Let $a$ and $b$ be nonnegative integers. prove that $\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1$

Question

Let $a$ and $b$ be nonnegative integers. prove that $\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1.$

So far I have tried using Linear Diophantine Equations, the GCD With Remainder theorem (replacing one input into the gcd with the remainder of the bigger term divided by the smaller term), and writing $d=gcd(a,b)$, let $a=ds, b=dr$, but I am still stuck.

Thanks.

$\gcd(8,4)=4\neq 2^{gcd(4,3)}-1=2^1-1=1$ are you sure you have written it correctly. — zwim, Dec 03 '17 at 04:29
Yep, it comes from the factorisation $X^n-1=(X-1)(1+X+...+X^{n-1})$ with $X$ a certain power of $2$. — zwim, Dec 03 '17 at 04:37

score 0 · Answer 1 · answered Dec 03 '17 at 04:51

Let $d = gcd(a, b), a > b, (x, y) := gcd(x, y)$. $$(2^a-1, 2^b-1) = (2^{a}(2^{b-a}-1), 2^a-1) = (2^{b-a}-1, 2^a-1) = \ldots = 2^d - 1 $$ Compare the above with the derivation of d = gcd(x, y).

For time reason I cannot write it in detail. I am sorry and I will edit this post latter.

Let $a$ and $b$ be nonnegative integers. prove that $\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1$

1 Answers1