<s>[INST] Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$ Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$.
I am rather weak in number theory so I hope the get some feedback if my solution is correct.
 [/INST] It's $\,(\color{#0a0}{n^4\!+\!1})(\color{#c00}{n^4\!-\!1})^2$ so $\,n\,$ odd $\,\Rightarrow\begin{align}&\ \,2\mid \color{#0a0}{n^4\!+\!1}\\ &2^4\mid \color{#c00}{n^4\!-\!1}\end{align}\,$ since $\,\begin{align} n&\equiv \pm1\ \ \pmod{\!4}\\\Rightarrow \color{#c00}{n^4}&\equiv \color{#c00}{(\pm1)^4}\!\!\!\!\pmod{\!4^2}\end{align}$
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