<s>[INST] 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.
 [/INST] Hint $\ $ Specialize $\ c,a,b = 4,3,2\ $ below
Lemma $\ \ 5\nmid n\pm 1,\,\ ab\mid n,\,\ \color{#c00}{b^2\!\mid 5c\!-\!4}\ \Rightarrow\ {\rm lcm}(5,a^2,b^4)\mid n^2(n^2\!+5c\!-\!4)$
$\begin{eqnarray}{\bf Proof}\quad\! &&a\mid n\,\Rightarrow\ a^2\!\mid n^2,\ \ \ {\rm and}\ \ \ \  b\mid n\,\Rightarrow\, b^2\mid n^2,\ n^2\!+\color{#c00}{5c\!-\!4}\\\ \\
&&5\nmid n\pm1\,\Rightarrow\, 5\mid n\ \ {\rm or}\ \  5\mid \color{#0a0}{n\pm 2}\,\Rightarrow\,5\mid n^2\ \ {\rm  or}\ \ 5\mid \!\!\!\!\underbrace{(n^2-4)}_{\large\color{#0a0}{(n-2)(n+2)}}\!\!\!\! + 5c\end{eqnarray}$
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