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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?

J. W. Tanner
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Lex_i
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    $(n-2)(n-1)n\cdot (n-1)$ is divisible by $3$. If $n$ is odd, then $(n-1)$ is even and thus, $(n-1)^2$ is divisible by $4$. If $n$ is even, $n$ and $(n-2)$ are even and thus divisible by $4$. – Evil Witch Feb 01 '23 at 00:29
  • @Lex_i Regarding divisibility by $3$, instead of what you did, note that the product includes $3$ consecutive numbers (i.e., $n-2$, $n-1$ and $n$), as you already stated, so one of those consecutive numbers is a multiple of $3$. – John Omielan Feb 01 '23 at 00:32
  • By the first dupe it's divisible by $3$ and also by $4$ being $(\color{#c00}{n(n!-!1)})^2!-2\color{#c00}{n(n!-!1})^2$ and $,2\mid \color{#c00}{n(n!-!1)},$ (or examine odd/even cases as in the 2nd dupe) $\ \ $ – Bill Dubuque Feb 01 '23 at 03:49

2 Answers2

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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.

Lourrran
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Suppose $(n,12)=1.$ We use Euler and lil' Fermat.

$$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 2(n-1)\pmod {2^2},$$ and $$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 0\pmod 3.$$

So by CRT, using Bezout, we have $1×4×2×0-1×3×2(n-1)\equiv6(n-1)\equiv0\pmod{12}.$

Suppose $(n,12)\neq1.$ Then either $2\mid n$ or $3\mid n.$

If $2\mid n,$ then if $3\not \mid n$, we are done by the previous calculation and CCRT.

If $3\mid n,$ then $n^4-4n^3+5n^2-2n\equiv 0\pmod3.$ So we're done by CCRT.

If $2\not\mid n$, then $3\mid n,$ so we're back to Bezout, since we have already done the modular arithmetic: $$1×4×0-1×3×2(n-1)\equiv 0\pmod {12}.$$

suckling pig
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