Primes of the Form of a Power of Two Plus One

Theorem

Let $n \in \N$ be a natural number.

Let $2^n + 1$ be prime.

Then $n = 2^k$ for some natural number $k$.

Proof

Suppose $n$ has an odd divisor apart from $1$.

Then $n$ can be expressed as $n = \left({2 r + 1}\right) s$.

But then:

$2^{\left({2 r + 1}\right) s} = \left({2^s + 1}\right) \left({2^{2rs} - 2^{\left({2r-1}\right) s} + 2^{\left({2r-2}\right) s} - \cdots - 2^s + 1}\right)$

as is apparent after some algebra. (Note that the result Sum of Odd Positive Powers‎ may also be invoked.)

Hence $2^n + 1$ can be prime only if $n$ has only even divisors.

That is, if $n = 2^k$ for some natural number $k$.

$\blacksquare$

Note

Primes of the form $2^{\left({2^k}\right)} + 1$ are called Fermat primes.