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 $\gcd(f_b,f_a)=1$?

Claim. $f_n=f_0\cdots f_{n-1}+2$.
The result holds for $f_1$: $f_0=2^{2^0}+1 = 2^1+1 = 3$, $f_1=2^{2}+1 = 5 = 3+2$.
Assume the result holds for $f_n$. Then
$$\begin{align*}
f_{n+1} &= 2^{2^{n+1}}+1\\
&= (2^{2^n})^2 + 1\\
&= (f_n-1)^2 +1\\
&= f_n^2 - 2f_n +2\\
&= f_n(f_0\cdots f_{n-1} + 2) -2f_n + 2\\
&= f_0\cdots f_{n-1}f_n + 2f_n - 2f_n + 2\\
&= f_0\cdots f_n + 2,
\end{align*}$$
which proves the formula by induction. $\Box$
Now, let $d$ be a common factor of $f_b$ and $f_a$. Then $d$ divides $f_0\cdots f_{a-1}$ (because it's a multiple of $f_b$) and divides $f_a$. That means that it divides
$$f_a - f_0\cdots f_{a-1} = (f_0\cdots f_{a-1}+2) - f_0\cdots f_{a-1} = 2;$$
but $f_a$ and $f_b$ are odd, so $d$ is an odd divisor of $2$. Therefore, $d=\pm 1$. So $\gcd(f_a,f_b)=1$. 
