362 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xil 



by a method first applied to equations of any degree n (see Lagrange 1 of 

 Ch. XXIII). His method for t 2 Ait 2 = A, where A is positive and not a 

 square, is as follows. First, consider solutions with u prime to A. Then 

 we can determine integers 6 and y such that t = 6u Ay, 6 < ^A. For 

 this value of t, the initial equation becomes, after division by A, 



E 1 u 2 -28uy+Ay 2 = l, 



where (0 2 - A) /A = E is an integer. Employ in turn each value of 6 for 

 which 6 2 = A (mod A) and solve the new equation by developing into a 

 continued fraction either root of the corresponding quadratic 



Second, for solutions with u = ru', A = r 2 A', whence t = rt f , with u f , A' 

 relatively prime, we have only to treat t' 2 Aw' 2 = A' as before. 



The same method applies to Bt 2 + Ctu + Du 2 = A, C 2 > 457). By 

 the same substitution we now get EiU 2 Quy + A By 2 = 1, where 

 E! = (B6 2 + C9 + D}IA, Q = 2B6 + C. 



He noted that a conjecture made by Euler 71 is false since 101 = x 2 79?/ 2 

 has no integral solutions, although 101 = 4-4-79 + 38 2 79. 



He applied (p. 719-723) the method of his former paper to deduce the 

 solution u = 34, t = 123, of 101 = t 2 13u 2 , chosen probably in view of 

 his correspondence with Euler next mentioned. 



Euler 77 stated he found trouble in applying Lagrange's 75 method of 

 solv'ng (13) to the case 101 = p 2 13g 2 . By that method we seek an 

 integer a < 101/2 such that a 2 13 is divisible by 101. This is true for 

 a = 35. Then (14) becomes A.\ = 12 = p\ \3q\. Since 12 is divisible 

 by the square 4, set the quotient 3 equal to t 2 I3u 2 . Then t = 4, u = 1, 

 whence p { = 8, qi = 2. By Lagrange's method, 



35-8T13-2 



q ~ 



A 1 12 Ai 12 ' 



As these are not integers, one should conclude that the problem is impossible. 

 However, p = 123, q = 34 are solutions, which fact led Euler to believe 

 that Lagrange's method is not sufficient. He noted that this solution 

 123, 34 is given by pi = 47, q l = 13: 



p = 123 = (35-47-13 -13)/12, g = 34 = (35-13-47)/12. 



But what reason leads us to suppose that p : = 47, q = 13? 



To test whether A = p 2 Bq 2 is possible or not, Euler gave for the 

 case A a prime the following rule, of which he had no proof: Subtract 

 from A any multiple of 45; if A 4nB is of the form ab 2 , where a is a 

 prime or unity, and if a = p 2 Bq 2 is solvable, then the proposed equation is 

 solvable. Thus, 101 = p 2 - I3q 2 is solvable since 101 - 4-13 = 7 2 and 

 1 = p 2 13g 2 is solvable. 



77 Letter to Lagrange, Jan., 1770; Euler's Opera postuma, 1, 1862, 571-3; Lagrange's 

 Oeuvres, XIV, 214-8. See Lagrange, 85 end. 



