Prove that $\gcd(a,b)=d \Rightarrow \gcd(p^a-1,p^b-1)=p^d-1$

Question

I have to prove the following:.

For $a,b,p,d \in \mathbb{N}$ and $p>1$, $\gcd(a,b)=d \Rightarrow \gcd(p^a-1,p^b-1)=p^d-1$.

So as far as I know I have to prove two things:
1. $p^d-1 \mid p^a-1$ and $p^d-1 \mid p^b-1$.
2. $(c \mid p^a-1$ and $c \mid p^b-1) \Rightarrow c \mid p^d-1$.

I already managed to prove the first one, but i am not sure how to solve the second. Any help would be appreciated!

Answer 1

Suppose $c \mid p^a-1$ and $c \mid p^b-1$.

Since $a,b$ are positive integers, and $d=\gcd(a,b)$, there exist positive integers $x,y$ such that $ax - by=d$.

\begin{align*} \text{Then}\;\;&c \mid p^a-1\;\text{and}\;c \mid p^b-1\\[4pt] \implies\;&p^a \equiv 1\;\,(\text{mod $c$})\;\;\,\text{and}\;\;\,p^b \equiv 1\;\,(\text{mod $c$})\\[4pt] \implies\;&p^{ax} \equiv 1\;\,(\text{mod $c$})\;\;\,\text{and}\;\;\,p^{by} \equiv 1\;\,(\text{mod $c$})\\[4pt] \implies\;&p^{by}p^{ax-by} \equiv 1\;\,(\text{mod $c$})\\[4pt] \implies\;&p^{by}p^d \equiv 1\;\,(\text{mod $c$})\\[4pt] \implies\;&p^d \equiv 1\;\,(\text{mod $c$})\\[4pt] \implies\;&c \mid p^d-1\\[4pt] \end{align*} as was to be shown.