5th Chinese 1990 Problem B1 Given a positive integer n, for what real u, v do the simultaneous equations a 2 + b 2 + c 2 = 169n 2 , a 2 (ua 2 + vb 2 ) + b 2 (ub 2 + vc 2 ) + c 2 (uc 2 + va 2 ) = (2u + v)(13n) 4 /4 have a solution in positive integers a, b, c? Solution
: there is a solution iff v = 2u. We can rearrange the second equation slightly as: u(a 4 + b 4 + c 4 ) + v(a 2 b 2 + b 2 c 2 + c 2 a 2 ) = (2u + v)/4 (13n) 4 . But a 2 b 2 + b 2 c 2 + c 2 a 2 = (a 2 + b 2 + c 2 ) 2 /2 - (a 4 + b 4 + c 4 )/2, so the equation becomes: (u - v/2) (a 4 + b 4 + c 4 ) + v(a 2 + b 2 + c 2 ) 2 /2 = (2u + v)/4 (13n) 4 . Using the other equation this gives: (2u - v)(a 4 + b 4 + c 4 )/2 = (2u - v)/4 (13n) 4 . So if 2u - v is non-zero we can derive the integer equation: 2(a 4 + b 4 + c 4 ) = (13n) 4 . It is unlikely that this has solutions. Indeed, we can use the standard descent argument. If it has a solution in positive integers, then take the solution with smallest n. The lhs is even, so n must be even. Hence a 4 + b 4 + c 4 = 0 mod 8. But fourth powers are all 0 or 1 mod 8, so a, b, and c must all be even. But that means a/2, b/2, c/2, n/2 is a solution with smaller n. Contradiction. On the other hand, if 2u - v is zero, then the second equation adds nothing. Indeed, dividing by u it becomes simply (a 2 + b 2 + c 2 ) 2 = (13n) 4 , which is just the square of the first equation. We can easily satisfy the first equation by taking a = 12n, b = 4n, c = 3n. 5th Chinese 1990 © John Scholes jscholes@kalva.demon.co.uk 1 May 2002