Wieferich prime is a prime number $p$ such that $p^2$ divides $2^{p - 1} - 1$.

There are only two Wieferich primes known and it is an open problem if
there are infinitely many non-Wieferich primes.

Mersenne numbers are $M_n=2^n-1$.

> When is a prime factor of Mersenne number Wieferich prime?

[Wikipedia claims](https://en.wikipedia.org/wiki/Wieferich_prime#Connection_with_Mersenne_and_Fermat_primes)

> A prime divisor $p$ of $M_q$, where $q$ is prime, is a Wieferich prime if and only if $p^2$ divides $M_q$.

Wikipedia reference doesn't appear to work, but we found it in paper:

[link](http://math.colgate.edu/~integers/t19/t19.pdf)
PRIME POWER DIVISORS OF MERSENNE NUMBERS AND WIEFERICH PRIMES OF HIGHER ORDER, Ladislav Skula


Confusion is possible, but we believe that Wikipedia's claim implies
infinitely many non-Wieferich primes.

Assume the set of non-Wieferich primes is finite and let $P$
their product.

Then for all primes $q$, $M_q=d u$ where $d$ is divisor of $P$
and $u$ is product of powers of Wieferich primes with exponents at least $2$.

If all exponents are $2$, this implies that the squarefree free part of $M_q$ is bounded,
which is easily shown to be impossible
e.g. see [this queston](https://mathoverflow.net/questions/149511/squarefree-parts-of-mersenne-numbers)

> What is wrong with this argument?