Are Mersenne prime exponents always odd? I have been researching Mersenne primes so I can write a program that finds them. A Mersenne prime looks like $2^n-1$. When calculating them, I have noticed that the $n$ value always appears to be odd. Is there confirmation or proof of this being true?
 A: Sorry everyone, but the question clearly states if Mersenne prime powers are always odd.
22 − 1 is a Mersennse prime which power is 2 which is an even number. 
Therefore the only possible answer is no. 
The only thing we can be certain is that if 2n − 1 is prime, then n is prime as it has been proved in other answers.
It just happens that 2 is the only even prime number.
A: Theorem.  If $2^n-1$ is prime then $n$ is prime.
Proof.  Suppose that $2^n-1$ is prime, and write $n=st$ where $s,t$ are positive integers.  Since
$$x^s-1=(x-1)(x^{s-1}+x^{s-2}+\cdots+1)\ ,$$
we can substitute $x=2^t$ to see that $2^t-1$ is a factor of $2^n-1$.  Since $2^n-1$ is prime there are only two possibilities,
$$2^t-1=1\quad\hbox{or}\quad 2^t-1=2^n-1\ .$$
Therefore $t=1$ or $t=n$.  We have shown that the only possible factorisations of $n$ are $n\times1$ and $1\times n$.  Hence, $n$ is prime.
Comment.  If $n$ is prime it is not always true that $2^n-1$ is prime.  For example, $2^{11}-1=23\times89$.
A: Yes, if $n$ is positive and even, then one can factor into integers:
$$2^n - 1 = (2^{n / 2} + 1)(2 ^ {n / 2} - 1).$$
Thus in this case $2^n - 1$ is composite if $2^{n / 2} - 1 > 1$, that is, if $n > 2$.
A: If $n$ is composite then $n=ab$, where $a\ne n$, $b\ne n$. So:
$$
2^{ab}-1 = (2^a)^b -1 = (2^a-1)(1+ 2^a + 2^{2a} + \ldots + 2^{a(b-1)})
$$
So $(2^{a}-1)$ divides $(2^{n}-1)$
A: Note that for $m\ge 2$
$$2^{2m}-1=(2^m)^2-1=(2^m-1)(2^m+1)$$
is not a prime number.
A: If $n=2k$ then we can show that $3|(4^k-1)$ by induction. When $k=1$, $3|3$ and we are done. Suppose $ 3|(4^m-1)$. Note that $$4^{m+1}-1=4(4^m-1)+3.$$ Which is divisible by $3$ by the induction hypothesis and the fact that $3|3$. 
So $2^{2k}-1$ is not prime when $k>1$.
Alternatively: $4^n-1=(4-1)(4^{n-1}+...+1)$ (used David's idea here).
A: Yes, the formula $2^n-1$ may only yield a prime if $n$ is prime, therefore the only even value of $n$ which yields a prime is $2$, which yields the Mersenne prime $2^2-1=3$. However for some prime values of $n$ the result is not prime, for example $2^{11}-1=23\cdot 89$.
In general $a^x-1$ is composite if $x$ is composite. This can be seen very easily by writing out the numbers in the base $a$. The number will have $x$ digits, and where $x$ is composite, possible factorisations involving $a^y-1$ (where $y$ is a factor of $x$) become obvious.
$2^4-1$ in binary is $1111$ which can be factored as $101\cdot 11$
$2^{15}-1$ in binary is $111 111 111 111 111$ which can be factored as $1001001001001 \cdot 111$ or $10000100001 \cdot 11111$

$10^8-1$ is $99999999$ in decimal which can be factored as $1010101 \cdot 99$ or $10001 \cdot 9999$. For $a$ greater than $2$ we can also separate out the factor $a-1$ (in this case, $9$): $9 \cdot 1010101 \cdot 11$ or $9 \cdot 10001 \cdot 1111$ 
As an aside, the presence of $a-1$ as a factor means that all $a^x-1$ with $a>2$ and $x>1$ are composite.
