proove that if a prime number is represented by this phormula p=(2^n)+1 with n>0 then n is a power of 2... it's all for you...

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- Oct 22nd 2006, 09:10 AMAglaiafermat...or not fermat?
proove that if a prime number is represented by this phormula p=(2^n)+1 with n>0 then n is a power of 2... it's all for you...

- Oct 22nd 2006, 09:16 AMThePerfectHacker
The important fact is that, we can factor:

$\displaystyle x^n+y^n$ where $\displaystyle n$ is odd.

Now, assume,

$\displaystyle 2^n+1$ is prime where $\displaystyle n$ is divisible by odd.

Then,

$\displaystyle 2^{(2k+1)j}+1$

Thus,

$\displaystyle (2^j)^{2k+1}+1$

Can be factored as,

$\displaystyle (2^j+1)(2^{2kj}-2^{(2k-1)j}+2^{(2k-2)j}-...+2^{2j}-2^j+1)$

It has a proper nontrivial factorization, thus it cannot be prime.

Thus, $\displaystyle n$ cannot be divisble by odd number that is, $\displaystyle 2^m$

Thus,

$\displaystyle 2^{2^m}+1$

Which were studied by Fermat (my favorite mathemation).