Show that if $n = \frac{a^{2p}-1}{a^2-1}$ where $a$ is an integer $>1$ and $p$ is an oddprime that doesn't divide $a(a^2-1)$, then $n$ is pseudoprime to the base $a$. Conclude that there are infinitely many pseudoprimes to any base $a$.
This seems to be similar to Fermat's Little Theorem: We would usually have to prove $a^{p-1} \equiv 1$ (mod p), but in this case it seems to replaced the p with n (which seems to contain p either way). My question is, how would I go about proving this?
So far, I have if $2p|(n-1)$ and $a^{2p}\equiv 1$ (mod n) , then we can conclude that $a^{n-1} \equiv 1$ (mod n).
Since $p$ is odd, $2p$ should make it even, we can also see that $(n-1) = \frac{a^{2p}-1}{a^2-1}-1$
I'm not entirely sure what path to take after this.