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.
 A: If we factor the power sum identity we find the problem equivalent to $$m^2 = \tfrac{1}{3}\left[\frac{n(n + 1)}{2}\right]^2(2n^2 + 2n - 1)$$ or $$m'^2 = \tfrac{1}{3}( 2n^2 + 2n - 1 )$$ this is an integer hence the $1/3$ implies  $n\equiv 1\pmod 3$ so let $n = 1 + 3n'$ to get $$m'^2 = 6n'^2 + 6n' + 1$$ but I don't know what now.
The RHS can be described by the recurrence relation: $a_0 = 1$, $a_{n+1} = a_n + 12 n$. Is there any theory about recurrences with '$n$' in them I might be able to use to tell when it takes on square values?
Completing the square like here gives $3(2n'+1)^2 - 2m'^2 = 1$ almost a pell equation with the condition one value is odd.
A: If $$a=1+2+3+\cdots+n$$ then
$$
1^5+2^5+3^5+\cdots+n^5 = \frac{4a^3-a^2}{3},
$$
so now the question is: when is that a square?
A: This is OEIS A031138, which lists some more and says $a(n) =11\cdot(a(n-1)-a(n-2)) + a(n-3) \\ a(n)=-1/2+((3-\sqrt 6)/4)\cdot(5+2\sqrt 6)^n+((3+\sqrt 6)/4)\cdot(5-2\sqrt 6)^n$
A: $$1^5 + 2^5 + ... + n^5 = \frac{1}{12} n^2 (n+1)^2 (2 n^2+2 n-1) $$
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
