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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 $k<n$, then $m$ divides $2$.

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

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If $m$ divides $F_i$ for some $i \le n-1$, then $m$ divides the left-hand side. If furthermore $m$ divides $F_n$, then $m$ divides the difference $(\prod_0^{n-1}F_k)-F_n$, so $m$ divides $-2$. –  André Nicolas Jan 4 '12 at 1:32
    
More generally, if $d|a$ and $d|b$ then $d|(a\pm b)$ (or more generally still, $d|ax+by$) –  Aaron Jan 4 '12 at 2:16
    
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2 Answers

up vote 6 down vote accepted

If $m$ divides $F_k$ and $F_n$, with $k<n$, 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. –  iDontKnowBetter Jan 4 '12 at 2:42
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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

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

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