<s>[INST] What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square? 
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?

Initially, this seemed like one could work it out with $AM-GM$, but it doesn't seem so.
From $AM-GM$ one gets that $$\frac{1^2+2^2+ \dots+n^2}{n} \geqslant \sqrt[\leftroot{-1}\uproot{2}n]{1^2\cdot2^2\dots\cdot n^2}
$$
is this of any help here?
Remark.  Thanks to Favst, the source of the problem is Problem 1 of 1994 British Mathematical Olympiad Round 2
 [/INST] Following a hint by Mostafa Ayaz, we have $(n+1)(2n+1)=6k^2$ for some integer $k$.  That is,
$$(4n+3)^2-3(4k)^2=1\,.$$
Hence, $(4n+3)+(4k)\sqrt{3}=(2+\sqrt{3})^m$ for some nonnegative integer $m$.  Therefore,
$$4n+3=\sum_{r=0}^{\left\lfloor\frac{m}{2}\right\rfloor}\,\binom{m}{2r}\,2^{m-2r}\,3^r\,.$$
If $m$ is odd, then
$$4n+3\equiv 2m\cdot 3^{\frac{m-1}{2}}\pmod{4}\,.$$
If $m$ is even, then
$$4n+3\equiv  3^{\frac{m}{2}}\pmod{4}\,.$$
Since $$4n+3\equiv 3\pmod{4},$$ we need $$m\equiv 2\pmod{4}\,.$$
That is, $m=4s+2$ for some nonnegative integer $s$
$$4n+3+(4k)\sqrt{3}=(7+4\sqrt{3})\,(97+56\sqrt{3})^s\,.$$
That is, $n=a_s$ and $k=b_s$, where
$$a_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s+(7-4\sqrt{3})\,(97-56\sqrt{3})^s-6}{8}$$
and
$$b_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s-(7-4\sqrt{3})\,(97-56\sqrt{3})^s}{8\sqrt{3}}\,.$$
Note that $a_0=1$, $a_1=337$, and
$$a_s=194\,a_{s-1}-a_{s-2}+144\text{ for }s=2,3,4,\ldots\,.$$
Furthermore, $b_0=1$, $b_1=195$, and
$$b_s=194\,b_{s-1}-b_{s-2}\text{ for }s=2,3,4,\ldots\,.$$
Therefore, the next smallest pair $(n,k)$ apart from the one given by Robert Israel is
$$(n,k)=(a_2,b_2)=(65521,37829)\,.$$
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