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



E. Fauquembergue 127 gave the general solution of (x 2 +y 2 )(2x 2 y 2 ) = 2z 2 . 

 A. Gerardin 128 gave x, y, z = 3f, 4f, 5/ 2 and h/2, 2h/3, 5/i 2 /36. 



4 +4 2 +l = y 2 is impossible in rational numbers. 129 Cf. Pietrocola. 131 



T. Pepin 130 treated x*-8x 2 y 2 +8y 4 = z 2 by the method of descent appli- 

 cable only if y is even ; then x = X 8 - 8 Y 8 , y = 2X YZ, z = Z 4 - 32X 4 F 4 . For 

 y odd the equation is reduced to the pair pq = rs, p 2 4<? 2 +4pg'+8s 2 r 2 = 0, 

 to which the method of descent is applicable. There exist only six sets of 

 solutions x, y, z, each =1=0, with ?/<10 10 . 



C. Pietrocola 131 discussed the equivalent equations 



z 4 +4tey+(2ft-l)V = 2 2 , (x 2 +2hy 2 +z)(x 2 +2hy 2 -z') = (4/i-l)2/ 4 . 

 From one solution he derived another and proved the equation impossible 

 if h = 1. The last result had been proposed as a problem by P. Tannery. 132 



A. S. Werebrusow 133 listed many values of m between 100 and +100 

 for which x 4 -\-mx 2 y 2 -\-y 4 = z 2 is impossible, and stated that it is impossible for 

 m positive or for ra = 8&+3 negative if m+2 and m2 are primes. 



A. Gerardin 134 noted that the last statement fails for m = 99. 



Gleizes and H. B. Mathieu 135 gave special expressions for m for which the 

 equation is solvable. 



A. Cunningham 136 noted that the equation is solvable for m = 60, 99, 

 72, 96, contrary to Werebrusow, 133 and for m = 91, 90, contrary to 

 Euler 108 (p. 495, p. 498); and corrected various misprints on pp. 496-8 of 

 Euler's paper. 



L. Aubry 137 stated that Werebrusow's 133 theorem is true for a positive 

 m=l, 5 or 7 (mod 8), and a negative m= (8&+5), but false for a positive 

 m = 8&+3. Aubry (pp. 57-9) treated x*-\-bx 2 y 2 +cy 4 = dz 2 , given 



d = p*+bp 2 q 2 +cq\ 



by setting X 2 = p 2 ucq 2 v, y 2 = q 2 u+(bq 2 +p 2 )v and deducing an equation of 

 the initial form, whence one solution leads to two new solutions. 



H. C. Pocklington 138 proved that x*x 2 y 2 +y*, x 4 +14o: 2 ?/ 2 +i/ 4 are neither 

 squares if x ^y. UN is not of the form 8n3 and is not divisible by any 

 prime 4n+l, and at the same time ./V^ 4 is an odd power of an odd prime 

 (including unity), then (x 2 +y 2 ) 2 =FNx 2 y 2 = z 2 is impossible in integers. For 

 N = l and the upper signs, we see that x 4 +x 2 y 2 -\-y*=z 2 is impossible. Also 

 x 4 14x 2 y 2 +y 4: = z 2 is impossible. There is a list of values of n<100 for 

 which x*nx 2 y 2 +y 4 =z 2 is impossible. The complete solution is given of 



127 L'intermSdiaire des math., 4, 1897, 70. 



128 Ibid., 16, 1909, 175. 



129 Ibid., 1897, 20,83, 203,229; 1898,89,128; 1900,87-90; 1903,158; 1905,109. 



130 Mem. Accad. Pont. Nuovi Lincei, 14, 1898, 71-85. 



131 Giornale di mat., 36, 1898, 77-80. 



132 L'intermddiaire des math., 1897, 20, 30, 203. 



133 Ibid., 15, 1908, 52, 282 (corrections); Mat. Sbornik, Moscow, 26, 1908, 599-617. 



134 L'interme'diaire des math., 16, 1909, 154. 

 Ibid., 15, 1908, 159. 



Ibid., 17, 1910,201. 

 lbid., 18, 1911,203. 

 " 8 Proc. Cambridge Phil. Soc., 17, 1914, 111-118. 



