<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] This can be solved by using the quadratic formula on $2n^2 + 3n + (1-6p^2)=0$ to convert to the Pell's equation $r^2 - 48p^2 = 1$.  Standard techniques can be used to generate infinitely many solutions, though we need to be careful to keep only those with $r \equiv 3\pmod{4}$.
For example, we can arrive at infinitely many solutions by writing $r+p\sqrt{48} = (7+\sqrt{48})^{2k+1}$, beginning from the primitive solution $r=7, p=1$.
Here is a proof that results from that process:
If $(n,p)$ is a positive solution to $(n+1)(2n+1)=6p^2$, then $(97n+168p+72,56n+97p+42)$ is a larger solution, which we can see by computing directly:
$$((97n+168p+72) + 1)(2(97n+168p+72) + 1) - 6(56n+97p+42)^2$$
$$ = (n+1)(2n+1)-6p^2$$
Beginning from the solution $(n,p)=(1,1)$, this generates infinitely many solutions by induction.
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