2^n-1 is prime

1. Nov 18, 2009

Char. Limit

First, remember that the OP is new to pure number theory.

We all know that not every number that follows the formula 2n-1 is prime. My question is, without using trial and error, how would you prove or disprove this statement?

"All numbers that obey the formula 2n-1, when n is a real integer number greater than 1, are prime."

2. Nov 18, 2009

Staff: Mentor

Re: 2^n-1

Find counterexample.

3. Nov 18, 2009

dodo

Re: 2^n-1

Another possibility is noting that, for an even n = 2h, we have that 2^n-1 = (2^h+1)(2^h-1), which makes the original number composite for h > 1.

4. Nov 18, 2009

VeeEight

Last edited by a moderator: Apr 24, 2017
5. Nov 18, 2009

Mensanator

Re: 2^n-1

For every even n, 2^n-1 is divisible by 3.

6. Nov 23, 2009

robert Ihnot

Re: 2^n-1

It doesn't have to be even. If ab is composite, then $$\frac{2^{ab}-1}{2^a-1}=1+2^a+2^{2a} +++2^{ab-a}$$

Last edited: Nov 23, 2009