Finite Summation = Perfect Square

By expanding out and using the well-known formulas for the sum of the first $n$ integers and the first $n$ squares, we get that the sum is $2n \sum k^2 - 2\left( \sum k \right)^2 = n^2(n^2-1)/6$ .

If this is a square, then $(n^2-1)/6$ is a rational square. But since the denominator has no square factors, this implies that $(n^2-1)/6$ is a square integer.

Solving $n^2-6m^2=1$ by whatever your favorite Pell's equation method is, we see that $n+m\sqrt{6}$ is a power of $5+2\sqrt{6}$ . This leads to solutions $1, 5, 49, 485, 4801, \ldots$ , so the answer is $\fbox{540}$ .

First math out the sum to get $\frac{n^2(n^2-1)}6=x^2$ . Since $\gcd(n^2,n^2-1)=1$ , we know that $2\not\mid n^2$ because then there is an odd number of even numbers dividing $x^2$ . Similarly, $3\not\mid n^2$ . Hence, $n\mid x$ . We can thus rewrite as $n^2-6\left(\frac{x}n\right)^2=1$ , where both $n$ and $\frac{x}n$ are integers. This is just Pell's equation! Since $\left(n,\frac{x}n\right)=(5,2)$ satisfies it, we have $n_i=10n_{i-1}-n_{i-2}$ with $n_0=1$ and $n_1=5$ . Hence, $n_2=49$ and $n_3=485$ , with no more solutions $n\le1000$ . Hence, $1+5+49+485=\boxed{540}$ . Very nice problem, I'm surprised Pell's equation came up. Love it!