Primes of form Power of Two plus One

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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 1

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

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

So:

\(\displaystyle 2^n + 1\) \(=\) \(\displaystyle 2^{\left({2 r + 1}\right) s} + 1\)
\(\displaystyle \) \(=\) \(\displaystyle \left({2^s}\right)^{\left({2 r + 1}\right)} + 1^{\left({2 r + 1}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({2^s + 1}\right) \left({2^{2rs} - 2^{\left({2r-1}\right) s} + 2^{\left({2r-2}\right) s} - \cdots - 2^s + 1}\right)\) Sum of Odd Positive Powers‎

and so $2^n + 1$ is not prime.


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

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

$\blacksquare$


Proof 2

A specific instance of Primes of form Power plus One:

$q^n + 1$ is prime only if:

$(1): \quad q$ is even

and

$(2): \quad n$ is of the form $2^k$ for some positive integer $k$.


As $2$ is even, the result applies.

$\blacksquare$


Also see


Historical Note

In $1640$, Pierre de Fermat wrote to Pierre de Fermat wrote to Bernard Frénicle de Bessy announcing this result.

From the fact that all integers of the form $2^n + 1$ such that $n = 2^k$ that he tested were prime, he went on to make his famous Fermat Prime Conjecture: that they are all prime.

This was refuted by Leonhard Paul Euler, who discovered that $2^{2^5}$ is composite in $1732$.