Not Pythagoras Theorem

This is actually the same as Average of the Squares . I'll just repost my solution here:

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 $n = \boxed {337}$

LOL I am lazy so I commanded the computer to give me an answer

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public class MainPanel{
    public static void main(String[] args) {
        for (int kdls = 0; kdls < 20000; kdls++) {
            for (int dfskjl = 0; dfskjl < 20000; dfskjl++) {
                if (getSigma(kdls) == Math.pow(dfskjl, 2)){
                    System.out.println(kdls);
                    break;
                }
            }
        }
    }
    public static double getSigma(int Sigma){
        double Square = 0;
        for (int dfsklj = 0; dfsklj < Sigma; dfsklj++) {
            Square += Math.pow(dfsklj+1, 2);
        }
        return Square / Sigma;
    }
}

When run, the program gave me

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1,
337