CHAP. XXII] ax 4 +by 4 = cz 2 , 629 



d, a, given by the last four equations, into 7/ 2 +r 2 a; 2 = a/3/s, we get 



2 



Denote the quantity in brackets by A. Evidently s is not negative. Ac- 

 cording as s is unity or the prime p, we get 2A 2 =f 4 +pg 4 or <7 4 +p/ 4 . Con- 

 versely, any solution of one of the latter equations leads to a solution of the 

 proposed equation with p = w 2 -f-r 4 , since/, g, A determine x, y, z. 



Next, let y, r be both even or both odd. The only modification needed 

 in the above case is to divide y 2 r 2 x 2 , nx 2 z, yrxby 2, and use d/2 = eg. 

 The result is B 2 = sf 4 +4:g 4 pfs, where 



nrB = -( n 2 / 2 4-r 2 # 2 ) (p+r^fg. 

 fi\ s / 



For s = l, we have 2 =/ 4 +4p0 4 , which implies 



.BT/ 2 Bf 2 



Hence the initial equation is reduced to a similar one pb 4 c 4 = d 2 , where 

 c, b are relatively prime. It thus remains to consider c 4 pb 4 = d 2 . First, 

 let one of c, d be even and the other odd. Then c 2 d = pe 4 , c 2z Fd = ft 4 , 

 b = eh, whence h 4 -\-pe 4 = 2c 2 . Next, let c and d be both odd or both even. 

 Then (c 2 d)/2 = 4py 4 or pv 4 , (c 2= Fd)/2 = u 4 or 4^ 4 . Then c 2 = u 4 +4py 4 or 

 c 2 = 4w 4 +pv 4 , which is reduced to the former type by multiplication by 4. 



0. Terquem 70 proved that neither x 4 +2y* nor z 4 4i/ 4 nor x 4 8y* is a 

 square if 2/40, and that zl/z is not a square. 



* J. Bertrand 71 treated ax 4 +by 4 = D. 



C. G. Sucksdorff 72 treated 2 m x*2 n y 4 = 2 p z 2 for x, y, z odd, positive and 

 relatively prime. It suffices to treat eight cases having n = p = 0, m = 4/z-fO, 

 1, 2, 3; four with the minus sign having m = p = 0, n = 4/x+0, 1, 2, 3; four 

 having m = n = 0, p = 2 M +0, 1. First, 2 4 "x 4 +y 4 = z 2 . The factors z 2 2 V 

 must be a 4 , /3 4 , where a/5 = y. By subtraction, 2 2 " +1 z 2 = a 4 /? 4 . Hence 



Eliminating a, , we get w 4 +v 4 = ^ 2 , of the given type. A like method of 

 descent applies to 2 4 " +l x*+y* = z' 2 , whence 



(lower sign excluded since the sum of two odd squares is not divisible by 8) ; 

 thus 8/3 4 = o: 4 -7/ 2 , a 2 y = 2y\ a 2z F?/ = 45 4 , whence cr = 7 4 +25 4 . For 



reference is made to Euler's 67 treatment of a+ex 4 = D, where t/ 4 is taken 

 as <z; various solutions result. The impossibility of 2 4M+2 z 4 +?/ 4 = z 2 follows 



70 Nouv. Ann. Math., 5, 1846, 75-78. 



71 Trait6 61em. d'algebre, Paris, 1850, 244. 



72 Disquisitio au et quatenus aequatio 2 m x 4 2"y 4 = 2 p 2 2 solutione gaudeat in integris. . . . 



Helsingfors, 1851, 16 pp. 



