Squarefree parts of Mersenne numbers The $n$-th Mersenne number is $M_n=2^n-1$. Write $M_n=a_n b_n^2$ where $a_n$ is positive and squarefree. 
Question 1: What lower bound can be proved for $a_n$?
Let $A$ be the set of all possible $a_n$. The natural density of $A$ is defined as
$$
\delta_A=\lim_{X \rightarrow \infty} \frac{\# \{a \in A | a \le X\}}{X}.
$$
Question 2: What can be proved about $\delta_A$? Is it possible to show that $\delta_A=0$?
Note: I am interested in unconditional answers to the above questions. It is easy to give answers conditional on the ABC conjecture. Indeed, the ABC conjecture shows that for any $\epsilon>0$ there is some $K_\epsilon>0$ such that
$$
a_n \ge K_\epsilon \cdot 2^{n(1-\epsilon)}.
$$
Thus $\# \{ a \in A | a \le X\}=O(\log(X))$, which gives $\delta_A=0$.
 A: Here's a simple proof using user43383's idea and a recent result of Andrew Granville: http://arxiv.org/abs/1212.6306
According to Theorem 1 of that paper, $2^n-1$ always has a primitive prime factor $p$ that occurs to an odd exponent, except when $n=1$ or $n=6$ (where there are no primitive prime factors at all). Primitive here means that $2$ has order $n$ modulo $p$. In particular, $p\equiv 1\pmod{n}$. Clearly, $p \mid a_n$. So if $q_n$ denotes the smallest prime congruent to $1$ modulo $n$, then $a_n \geq q_n$ for every $n \neq 1, 6$. And in fact, by a direct check, $a_6 = 7 = q_6$, so $a_n \geq q_n$ for every $n > 1$.
Trivially, $q_n \geq n+1$ for every $n$. In fact, it is usually much larger. The simple fact that the primes have density zero implies that for every positive integer $K$, one has $q_n > Kn$  apart from a set of $n$ of density zero. (To see this, note that if this inequality fails, then one of $n+1$, $2n+1$, $\dots$, or $(K-1)n+1$ is prime, and each of these conditions puts $n$ in a set of density zero.) 
Hence: $a_n \geq n+1$ for all $n > 1$, and for every fixed $K$, $a_n > Kn$ for all $n$ outside of a set of density zero. These two facts are enough to imply that $\{a_n\}$ itself has density zero.
(Proof of the last bit: Fix $K$. Given a large $x$, if $a_n \leq x$, then the inequality $a_n \geq n+1$ forces $n \leq x$. If $n \leq x$ and $a_n \leq x$, then either $n < x/K$ or $a_n \leq Kn$. The former holds for at most $x/K$ values of $n$, and the latter holds only for $o(x)$ values of $n$, as $x\to\infty$. Thus, the number of distinct $a_n$ with $a_n \leq x$ is at most $x(1/K+o(1))$; so the upper density of $\{a_n\}$ is at most $1/K$. But this holds for all $K$.) 
A: I think I can prove the bound $a_n> \prod 2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ if $n=\prod p_i^{\alpha_i}$. It is certainly very far of the real $a_n$, but it is more than enough to prove $\delta_A=0$.
First a lemma:
$\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ is never a square
Proof: If $p=2$ it is obviously true. Wlog $p$ is odd. 
We actually want to analyze the polynomial $f(x)=\frac{x^p-1}{x-1}$ for $x=2^{p^{d-1}}$. We will prove actually that this polynomial is never a square for "big" $x$. The idea is approximating $\sqrt{f(x)}$ by a polynomal with rational coefficients.
Notice that $\sqrt{f(x)}<\sqrt{\frac{x^p}{x-1}}=x^{\frac{p-1}{2}}\sqrt{\frac{x}{x-1}}=x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}$ (by the generalized binomal theorem).
So $\sqrt{f(x)}-x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k} < x^{\frac{p-1}{2}}\sum\limits_{k=\frac{p+1}{2}}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}<\frac{\binom{p+1}{\frac{p+1}{2}}}{2^{p+1}}\frac{1}{x-1}$ (substituting all denominators for the first one, which is greater).
Also, the left side can't be zero, because all the coefficients of $(x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k})^2$ are at most $1$ (the first ones are exactly $1$, while the last ones are strictly less than $1$).
So if $\sqrt{f(x)}$ is an integer, the left side will be at least $\frac{1}{2^{p-1}}$ (because all denominators divide $2^{p-1}$). So, if $f(x)$ is a perfect square,  $x-1<\frac{\binom{p+1}{\frac{p+1}{2}}}{2}<2^{p}-1$.
It indeed doesn't happen for $x=2^{p^{d-1}}$ for $d>1$. For $x=2$, $f(2)$ is clearly not a square because $f(2) \equiv 3$ mod $4$.
Now back to the bound.
