# Are Wieferich primes Wieferich numbers?

In the article 'On a conjecture of Crandall concerning the $$qx+1$$ problem' by Franco and Pomerance, they define Wieferich primes to be prime numbers $$p$$ for which $$p^2|2^{p-1}-1$$ and Wieferich numbers to be odd positive integers $$q$$ for which $$q^2|2^{l(q)}-1$$, where $$l(q)$$ is the order of $$2$$ in the group $$(\mathbb{Z}/q\mathbb{Z})^*$$. They then mention that Wieferich primes are Wieferich numbers. But I don't see how $$p^2|2^{p-1}-1$$ implies that $$p^2|2^{l(p)}-1$$. It would make sense if it was $$\varphi(q)$$ instead of $$l(q)$$, which is an alternative definition, but this is not the one they gave. Is their statement true?

$$l(p)$$ is the least positive number $$l$$ that $$p|2^l-1$$.
Assume $$p^2\nmid 2^l-1$$. Then, $$p|2^{l(\frac{p-1}{l}-1)}+...2^l+1\Rightarrow p|1+...1=\frac{p-1}{l}$$ since $$2^l\equiv 1$$ mod $$p$$ and this leads to contradiction.