Problem:

Show that $\displaystyle 8^{n}+1$ is composite for all $\displaystyle n\geq1$.

What I have so far:

In general,

$\displaystyle a^{n}+1 = (a+1)(a^{(n-1)} - a^{(n-2)} + a^{(n-3)} - \cdots + 1)$

but this only holds for $\displaystyle n$ odd.

Is there a way to factor when $\displaystyle n$ is even? Or would I have to find a solution specific to $\displaystyle a=8$?