CHAP, xvi] CONCORDANT FORMS. 473 



Euler 77 called x 2 +my 2 and x 2 +ny 2 concordant forms if they can both be 

 made squares by choice of integers x, y each not zero ; otherwise, discordant 

 forms. He treated the problem: Given an integer m, to find all integers n 

 for which the two forms are concordant. Set m = nv, where one factor may 

 be unity. Then x 2 +my 2 = (y,p~ -\-vq~Y\i x = d=(/x;; 2 vq 2 ), y = 2pq. To make 

 x 2 -{-ny 2 = w~, we must take n = (w 2 x 2 } / (4p 2 q 2 ) , where x has the preceding 

 value. Then both factors wx must be even. Set p 2 q 2 = r 2 s 2 , where 2r 2 

 divides w-\-x, and 2s 2 divides w x, the respective quotients being/ and g. 

 Then n =fg. Hence we consider x, r, s as known and seek /, g such that 

 fr 2 gs 2 = x. The latter is satisfied by /=/is 2 =b<rx, g = hr 2 px, where p/<r 

 is the convergent preceding r 2 /s 2 for the continued fraction for the latter. 

 A table ( 10) gives the values of p, a for r 2 ^12 2 , s 2 ^r 2 . For ra = l, Euler 

 found ( 12) those values numerically < 100 of n which result from small 

 values of r, s. It is known that x 2 y 2 are discordant; also x 2 -\-y 2 , x 2 +2y 2 . 

 Proof is given ( 15-19) that x 2 +y 2 , x 2 +3y 2 are discordant; also (20-23, 

 31) that x 2 -\-y 2 =z 2 , x 2 +4y 2 = v 2 are impossible. Hence 



and v 2 4?/ 2 = x 2 , v~ 3?/ 2 = z 2 are discordant. But x 2 -\-y 2 , x 2 -\-7y 2 are squares 

 for x = 3, 2/ = 4. In papers 109, 110, 113-4 of Ch. XVI it is noted that 

 Z 2 db7/ 2 and 2 4?/ 2 are not both squares; also, x 2 y 2 and x 2 T3?/ 2 . 



Euler 78 satisfied x--\-my 2 = D as in the last paper, 77 and noted that then 

 x 2 +ny 2 =(/jip 2 -vq 2 +2Mp 2 q-} 2 if n = M 2 p 2 <?+M(np 2 -vq' 2 ), where M is arbi- 

 trary. If M^N+v/p 2 , n=(Np 2 +v}(Nq 2 +fj.}. 



Euler 79 noted that x 2 +aby 2 is a square for x = (ap 2 bq-}, y = 

 that x- + cdy 2 = n for x = 77 (cr 2 ds 2 ) , y = 2-rjrs. Hence set 



pq = rirs = trifghk, p = rjfg, q = hk, r=tfh, s = gk. 

 By the values of x, 



h 2 rj 



Set 6=^. Hence must d(ec/ 2 +bk 2 )(eaf 2 +dk 2 ) = D. But this condition 

 was not discussed. 



Euler 80 had previously treated the more special problem to find all 

 integers N such that A 2 +B 2 and A 2 +NB 2 are both squares for AB^Q. 

 Take A = x 2 - y 2 , B = 2xy. Then 



A 2 +B 2 =(x 2 +y 2 ) 2 , A 2 +NB 2 =(x 2 -y-y-+mx 2 y 2 = z 2 . 



The last gives for 2V an expression which is an integer if z = x 2 +2ax 2 y 2 y 2 . 

 According as the upper or lower sign is chosen, we have 



N=(ax 2 +l)(ay 2 +l) or (aZ 2 -l)(a7/ 2 +l) + l. 



To investigate the rational en's for which the first N is integral, when x and 

 y are integral, set a = af(q 2 s 2 ), x = pq, y = rs, where a is an integer, while p, 



77 Mem. Acad. Sc. St. Petersb., 8, 1817-8 (1780), 3; Comm. Arith., II, 406. 



78 Opera postuma, 1, 1862, 253 (about 1769). 

 n Ibid., 256 (about 1782). 



80 Nova Acta Acad. Petrop., 11, 1793 (1777), 78; Comm. Arith., II, 190-7. 



