<s>[INST] How can I find integers $n \gt 1$ such that the average of $1^2,2^2,3^2...n^2$ is itself a perfect square. 
$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$

so we would like to solve 

$6k^2=(n+1)(2n+1)$

here we see that $6|(n+1)(2n+1)\implies 2|n+1$
hence we can set $n=2j-1$
$2j(4j-1)=6k^2\implies j(4j-1)=3k^2$
how can  i find $j$ ,beside trial-and-error
$n=1$ is trivial
 [/INST] We are looking for $n,k\in\mathbb{N}$ such that $k^2=\frac{1}{n}\sum^n_{i=1}i^2=\frac{(n+1)(2n+1)}{6}$. Set $n_0=n+1$. 
The equation becomes $n_0(2n_0-1)=6k^2$. So $n_0$ is even. 
Set $n_0=2n_1$ to get $n_1(4n_1-1)=3k^2$. Now suppose $3|n_1$ and set $n_1=3n_2$. Then $n_2(12n_2-1)=k^2$. Since $n_2$ and $12n_2-1$ are coprime, both have to be squares. But $12n_2-1$ is never a square because $-1$ is not a quadratic residue modulo $4$. 
Hence we must have $3\nmid n_1$. But then $3|4n_1-1$, so $n_1\equiv 1\mod{3}$. Set $n_1=3m+1$. 
The equation is now $(3m+1)(4m+1)=k^2$. Since $3m+1$ and $4m+1$ are coprime, $3m+1=k_1^2$ and $4m+1=k_2^2$. 
Now just go through the first few squares to find some solutions. The smallest solution is $k_1=13$, $k_2=15$, $m=56$ which gives $n=337$.
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