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Let $M_n$ denote the Mersenne numbers $M_n=2^n-1$.

As $n$ varies, must $M_n$ be divisible by arbitrary large prime $p$ with exponent one, i.e. $p \mid M_n, p^2 \nmid M_n$?

In other words, must the exponential diophantine equation $2^n-1 = A x^2 y^3$ for constant $A$ has only finitely many solutions $(n,x,y)$ and $n >1$?

Related question is this.

Also related question that might show there are infinitely many non-Wieferich primes is here.

Looking for unconditional results, abc easily implies it.

Added

The paper Remarks on Exponential Congruences and Powerful Numbers P. RIBENBOIM on p7.

(M') There exist infinitely many Mersenne numbers which are not powerful.

(M') implies B_2 implies infinitely many non-Wieferich primes.

One easy way to construct infinitely many non-powerful Mersenne numbers is to observe that 3 divides $M_{6n+2}$ with exponent one, so $M_{6n+2}$ is not powerful.

Doesn't this approach give infinitely many non-Wieferich primes?

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    $\begingroup$ Regarding the latest addition on (M'): I think, the author defines Mersenne numbers as $2^p-1$ for some prime $p$, because the paper states that every prime factor of Mersenne number is primitive, which is true for prime exponents and false for arbitrary exponents. So no, the problem of proving the infinitude of non-Wieferich prime is still hard $\endgroup$ Jun 6 at 13:26
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This is at least as strong as the infinitude of non-Wieferich primes. Indeed, if there are only finitely many non-Wieferich odd primes and they are equal to $p_1, \ldots, p_k$, then one can choose $n$ to be divisible by $p_1(p_1-1)\ldots p_k(p_k-1)$. Then $2^n-1$ is automatically divisible by $p_i^2$ for $i\leq k$ and if $2^n-1$ is divisible by any other prime $q$, then $q$ is a Wieferich prime, so $q^2\mid 2^n-1$. Hence, for any such $n$ the number $2^n-1$ would be of the form $x^2y^3$.

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  • $\begingroup$ Thanks. Did you reduce the problem to $2^n-1=x^2 y^3$ with no other constants? $\endgroup$
    – joro
    Jun 4 at 15:09
  • $\begingroup$ @joro No, I did not. It's more of a special case: if your statement above holds for $A=1$, then there are infinitely many non-Wieferich primes. It is not obvious, however, if one can prove even that $2^n-1=x^2y^3$ has finitely many solutions, assuming that there are infinitely many non-Wieferich primes $\endgroup$ Jun 4 at 15:39
  • $\begingroup$ I edited with a subsequence of Mersenne numbers. $\endgroup$
    – joro
    Jun 4 at 16:36
  • $\begingroup$ @joro I don't get it, my sequence is just an infinite arithmetic progression $a(n)=n\prod p_i(p_i-1)$. Do you want $a(n)$ to have some other properties? $\endgroup$ Jun 4 at 17:36
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    $\begingroup$ @joro Isn't it obvious? $M_{6n+2}$ is divisible by $3$, but not by $9$ $\endgroup$ Jun 6 at 9:07

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