450 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xv 



It remains to make y z a square. Since d a, 



Choose the lower signs. Then 3(n 2 1) is the square of (n+V)flg if 



_/ 2 +3g 2 

 30 2 -/ 2 ' 



Multiply the resulting values of a, b, c by (3</ 2 / 2 ) 2 ; we get 

 a = d = 4/V, 6, c = 



r=/ 8 +30/V+810 8 , s = 16/y. 



For/=<7=l, we get p = q = 5, r = 7, s = l, whence z = ?/ = 50, 2 = 14. From 

 one solution z, ?/, 0, we get ( 15) a second solution 



(5) X = , Y= , . 



In the " additamentum ' (16), Euler treated the problem to find 

 three squares z 2 , y 2 , z 2 whose differences are squares. Using (3) and (4), 

 we have 



x 2 -y 2 =(p' 2 -q z ) 2 ) z 2 -z 2 =(r 2 -s 2 ) 2 , 7/ 2 -z 2 = 4(j>y-r 2 s 2 ), 

 the last being a square if dbcd(a? W)(d- c 2 ) = D. This is satisfied if 



From one solution we get a second by (5) . 



E. Waring 34 noted that, in the problem to find three numbers the sum 

 and difference of any two of which are squares, four of the conditions are 

 satisfied if we employ either of Landen's 23 notations for the numbers or 

 the notation a 2 rc 2 +& 2 2/ 2 > 2abxy, a 2 t/ 2 +6 2 x 2 , but gave no discussion. He 

 recalled Rolle's 18 values A, B, C. 



Euler 35 treated the problem to find four positive numbers in arith- 

 metical progression such that the sum of any two is a square: 



Hence all are expressible in terms of p, q, r, subject to two conditions 



2r 2 = p 2 +J 2 = g 2 +s 2 . 

 Thus r = m = z 2 +?/. We get 2r 2 = El and satisfy 2r 2 = p 2 + 2 by taking 



the first term being positive, whence p<t. Similarly, we satisfy 2r 2 = 

 by taking r = x\+y\ and 



Then x-+y- = x 2 l -\-y 2 l is satisfied by taking 



x=fz+l, xi=fz-l t y = z-f, yi = 

 as may be done without loss of generality by removing a common square 



34 Meditationes Algebraicae, ed. 3, 1782, 328. 



35 Posthumous paper, 1781, Comm. Arith., II, 617-25; Opera postuma, 1, 1862, 119-127. 



Reprinted, Sphinx-Oedipe, 4, 1909, 33-42. 



