Find all positive integers m and n such that [m2/n] + [n2/m] = [m/n + n/m] + mn.
Solution
Answer: (m, n) = (k, k2 + 1) or (k2 + 1, k) for any positive integer k.
Put m = n2 + 1. If n > 1, then lhs = [n3 + 2n + 1/n] + [n2/(n2 + 1] = n3 + 2n, and rhs = [n + 1/n + n/(n2 + 1)] + n3 + n = n3 + 2n, so we have a solution. It is easy to check that n = 1 also gives a solution. Obviously there is symmetry between m and n, so n = m2 + 1 is also a solution. We now show that there are no other solutions. Without loss of generality we may assume n ≥ m.
Now regard m as fixed and put f(n) = [m2/n] + [n2/m] - [m/n + n/m] - mn. Put g(n) = m2/n + n2/m - m/n - n/m - mn. So |g(n) - f(n)| < 3.
g(n) = (1/m)(n - (m2+1)/2)2 - (1/4m)( (m2+1)/2)2 + (m2-m)/n. So if n ≥ m2 + 2, then g(n) ≥ m + 2/m + (m2-m)/(m2+2) > m. Hence f(n) > m-3. So for m > 2 there are no solutions for n > m2 + 1.
Also if m ≤ n ≤ m2 - 1, then (n - (m2+1)/2)2 ≤ (m2 - 3)2/4 and (m2-m)/n ≤ (m-1), so g(n) ≤ -m + 2/m - 1 < 0. Hence f(n) < 0 for m > 2. So for m > 2 there are no solutions for n < m2 (and ≥ m). It is easy to check directly that f(m2) = -(m-1) < 0.
So it remains to check m = 1 and m = 2. If m = 1, then f(1) = -1. If n > 1, then f(n) = 0 + n2 - n - n = n(n - 2). So for m = 1, the only solution is n = 2.
So finally take m = 2 and consider n >= 2. It is easy to check that f(2) = -2, f(3) = -3, f(4) = -1. For n ≥ 6, we have f(n) = 0 + [n2/2] - [n/2] - 2n ≥ n2/2 - 1/2 - 5n/2 > 0.
© John Scholes
jscholes@kalva.demon.co.uk
16 Dec 2002