Goldbach's Theorem/Proof 1
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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 $k \in \Z_{>0}$.
\(\ds F_m - 1\) | \(\equiv\) | \(\ds -1\) | \(\ds \pmod p\) | 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.