TITLE: Can a Mersenne number ever be a Carmichael number?
QUESTION [36 upvotes]: Can a Mersenne number ever be a Carmichael number?
More specifically, can a composite number $m$ of the form $2^n-1$ ever pass the test: $a^{m-1} \equiv 1 \mod m$ for all intergers $a >1$ (Fermat's Test)?
Cases potentially proved so far: (That are never Carmichael numbers)

where $n$ is odd
where $n$ is prime

Work using "main" definition:
First off take the definition of a Carmichael number:

A positive composite integer $m$ is a Carmichael number if and only if  $m$
  is square-free, and for all prime divisors $p$ of $m$, it is true that $p -
1 \mid m - 1$.

Let's assume $m=2^n-1$ is squarefree. (Best case, and I believe it always is for $2^p-1$)
Take the case where $n$ (in $2^n-1$) is a prime $p$. All factors of $2^p-1$ must of the form: $2kp+1$ for some constant $k$. So will $2kp$ ever divide $2^p-2$? Factoring a $2$ out gives us $kp \mid 2^{p-1}-1$, or split into two: $k \mid 2^{p-1}-1$ and $p \mid 2^{p-1}-1$ must both be true. By Fermat's little theorem, $2^{p-1} \equiv 1 \mod p$, so $p \mid 2^{p-1}-1$ is always true. 
So if $k \mid 2^{p-1}-1$ for $k = {q-1 \over p}$, is false for at least one factor $q$ of $2^p-1$, no Carmichael numbers can exist of form $2^p-1$. 
Now for other cases where $n$ is composite, lets say $n=cp$, for some prime $p$, and some number $c$:
$\begin{align}2^{cp}-1&=(2^p-1)\cdot \left(1+2^p+2^{2p}+2^{3p}+\cdots+2^{(c-1)p}\right)\end{align}$
Thus: $2^{n-1} \mid 2^p-1$
Because of that, we must look at the factors of $2^p-1$ when considering if $2^{cp}-1$ is a Carmichael number. So we know those factors are already of form $2kp+1$, and then $kp \mid 2^{cp-1}-1$.
This is where I'm left. on an incomplete proof.
Using Bernoulli definition: 

An odd composite squarefree number $m$ is a Carmichael number iff $m$
  divides the denominator of the Bernoulli number $B_{n-1}$.

Using the Von Staudt–Clausen theorem, there may be a way to proof that that factors of the Bernoulli number denominators may never divide a mersenne number.

REPLY [3 votes]: Just some initial observations:

Suppose  $m=2^n-1$ and suppose $m$ is Carmichael. If $p$ is prime and $p \mid m$, then $p -1\mid m-1=2(2^{n-1}-1)$. Since $2^{n-1}-1$ is odd, we must have $p \equiv 3 \pmod 4$ for all $p \mid m$.
For $n \ge 2$, $m \equiv 3 \pmod 4$. $m$ is Carmichael and hence square free. If  $$m = \prod_{i=1}^kp_i\qquad\text{for $p_i$ distinct primes}$$
then
$$\begin{align}
m&\equiv \prod_{i=1}^k3 \pmod 4\\
&\equiv \prod_{i=1}^k(-1) \pmod 4\\
&\equiv (-1)^k \pmod 4
\end{align}$$
So $k$ must be odd, and $m$ is the product of an odd number of distinct primes with $p_i \equiv 3 \pmod 4$.

Certainly $2^n \equiv 1 \pmod m$, and since $2^k<m$ for $m<n$, $2$ has order $n$ modulo $m$. But $$2^{m-1} \equiv 1 \pmod m$$
so $n \mid m-1$. In particular, $2^n \equiv 2 \pmod n$, so $n$ is either prime, a pseudoprime to the base $2$ or $n$ is even and $2^{n-1} = 1 \pmod{\frac n2}$.

None of these conclusions are that restrictive, since we know that a Mersenne number can be prime! I'll try to post more as I think of it.