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.

With $n:=2m+1$, the polynomial is
$$(2m)^2(2m+2)^2(4m^2+4m+2)^2(16m^4+32m^3+24m^2+8m+2)
\\=2^7m^2(m+1)^2(2m^2+2m+1)(8m^4+16m^3+12m^2+4m+1).$$
The last two factors are always odd and the remaining ones form at least a multiple of $2^9$.

More precisely, if the multiplicity of the factor $2$ in the prime factorization of $\dfrac{n-1}2$ is $p>0$, then $k=2p+7$, otherwise $k=2q+7$ where $q$ is the multiplicity of $2$ in $\dfrac{n+1}2$.
E.g., with $n=15$, $p=0,q=3$ and $k=13$. Indeed, $15^{12}-15^8-15^4+1=2^{13}\,15837863153$, and with $n=17$, $p=3$ and $17^{12}-17^8-17^4+1=2^{13}\,71120027025$.
