Proving $n(n-2)(n-1)^2$ is divisible by 12 I want to show that $n^4 - 4n^3 + 5n^2 -2n$ is divisible by $12$ whenever $n>0$.
I reduced this to $n(n-2)(n-1)^2$. My approach has been to check divisibility by $3$ and $4$.
In both cases, squares are always congruent to $0$ or $1$. So
\begin{align*}n^2 &\equiv_3 0, 1 \\ \implies (n-1)^2 &\equiv_3 n-2, n-1\end{align*}
Given $-1\equiv_3 2$, then $(n-1)^2\equiv_3 n-2$ or $(n-1)^2\equiv_3 n-1$. But now I'm stuck here.
I know the polynomial is made up of a product of 3 consecutive numbers, so either two of them are even and one odd, or two of them are odd and one even. When I made the assumption $n$ is even, then $n-1$ is odd and $n-2$ is even, eventually landing at $n(n-2)(n-1)^2 \equiv_4 n(n-2)$ but so far nothing helpful has come up.
What am I missing?

If $n$ is even, $(n-2)$ is also even, and $n(n-2)$ is divisible by 4
If $n$ is odd, $(n-1)$ is even, and $(n-1)^2$ is divisible by 4.
