if $(m,n)=1$ is this true that $(2^m-1,2^n-1)=1$?

Question

if $(m,n)=1$ is this true that $(2^m-1,2^n-1)=1$ ?

Observing $2^i-1$'s shows that it seems true! But how to prove it?

$2 \rightarrow 3$

$3 \rightarrow 7$

$4 \rightarrow 15$

$5 \rightarrow 31$

$6 \rightarrow 63$

Yes it is true, see here: http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 — pritam, Dec 01 '13 at 13:04

Konstantinos Gaitanas · Answer 1 · 2013-12-01T14:00:42.887

4

Suppose the contrary:
Let $n$ be the smallest natural number for whom there exists a natural number $m$ with $n<m$ and $(m,n)=1$ but $(2^m-1,2^n-1)=d>1$

Let $m=kn+a$ with $1<a\leq n-1$. If $d|2^m-1$ and $d|2^n-1$ then $d|2^a-1$.
If $a=1$ then $d|1$ which means $d=1$ a contradiction.
If $a>1$ then $2^a-1<2^n-1$ has a common divisor with $2^m-1$ which contradicts that $n$ was the smallest proper number for our claim.

edited Dec 01 '13 at 14:00

answered Dec 01 '13 at 13:21

Konstantinos Gaitanas

9,115

What if $k=0$ and $a=n$ ? Shouldn't be $n=km+a$ and by minimality $k>0$ ? – Xoff Dec 01 '13 at 13:27
if $k=0$ then $m=a$ – Konstantinos Gaitanas Dec 01 '13 at 13:32
There is some confusion. At first $m$ is the smallest number, and in the last sentence, it's $n$. – Xoff Dec 01 '13 at 13:56
@Xoff it needed a small change, see now. – Konstantinos Gaitanas Dec 01 '13 at 14:01
1

ok, it's a nice answer :) – Xoff Dec 01 '13 at 15:07
But $n,m$ are arbitrary! Does your answer apply to all $m,n$ ? – Mahdi Khosravi Dec 01 '13 at 15:08
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@Mahdi Khosravi arbitrary?No my friend. $m$ seems to be arbitrary but it is not. $n$ is defined in a unique way and $m$ comes in front because of $n$ -and so neither $n$ nor $m$ are not arbitrary- – Konstantinos Gaitanas Dec 01 '13 at 16:03
@KonstantinosGaitanas I mean the only restriction on $m$ and $n$ is that they are relatively prime, nothing else. – Mahdi Khosravi Dec 01 '13 at 16:29
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@MahdiKhosravi what else do you need?and $n$ is the s m a l l e s t one with the mentioned property(another restriction) – Konstantinos Gaitanas Dec 01 '13 at 19:37

hmakholm left over Monica · Answer 2 · 2013-12-01T13:01:36.723

Euclid's algorithm on $(m,n)$ ends with $1$. Apply the map $x\mapsto 2^x-1$ to each of the intermediate values in the algorithm, and you get a way to make $1$ from $2^m-1$ and $2^n-1$ by successively subtracting multiples of one from the other.

If $m \ge an \ge 1$, then $$2^{m-an}-1 = (2^m-1) - \left(\sum_{j=1}^{a} 2^{m-jn}\right)(2^n-1) $$

In fact, the resulting sequence is exactly what the Euclidean algorithm will produce when started in $(2^m-1,2^n-1)$.

if $(m,n)=1$ is this true that $(2^m-1,2^n-1)=1$?

2 Answers2