Is it true that if $(a,b) = 1$, then $(2^a - 1, 2 ^ b -1) = 1$?

Question

I played around with some values and so far the conjecture seems to be true, however I haven't come up with a proof. I am not looking for a complete proof, only a couple of hints.

Answer 1

Suppose $a,b$ are positive integers with $(a, b) = 1$. Let us choose a Bezout pair $m,n \in \mathbb{Z}$ such that $ma + nb = 1$. Now, suppose for the sake of illustration that $m > 0$ and $n < 0$. Then $ma = 1 + (-n) b$, so $(2^a)^m = 2 \cdot (2^b)^{-n}$. So, if $d \mid 2^a - 1$ and $d \mid 2^b - 1$, then $2^a \equiv 1 \pmod{d}$ and $2^b \equiv 1 \pmod{d}$...