When is $1^5 + 2^5 + \ldots + n^5$ a square? When is $1^5 + 2^5 + \ldots + n^5$ a square? I found that this happens sometimes: $n=13$ gives $1001^2$, $n=133$ gives $9712992^2$ and $n=1321$ gives $942162299^2$.
I feel that the identity$$\displaystyle\sum_{i=1}^n i^5 = \tfrac{1}{12}[2n^6+6n^5+5n^4-n^2]$$ will be useful, since it's all square powers except one.. but I see no way to connect that.

$$1^5 + 2^5 + ... + n^5 = \frac{1}{12} n^2 (n+1)^2 (2 n^2+2 n-1) $$
Now we need to solve $$\frac{1}{12} n^2 (n+1)^2 (2 n^2+2 n-1) $$ is a perfect square
$$\iff\frac{2 n^2+2 n-1}{12} = k^2$$
[do some thing here] I don't know how to solve more
