Odd divisibility induction proof Prove that for odd $n>3$ 
$$64\ | \ n^4-18n^2+17$$
I checked that for $n=5$ it works. I think I need to assume that for $2n+1$ it holds and show that $2n+3$ also holds. Any ideas?

If $n=2k+1$, we have
$$
\begin{align}
n^4-18n^2+17
&=(2k+1)^4-18(2k+1)^2+17\\[6pt]
&=16k^4+32k^3-48k^2-64k\\
&=384\binom{k}{4}+768\binom{k}{3}+320\binom{k}{2}-64\binom{k}{1}\\
&=64\left[6\binom{k}{4}+12\binom{k}{3}+5\binom{k}{2}-\binom{k}{1}\right]
\end{align}
$$
This indicates that $64$ divides $n^4-18n^2+17$ for all odd $n$.
