<s>[INST] Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.

I was thinking of using induction, but wasn't really sure how to do it.
 [/INST] Here's a proof based on the fact that among any five consecutive numbers, one of them must be divisible by $5$.
Suppose that $n^2(n^2+1)(n^2-1)$ is not divisible by $5$.  Then either $n^2+2$ or $n^2-2$ is divisible by $5$.  Let's write this as $n^2+2\sigma=5k$, where $|\sigma|=1$.
But if $n^2(n^2+1)(n^2-1)$ is not divisible by $5$, then neither is $n(n^2-1)=n(n+1)(n-1)$, and therefore $5$ divides either $n-2$ or $n+2$.  Let's write this as $n+2\tau=5h$ with $|\tau|=1$.
Combining these cleverly, we can conclude $n^2-\sigma\tau n=5(k-\sigma\tau h)$, which implies $5$ divides $n(n-\sigma\tau)$, where $|\sigma\tau|=1$.  But that implies $5$ divides $n(n-1)(n+1)$, which is a contradiction.  Therefore $5$ must divide $n^2(n^2+1)(n^2-1)$.
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