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Given that the Fermat numbers $F_m$ are pairwise relatively prime.

Prove that there are infinitely many primes.

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    $\begingroup$ So, each Fermat number is either prime or is divisible by a unique prime number which does not divide any other Fermat number $\endgroup$ – lab bhattacharjee Jan 17 '15 at 18:45
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    $\begingroup$ Let $p_m$ be the smallest prime divisor of $F_m$. Since the $F_k$ are pairwise relatively prime, the $p_m$ are distinct. So the set of all $p_m$ is infinite. (The relative primality is not hard to prove, but it looks as if you are not expected to give a proof.) $\endgroup$ – André Nicolas Jan 17 '15 at 18:46
  • $\begingroup$ @AndréNicolas If I wanted to prove its primality how would I be able to prove it? $\endgroup$ – Alexis Dailey Jan 17 '15 at 18:48
  • $\begingroup$ Let $m\lt n$. Note that $2^n=(2^m)^{n-m}$ So if $x=F_m$ then $F_n=(x-1)^{2^{n-m}}+1$. Using the binomial theorem to expand, we find that $x$ divides $F_{n}-2$. So any divisor $d$ of $F_m$ divides $F_n-2$. If $d$ also divides $F_n$, then $d$ divides $2$. But $d$ is odd, so $d=1$. (You will find variant proofs on MSE, the question has been asked repeatedly. Answers are likely to be less terse than this comment.) $\endgroup$ – André Nicolas Jan 17 '15 at 19:05
  • $\begingroup$ @AndréNicolas. For answers on MSE see here. So the OP already knows it. $\endgroup$ – Dietrich Burde Jan 17 '15 at 19:30

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