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.
Hint:
$$
n^2(n^2+1)(n^2−1)\cong n^2(n^2-4)(n^2−1) = (n-2)(n-1)(n^2)(n+1)(n+2)
$$
Lemma:

For any polynomial of integer coefficients $P(n)$, $n=0,1,2,3,4$ exhaust the possible values of $P(n)\bmod 5$.

Indeed for any $a$, $(n+a)\bmod 5=(n\bmod5+a)\bmod5$, and $(na)\bmod5=((n\bmod5)a)\bmod5$, so that $P(n)\bmod5=P(n\bmod 5)\bmod5$. QED.

Now,
$$0^2(0^4-1)\bmod5=0\\
1^2(1^4-1)\bmod5=1\cdot0\bmod5=0\\
2^2(2^4-1)\bmod5=4\cdot15\bmod5=0\\
3^2(3^4-1)\bmod5=9\cdot80\bmod5=0\\
4^2(4^4-1)\bmod5=16\cdot255\bmod5=0\\$$