Prove $\gcd(k, l) = d \Rightarrow \gcd(2^k - 1, 2^l - 1) = 2^d - 1$

Question

This is a problem for a graduate level discrete math class that I'm hoping to take next year (as a senior undergrad). The problem is as stated in the title:

Given that $\gcd(k, l) = d$, prove that $\gcd(2^k - 1, 2^l - 1) = 2^d - 1$.

The problem also says "hint: use Euclid's lemma," although I'm not sure what part of the problem should be applied to. I've been thinking about it for a few days and I'm completely stumped.

I'm not really even sure how to show that it divides either $2^k - 1$ or $2^l - 1$. From the given, we know that $\exists c: dc = k$. We need to show that $\exists c': (2^d - 1)c' = 2^k - 1$. Obviously $c' = 2^c$ gives you $(2^d - 1)c' = 2^k - 2^c$, but I can't figure out how to get rid of the extra terms that the $- 1$ brings in in various places.

From Euclid's lemma on the left side, you know $\exists i: di = k - l$, and applying it on the right side, you know it suffices to show that $\gcd(2^k - 2^l, 2^l - 1) = 2^d - 1$. And by Bezout's identity, it's enough to show that $2^d - 1$ can be written as a linear combination of either of those things.

Can anyone give me a hint?

Answer 1

Hint $\ $ By below $\rm\,\ a^M\!-\!1,\:a^N\!-\!1\ $ and $\rm\,\ a^{\,(M,N)}\!-\!1\ $ have the same set $\rm\,S\,$ of common divisors $\rm\,d,\, $ therefore they have the same greatest common divisor $\rm\ (= \max\ S).$

$$\begin{eqnarray}\rm\ \ mod\,\ d\!:\ \ a^M,\:a^N\equiv 1&\iff&\rm ord(a)\ |\ M,N\color{#c00}\iff ord(a)\ |\ (M,N)\iff a^{\,(M,N)}\equiv 1\\ \rm i.e.\ \ \ d\ |\ a^M\!-\!1,\:a^N\!-\!1\ &\iff&\rm\ d\ |\ a^{\,(M,N)}\!-\!1,\qquad\ \ \, where \rm\quad\! (M,N)\, :=\, gcd(M,N) \end{eqnarray}$$

Remark $\ $ Above we used $\ a\mid b,c \color{#c00}\iff a\mid (b,c),\ $ the fundamental universal property of the gcd. Compare $\, a<b,c \!\iff\! a< \min(b,c),\ $ and $\ a\subset b,c\iff a\subset b\cap c.\,$ Exploiting such universal "iff" definitions allows us to quickly and easily simultaneously prove both directions of the equivalence.

The conceptual structure that lies at the heart of this simple proof is the ubiquitous order ideal. $\ $ See my post here for more on this and the more familiar additive form of a denominator ideal.

More generally: $\rm\ gcd(f(m), f(n))\, =\, f(gcd(m,n))\ \ \, if\, \ \ f(n) \equiv f(n\!-\!m)\,\ \ (mod\ f(m)),\, $ and $\rm\, f(0) = 0.\, $ See my post here for a simple inductive proof.

In fact there is a q-analog: the result also holds true for polynomials $\rm \ f(n) = (x^n\!-\!1)/(x\!-\!1),\, $ and $\rm\ x\to 1\ $ yields the integer case (Bezout identity) - see my post here for a simple proof.