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(A Fermat number $F_n$ is such that $F_n = 2^{2^n} + 1, \; \; n=0,1,2,3...$.)

We will show that any two Fermat numbers are relatively prime; hence there must be infinitely many primes. We verify the recursion

$\prod_{k=0}^{n-1} F_k = F_n - 2 $

from which our assertion follows immediately. Indeed, if $m$ is a divisor of, say, $F_k$ and $F_n$, where $k, then $m$ divides $2$.

Where did the "then $m$ divides $2$" assertion come from?

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    Related: http://math.stackexcha$n$ge.com/questions/68653/fermat-numbers-and-gcd2012-01-04

2 Answers 2

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If $m$ divides $F_k$ and $F_n$, with $k, then $m$ divides $\prod_{k=0}^{n-1} F_k = F_n - 2$, and so $m$ divides the difference $ F_n-\prod_{k=0}^{n-1} F_k = 2$.

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    ha, simple enough. thanks.2012-01-04
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HINT $\rm\ \ m\ |\ F_n\ $ and $\rm\ m\ |\ F_k\ |\ F_n-2\ \ \Rightarrow \ \ m\ |\ F_n - (F_n-\:2)\ =\ 2 $

NOTE $\ $ An alternative simpler proof of the main result follows by specializing $\rm\:c\:,\:k\:$ in

$\rm\ \gcd(c+1,\ c^{2\:k}+1)\ =\ gcd(c+1,\:2)$