Problem 681

Computing and completing the square, we find
$$\begin{aligned} \frac{1}{n} \sum_{k=m}^{m+n-1} k^2 &= m^2 + (n-1)m + \frac{(n-1)(2n-1)}{6} \\ &= \left(m + \frac{n-1}{2}\right)^2 + \frac{n^2-1}{12} \end{aligned}$$
is a perfect square.

Since $n^2 - 1$ modulo 8 is either 0 or odd, $\frac{n^2-1}{12}$ must be even. Also, from the above equation we know that $\frac{n^2-1}{12}$ is the difference of two perfect squares, and since it is even, it must therefore be a multiple of 4. Hence we obtain $n^2 \equiv 1 \pmod{48}$, so $n$ can only be of the form $24p \pm 1, 24p \pm 7$.

Conversely, if $n$ is a positive integer of the form $24p \pm 1, 24p \pm 7$, then $n^2 \equiv 1 \pmod{48}$. There exists a nonnegative integer $a$ such that $\frac{n^2-1}{12} = 4a$. Take $m = a - 1 - \frac{n-1}{2}$, then
$$\begin{aligned} \frac{1}{n} \sum_{k=m}^{m+n-1} k^2 &= \left(m + \frac{n-1}{2}\right)^2 + \frac{n^2-1}{12} \\ &= (a-1)^2 + 4a \\ &= (a+1)^2. \end{aligned}$$