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



Thus Z(b+eZ) =P 2 f 2 . For a suitably chosen rational A, we may set 



b+eZ = A(P+f), Z = (P-f)/A. 



Eliminating P, we get Z=(2fA b)[(eA 2 ). In our case, e=/=l, b = k, 

 whence Z = (k 2A)/(A 2 1) is to be a square z 2 . Thus k 2A = mp 2 , 

 A 2 l = mq 2 * Of the solutions of the latter Pell equation, those are to be 

 selected which satisfy the first equation (a " solution " which he admitted 

 was imperfect). By eliminating m and setting 2A=a, p/q = 2n, we get 

 k = a+(a 2 4)n 2 , the case treated by Euler 108 at the end of his second paper. 

 Kausler treated at length (pp. 219-236) the problem to make k integral 

 by choice of rational values of a, n. 



N. Fuss 112 required integers m such that x 4 +mx 2 y 2 +y i = z 2 . Set 



Then z~ (x 2 + y~] 2 = afix 2 y 2 , z 2 (x 2 y 2 ) 2 = 7 5x 2 y 2 . For x = pq,y = rs, we have 



z-x 2 +y 2 = 5q 2 r 2 . 



Eliminating z and replacing x, y by their values, we get three linear equa- 

 tions between a, /3, 7, 5, which give 



_aq 2 -2r 2 as 2 -2p 2 aq 2 s 2 -2p 2 q z -2r 2 s 2 



p 2 ' r 2 ' pV 



of which the last may be replaced by 75 = a/3 +4. If p = r=l, then 

 yd = (aq 2 2) (as 2 2), and a, q, s may be given any values; as the values 

 of ra<100 we get 2, 8, 12, 16, 17, 22, 23, 26, 31, -, 94. 



R. Adrain 113 proved by descent that x*+x 2 y 2 +y 4 ^n. He and T. 

 Strong (p. 151) also noted that (x 2 +y 2 ) 2 x 2 y 2 = a 2 requires that a 2 +x 2 y 2 = D 

 and a 2 3x 2 y 2 = D = (x 2 y 2 ) 2 , whereas a 2 -{-q 2 and a 2 3q 2 are not both 

 squares (Euler's Algebra, Second English transl., II, 481). H. J. Anderson 114 

 noted that we may take x and y positive and relatively prime. If x and y 

 are both odd, x 4 +x 2 y 2 +y* = 8n+3 =|= D . Hence we may take x even, y odd. 

 Thus (x 2 +?/ 2 ) 2 x 2 y 2 is an odd square, whence x 2 +y 2 = p 2 -\-q 2 , xy = 2pq. 

 By an argument like that in Euler's Algebra, II, Art. 230, we conclude that 

 r 2 s 2 and r 2 4s 2 are odd squares, where s is even, and r, s are divisors of 

 x, y, and similarly that t 2 u 2 and t 2 4v? are odd squares, where u is even, 

 and t, u are divisors of r, s. Finally, we would reach odd squares v 2 w 2 

 and v 2 4;W 2 , where %w no longer has divisors. Hence the problem is 

 impossible. 



A. M. Legendre 115 found only two solutions of m 4 4m 2 n 2 +n 4 = p 2 , viz., 

 (m, n, p} = (15, 4, 191), (442, 161, 364807). The complete solution, includ- 

 ing (2, 1, 1), was given by E. Lucas. 116 



112 M<m. Acad. Sc. St. P<tersbourg, 9, 1824 (1820), 159. 



118 The Math. Diary, New York, 1, 1825, 147-150. Cf. Genocchi 119 and Pocklington 158 ; 

 also Beha-Eddin 60 of Ch. XIV and Kausler 10 of Ch. XXVI. 



114 Ibid., 150-1. 



115 Th6orie des nombres, ed. 3, 2, 1830, 127; Maser, II, 124. See Legendre 47 of Ch. XIX. 

 118 Recherches sur 1'analyse inde'termine'e, Mouline eur Allier, 1873, p. 67. Bull. Bibl. Storia 



Sc. Mat. Fis., 10, 1877, 291-2. 



