Find all positive integer $m$ such $$2^{m}+1\mid5^m-1\,.$$

It seem there no solution, I think it might be necessary to use quadratic reciprocity knowledge to solve this problem.

Let $M=2^m+1$, so we have $$5^{m}\equiv 1\pmod{M}. $$ If $m$ is odd, then we have $$\left(5^{\frac{m+1}{2}}\right)^2=5\pmod {M}\Longrightarrow \left(\dfrac{5}{M}\right)=1,$$ so $$\left(\dfrac{M}{5}\right)(-1)^{\frac{(5-1)(M-1)}{4}}=\left(\dfrac{M}{5}\right)(-1)^M=-\left(\dfrac{M}{5}\right).$$


closed as off-topic by davidlowryduda Dec 15 '18 at 16:47

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    $\begingroup$ What did you try? $\endgroup$ – Robert Z Dec 13 '18 at 8:15
  • $\begingroup$ @RobertZ I have post my some try,Following step I can't do it $\endgroup$ – function sug Dec 13 '18 at 8:22
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    $\begingroup$ Are we assuming $(2^m+1)\mid(5^m-1)$? $\endgroup$ – manooooh Dec 13 '18 at 8:26
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    $\begingroup$ @manooooh,yes. it's right $\endgroup$ – function sug Dec 13 '18 at 8:27
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    $\begingroup$ If $m$ is odd then $2^m +1$ is divisible by 3 but $5^m - 1$ is not. No need for quadratic reciprocity $\endgroup$ – user420261 Dec 13 '18 at 8:34