Goldbach's Theorem/Proof 2

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:

\(\ds d\)

\(\divides\)

\(\ds F_n\)

Definition of Common Divisor of Integers

\(\, \ds \land \, \)

\(\ds d\)

\(\divides\)

\(\ds F_m\)

(where $\divides$ denotes divisibility)

\(\ds \leadsto \ \ \)

\(\ds d\)

\(\divides\)

\(\ds F_n - 2\)

as $F_m \divides F_n - 2$

\(\ds \leadsto \ \ \)

\(\ds d\)

\(\divides\)

\(\ds F_n - \paren {F_n - 2}\)

\(\ds \leadsto \ \ \)

\(\ds d\)

\(\divides\)

\(\ds 2\)

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.