What is the smallest n > 1 for which the average of the first n (non-zero) squares is a square?
Solution
Answer: 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 = a2 and 4m+1 = (a+k)2, then (subtracting) m = 2ak + k2, so a2 - 6ak - (3k2 + 1) = 0, so a = 3k + √(12k2+1). By inspection the smallest solution of this is k = 2, giving a = 13 and hence m = 56 and n = 337.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002