# Chances to refute : Mersenne numbers with prime exponents are squarefree$\$?

Suppose, a third Wieferich prime will be found.

Can we estimate the chance that this new Wieferich-prime will refute the conjecture that a Mersenne number $$\ 2^q-1\$$ with $$\ q\$$ prime is always squarefree ?

It is well known that $$\ p^2\mid 2^q-1\$$ with primes $$\ p,q\$$ implies that $$\ p\$$ must be a Wieferich prime and that the two known Wieferich primes cannot satisfy $$\ p^2\mid 2^q-1\$$.

If $$\ p^2\mid 2^q-1\$$ , then the order of $$\ 2\$$ modulo $$\ p^2\$$ must be $$\ q\$$.

Hence a third Wieferich prime $$\ p\$$ would refute the above conjecture if and only if the order of $$\ 2\$$ modulo $$\ p^2\$$ is a prime number , which is not the case for the known Wieferich-primes. What is the chance this is the case , if the third Wieferich prime has , lets say , $$\ 20\$$ digits ?

Also : How is the situation in the case of Fermat-numbers ? Can we hope that a third Wieferich prime will refute that Fermat-numbers are always squarefree ?

• Don't add additional parts of the question to comments - add it to the question. – Thomas Andrews Apr 11 at 15:54
• @ThomasAndrews Done – Peter Apr 11 at 15:55
• >This also implies that Wieferich primes can be defined as primes $p$ such that the multiplicative orders of 2 modulo $p$ and modulo $p^2$ coincide -wikipedia – Roddy MacPhee Apr 12 at 12:48
• It would be nice if the downvoter would give a reason for the downvote. – Peter Apr 13 at 7:49