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

Prove that there are infinitely many primes.

Alexis Dailey
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    So, each Fermat number is either prime or is divisible by a unique prime number which does not divide any other Fermat number – lab bhattacharjee Jan 17 '15 at 18:45
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    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.) – André Nicolas Jan 17 '15 at 18:46
  • @AndréNicolas If I wanted to prove its primality how would I be able to prove it? – Alexis Dailey Jan 17 '15 at 18:48
  • 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.) – André Nicolas Jan 17 '15 at 19:05
  • @AndréNicolas. For answers on MSE see here. So the OP already knows it. – Dietrich Burde Jan 17 '15 at 19:30
  • @AlexisDailey: I assumed your "prove its primality" was intended to mean relative primality. We do not need to identify the smallest prime factor of $F_m$ to know there is one. But if you are looking for efficient algorithms, there are methods for Fermat numbers that are much more efficient than the "general" methods for numbers of the same size. – André Nicolas Jan 17 '15 at 21:48

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