I have been researching Mersenne primes so I can write a program that finds them. A Mersenne prime looks like $2^n1$. When calculating them, I have noticed that the $n$ value always appears to be odd. Is there confirmation or proof of this being true?
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10The value of $n$ will in fact always be prime, and therefore odd, except for the single isolated case $n=2$. The proof of this fact is not difficult, but a bit too long for a comment. – David Sep 13 '14 at 12:58

6What about $2^21$? – illysial Sep 13 '14 at 12:58

@illysial Well, that is prime, but remember that two is a bit of an outcast in the prime numbers. – Progo Sep 13 '14 at 12:59

Ah of course.. Sorry for the trivial example – illysial Sep 13 '14 at 12:59

@illysial Don't feel sorry. It was good to point out. – Progo Sep 13 '14 at 13:01

@Progo Why is $2$ an outcast? Since it is the only even prime number? – Antoine Sep 13 '14 at 16:15

@Antoine That is why id is an outcast. All the other primes are odd, two is special. – Progo Sep 13 '14 at 17:04

8@Progo Well, $97$ is then special as well. It is the only prime which is divisible by $97$ (being even means being divisible by $2$). – Antoine Sep 13 '14 at 18:42

@Antoine This is true. – Progo Sep 13 '14 at 18:53

6@Progo: Well, all prime numbers are odd. The fact that $2$ is the unique even prime makes him quite the odd prime number, I'd say. – Asaf Karagila Sep 14 '14 at 04:00

5@AsafKaragila There's a saying I like, "Two is the oddest prime." – Akiva Weinberger Sep 14 '14 at 13:27

Another question to ask would be "What is the density of primes such that 2^n  1 is also prime?" – Ryan Sep 15 '14 at 03:51

Why is this question generating so many upvotes? – Christian Chapman Sep 16 '14 at 19:29

I believe that one of the reasons is that it was a 'Hot network question'. – Antoine Sep 17 '14 at 15:21
8 Answers
Theorem. If $2^n1$ is prime then $n$ is prime.
Proof. Suppose that $2^n1$ is prime, and write $n=st$ where $s,t$ are positive integers. Since $$x^s1=(x1)(x^{s1}+x^{s2}+\cdots+1)\ ,$$ we can substitute $x=2^t$ to see that $2^t1$ is a factor of $2^n1$. Since $2^n1$ is prime there are only two possibilities, $$2^t1=1\quad\hbox{or}\quad 2^t1=2^n1\ .$$ 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 $2^n1$ is prime. For example, $2^{11}1=23\times89$.
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Thank you for providing proof, it really helps me see why this is true. Sadly, I have to wait three minutes to accept your answer. – Progo Sep 13 '14 at 13:06

18@Progo Don't be sad about having to wait to accept an answer. It's usually best to wait a couple of days anyway, just in case an even better answer is posted. – David Richerby Sep 13 '14 at 13:51

