# Goldbach's Theorem/Proof 1

Jump to navigation
Jump to search

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

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

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

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

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

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

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

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.

$\blacksquare$

## Source of Name

This entry was named for Christian Goldbach.