### IMO 2003

Problem A2

Find all pairs (m, n) of positive integers such that m2/(2mn2 - n3 + 1) is a positive integer.

(m, n) = (2k, 1), (k, 2k) or (8k4 - k, 2k)

Solution

Thanks to Li Yi

The denominator is 2mn2 - n3 + 1 = n2(2m - n) + 1, so 2m >= n > 0. If n = 1, then m must be even, in other words, we have the solution (m, n) = (2k, 1).

So assume n > 1. Put h = m2/(2mn2 - n3 + 1). Then we have a quadratic equation for m, namely m2 - 2hn2m + (n3 - 1)h = 0. This has solutions hn2 +- N, where N is the positive square root of h2n4 - hn3 + h. Since n > 1, h ≥ 1, N is certainly real. But the sum and product of the roots are both positive, so both roots must be positive. The sum is an integer, so if one root is a positive integer, then so is the other.

The larger root hn2 + N is greater than hn2, so the smaller root < h(n3 - 1)/(hn2) < n. But note that if 2m - n > 0, then since h > 0, we must have the denominator (2m - n)n2 + 1 smaller than the numerator and hence m > n. So for the smaller root we cannot have 2m - n > 0. But 2m - n must be non-negative (since h is positive), so 2m - n = 0 for the smaller root. Hence hn2 - N = n/2. Now N2 = (hn2 - n/2)2 = h2n4 - hn3 + h, so h = n2/4. Thus n must be even. Put n = 2k and we get the solutions (m, n) = (k, 2k) and (8k4 - k, 2k).

We have shown that any solution must be of one of the three forms given, but it is trivial to check that they are all indeed solutions.