3Man. I which my undergrad textbooks had had proofs as well written as this. +1 good sir; +1. – imallett Sep 14 '14 at 07:40
Note that for $m\ge 2$ $$2^{2m}1=(2^m)^21=(2^m1)(2^m+1)$$ is not a prime number.
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Yes, the formula $2^n1$ 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 $2^21=3$. However for some prime values of $n$ the result is not prime, for example $2^{11}1=23\cdot 89$.
In general $a^x1$ 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 $a^y1$ (where $y$ is a factor of $x$) become obvious.
$2^41$ 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 $10000100001 \cdot 11111$
$10^81$ 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 $a1$ (in this case, $9$): $9 \cdot 1010101 \cdot 11$ or $9 \cdot 10001 \cdot 1111$
As an aside, the presence of $a1$ as a factor means that all $a^x1$ with $a>2$ and $x>1$ are composite.
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Sorry everyone, but the question clearly states if Mersenne prime powers are always odd.
2^{2} − 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 2^{n} − 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.
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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 $n > 2$.
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If $n$ is composite then $n=ab$, where $a\ne n$, $b\ne n$. So: $$ 2^{ab}1 = (2^a)^b 1 = (2^a1)(1+ 2^a + 2^{2a} + \ldots + 2^{a(b1)}) $$
So $(2^{a}1)$ divides $(2^{n}1)$
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If $n=2k$ then we can show that $3(4^k1)$ by induction. When $k=1$, $33$ and we are done. Suppose $ 3(4^m1)$. Note that $$4^{m+1}1=4(4^m1)+3.$$ Which is divisible by $3$ by the induction hypothesis and the fact that $33$.
So $2^{2k}1$ is not prime when $k>1$.
Alternatively: $4^n1=(41)(4^{n1}+...+1)$ (used David's idea here).
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How does (lack of) divisibility by $3$ prove it to not divisible by all other primes till $2^n1$? Or do you mean to show similarly for all primes till $2^n1$? If so, what would you take as $a$ in $n=ak$, for showing for (lack of ) divisibility by prime $5$? – jiten Nov 26 '21 at 08:38

1A mersenne prime is a prime. The answer proved that $3$ divides $2^n1$ if $n$ is even. – Siong Thye Goh Nov 26 '21 at 14:03

@SiongThyeGoh why only for even $n$, it could be a multiple of any other prime like $5,7$. Is there no need to show for such cases, and showing just failure for even $n$ suffices? – jiten Nov 26 '21 at 14:24

1the question is prove that for $n$ even, $2^n1$ is not a prime. It suffices to show that it is divisible by $3$ for even $n$. The question is about even $n$. – Siong Thye Goh Nov 26 '21 at 16:52
There is an extremely simple proof of this fact if you simply write the number in binary and realize that $2^n  1$ is written as $n$ $1$'s. If $n$ is a composite number (i.e. non prime) then it can be grouped into a subset of $1$'s. Here's an example, $2^{10}  1$:
\begin{align} 2^{10}  1 =&\ 1111111111_2 = 1023 \\=&\ 1111100000_2 + 0000011111_2 \\ =&\ (2^5  1)*2^5 + (2^5  1)\\ =&\ 31*2^5 + 31 \\ =&\ 31(2^5 + 1)\\ =&\ 31 * 33 \end{align}
or
\begin{align} 2^{10}  1 =&\ 1111111111_2 = 1023 \\=&\ 1100000000_2 + 0011000000_2 + 0000110000_2 + 0000001100_2 + 0000000011_2 \\ =&\ 3*2^8 + 3*2^6 + 3*2^4 + 3*2^2 + 3*2^0 = 1023\\ =&\ 3*(2^8 + 2^6 + 2^4 + 2^2 + 2^0) = 1023 \\ =&\ 3*341 = 1023 \end{align}
If $n$ is composite (i.e. nonprime) then the $1$'s can be grouped into sets which will make $2^n  1$ nonprime. Basically, assume that $n$ is nonprime such that $n = \alpha \beta$, then $2^n  1$ will be divisible by $2^\alpha  1$ and $2^\beta  1$.
It's worth pointing out that (obviously) $n$ prime isn't enoughbecause all we've proved is that "if $n$ isn't prime $2^n  1$ definitely isn't" (I just showed how to find a factor for any $2^n  1$ where $n$ isn't prime). And you probably already realize things like $2^{11}  1$ exist where that's not a prime number (also definitely not divisible by $11$ btw) and thus not a Mersenne's Prime.
A couple of quick examples:
\begin{align*} 2^4  1 &\ \leadsto 4 = 2*2 \leadsto 2^4  1 \text{ is divisible by } 2^2  1 = 3& 15 =&\ 3*5 \\ 2^{12}  1 &\ \leadsto 12 = 4*3 \leadsto 2^4  1 = 15 \text{ and } 2^3  1 = 7 & 4095 =&\ 7*582, 15*273\end{align*}
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