# Noob[est] : Proofs from the Book : Existence of Infinite Primes : Part 2 : Fermat Numbers

Fermat numbers are numbers of the form : $F_n = 2^{2^n} +1$. Kind of obvious that the $F_n$ and $F_r$ with $n \ne r$ are relatively prime. Not a problem. But how that implies there are infinite primes?

Here is the problem - there is a sequence $\mathbb{F} = <F_n>$ where every element is relatively prime to one another. How that implies that there are infinite prime numbers?

• $F_1$ is a prime.$F_2$ has a prime factor that $F_1$ doesn't have. $F_3$ has a prime factor that $F_1$ and $F_2$ don't have and so on. – pointer Jun 5 '14 at 8:24
• Oh man! How much dumb I can get! Please add this as an answer so that I can upvote :) – Noga Tailcutter Jun 5 '14 at 8:26
• For two numbers to be relatively prime it means that they don't have any prime divisor in common. Since every number can be written as a product of primes, a sequence of pairwise relatively prime numbers $(x_i)_{i \in \Bbb N}$ ensures that for each $n$ their is a prime dividing $x_n$ but not dividing any $x_m$ for $n \neq m$. – Stefan Mesken Jun 5 '14 at 8:28
• You can up vote comments too. – pointer Jun 5 '14 at 8:40

As the sequence $<x_n>$ is such that $x_i,x_j$ are relatively prime for $i \ne j$ - they must have no common prime factor. That would mean all $x_n$ are having different prime factors.