# Goldbach's Theorem/Proof 2

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## Theorem

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

Then $F_m$ and $F_n$ are coprime.

## Proof

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

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

From the corollary to Product of Sequence of Fermat Numbers plus 2:

- $F_m \divides F_n - 2$

But then:

\(\displaystyle d\) | \(\divides\) | \(\displaystyle F_n\) | Definition of Common Divisor of Integers | ||||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle d\) | \(\divides\) | \(\displaystyle F_m\) | (where $\divides$ denotes divisibility) | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle d\) | \(\divides\) | \(\displaystyle F_n - 2\) | as $F_m \divides F_n - 2$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle d\) | \(\divides\) | \(\displaystyle F_n - \paren {F_n - 2}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle d\) | \(\divides\) | \(\displaystyle 2\) |

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.