CHAP, xxii] SINGLE EQUATIONS OF DEGKEE FOUR. 661 



x = p+t, y = q-\-tz. After dividing by t, we obtain B+Cz+tQ = Q, where Q 

 is quadratic in t and z, while B and C are constants. From the solution 

 t = Q, z B/C of this cubic, we obtain a second by the tangent method. 

 Euler 253 treated as two separate problems the solution of 



Then 



jcV=F2V= I(x 2 +y 2 -z 2 v 2 ) 2 , 



The left members will be squares if 



whence 



2 2pr 2 2qs 



From (1) we obtain z 2 /v 2 and y 2 /z 2 by multiplication and division. Hence 

 we have values a, b, c, d for which x = at,v = bt,y = cu, z = du. Then (2) give 



(a?b 2 )t 2 +(c 2 -d 2 ')u 2 = 2mbdtu, (a 2 b 2 )t 2 -(c 2 -d 2 ')u 2 = 



2 



> o - 



2qs 

 By subtraction, we get t/u. Hence we take 



t = c* d 2 , u = b(mdnc) 



and obtain x, y, v, z. Changing the sign of n, we obtain a second set of 

 solutions. Rational solutions result only when the product of the right 

 members of (1) is a rational square. For the upper signs, take p = 2fg, 

 r=f 2 g 2 , q = 2hk, s = h 2 k 2 . Then the condition is 



It is the square of 3mnfg(fg) for 

 h = g, k=fg, f 



See Euler 81 of Ch. XVI. 



Euler 254 used the preceding V+ = F to find x 2 , - , v 2 such that 



a = x 2 y 2 z 2 v 2 , (3=z 2 x 2 y 2 v 2 , y = y 2 z 2 x 2 v 2 

 shall be squares. We have 



F+4a=(x 2 +y 2 -z 2 +v*) 2 , F+4:p=(x 2 +z 2 -y 2 +v 2 ) 2 , 



Hence we seek solutions of F=0. Solving the latter for x 2 we get 



x 2 = y i+ z 2_ v 2 +2 T, T 2 = y 2 (z 2 -v 2 }-z 2 v 2 . 

 Now 2 2 -v 2 = D for z = 5, v = 3, whence T 2 = 16?/ 2 -225 = (42/-Z) 2 if 



263 Acta Acad. Petrop., 2, II, 1781 (1778), 85; Comm. Arith., II, 366; Op. Om., (1), III, 429. 

 254 Opera postuma, 1, 1862, 257-8 (about 1782). For sums, instead of differences, see Euler 81 

 of Ch. XVI. 



