Goldbach's Theorem/Proof 2

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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:

\(\displaystyle d\) \(\divides\) \(\displaystyle F_n\) Definition of Common Divisor of Integers
\(\, \displaystyle \land \, \) \(\displaystyle d\) \(\divides\) \(\displaystyle F_m\) (where $\divides$ denotes divisibility)
\(\displaystyle \leadsto \ \ \) \(\displaystyle d\) \(\divides\) \(\displaystyle F_n - 2\) as $F_m \divides F_n - 2$
\(\displaystyle \leadsto \ \ \) \(\displaystyle d\) \(\divides\) \(\displaystyle F_n - \paren {F_n - 2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle d\) \(\divides\) \(\displaystyle 2\)

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.