Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$

Question

Somewhere on Stack Exchange I saw the equation

$$\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1.$$

I had never seen this before, so I started trying to prove it. Without success...

Can anyone explain me (so actually prove) why this equation is true?

And can we say the same when replacing the '$2$' by any integer number '$a$'?

Answer 1

In general, if $p=\gcd(m,n)$ then $p=mx+ny$ for some integers $x,y$.

Now, if $d = \gcd(2^m-1,2^n-1)$ then $2^m \equiv 1 \pmod d$ and $2^n \equiv 1\pmod d$ so $$2^p = 2^{mx+ny} = (2^m)^x(2^n)^y \equiv 1 \pmod d$$

So $d\mid 2^p-1$.

On the other hand, if $p\mid m$ then $2^p-1\mid 2^m-1$ so $2^p-1$ is a common factor.

And yes, you can replace $2$ with any $a$.

Answer 2

Let $(m,n)=p$, then $p\mid m$, and $p\mid n$, then $m=m_1p$, $n=n_1p$, $(m_1,n_1)=1$, then $(2^{m_1p}-1,2^{n_1p}-1)=((2^{p})^{m_1}-1,(2^{p})^{n_1}-1)=((2^{p}-1)(.....),(2^{p}-1)(.....))=(2^{p}-1)$

Answer 3

Suppose $x$, $m$ and $n$ are positive integers with $m$ and $n$ coprime. First let us show that $$r = 1 + x + {x^2} + \ldots + {x^{m - 1}}$$ and $$s = 1 + x + {x^2} + \ldots + {x^{n - 1}}$$ are relatively prime. If $d$ is a common divisor of $r$ and $s$, then $d$ is relatively prime to $x$ because $r$ and $s$ are one more than a multiple of $x$. Let $m$ be greater than $n$ (or vice versa) and consider $$r - s = {x^n} + {x^{n - 1}} + \ldots + {x^{m - 2}} + {x^{m - 1}} = {x^n}(1 + x + \ldots + {x^{m - n - 1}})$$ and notice that $d$ divides $r - s$ and so must be a divisor of $1 + x + \ldots + {x^{m - n - 1}}$. Observe that $m - n$ is relatively prime to both $m$ and $n$, so we can likewise use geometric sums which eventually becomes shorter and shorter until we conclude that $d$ must divide 1 i.e. $d = 1$. Now if we let $$d' = \gcd (m',n')$$ with $m' = md'$ and $n' = nd'$, then $m$ and $n$ are coprime and $${2^{m'}} - 1 = ({2^{d'}} - 1)(1 + {2^{d'}} + {2^{2d'}} + \cdots + {2^{(m - 1)d'}})$$ $${2^{n'}} - 1 = ({2^{d'}} - 1)(1 + {2^{d'}} + {2^{2d'}} + \cdots + {2^{(n - 1)d'}})$$ which are geometric sums with $x = {2^{d'}}$ and we showed that $\gcd (r,s) = 1$. This completes the proof.

Answer 4

Yes, you can say the same when replacing $2$ with an integer $a \geqslant 2$.

Lemma. Suppose that $a \geqslant 2$, $m, n \in \mathbb{N}$ and $\gcd(m, n)=1$. Then $\gcd(a^m-1, a^n-1)=a-1$.

Proof. It is obvious that $(a-1) | \gcd(a^m-1, a^n-1)$. So, we only need to prove that $\gcd(a^m-1, a^n-1) | (a-1)$.

It is well known that if $\gcd(m, n)=1$, then there exist $k, l \in \mathbb{N}$ such that $mk-nl=1$. If is obvious that $(a^n-1)|(a^{nl}-1)$, therefore $$ \gcd(a^m-1, a^n-1) | (a^{nl}-1), $$ and for the same reason $$ \gcd(a^m-1, a^n-1) | (a^{mk}-1). $$

Now we just observe that $$ (a^{mk}-1)-a\cdot(a^{nl}-1) = (a^{nl+1}-1)-(a^{nl+1}-a) = a - 1, $$ therefore $$ \gcd(a^m-1, a^n-1) | (a-1), $$ QED.

Now we can prove the main statement: for $b \geqslant 2$ we have: $$ \gcd(b^m-1, b^n-1) = b^{\gcd(m,n)}-1. $$ Proof. Set $a = b^{\gcd(m, n)}$, $m'=m/\gcd(m,n)$ and $n'=n/\gcd(m,n)$. Clearly, $\gcd(m',n')=1$, and by the lemma we have $$ \gcd(a^{m'}-1,a^{n'}-1) = a-1, $$ which is exactly what we need, QED.

Answer 5

Hint $\rm\bmod d\!:\ a^{\large m}\!\equiv 1\equiv a^{\large n}\!\iff order(a)\mid m,n\iff order(a)\mid(m,n)\iff a^{\large (m,n)}\!\equiv 1$

Therefore $\rm\ \ d\mid a^{\large m}\!-1,\,a^{\large n}\!-1$ $\iff$ $\rm\:d\mid a^{\large (m,n)}\!-1,\ \,$ hence $\rm\,\ \ (a^{\large m}\!-1,\,a^{\large n}\!-1)\, =\, a^{\large (m,n)}-1$

Answer 6

This problem can be solved in a very simple manner by throwing everything wisely into induction. Rephrase the statement and do strong induction on $a$. The base case of $a=1$ is trivial to verify. Now let $a\geq1$ be arbitrary and assume that the statement is true for any $k\leq a-1$, that is, for any such $k$, $\gcd(2^k-1,2^b-1)=2^{\gcd(k,b)}-1$ for any $b$. Now, if $b<a$ we are done by the induction hypothesis, so let us focus on the case $b\geq a$. Assume that $b=aq+r, 0\leq r< a$; then, (using the binomial theorem) $$2^b-1=((2^a-1)+1)^q 2^r-1=(\{Integer\; Expression\} (2^a-1)+1)\;2^r-1=\{Integer\; Expression\}\;2^r\; (2^a-1)+(2^r-1).$$ This transforms the problem into calculating $gcd(2^a-1,2^r-1)$, which, by the induction hypothesis is $2^{\gcd(a,r)}-1=2^{\gcd(a,b)}-1$, by the Euclidean Algorithm.