# Goldbach's Theorem

## Contents

## Theorem

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

Then $F_m$ and $F_n$ are coprime.

## Proof

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 $k \in \Z_{>0}$.

\(\displaystyle F_m - 1\) | \(\equiv\) | \(\displaystyle -1\) | \(\displaystyle \pmod p\) | $\quad$ as $p \divides F_m$ | $\quad$ | ||||||||

\(\displaystyle F_n - 1\) | \(\equiv\) | \(\displaystyle -1\) | \(\displaystyle \pmod p\) | $\quad$ as $p \divides F_n$ | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {F_n - 1}^{2^k}\) | \(\equiv\) | \(\displaystyle -1\) | \(\displaystyle \pmod p\) | $\quad$ Fermat Number whose Index is Sum of Integers | $\quad$ | |||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {-1}^{2^k}\) | \(\equiv\) | \(\displaystyle -1\) | \(\displaystyle \pmod p\) | $\quad$ Congruence of Product | $\quad$ | |||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 1\) | \(\equiv\) | \(\displaystyle -1\) | \(\displaystyle \pmod p\) | $\quad$ Congruence of Powers | $\quad$ | |||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(\equiv\) | \(\displaystyle 2\) | \(\displaystyle \pmod p\) | $\quad$ | $\quad$ |

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.

$\blacksquare$

## Source of Name

This entry was named for Christian Goldbach.

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 2.4$: The rational numbers and some finite fields: Further Exercises $9$ - 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $257$ - 2001: Michal Křížek, Florian Luca and Lawrence Somer:
*17 Lectures on Fermat Numbers*: Theorem $4.1$