15th USAMO 1986 Problem 3 What is the smallest n > 1 for which the average of the first n (non-zero) squares is a square? Solution
: 337. We need 1/6 (n+1)(2n+1) a square. We need n = 6m+1 for it to be an integer. So (3m+1)(4m+1) must be a square. But 3m+1 and 4m+1 are relatively prime, so each must be a square. Suppose 3m+1 = a 2 and 4m+1 = (a+k) 2 , then (subtracting) m = 2ak + k 2 , so a 2 - 6ak - (3k 2 + 1) = 0, so a = 3k + √(12k 2 +1). By inspection the smallest solution of this is k = 2, giving a = 13 and hence m = 56 and n = 337. 15th USAMO 1986 © John Scholes jscholes@kalva.demon.co.uk 11 May 2002