# fermat...or not fermat?

• Oct 22nd 2006, 09:10 AM
Aglaia
fermat...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 AM
ThePerfectHacker
Quote:

Originally Posted by Aglaia
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...

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).