Proving $n^2(n^2+16)$ is divisible by 720 Given that $n+1$ and $n-1$ are prime, we need to show that  $n^2(n^2+16)$ is divisible by 720 for $n>6$.
My attempt:
We know that neither $n-1$ nor $n+1$ is divisible by $2$ or by $3$, therefore $n$ must be divisible by both $2$ and $3$ which means it must be divisible by $6$.
So, $n = 6k$ and since $n>6$ we must have $k>3$. So the expression becomes $36k^2(36k^2+16) = 144k^2(9k^2+4)$
...
But then I get stuck.
Could someone please guide me towards a solution.Thanks.

$n=6k$ implies $n\equiv k \bmod 5$. If $k\equiv 0 \bmod 5$, then you're done.
Othwerwise, $n-1$ and $n+1$ prime implies $k\equiv n\equiv \pm 2 \bmod 5$ and so $9k^2+4\equiv -k^2+4 \equiv 0 \bmod 5$.
