1981 IMO Problems/Problem 3

Problem

We first observe that since $\gcd(m,n)|1$ , $m$ and $n$ are relatively prime. In addition, we note that $n \ge m$ , since if we had $n < m$ , then $n^2 +nm -m^2 = n(n-m) - m^2$ would be the sum of two negative integers and therefore less than $-1$ . We now observe

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i.e., $(m,n) = (a,b)$ is a solution iff. $(b, a+b)$ is also a solution. Therefore, for a solution $(m, n)$ , we can perform the Euclidean algorithm to reduce it eventually to a solution $(1,n)$ . It is easy to verify that if $n$ is a positive integer, it must be either 2 or 1. Thus by trivial induction, all the positive integer solutions are of the form $(F_{n}, F_{n+1})$ , where the $F_i$ are the Fibonacci numbers. Simple calculation reveals $987$ and $1597$ to be the greatest Fibonacci numbers less than $1981$ , giving $987^2 + 1597^2=3524578$ as the maximal value.