Square of Wieferich prime is Poulet number

Statement

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

$2^{p-1}\equiv 1(\mathrm{mod}{p}^2)$

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 $2^{p-1}\equiv 1(\mathrm{mod}{p}^2)$

To prove: $2^{p^2-1}\equiv 1(\mathrm{mod}{p}^2)$

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 $2^{p^2-1}\equiv 1(\mathrm{mod}{p}^2)$.