<s>[INST] $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square 
Prove that there exist infinitely many positive integers $n$ such that $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers with this property. 

The given condition is equivalent to $(2n+2)(2n+1) = 12p^2$ where $p$ is a positive integer. Then since $\gcd(2n+2,2n+1) = 1$, we have that $2n+2 = 4k_1$ and $2n+1 = k_2$. We must have that $k_1$ is divisible by $3$ or that $k_2$ is divisible by $3$. If $k_1$ is divisible by $3$ and $k_2$ is not, then we must have that $k_1$ is divisible by $9$ and so $2n+2 = 36m$. Then we need $3mk_2$ to be a perfect square where $k_2+1 = 36m$. Thus if $3mk_2 = r^2$, we get $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$.
 [/INST] Using this:
$$
\frac{\sum_{k=1}^N k^2}{N}= \frac{1}{6} (N+1) (2 N+1)=K^2
$$
the condition can be written as (as you pointed out): 
$$(N+1) (2 N+1)=6K^2$$
through numerical search, the last part of the problem is solved
$$N=1,337,65521$$
Not really a satisfying answer i know.
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