Let $p^d$ a prime power factor of $n$. Notice that $gcd (\frac{2^{p^d}-1}{2^{p^{d-1}}-1},2^{p^{d-1}}-1)$ must divide $p$, (in general $gcd(\frac{x^p-1}{x-1},x-1)|p$, because $\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+...+1 \equiv p \,(mod \,(x-1))$) therefore this gcd is $1$ (because clearly $p \nmid 2^{p^{d-1}}-1$).
Now we take a prime $q$ with odd exponent in $\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ (such prime exists because this expression is not a square). Since $q$ doesn't divide $2^{p^{d-1}}-1$ (by the earlier $gcd$ condition), $q$ satisfies $ord_q(2)=p^d \Rightarrow 2p^d|q-1 \Rightarrow q>2p^d$. Notice that this prime $q$ is a factor of $a_{p^d}$.
Repeating for $d-1$, $d-2$, ..., $1$ we get that $2^{p^d}-1$ has at least $d$ distinct prime factors that divide it with odd exponent, whose product is at least $2^d p^{\frac{d(d+1)}{2}}$.
Since $2^{p_i^{\alpha_i}}-1|2^n-1$ and all $2^{p_i^{\alpha_i}}-1$ are pairwise coprime, we proved the bound previously stated. (notice that the factor $2^{\alpha_1}$ actually doesn't appear for $i=1$ ($p_1=2$), but it won't make a big difference).
Now, for the proof that $\delta_A=0$, we will use that the bound we proved gives $a_n>n 2^{\omega(n)-1}$. This is because $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}\ge 2 p_i^{\alpha_i}$ (for $i=1$, since we don't have the factor $p_i^{\alpha_i}$ we use only $p_1^{\frac{(\alpha_1)(\alpha_1+1)}{2}}\ge p_1^{\alpha_1}$, what explains the $-1$ after $\omega(n)$)
Now, set some fixed $t$:
\delta_A=$\lim\limits_{n \to \infty} \frac{\#\{ a_k \in A | a_k<n\}}{n} = \lim\limits_{n \to \infty} \frac{\#\{a_k \in A | a_k<n, \omega(k)<t\}}{n} + \frac{\#\{a_k \in A | a_k<n, \omega(k)\ge t\}}{n}$
By http://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem, $\frac{\#\{a_k \in A | a_k<n, \omega(k)<t\}}{n}$  goes to zero, because by our bound, if $a_k<n$, then $k<n$, and one of the consequences of the theorem presented is that the set of numbers with $\omega(k)<t$ has natural density zero.
By our bound $\#\{a \in A | a<n, \omega(a)\ge t\}$
 is at most $\frac{n}{2^{t-1}}$, because it has no elements $a_k$ with $k>\frac{n}{2^{t-1}}$ otherwise $a_k>k2^{\omega(k)-1}>\frac{n}{2^{t-1}}2^{t-1}=n $.
Therefore for each fixed $t$, $\delta_A\le\frac{1}{2^{t-1}}$, so we prove that indeed $\delta_A=0$.
(notice that the factor $2^{\alpha_i}$ in $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ is very important, otherwise we wouldn't be able to handle squarefree $n$)
A: Le Maohua, “On Mersenne Numbers” [in Chinese], Journal of Jishou University (Natural Science Edition) 20(1) (March 1999): 17–19, shows that the squarefree part of $2^p - 1$, with $p \ge 11$ prime, is greater than $(\pi{}p/log\thinspace{}p)^2$.
(From here to end added in 2023:) Since Le’s paper is not widely available, I will attempt some explanation of it, working from notes in English that were provided to me privately and which I am not in a position to share.
Le was apparently the first to notice the relevance to this problem of a paper of Wilhelm Ljunggren, “Über die Gleichungen $1 + Dx^2 = 2y^n$ und $1 + Dx^2 = 4y^n$,” Det Kongelige Norske Videnskabers Selskab Forhandlinger 15(30) (1942): 115–118, on the Diophantine equation $1 + Dx^2 = 4y^n$. The case $D = a$, $x = b$, $y = 2$, and $n - 2 = p$ (with $p$ prime) corresponds to a Mersenne number factorization in the form proposed in the original post above, $M_p := 2^p - 1 = a \cdot b^2$. Now if $h(\cdot)$ represents the class number of the imaginary quadratic number field $\mathbb{Q}\left(\sqrt{\cdot}\right)$, then by Gauss’s theory of classes, $h_p(-M_p) = h(-b^2a) = h(-a)$. Ljunggren showed that $p - 2$ divides the class number $h(-a)$, which is in itself a notable result.
After verify the nonexistence of a square divisor in published factorizations of $M_p$ for the small cases $11 \le p < 101$, Le uses Ljunggren’s result and estimates of the class number by Louboutin and others to establish a lower bound on the squarefree part of $M_p$ in the remaining cases.
