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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?

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    $\begingroup$ $F_1$ 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. $\endgroup$ – pointer Jun 5 '14 at 8:24
  • $\begingroup$ Oh man! How much dumb I can get! Please add this as an answer so that I can upvote :) $\endgroup$ – Noga Tailcutter Jun 5 '14 at 8:26
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    $\begingroup$ 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 $n \neq m$. $\endgroup$ – Stefan Mesken Jun 5 '14 at 8:28
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    $\begingroup$ You can up vote comments too. $\endgroup$ – pointer Jun 5 '14 at 8:40
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The comments actually do answer my question. user121270 - said it right - so did user155124.

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.

As the sequence is infinite, then, there must be infinite primes to continue creating the sequence. Hence, QED.

and 100 points for me for being dumbo here.

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