IMO 2003 Problem A2 Find all pairs (m, n) of positive integers such that m 2 /(2mn 2 - n 3 + 1) is a positive integer.
(m, n) = (2k, 1), (k, 2k) or (8k 4 - k, 2k) Solution Thanks to Li Yi The denominator is 2mn 2 - n 3 + 1 = n 2 (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 = m 2 /(2mn 2 - n 3 + 1). Then we have a quadratic equation for m, namely m 2 - 2hn 2 m + (n 3 - 1)h = 0. This has solutions hn 2 +- N, where N is the positive square root of h 2 n 4 - hn 3 + 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 hn 2 + N is greater than hn 2 , so the smaller root  0, then since h > 0, we must have the denominator (2m - n)n 2 + 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 hn 2 - N = n/2. Now N 2 = (hn 2 - n/2) 2 = h 2 n 4 - hn 3 + h, so h = n 2 /4. Thus n must be even. Put n = 2k and we get the solutions (m, n) = (k, 2k) and (8k 4 - 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. 44th IMO 2003 © John Scholes jscholes@kalva.demon.co.uk 24 Jul 2003 Last corrected/updated 24 Jul 2003