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In the course of considering this Diophantine equation, I convinced myself that the following question is interesting:

If $n$ is large, must it be the case that the square-free part of $2^n-1$ is also large?

Let me elaborate on the setup of the question to remove any ambiguities.

Recall the fact that every natural number $n$ can be written uniquely as the product of some square-free number $k$ and some square $m^2$, i.e. $n=km^2$ where $p|k\Rightarrow p||k$.

(The following notation is not standard, but can be found at least on Wikipedia.)

Since every natural can be factored uniquely in this way, we can define a function sending naturals to their square-free parts. Define $\text{core}_2(n)=k$ when $n=km^2$ is the factorization into square-free and square parts. The question posed at the beginning of the post can now be rephrased as:

Is it true that $\lim_{n\rightarrow\infty}\text{core}_2(2^n-1)=\infty$?

To the best of my knowledge this is an open problem. Experimental evidence suggests that it should be true; it actually looks like $2^n-1$ itself should be square-free infinitely often, as shown by the following low quality graph:

Graph of ln(core_2(2^n-1))

It is also easy enough to prove that there is a subsequence whose square-free parts grow without bound ($2^{2^n}-1$ suffices for this), so we can say that the $\limsup \text{core}_2(2^n-1)$ tends towards infinty. But a simple proof for the $\liminf \text{core}_2(2^n-1)$ seems harder.

My two questions for you are as follows:

1) Does anyone know of a general theory or collection of results that can give us control over the square-free parts of integer sequences defined by an expression containing an exponential term? Everything I have managed to find is rather ad hoc or specifically about polynomial values.

2) Does anyone have a nice and simple argument to turn my factoid into a proper fact?

EDIT: Major revisions to the statement of the problem for the sake of clarification. Apologies to anyone who spent time thinking in a different direction than intended.

EDIT 2: As per the comments of Wojowu and Pasten, the full strength $abc$ conjecture is enough to resolve this question (as well as generalizations to other bases besides $2$). As per Gerry Myerson's comment, I have accidentally re-asked a question that has already been resolved on this site! You may find the relevant question and answer here.

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  • $\begingroup$ $2^n-1$ will be squarefree whenever $n$ is not divisible by $\text{ord}_{p^2}(2)$ for $p$ an odd prime. I would guess you can do some approximations with replacing $2$ with an arbitrary unit mod $p^2$ which is not $1$ mod $p$. $\endgroup$
    – user44191
    Commented May 1, 2018 at 21:40
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    $\begingroup$ This is very closely related to the $abc$ conjecture (for $a=2^n-1,b=1,c=2^n$). This gives very strong conditional lower bounds. I don't know much about unconditional bounds, but just showing that the squarefree parts go to infinity might have been proven. $\endgroup$
    – Wojowu
    Commented May 1, 2018 at 21:47
  • $\begingroup$ @Wojowu I'm not sure how to use the conclusion of the $abc$ conjecture to get at the lower bounds you are describing; can you elaborate on how the radical of $abc$ (which is really just the radical of $a$) can tell us about the square-free part of $a$? $\endgroup$
    – Richard Voepel
    Commented May 1, 2018 at 21:56
  • $\begingroup$ The crux is that in this case, the radical of $abc$ is just twice the radical of $a$. $\endgroup$
    – Wojowu
    Commented May 1, 2018 at 21:57
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    $\begingroup$ Previous question on this topic: mathoverflow.net/questions/149511/… See also mathoverflow.net/questions/203502/… although that one is just about $2^p-1$, $p$ prime. $\endgroup$
    – Gerry Myerson
    Commented May 1, 2018 at 22:38

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