Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(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?

share|cite|improve this question
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
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$.

share|cite|improve this answer
ha, simple enough. thanks. – iDontKnowBetter Jan 4 '12 at 2:42

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.