Square of Wieferich prime is Poulet number

Statement

Suppose $p$ is a Wieferich prime, i.e., a prime number such that:

$2^{p-1}\equiv 1{\pmod {p^{2}}}$

Then, $p^{2}$ is a Poulet number (also called Sarrus number), i.e., a Fermat pseudoprime to base 2.

Proof

Given: $p$ is a prime such that $2^{p-1}\equiv 1{\pmod {p^{2}}}$

To prove: $2^{p^{2}-1}\equiv 1{\pmod {p^{2}}}$

Proof: We have:

$\!p^{2}-1=(p-1)(p+1)$

Thus, $p-1$ divides $p^{2}-1$. This gives that:

$2^{p-1}-1\mid 2^{p^{2}-1}-1$

Combining this with the given information, we get that $2^{p^{2}-1}\equiv 1{\pmod {p^{2}}}$.