On the GCD of a Pair of Fermat Numbers I've been working with the Fermat numbers recently, but this problem has really tripped me up. If the Fermat theorem is set as $f_a=2^{2^a}+1$, then how can we say that for an integer $b<a$, the $\gcd(f_b,f_a)=1$?

This is because the Fermat numbers belong to the companion Lucas sequence $V(3,2) = 2^{k} + 1$. Hence, all the prime factors of either $V_{p}$ where $p \neq 3$ is a prime, or $V_{2^{n}}$, are primitive; that is, they enter the sequence for the first time as factors at that very term.
So, as every prime factor of $f_{a}$ is primitive, it follows that $gcd(f_{a},f_{b}) = 1$  when $b < a$.
