16th USAMO 1987 Problem 1 Find all solutions to (m 2 + n)(m + n 2 ) = (m - n) 3 , where m and n are non-zero integers. Solution
: (m, n) = (-1, -1), (8, -10), (9, -6), (9, -21). If m, n > 0, then lhs is certainly positive, so we must have m > n to make the rhs positive. But then m 2 + n > m 2 and n 2 + m > m, so lhs > m 3 > rhs. Contradiction. So there are no solutions with m and n both positive. Put M = |m|, N = |n|. Consider next m = M, n = -N. So we have (M 2 - N)(N 2 + M) = (M + N) 3 . Hence M 2 N 2 + M 3 - N 3 - MN = M 3 + 3M 2 N + 3MN 2 + N 3 . N is non-zero, so we can divide to get: 2N 2 + N(3M - M 2 ) + 3M 2 + M = 0. Regarding this as a quadratic in N we solve, getting N = ( (M 2 - 3M) ± √(M 4 - 6M 3 + 9M 2 - 24M 2 - 8M) )/4 . Thus M 4 - 6M 3 - 15M 2 - 8M = M(M - 1) 2 (M - 8) must be a square. Hence M(M - 8) must be a square. We have (M - 4) 2 = M(M - 8) + 16 and for M ≥ 13, we have (M - 5) 2 = M 2 - 10M + 25  M. Finally, consider the case m = -M, n = -N. Then we get (M 2 - N)(N 2 - M) = (N - M) 3 , so 2N 2 - (M 2 + 3M)N + (3M 2 - M) = 0. Solving for N, we get N = ( (M 2 + 3M) ± √(M 4 + 6M 3 + 9M 2 - 24M 2 + 8M) )/4. So we require M(M 3 + 6M 2 - 15M + 8) = M(M + 8)(M - 1) 2 to be a square. Hence M(M + 8) is a square. We have (M + 4) 2 > M(M + 8) and for M ≥ 5, we have (M + 3) 2 = M 2 + 6M + 9 < M 2 + 8M. So we need only check M = 1, 2, 3, 4. We find M = 1 gives the only square, and that gives the solution (m, n) = (-1, -1). 16th USAMO 1987 © John Scholes jscholes@kalva.demon.co.uk 26 Aug 2002