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In this page of Caldwell, after the proof of the link with Wieferich primes and pointing out that the two known ones can't be divisors of any Mersenne numbers, in the Comment he asserts «... so $M_q$ is square-free for all primes less than $4 \cdot 10^{12}$». I don't get this latter number, since previously it's said the Wieferich primes are checked up to $6.7\cdot 10^{15}$.

Falcon
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  • Maybe because the prime exponent must be smaller than a possible prime divisor. – Peter Jul 16 '23 at 19:24
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    Additionally , I read in wikipedia that the range upto $10^{17}$ is covered in the mean time. Anyway , we can say , if $p^2\mid 2^q-1$ with primes $p,q$ , then we have $p>6\cdot 10^{15}$. A similar statement is true for the Ferrmat numbers. Hence both Mersenne numbers with prime exponent and Fermat numbers have very little scope to be non-squarefree ! – Peter Jul 16 '23 at 19:26
  • The last update seems to be that there is neither a Wall sun sun prime nor a third Wieferich prime upto $2^{64}$. What this means for the prime exponent ? I do not know. I wonder whether the search limit for Wieferich primes to other bases (like $10$) has also be extended. – Peter Jul 20 '23 at 14:13
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    I really wonder how another Wieferich prime with an order with respect to base $2$ being a prime less than $4\cdot 10^{12}$ can be ruled out. – Peter Jul 20 '23 at 14:16

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