Find all pairs m, n of positive integers such that m2 - n divides m + n2 and n2 - m divides m2 + n.
Solution
Assume n ≥ m.
(m+1)2 - m = (m+1) + m2. Clearly n2 increases faster than n, so n2 - m > n + m2 for n > m+1 and hence there are no solutions with n > m+1. It remains to consider the two cases n = m and n = m+1.
Suppose n = m. Then we require that n2 - n divides n2 + n. If n > 3, then n2 > 3n, so 2(n2 - n) > n2 + n. Obviously n2 - n < n2 + n, so if n > 3, then n2 - n cannot divide n2 + n. It is easy to check that the only solutions (with n = m) less than 3 are n = 2 and n = 3.
Finally suppose n = m+1. We require m2 - m - 1 divides m2 + 3m + 1. If m >= 6, then m(m - 5) > 3, so 2(m2 - m - 1) > m2 + 3m + 1. Obviously m2 - m - 1 < m2 + 3m + 1, so m2 - m - 1 cannot divide m2 + 3m + 1 for m >= 6. Checking the smaller values, we find the solutions less than 6 are m = 1 and m = 2.
Summarising, the only solutions are: (n, m) = (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2).
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002