668 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxii 



It suffices to treat the case in which x and y are relatively prime, also t 

 and u. For problem 305 (i), AB is the square of Axy(p 2 -\-q 2 ) if 



t = xy(p 2 q 2 }-{- 2y 2 pq, u = xy (p 2 <f) 2x 2 pq. 

 Then C is found to have the factor A, so that AC= D if 

 4p2q*x*-4pq(p* - q 2 )x 3 y+ (p 4 - 6p 2 q 2 +q*)x 2 y 2 +4pq(p 2 - q 2 )xy*+4p 2 q 2 y* = Q 2 - 

 Taking Q = 2pqx 2 (p 2 q 2 )xy 2pqy 2 + ay 2 , we have 



a (a 4pq) y 2 2a (p 2 q 2 } xy + pq (a. pq) x 2 = . 

 For a = Ipq, we obtain the solution 



x = 2(p 2 -q 2 ), y = 3pq, t = 3(p*+p 2 q 2 +q*), u=(p 2 -q 2 )*. 

 For a = pq, we obtain a similar solution. For = =F2p 2 , we get 



x = p(p2q), t = p(2p^q)(p z ^2pq+3q i ), 

 y= q(2pq-), u = q(p2q}(q*2pq+3p 2 ). 



For problem (ii), BC is a square if a: = 3, y = 5, t= 11, u =45, or if 



= 3n 4 +6ra 2 n 2 m 4 , 2/ = 3ra 4 +6w 2 ft 2 n 4 , t = mx, u = ny. 



For problem (iii), we apply the last solution with w 2 +n 2 = D. 



Euler 306 required four numbers the four elementary symmetric functions 

 of which are squares. For the numbers Mab, Mbc, Mcd, Mda the conditions 

 reduce to 



dbcd= D, bd(a 2 +c 2 )+ac(b+d) 2 = D, M = (ab+bc+cd+dd)ff 2 . 



Finding the second condition impossible if b/d = 2or 3, Euler took b/d = p 2 /q 2 . 

 Then must p 2 q 2 (a 2J [-c 2 )-{-ac(p 2 -\-q 2 ) 2 be a square, say that of pqa+cm/n. 

 Thus a/c is found, and we readily form the condition that ac and hence 

 abed shall be a square. By trial Euler found the two solutions 307 a = 64, 

 5 = 9, d = 4, c = 49 or 289, M=1469 or 4589; also one of another type: 

 a = 16, 6 = 5, c = 5, d = 4, / = 3, M = 2l. He discussed at length the problem 

 to find b, d such that the initial second condition can be satisfied by choice 

 of a, c. 



Euler 308 treated x 2 +y 2 +z 2 = D, X 2 y 2 +x 2 z 2 +y 2 z 2 = D. The first is satis- 

 fied if x = p 2 -\-(f r 2 , y = 2pr, z = 2qr. The second then becomes 



(!) (p 2 +q 2 )(p 2 +q 2 -r 2 ') 2 +4p 2 q 2 r 2 = D. 



Set n=(p r')/q and eliminate r. Then shall 



(p 2 +q 2 } {2np+(l-n 2 )q} 2 +4p 2 (p-nq') 2 = D =R 2 . 

 Set R = (1 n 2 }q 2 +2npq+ap 2 . The terms in p 2 q 2 cancel if 



306 F. van Schooten had proposed to find rational sides of a triangle given the base a, altitude 



6 and ratio m : n of the other sides (mz, nz). Thus b=2mnxy, a = (m 2 ri 2 )(x*+y 2 ), 

 2 2 = (a; 2 +2/ 2 )[(mn) 2 a; 2 +(TOTn)V]> falling under problem (i). The simplest solution 

 isx = 3, y = 5, m = 28, w = 17, a = 33, 6=28. 

 3Novi Comm. Acad. Petrop., 17, 1772, 24; Comm. Arith., I, 450; Op. Om., (1), III, 172. 



307 Reproduced by A. Ge'rardin, Pintermediaire des math., 16, 1909, 105-6. 



308 Acta Acad. Petrop., 3, I, 1782 (1779), 30; Comm. Arith., II, 457; Op. Om., (1), III, 453. 



