If $2^n+1$ is prime, why must $n$ be a power of $2$?

A little bird told me that if $2^n+1$ is prime, then $n$ is a power of $2$. I tend not to trust talking birds, so I'm trying to verify that statement independently.

Suppose $n$ is not a power of $2$. Then $n = a \cdot 2^m$ for some $a$ not a power of $2$ and some integer $m$. This gives $2^n+1 = 2^{a \cdot 2^m}+1$. Now I suspect there's a way to factor that, but I don't see how. Can someone give me a hint?

Solution 1:

Hint. For any odd natural number $a$, the polynomial $x+1$ divides $x^a+1$ evenly.

In particular, we have $$ \frac{x^a+1}{x+1}=\frac{(-x)^a-1}{(-x)-1}=1-x+x^2-\cdots+ (-x)^{a-1}$$ by the geometric sum formula. In this case, specialize to $x=2^{2^{\large m}}$ and we have a nontrivial divisor. (Also, $x^a+1\equiv(-1)^a+1\equiv-1+1\equiv0\mod x+1$ inside $\Bbb Z[ x]$ is pretty slick.)