Can 2^n-1 Be Proven to Always Be Prime for Real Integer n Greater Than 1?

[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."

 

[Borek]

Find counterexample.

 

[dodo]

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.

 

[VeeEight]

You may also wish to check out http://en.wikipedia.org/wiki/Mersenne_prime#Theorems_about_Mersenne_numbers"

 

[Mensanator]

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

 

[robert Ihnot]

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}