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.
 A: HINT (since this is worth solving yourself): You need to use that $n-1, n+1$ are prime. If either of these are equal to $5$, you have the excluded values $n=6$ or $n=4$.
Otherwise, note that the product of any five successive integers is divisible by $5$. You are given $n+1$ and $n-1$ - can you see which five successive integers you might use?

Further to comments below, and to fill out a solution. We note that the conditions mean that $(n+1)$ and $(n-1)$ have no factors $2,3,5$ and that $720 = 16\times 9 \times 5$.
The product of three successive integers $(n-1)n(n+1)$ is divisible by $3$, so $n$ is divisible by $3$ and $n^2$ is divisible by $9$.
Also $n$ must be even, so $n^2$ and $n^2+16$ are both divisible by $4$ and we have a factor $16$.

 One of the five consecutive numbers $n-2, n-1, n, n+1, n+2$ is divisible by $5$. It is not $n-1$ or $n+1$. If it is $n$ we are done - we have the factor $5$ we need. Else $(n+2)(n-2)=n^2-4=n^2+16-20$ is divisible by $5$, whence the same is true of $n^2+16$ and again we are done.


It is also possible to do this with congruences, of course, but I have always liked this kind of method, so I look out for it.
A: Hint $\ $ Specialize $\ c,a,b = 4,3,2\ $ below
Lemma $\ \ 5\nmid n\pm 1,\,\ ab\mid n,\,\ \color{#c00}{b^2\!\mid 5c\!-\!4}\ \Rightarrow\ {\rm lcm}(5,a^2,b^4)\mid n^2(n^2\!+5c\!-\!4)$
$\begin{eqnarray}{\bf Proof}\quad\! &&a\mid n\,\Rightarrow\ a^2\!\mid n^2,\ \ \ {\rm and}\ \ \ \  b\mid n\,\Rightarrow\, b^2\mid n^2,\ n^2\!+\color{#c00}{5c\!-\!4}\\\ \\
&&5\nmid n\pm1\,\Rightarrow\, 5\mid n\ \ {\rm or}\ \  5\mid \color{#0a0}{n\pm 2}\,\Rightarrow\,5\mid n^2\ \ {\rm  or}\ \ 5\mid \!\!\!\!\underbrace{(n^2-4)}_{\large\color{#0a0}{(n-2)(n+2)}}\!\!\!\! + 5c\end{eqnarray}$
A: $n=6k$ 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 $9k^2+4\equiv -k^2+4 \equiv 0 \bmod 5$.
