CHAP. XXII] ax 4 + by 4 = cz 2 . 633 



where \ = ax' 2 x' 2 by' 2 y f2 , n = xyz'-{-zx'y r . For a+b = c, we may set 



and deduce his 159 and Pepin's 87 formulas. For a = c = l, x' = z f = l, 2/ = 0, 

 we get Lagrange's formula. He announced the empirical result that the 

 complete solution of (1) in integers is given by as many systems (6) as 

 (1) has primitive solutions (x f , y', z'). For 8x* 3?/ 4 = 5z 2 , Pepin's ten 

 systems reduce to the two systems (6) with (x r , y', z'} = (1, 1, 1), (2, 1, 5). 

 For 90 the case c = a+b, set x = y = z = l in (6) and drop the accents; we get 



X = a(a-b)x*-b(3a+b)xy' 2 -2bcyz, 



while Y is derived from X by interchanging a, b and x, y. He gave another 

 set of formulas of like degree. By finding a relation 



EX 2 x*+GY 2 y 2 -2LXYxy-H(X 2 y 2 +YW) = 



such that Y/X is a function of y, x, involving only the irrationality 

 (ax 4 -\-by*y 12 , he obtained the quadratic formulas 



X= -(a-6) 2 z 2 +46q/ 2 , F=[2c 2 -(a-6) 2 >?/+2c(a-fc)z, 



and stated that a like discussion may be made for 



ax*+by 4 +dx 2 y* = cz 2 , c = a-\-b+d. 



Desboves 91 noted that, if (1) is solved completely by (6) when (x f , y', z'} 

 is replaced by (#, y' if z\) for i = l, 2, 3, in succession, then anyone of these 

 solutions is called primitive if one does not get it when one determines all 

 solutions given by the other two and continues the calculations with them. 



Desboves 92 stated that we can find, by a single system of formulas (not 

 given), the complete solution of arc 4 by 4 = 2z 2 when a and b are consecutive 

 primes 8n+7 and 8n+5 or 8n+5 and 8n+3. 



T. Pepin 93 treated x*+2 k >7y* = z* for k = 2a and 4a+3. He 94 gave a 

 detailed discussion of 



5z 4 - 3?/ 4 = 2z 2 , 5x* - 2y* = 3z 2 , 3;r 4 + 5 y* = 8z 2 , 8z 4 - 5y 4 = 3z 2 . 



E. B. Escott 95 noted that if in x 4 +y i = az 2 we set x = zk/l we obtain a 

 quadratic for z~ which will be rational if (alY (2a) 2 (ky} 4 = (aml 2 } z , so that 

 the problem reduces to the pair of equations p 2 db2a<? 2 = D (Ch. XVI). 



Axel Thue 96 proved that x 4 2 m y 4 = l has no integral solutions. 



Escott 97 solved 4A 4 +1 = 5 2 C by noting that the left member has the 

 factors 2A 2 2A + 1, whence (2A1) 2 +1 = (mod B 2 ). 



A. Gerardin 98 noted that if (a, |8, 7) and (A, B, C) are two solutions of (1), 

 x = a+ Au, y = (3-}-Bu, z = y -\-Su-\-Cu? give a new solution provided a certain 



90 Comptes Rendus Paris, 104, 1887, p. 1832. 



91 Ibid., 1602-3. 



92 Assoc. frans. av. sc. f 16, 1887, I, 175 (in full). 



93 Mem. Ace. Pont. Nuovi Lincei, 4, 1888, 227. 



94 Ibid., 9, I, 1893, 247-284. 



95 L'interm6diaire des math., 7, 1900, 199 (reply to 3, 1896, 130). 

 98 Archiv for Math, og Naturvidenskab, 25, 1903, No. 3. 



97 L'interm6diaire des math., 12, 1905, 155-6. 



"Bull. Soc. Philomathique, (10), 3, 1911, 234-6; Sphinx-Oedipe, 6, 1911, 101-2. 



