Find all pairs (m, n) of positive integers such that m^{2}/(2mn^{2} - n^{3} + 1) is a positive integer.

**Answer**

(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 < h(n^{3} - 1)/(hn^{2}) < n. But note that if 2m - n > 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.

© John Scholes

jscholes@kalva.demon.co.uk

24 Jul 2003

Last corrected/updated 24 Jul 2003