Prove that $\gcd(g_a,g_b) = 1$ given that for $n \in Z^{\geq 0}$, define $g_n = 2^{2^n} + 1$

Question

Prove that $\gcd(g_a,g_b) = 1$ given that for $n \in Z^{\geq 0}$, define $g_n = 2^{2^n} + 1$. I have already proved that $g_0\cdot g_1\cdots g_{n-1} = g_n -2$ if this hint is useful in this proof.

Answer 1

If $g_n=2^{2^n}+1$ and $g_n=g_0\cdot g_1 \cdots g_{n-1}+2$.

The gcd must divide 2, but it cannot be 2 as $g_n$ cannot be divided by 2. Hence $\gcd(g_a,g_b)=1$

Answer 2

All the $g_i$ are odd.
If $(g_a,g_b)=d>1$ then there must be an odd prime $p$ which divides both $g_a$ and $g_b$.
Assume $a>b$. Then $g_a=g_1\cdots g_b\cdots g_{a-1}-2$ which means that if $p\mid g_a ,p\mid g_b$ then $ p\mid g_a- g_1\cdots g_b\cdots g_{a-1}$ which means $p\mid 2$ a contradiction.