Intereting Posts

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Let $(2^m-1,2^n+1)=1$ and suppose $m$ is even. Also, $m,n,k\in \mathbb{N}$. We have

$$(2^{m}-1)x+(2^n+1)y=1$$

$$(2^{2k}-1)x+(2^n+1)y=1$$

$$(2^{2k}-1)y\equiv 1\pmod{2^n+1}$$

$$(2^k+1)(2^k-1)y\equiv 1\pmod{2^n+1}$$

I don’t konw now to procede from here….

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$\overbrace{p\mid\color{#0a0}{2^{\large m}\!-\!1}}^{\color{#08f}{\Large\Rightarrow\,\ p\: \rm\ odd}},\color{#c00}{2^{\large n}\!+\!1}\,$

$\Rightarrow\,

{\rm mod}\ p\!:\ \color{#0a0}{2^{\large m}\! \equiv}\!\!\!\!\!\!\!\!\overset{\Large\ [\,\color{#c00}{ -1\ \equiv\ 2^{\LARGE n}}]^{\LARGE\color{#90f}2}\ \ \ \ }{ 1\color{#90f}{\equiv {2^{\large 2n}}}}\!\!\!\!\!\!\!\!\!$

$\Rightarrow\,

{\rm ord}\,2\mid\overbrace{(\color{#0a0}m,\color{#90f}{2n})= (\color{#0a0}m,\color{#90f}n)}^{\Large\rm\ \ by\ \color{#0a0}m\ odd}\,$

$\Rightarrow

\overbrace{\color{#90f}{\bf 1}\equiv\color{#c00}{2^{\large\color{#90f} n}\!\equiv -1}}^{\color{#08f}{\Large\Rightarrow\,\ p\ \mid\ 2\quad\:\ }}$

**Remark** $\ $ Above is special case $\,m$ odd, $\,k=2\,$ of the following

**Lemma** $\,\ \ a^{\large m}\equiv 1\equiv a^{\large kn},\ $ $(m,k) = 1\,$ $\Rightarrow\, a^{\large (m,n)}\equiv 1\,\Rightarrow\,a^{\large n}\equiv 1$

**Proof** $\ \ {\rm ord}\ a\mid m,kn\, \Rightarrow\, {\rm ord}\ a\mid \underbrace{(m,kn)\!=\!(m,n)}_{\large (m,k)\,=\,1}\mid n\,\Rightarrow\, a^{\large n}\equiv 1$

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