If gcd$(a,2)=1,$ and $p$ is an odd prime, then $(a^{p-1})-1$ is divisible by $8$.

Question

Prove that If gcd$(a,2)=1,$ and $p$ is an odd prime, then $(a^{p-1})-1$ is divisible by $8$.

Answer 1

This is true if $p$ is ANY odd integer infact, $p$ does not have to be prime.

Note that $c^2 \equiv_8 1$ if $c$ is odd; it is easy to check, check $c \equiv_8 1,3,5,7$. Note that $a^{p-1}$ is an odd square for any odd $p$ and thus even $p-1$ i.e., $a^{p-1} = c^2$ for some odd $c$. So what must $a^{p-1}$ be mod 8. Can you finish from here.