IMO 1981 SL 12

Origin: NET

Problem

Determine the maximum value of m2 + n2 where m and n are integers satisfying m, n \in{1, 2, . . ., 100} and (n2 −mn −m2)2 = 1.

Solution

We will solve the contest problem (in which m, n \in{1, 2, . . ., 1981}). For m = 1, n can be either 1 or 2. If m > 1, then n(n −m) = m2 \pm 1 > 0; hence n −m > 0. Set p = n −m. Since m2 −mp −p2 = m2 −p(m + p) = −(n2 −nm −m2), we see that (m, n) is a solution of the equation if and only if (p, m) is a solution too. Therefore, all the solutions of the equation are given as two consecutive members of the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, . . . . So the required maximum is 9872 + 15972.