Prove that $p^2$, with $p$ prime, is a pseudoprime to the base $b$ if and only if $b^{p-1} \equiv 1 \pmod{p^2}$.

Question

I have to prove that $p^2$, with $p$ prime, is a pseudoprime to the base $b$ iff $b^{p-1} \equiv 1\pmod{p^2}$.

I don´t have any idea, so ¿could someone helps me please?

Answer 1

I have just tried the following:

By definition, $p^2$ is a pseudoprime to base $b$ iff $b^{p^{2}-1} \equiv 1 \pmod{p^2}$. On the other hand, by Euler's theorem $b^{p^{2}-p} \equiv 1 \pmod{p^2}$ for all $b$ with $(b,p^2)=1$. So $p^2$ is a pseudoprime to base $b$ iff $ 1 \equiv b^{p^{2}-1} \equiv b^{p^{2}-1}(b^{p^2-p})^{-1} \equiv b^{p-1} \pmod{p^2}. $