# Divisor of Fermat Number/Euler's Result

## Theorem

Let $F_n$ be a Fermat number.

Let $m$ be divisor of $F_n$.

Then $m$ is in the form:

$k \, 2^{n + 1} + 1$

where $k \in \Z_{>0}$ is an integer.

## Proof

It is sufficient to prove the result for prime divisors.

The general argument for all divisors follows from the argument:

$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$

So the product of two factors of the form preserves that form.

Let $n \in \N$.

Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$.

Then we have:

$2^{2^n} \equiv -1 \pmod p$

Squaring both sides:

$2^{2^{n + 1}} \equiv 1 \pmod p$

From Integer to Power of Multiple of Order, the order of $2$ modulo $p$ divides $2^{n + 1}$ but not $2^n$.

Therefore it must be $2^{n + 1}$.

Hence:

 $\ds \exists k \in \Z: \ \$ $\ds \map \phi p$ $=$ $\ds k \, 2^{n + 1}$ Corollary to Integer to Power of Multiple of Order $\ds p - 1$ $=$ $\ds k \, 2^{n + 1}$ Euler Phi Function of Prime $\ds p$ $=$ $\ds k \, 2^{n + 1} + 1$

$\blacksquare$

## Historical Note

In $1747$, Leonhard Paul Euler proved that a divisor of a Fermat number $F_n$ is always in the form $k \, 2^{n + 1} + 1$.

This was later refined to $k \, 2^{n + 2} + 1$ by François Édouard Anatole Lucas.