If Wieferich primes are finite...Then what? I am wondering if $1093$ and $3511$ are the only Wieferich Primes, then what would it imply? (A Wieferich Prime is a prime satisfying the congruence $2^{p-1}\equiv 1\pmod {p^2}$). I know of 3 cases:


*

*There are infinitely many "non-Wieferich" primes...

*Fermat Numbers are square-free

*Mersenne Numbers having a prime exponent are square-free.


Does the finiteness have more consequences?
Thank you.
 A: A good survey about this question is the article Wieferich Past and Future by Nicholas M. Katz. In section $1$ the consequences regarding FLT are discussed. However, since there is a proof of FLT the motivation for going further there has disappeared. However, other aspects related to elliptic curves, abelian varieties and semiabelian varieties are still of interest.
A: Let $M_n=2^n-1$ be the $n$th Mersenne number, and $F_n=2^{2^n}+1$ be the $n$th Fermat number.
$$M_{2^{n}}=2^{2^{n}}-1=\prod_{k=0}^{n-1}(2^{2^{k}}+1)=\prod_{k=0}^{n-1}F_{k}.\tag{1}$$
$$F_{n}=\frac{M_{2^{n+1}}}{M_{2^{n}}}=\frac{2^{2^{n+1}}-1}{2^{2^{n}}-1}=2^{2^{n}}+1.\tag{2}$$
The Mersenne numbers form a divisibility sequence, i.e., if $M_m\mid M_n$ then $m\mid n$. This implies, since $2$ is prime, that Mersenne numbers $M_{2^k}$, with a subscript a power of $2$ only inherit factors from those preceding Mersenne numbers, $M_{2^{k-j}}$, $1\leqslant j\leqslant k-1$, having subscripts being a lesser power of $2$.  These inherited factors are seen by (1) to be the Fermat numbers. 
Now (2) shows us the primitive prime factors of $M_{2^{n+1}}$ (by primitive we mean the primes that have never appeared before in the sequence ($M_n$)) are exactly those that make up the prime factorisation of $F_n$. 
Hence if $1093$  and $3511$ are the only Wieferich primes, then Fermat numbers are square-free which immediately implies Mersenne numbers of the form $M_{2^n}$ are also square-free.
