Exercise 10.2.1

Proof. Let $N=2^k+1,k\ge 1$, an odd prime. Suppose that there exists an odd divisor $l$ of $k$, with $l>1$, so $k=\mathit{ql},q\in \mathbb{N}$, and $1\le q<k$. Since $l$ is odd,

so $2^q+1$ divides $2^k+1$, and $1\le q<k$, hence $1<3\le 2^q+1<2^k+1$, so $2^q+1$ is a nontrivial divisor of $N=2^k+1$. Therefore $N$ is composite: this is a contradiction.

Thus $k$ has no odd non trivial divisor, so $k=2^n$ for some integer $n$, and $N=2^{2^n}+1$ is a Fermat prime. □