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.

In case $n$ is a multiple of $5$ the $n^2$ on the rhs is also.
If $n=2,3\pmod 5$ then $n^2\equiv4\pmod5$ , hence $n^2+1$ would be a multiple of $5$. 
In the remaining cases, i.e.,$n=1,4\pmod 5$ then in $n^2-1= (n+1)(n-1)$ one of the term on the right will be a multiple of $5$. 
So $n^2(n^2+1)(n^2-1)$ is always a multiple of $5$.
