A number theory problem by Priyanshu Mishra

Relevant wiki: Pell's Equation

We want to find positive integers $n$ such that $\begin{aligned} \frac{n(n+1)(2n+1)}{6n} & = \; m^2 \\ (n+1)(2n+1) & = \; 6m^2 \\ (4n+3)^2 - 48m^2 & = \; 1 \end{aligned}$ for some integer $m$ . The positive integer solutions of the equation $x^2 - 48y^2 = 1$ are given by $x + y\sqrt{48} \; = \; (7 + \sqrt{48})^n \hspace{2cm} n \in \mathbb{N}$ The first three of these powers is $7 + \sqrt{48}$ , $97 + 14\sqrt{48}$ and $1351 + 195\sqrt{48}$ . The first of these gives $n=1$ , so we ignore it. The second one is no good, because $97 \not\equiv 3 \pmod{4}$ . Thus we want the third solution, with $4n+3 = 1351$ , and hence $n=\boxed{337}$ . The root mean square of the first $337$ squares is $195$ .