Goldbach's Theorem

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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

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.


Sources