Average of the Squares

Using sum of squares formula, $\frac {\displaystyle\sum_{i=1}^n i^2}{n}$ $=\frac {n (n+1)(2n+1)}{6n}$ $=\frac {2n^2+3n+1}{6}$ Let this be equal to $y^2$ for a positive integer y. Rearranging and using the quadratic formula to solve for n, we get $n = \frac {-3 \pm \sqrt {48y^2+1}}{4}$ We reject the negative solution as n is a positive integer. Let $48y^2+1=x^2$ for some positive integer x. Rearranging, $x^2-48y^2=1$ This is Pell's equation and has solutions $(x,y)=(7,1), (97,14), (1351,195),\ldots$ (7,1) gives n=1, (97,14) gives a fractional n and (1351,195) yields the smallest positive integer value of $n = \boxed {337}$