---
tag: 0VN4H8M12X
kind: statement
role: proposition
name: If 2^n + 1 is prime then n is a power of 2
msc: [11A41]
uses: [0R03B54V4Z, 0SN0MAEKYX]
sources:
  - url: https://en.wikipedia.org/wiki/Fermat_number
    locator: lead section
---

## Statement

Let $n$ be a positive integer.
If $2^n + 1$ is a [[prime number|0SN0MAEKYX]], then $n = 2^k$ for a natural number $k$.
So every prime of the form $2^n + 1$ with $n \ge 1$ is a [[Fermat number|0R03B54V4Z]].

## Remarks

The proof: if $n = k m$ with $m > 1$ odd, then $x + 1$ divides $x^m + 1$ for $x = 2^k$. So $2^k + 1$ is a proper divisor of $2^n + 1$.
For example, $2^6 + 1 = 65 = 5 \cdot 13$, and $5 = 2^2 + 1$.
The converse is false, because $F_5 = 2^{32} + 1$ is composite.

## Explanations

### L1 human

An odd factor of the exponent gives a factorization of $2^n + 1$.
So the only exponents that can give primes are the powers of $2$, which have no odd factor except $1$.