with 2v, the first period will produce a convergent /p such that q3- zp2+1; 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, qa – zp2 = − 1, . 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 q2- sp2=+1. Or, says Euler, instead of going on and completing the second period, we can satisfy the latter equation more easily thus. |