Find all solutions to (m2 + n)(m + n2) = (m - n)3, where m and n are non-zero integers.
Solution
Answer: (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 m2 + n > m2 and n2 + m > m, so lhs > m3 > 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 (M2 - N)(N2 + M) = (M + N)3. Hence M2N2 + M3 - N3 - MN = M3 + 3M2N + 3MN2 + N3. N is non-zero, so we can divide to get: 2N2 + N(3M - M2) + 3M2 + M = 0. Regarding this as a quadratic in N we solve, getting N = ( (M2 - 3M) ± √(M4 - 6M3 + 9M2 - 24M2 - 8M) )/4 . Thus M4 - 6M3 - 15M2 - 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 = M2 - 10M + 25 < M2 - 8M. So we need only consider M ≤ 12. But obviously we cannot have M < 8, or M(M - 8) is negative. Checking the remaining values, we find M = 8 and 9 are the only solutions. They give the solutions (m, n) = (8, -10), (9, -6), (9, -21).
Next, consider the case m = -M, n = N. That clearly does not work. We get (M2 + N)(N2 - M) = - (M + N)3. So 2N2 + (M2 + 3M)N + 3M2 - M = 0, which has no solutions because 3M2 > M.
Finally, consider the case m = -M, n = -N. Then we get (M2 - N)(N2 - M) = (N - M)3, so 2N2 - (M2 + 3M)N + (3M2 - M) = 0. Solving for N, we get N = ( (M2 + 3M) ± √(M4 + 6M3 + 9M2 - 24M2 + 8M) )/4. So we require M(M3 + 6M2 - 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 = M2 + 6M + 9 < M2 + 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).
© John Scholes
jscholes@kalva.demon.co.uk
26 Aug 2002