Question

If 2^n + 1 is an odd prime for some integer n, prove that n is a power of 2. (H

Answer 1

Step-by-step explanation:

We will prove by contradiction. Assume that $2^n + 1$ is an odd prime but n is not a power of 2. Then, there exists an odd prime number p such that $p\mid n$. Then, for some integer $k\geq 1$,

$n=p\times k.$

Therefore

1. $2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p.$

Here we will use the formula for the sum of odd powers, which states that, for $a,b\in \mathbb{R}$ and an odd positive number $n$,

$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...+b^{n-1})$

Applying this formula in 1) we obtain that

$2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p=(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^{k}+1)$.

Then, as $2^k+1>1$ we have that $2^n+1$ is not a prime number, which is a contradiction.

In conclusion, if $2^n+1$ is an odd prime, then n must be a power of 2.