why must n be a prime (odd)?
 

I´m thinking:

Why must n from 2 exp n be a prime. After wikipedia it says:

Some definitions of Mersenne numbers require that the exponent [I]n[/I] be prime.

A number in the format 2 n - 1, where n is even could be a prime as well?

[quote=joblack;145609]I´m thinking:

Why must n from 2 exp n be a prime. After wikipedia it says:

Some definitions of Mersenne numbers require that the exponent [I]n[/I] be prime.

A number in the format 2 n - 1, where n is even could be a prime as well?[/quote]
Read the top of the Searching for Mersenne Primes section of the Wikipedia article on it.
[URL]http://en.wikipedia.org/wiki/Mersenne_number#Searching_for_Mersenne_primes[/URL]
It says:
[tex]2^{ab}-1=(2^a-1)\cdot \left(1+2^a+2^{2a}+2^{3a}+\cdots+2^{(b-1)a}\right)[/tex]
Which means that if the exponent, usually referred to as n or p, is composite, i.e. can be split into an a and b as that equation requires, then the related Mersenne number is composite.
The only time that an even n produces a Mersenne prime is n=2 (2^2-1=4-1=3).

[quote=joblack;145609]A number in the format 2 n - 1, where n is even could be a prime as well?[/quote]
A number 2^n-1 with even n will always be divisible by 3.

[QUOTE=jrk;145613]A number 2^n-1 with even n will always be divisible by 3.[/QUOTE]Writing a number in base 4 would clearly show the divisibility by 3.

[quote=retina;145616]Writing a number in base 4 would clearly show the divisibility by 3.[/quote]
2^4-1=4^2-1=15
I'm guessing since 4=3+1, 4^n-1 is divisible by 3. Is this so? If so, why is this? If not, what's the reason?

Well one way to show it with induction:

assume [tex]2^n-1 \equiv 0 (mod 3)[/tex]
[tex]2^n \equiv 1 (mod 3)[/tex]
[tex]2^{(n+2)} = 4*2^n \equiv 1 (mod 3)[/tex]
[tex]2^{(n+2)}-1 \equiv 0 (mod 3)[/tex]
[tex]2^0-1 \equiv 0 (mod 3)[/tex]

So if it is true for n then it is true for n+2, it is true for n=0, therefore it is true for all positive even n.

[QUOTE=Mini-Geek;145618]2^4-1=4^2-1=15
I'm guessing since 4=3+1, 4^n-1 is divisible by 3. Is this so? If so, why is this? If not, what's the reason?[/QUOTE]Writing any number of the form 2[sup]2n[/sup]-1 in base 4 gives: 3333...333 (a total of n 3's).

Basic result:

1. (a^n - b^n) is divisible by (a-b).
2. a^(mn) = (a^m)^n = (a^n)^m

From these two basic results, everything else follows.

Yep. My induction method for even n was pointless :smile:

[quote=Mini-Geek;145611]Read the top of the Searching for Mersenne Primes section of the Wikipedia article on it.
[URL]http://en.wikipedia.org/wiki/Mersenne_number#Searching_for_Mersenne_primes[/URL]
It says:
[tex]2^{ab}-1=(2^a-1)\cdot \left(1+2^a+2^{2a}+2^{3a}+\cdots+2^{(b-1)a}\right)[/tex]
Which means that if the exponent, usually referred to as n or p, is composite, i.e. can be split into an a and b as that equation requires, then the related Mersenne number is composite.
The only time that an even n produces a Mersenne prime is n=2 (2^2-1=4-1=3).[/quote]
This is bugging me:
If 2^a-1 is a factor, then surely 2^b-1 must be as well.

[QUOTE=davieddy;145649]This is bugging me:
If 2^a-1 is a factor, then surely 2^b-1 must be as well.[/QUOTE]

Yes:

2[sup]ab[/sup] - 1 = (2[sup]b[/sup]-1)*(1+2[sup]b[/sup]+2[sup]2b[/sup]+2[sup]3b[/sup]+ ... +2[sup](a-1)b[/sup])