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

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

[tex]n=p\times k.[/tex]

Therefore

[tex]2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p.[/tex]

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

[tex]a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...+b^{n-1})[/tex]

Applying this formula in 1) we obtain that

[tex]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)[/tex].

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

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