# Fermat numbers

Fermat numbers are these:

$F_n=2^{2^n}+1$

Are they all prime? If so, we’d have a truly incredible proof. Unfortunately, they are not. While $F_0,F_1,F_2,F_3,F_4$ are prime, $F_5$ is not (it’s divisible by $641$). Up through around $F_{32}$ there are no more Fermat numbers which are prime. It’s weird though, because the primes which do divide Fermat numbers seem to get big fast. The smallest prime dividing $F_{11}$ is $319,489$. The smallest prime dividing $F_{10}$ is $45,592,577$. Anyway, this isn’t relevant to us.

Here’s a fun fact I won’t prove. You can prove it via induction.

$F_n=F_0\cdot F_1\cdot\dots\cdot F_{n-1}+2$.

So suppose $p$ is some prime dividing $F_j$, and $n>j$. Obviously $p$ is odd, and since $F_j$ shows up in the product, $F_n$ cannot be divisible by $p$. This means that once a Fermat number is divisible by a prime, no other Fermat number will be ever again (in some sense, this is why the primes dividing Fermat numbers grow so quickly). Anyway, this means that if I have a collection of $k$ Fermat numbers, I also have a collection of at least $k$ primes. Since there are infinitely many Fermat numbers, there must be infinitely many primes.