# Goldbach's Theorem

## Theorem

Let $F_m$ and $F_n$ be Fermat numbers such that $m \ne n$.

Then $F_m$ and $F_n$ are coprime.

## Proof 1

Aiming for a contradiction, suppose $F_m$ and $F_n$ have a common divisor $p$ which is prime.

As both $F_n$ and $F_m$ are odd, it follows that $p$ must itself be odd.

Without loss of generality, suppose that $m > n$.

Then $m = n + k$ for some $k \in \Z_{>0}$.

 $\ds F_m - 1$ $\equiv$ $\ds -1$ $\ds \pmod p$ as $p \divides F_m$ $\ds F_n - 1$ $\equiv$ $\ds -1$ $\ds \pmod p$ as $p \divides F_n$ $\ds \leadsto \ \$ $\ds \paren {F_n - 1}^{2^k}$ $\equiv$ $\ds -1$ $\ds \pmod p$ Fermat Number whose Index is Sum of Integers $\ds \leadsto \ \$ $\ds \paren {-1}^{2^k}$ $\equiv$ $\ds -1$ $\ds \pmod p$ Congruence of Product $\ds \leadsto \ \$ $\ds 1$ $\equiv$ $\ds -1$ $\ds \pmod p$ Congruence of Powers $\ds \leadsto \ \$ $\ds 0$ $\equiv$ $\ds 2$ $\ds \pmod p$

Hence $p = 2$.

However, it has already been established that $p$ is odd.

From this contradiction it is deduced that there is no such $p$.

Hence the result.

$\blacksquare$

## Proof 2

Let $F_m$ and $F_n$ be Fermat numbers such that $m < n$.

Let $d = \gcd \set {F_m, F_n}$.

$F_m \divides F_n - 2$

But then:

 $\ds d$ $\divides$ $\ds F_n$ Definition of Common Divisor of Integers $\, \ds \land \,$ $\ds d$ $\divides$ $\ds F_m$ (where $\divides$ denotes divisibility) $\ds \leadsto \ \$ $\ds d$ $\divides$ $\ds F_n - 2$ as $F_m \divides F_n - 2$ $\ds \leadsto \ \$ $\ds d$ $\divides$ $\ds F_n - \paren {F_n - 2}$ $\ds \leadsto \ \$ $\ds d$ $\divides$ $\ds 2$

But all Fermat numbers are odd, so:

$d \ne 2$

It follows that $d = 1$.

The result follows by definition of coprime.

$\blacksquare$

## Source of Name

This entry was named for Christian Goldbach.