Question on Fermat numbers and Wiefrich primes Theorem 6.25. in the book 17 Lectures on 
Fermat NUmbers From Number Theory to Geometry states:
If there exist only finitely many Wieferich primes, then there 
exist infinitely many Fermat numbers that are not powerful. 
Does that mean if someone proves that Fermat Numbers are square free then it is proven that  there are only finitely many Wieferich primes?
Thank you
 A: No, it doesn't. The proposition
"If there exists only finitely many Wieferich primes, then there exists infinitely many Fermat numbers that are not powerful"
is still true when the conclusion is true but the premise is false.
A: Fermat numbers are pairwise coprime. So given one prime number $p$ (either Wieferich or non-Wieferich), there are only two possibilities: Either:


*

*$p$ divides one Fermat number $F_m$ but divides no other Fermat numbers, or

*$p$ divides no Fermat numbers at all


For example $114689$ (non-Wieferich) divides $F_{12}$ and no other Fermat numbers, while $114691$ (non-Wieferich) divides no Fermat number.
(The two known Wieferich primes are in case (2) above, so they do not divide Fermat numbers, but it is conceivable that other Wieferich primes exist that belong to case (1).)
Now consider a prime $p$ and a Fermat number $F_m$ that are so related (i.e. we assume $p|F_m$), then the following two statements are equivalent:


*

*$p^2|F_m$, the square of $p$ divides the associated Fermat number

*$p$ is a Wieferich prime



It is thought that infinitely many non-Wieferich primes exist, and it is likewise thought that infinitely many Wieferich primes exist.
Suppose the first of these conjectures is false. Then all primes, with a finite set of exceptions, are actually Wieferich. It follows that all but finitely many Fermat numbers are entirely composed of Wieferich primes. That means that all but finitely many Fermat numbers are powerful, i.e. on the form $a^2b^3$ for some integers $a,b$. This seems very unlikely, but nobody has disproved it. It would also follow that almost all (only finitely many excpetions) pernicious Mersenne numbers are powerful (in particular composite).
Until now, no Fermat number is known that is not square-free, so just one single instance of a powerful Fermat number would be a tremendous surprise.
Let us turn to the other conjecture, on the infinitude of the Wieferich primes. Suppose that conjecture turns out to be false. Then all, with the possible exception of finitely many, Fermat numbers are square-free. In particular, if $\{1093,3511\}$ is the full set of Wieferich primes, then absolutely all Fermat numbers are square-free (since $1093$ and $3511$ belong to case (2) above).
Remember that even if infinitely many Wieferich primes exist, as conjectured, that does not say anything about how many of them will divide Fermat numbers (or divide pernicious Mersenne numbers).
You:

Does that mean if someone proves that Fermat Numbers are square free
  then it is proven that there are only finitely many Wieferich primes?

No. If someone proves that all Fermat numbers are square-free, the implication is only that every Wieferich prime fails to divide a Fermat number. I.e. it would follow that every Wieferich prime is in case (2) above.
A: After doing some research I have found http://en.wikipedia.org/wiki/Wieferich_prime#Connection_with_Mersenne_and_Fermat_primes. There it is stated that 

" if there are only finitely many Wieferich primes, then there will be at most finitely many Mersenne numbers that are not square-free. Rotkiewicz showed that the converse is also true, that is, if there are infinitely many square-free Mersenne numbers, then there are infinitely many non-Wieferich primes.[37]
  "

and Theorem 6.26 from the book states if Fermat numbers are square free, so are Mersenne Numbers (or at least this is what I Understood).
 So I take it as: "If Fermat Numbers are square free, then Mersenne Numbers are square free, then there are infinitely many non-Wieferich primes!
Is my understanding valid?
