0

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!

user26857
  • 52,094
Johny Hunter
  • 327
  • 1
  • 2
  • 11
  • 1
    It turns out this already has been asked several years ago by someone else: https://math.stackexchange.com/q/7473/101420. Apparently it is true what they say: great minds think alike! – Vincent Jul 06 '17 at 09:03
  • Let $b=qa+r : 0\le r<a$ then $GCD(p^a - 1, p^{qa+r}-1)=GCD(p^a-1, (p^{qa}-1)p^r +(p^r-1))=GCD(p^a-1, p^r-1) = ... = p^d-1$ by Euclid's lemma for integers. – Siddharth Joshi Jul 06 '17 at 09:15

1 Answers1

2

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.

quasi
  • 58,772