Goldbach's Theorem/Proof 1

Proof
$F_m$ 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.

, suppose that $m > n$.

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

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.