30th BMO 1994 Problem 1 Find the smallest integer n > 1 such that (1 2 + 2 2 + 3 2 + ... + n 2 )/n is a square.
We want (n+1)(2n+1)/6 = m 2 . But n+1 and 2n+1 are coprime. Also 2n+1 is odd, so we must have either (1) 2n+1 = a 2 , n+1 = 6b 2 or (2) 2n+1 = 3a 2 , n+1 = 2b 2 . But 2n+1 = 2(n+1) - 1, so in case (1) we have a 2 + 1 = 12b 2 . But squares are 0 or 1 mod 3, so a 2 + 1 is 1 or 2 mod 3, whereas 12b 2 is 0 mod 3. So case (1) is not possible. In case (2) we have 4b 2 - 1 = 3a 2 , so either 2b + 1 = c 2 , 2b - 1 = 3d 2 or 2b + 1 = 3d 2 , 2b - 1 = c 2 , so c 2 = 3d 2 ± 2. We now try successively, d = 1, 2, ... . d = 1 gives c = 1, b = 1, n = 1. But we are given n > 1. d = 2 fails, d = 3 gives a = 15, b = 13, n = 337. Checking: (337 + 1)(2.337 + 1)/6 = 169.225 = (13.15) 2 . 30th BMO 1994 © John Scholes jscholes@kalva.demon.co.uk 14 Aug 2002