What is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer? What is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer?
The root mean square is defined as:
$$\sqrt{\left(\frac{a_1^2 + a_2^2 + a_3^2+\dots+a_n^2}{n}\right)}$$
I started off by the relation:
$$\sum^n_{i = 1}i^2 = \frac{n(n+1)(2n+1)}{6}$$
We need an $n$ such that:
$$\sqrt{\frac{n(n+1)(2n+1)}{6n}} = k$$
$$\implies (n+1)(2n+1) = 6k^2$$
$$\implies 2n^2 + 3n + (1-6k^2) = 0$$
$$\implies n = \frac{-3 \pm \sqrt{9 - 8(1-6k^2)}}{4} = \frac{-3 \pm \sqrt{1 + 48k^2}}{4}$$
For the equation to have rational solutions $1 + 48k^2$ must be a perfect square. So, the problem boils down to solving the smallest $k$ such that $1 + 48k^2$ is a perfect square. 
I have no idea how to proceed. 

Starting with $1+48k^2= m^2$ or $m^2-48k^2 = 1$, it makes sense to try convergents to $\sqrt{48}$, which are $\left\{6,7,\frac{90}{13},\frac{97}{14},\frac{1254}{181},\frac{1351}{195},\frac{17466}{2521}, ...\right\}$. 
We spot solutions $(m, k) = (7, 1), (97, 14), (1351, 195), \;\dots $   corresponding to $n = 1,  \frac{47}2, 337, ...$
