Goldbach's Theorem/Proof 1

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

\(\ds F_m - 1\)

\(\equiv\)

\(\ds -1\)

\(\ds \pmod p\)

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.

$\blacksquare$

Source of Name

This entry was named for Christian Goldbach.