Let $m,n$ be positive integers, and $m>2$, show that $$(2^m-1)\nmid(2^n+1)$$
if $m=n$ it is clear hold, because $\dfrac{2^n+1}{2^n-1}=1+\dfrac{2}{2^n-1}\notin N$
and if $m>n$, then we have $$2^m-1\ge 2^{n+1}-1>2^n+1$$ then this case also hold. But $m<n$ it seems hard to prove it
and maybe this problem have other methods,Thanks
We'll use the following result a few times:
${\;\;\;\;\;\;}$ Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$
Suppose that $2^m-1\mid2^n+1$.
Then $$\begin{align*}2^m-1\mid2^{2n}-1&\implies\gcd(2^m-1,2^{2n}-1)=2^m-1\\ &\implies2^{\gcd(m,2n)}-1=2^m-1\\ &\implies \gcd(m,2n)=m\\ &\implies m\mid2n.\qquad(1)\end{align*}$$
On the other hand $$\begin{align*}&\gcd(2^m-1,2^n-1)\mid\gcd(2^n+1,2^n-1)=\gcd(2^n+1,2)=1\\ &\implies2^{\gcd(m,n)}-1=1\\ &\implies\gcd(m,n)=1.\qquad(2)\end{align*}$$
Combining $(1)$ and $(2)$ gives $m\mid2$.
Note: In exactly the same way we can prove that $3^m-1\mid3^n+1\implies m\mid2$. For $a>3$, $a^m-1\mid a^n+1$ is impossible because the RHS is not divisible by $a-1$.
Side remark: You may note that the deduction of $(1)$ generalizes to:
${\;\;\;\;\;\;}$ Showing that $a^n - 1 \mid a^m - 1 \iff n \mid m$
The quesion is whether it is possible that $$ 2^m-1 | 2^n +1 $$
Of course, in general this requires that $m \lt n$ (except possibly $m,n$ small numbers, see below for this).
So let $n=am+r$ where $0 \le r \lt m$ .
Then we can derive
$\qquad 2^{am+r} +1 = 2^{am}2^r +1 \\ \qquad \qquad \qquad = (2^{am}-1)2^r +2^r+1 \\ \qquad \qquad \qquad = (2^{am}-1)2^r +(2^r+1) $
It is well known, that $2^{am}-1$ is divisible by $2^m-1$ , so if indeed $(2^m-1) | (2^n-1)$ we have
at one hand that $(2^m-1) | (2^{am}-1)$
and on the other hand it must thus also be true, that $2^m-1 | 2^r+1$
But this can only be true if $m=2,r=1$ and $2^2-1 = 3 |2^1+1 = 3 $ and never else because $r<m$ by definition.
Deny. $ $ Then $\ \color{darkorange}{m\mid 2n}\ $ by $\,\ 2^{\color{darkorange}m}\!-\!1\mid 2^n\!+\!1\mid 2^{\color{darkorange}{2n}}\!-\!1.\ $ Note that $\ \color{#c00}2 = {\color{#0a0}{2^n\!+\!1-(2^n\!-\!1)}}\ $ so
$2\nmid m\ \ \,\Rightarrow\,\ \ \color{darkorange}{m\mid n} \ \,\Rightarrow\,\ 2^m\!-\!1\mid \color{#0a0}{2^n\!+\!1,\,2^n\!-\!1} \ \Rightarrow\,\ 2^m\!-\!1\mid\color{#c00}2 \,\Rightarrow\, m=1,\ $ else
$2\mid m\,\Rightarrow\, \color{darkorange}{m/2\mid n}\, \Rightarrow\!\!\!\!\! \underbrace{2^{m/2}\!\!-\!1\mid \color{#0a0}{2^n\!+\!1}}_{\Large 2^{\Large m/2}-1\mid 2^{\Large m}-1\mid 2^{\Large n}+1 }\!\!\!\!\!\!, \color{#0a0}{2^n\!-\!1} \Rightarrow 2^{m/2}\!-\!1\mid\color{#c00}2 \,\Rightarrow\, m=2,\: $ contra $\ m>2$.
