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 ?

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    $\begingroup$ Don't add additional parts of the question to comments - add it to the question. $\endgroup$ – Thomas Andrews Apr 11 at 15:54
  • $\begingroup$ @ThomasAndrews Done $\endgroup$ – Peter Apr 11 at 15:55
  • $\begingroup$ >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 $\endgroup$ – Roddy MacPhee Apr 12 at 12:48
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    $\begingroup$ It would be nice if the downvoter would give a reason for the downvote. $\endgroup$ – Peter Apr 13 at 7:49

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