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.

If $n$ is divisible by five, then $n^2(n^2+1)(n^2-1)$ is divisible by $5$, since $n^2$ is a factor.
Now suppose that $n$ is not divisble by $5$. Then $n=5m+r$ for some $r=1,2,3,4$. Note that $(n^2 + 1)(n^2-1) = n^4-1$.
$$n^4 = (5m+r)^4 = (5m)^4 + 4 (5m)^3r + 6 (5m)^2 r^2 + 4 (5m) r^3 + r^4$$ by the binomial theorem. We see then that $n^4 = 5\cdot M + r^4$ for some $M$. Now we just need to show that $r^4-1$ is divisible by $5$ to establish that $n^4-1$ is divisible by $5$.
This does follow by Fermat's little theorem. If that sounds foreign to you, it's easy enough to check by hand. $r$ can take four different values, $1,2,3,4$ (remember we already assumed that $r\neq 0$ since we are assuming $n$ is not divisble by $5$).
$$1^4 -1 = 0$$
$$2^4 -1= 15$$
$$3^4 - 1 = 80$$
$$4^4 - 1 = 255$$
Thus we see that, if $n$ is not divisible by $5$, then $$n^4 -1 = 5 M + r^4 - 1$$ and both $5M$ and $r^4 - 1$ are divisible by five. Therefore, $n^4-1$ is divisible by five, and $$n^2(n^2+1)(n^2-1) = n^2(n^4-1)$$ is divisible by $5$.
