Find a criterion such that $\displaystyle\sum_{i=1}^ni$ divides $\displaystyle\prod_{i=1}^ni^2$ for $n\in\mathbb N$.
What I have done so far, $\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}{2}$ and $\displaystyle\prod_{i=1}^ni^2=(n!)^2$. So the problem reduces to for which $n\in\mathbb{N}$, $\frac{n(n+1)}{2}| (n!)^2$.
$\displaystyle\frac{n(n+1)}{2}| (n!)^2\Leftrightarrow \frac{n+1}{2}\mid (n-1)!n!\Leftrightarrow (n+1)\mid (n-1)!n!$ (because $n$ or $(n-1)$ is even)
But I'm stuck at this point, I don't know if the $n$ could be charactherized in a more "familiar" form. The only thing that I can say so far is if $n+1$ is prime, there is no divisibility. Are there another interpretation with my result? Is there an improvement to my result? Is another way to prove the wanted criterion?
$n+1$ has a unique prime representation, say: $n+1 = \prod_{i=1}^t p_i $ (the $p_i$'s aren't necessarily distinct). Certainly none of the $p_i$'s can be $n$ or $n-1$. They must all be smaller than that; but this just means that...