632 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxn 



type (1) in which a, b, c contain only the prime factors 2 and 3 [^erroneous 

 for 4x 4 -3y 4 = z 2 , Desboves 91 ]. 



Desboves 82 again gave (2) and, by replacing y by v x and then x 2 by u, 

 deduced (5). He noted that (3) is solvable in the further cases a = x(y 2 x), 

 xy 2 (x+y), x(x+y 2 ), X 2 y 2 (x 2 y 2 } 2 . He again (ibid., p. 440) gave (4). 

 He noted (ibid., p. 436-7) that (1) has the solutions 



X = Sax 4 - by 4 , Y = 4ax 3 y, Z = ax 4 +by 4 



if c = 81a 3 6 I4a?bx 4 y 4 -\-atfy 8 , and gave a simpler derivation of Lagrange's 

 solution of (3). For ax 4 +by 4 = cz 4 , see Desboves. 262 



Solutions of x 4 +y 4 = 17z 2 are 1, 2, 1 and 13, 2, 41, neither of which can 

 be obtained (ibid., p. 495) from a solution x, y, z by the formulas (4). 



T. Pepin 83 gave the complete solution of 7x 4 5y 4 = 2z 2 in integers. 

 Then 84 X = z, Y=xy,Z = (7x 4 +5y 4 )/2 give all the solutions of Z 4 +35F 4 = Z 2 

 in which Y is odd; while those with Y even are all obtained by the method 

 of descent. 



S. Re"alis 85 noted that x 4 3?/ 4 = 13z 2 has the solution 



2 /3- 96a/3 2 - 



if a 4 3/3 4 = 137 2 , and asked for the value of z. 



Pepin 86 noted that in Euler's 144 method of making a quartic V = P 2 +QR 

 a square, not only a rational root of 72 = or Q = or = or T = Q leads 

 to an infinity of solutions of V= D, but this may be true of further roots. 

 The latter happens for Ux 4 -7y 4 = z 2 , whence V = ll-7? = P 2 -i-QR, P = 2, 

 Q = ll+7 2 , R = l 2 . The complete solution is obtained by descent to 

 two irreducible solutions 1, 1, 2 and 2, 1, 13 by four sets of formulas, among 

 them being an infinity of solutions which escape the methods of Fermat 

 and Euler. To obtain (pp. 42-48) the complete solution of x 4 -\-20y 4 = z 2 , 

 that of 5n 4 ra 4 = 4Z 2 is found by descent. From one set of solutions x, y, z 

 of (1) for c = a+6 is derived, 87 by special assumptions, the new solutions 



X = \ 2 x 2 -bcvY, Y = \ 2 y 2 -ac f j 2 x 2 , Z= Y 2 -4a\p,xy(\ 2 x 2 +bc[ji 2 y 2 ), 

 where ju : \ = xyz : ax 2 bz 2 . 



Pepin 88 obtained by descent all solutions of 13x 4 Ily 4 = 2z 2 and all of 

 Sx 4 3y 4 = 5z 2 , whereas Euler's 144 method to make 40 4 15 =D does not 

 give all solutions. 



A. Desboves 89 proved that, if (x, y, z} and (x f , y', z'} are solutions of (1), 

 a new solution is given by 



x'X = x 2 \ 2 - 6c?/y , y'Y = 

 * '*' 2 



82 Nouv. Ann. Math., (2), 18, 1879, 434. 



83 Jour, de Math., (3), 5, 1879, 405-24. 



84 Ibid., (5), I,* 1895, 351-8. 



86 Nouv. Correep. Math., 6, 1880, 479. 



88 Atti Accad. Pont. Nuovi Lincei, 36, 1882-3, 49-67. 



87 Ibid., 67-70. Cf. Lucas. 81 

 **Ibid., 38, 1884-5,20-42. 



88 Comptes Rendus Paris, 104, 1887, 846-7. 



