<s>[INST] 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?
 [/INST] we have:
$$n^4-18n^2+17+64=(n^2-9)^2 $$
and because $(n^2-9)=(n-3)(n+3)$ is divisible by $8$ for odd numbers we can conclude.

By induction:
Assume that $n^4-18n^2+17=(n^2-9)^2-64$ is divisible by $64$ for an odd number $n$ we want to prove that this still true for $n+2$, we have
$$\begin{align}(n+2)^4-18(n+2)^2+17+64&= ((n+2)^2-9)^2\\
&=(n^2-9+4(1+n))^2 \\
&=(n^2-9)^2+2.4(1+n)(n^2-9)+2.16.(1+n)^2\end{align}$$
and because $n$ is odd we have $2$ divides $(1+n)$, and by induction hypothesis $8$ divides $(n^2-9)$ and $64$ divides $(n^2-9)^2$ which implies the result.
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