IMO 1981 Problem A3 Determine the maximum value of m 2 + n 2 , where m and n are integers in the range 1, 2, ... , 1981 satisfying (n 2 - mn - m 2 ) 2 = 1.
Experimenting with small values suggests that the solutions of n 2 - mn - m 2 = 1 or -1 are successive Fibonacci numbers. So suppose n > m is a solution. This suggests trying m+n, n: (m+n) 2 - (m+n)n - n 2 = m 2 + mn - n 2 = -(n 2 - mn - m 2 ) = 1 or -1. So if n > m is a solution, then m+n, n is another solution. Running this forward from 2,1 gives 3,2; 5,3; 8,5; 13,8; 21,13; 34,21; 55,34; 89,55; 144,89; 233,144; 377,233; 610,377; 987,610; 1597,987; 2584,1597. But how do we know that there are no other solutions? The trick is to run the recurrence the other way. For suppose n > m is a solution, then try m, n-m: m 2 - m(n-m) - (n-m) 2 = m 2 + mn - n 2 = -(n 2 - mn - m 2 ) = 1 or -1, so that also satisfies the equation. Also if m > 1, then m > n-m (for if not, then n >= 2m, so n(n - m) >= 2m 2 , so n 2 - nm - m 2 >= m 2 > 1). So given a solution n > m with m > 1, we have a smaller solution m > n-m. This process must eventually terminate, so it must finish at a solution n, 1 with n > 1. But the only such solution is 2, 1. Hence the starting solution must have been in the forward sequence from 2, 1. Hence the solution to the problem stated is 1597 2 + 987 2 . s are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 22nd IMO 1981 © John Scholes jscholes@kalva.demon.co.uk 14 Oct 1998