Let $a$ be a positive integer.

  1. Does there exist a positive integer $m$ other than 1 such that: $$a+m \ |\ a^m+1 \ $$ If not, what are the conditions of $a$ for $m$ to exist?

  2. If there exist such number $m$, are there infinitely many integers $m$ that satisfy $$a+m \ |\ a^m+1 \ $$ If not, what are the conditions of $a$ so that there are infinitely many integers $m$ that satisfy the question?

This problem I think might be similar to: Given $a$, does there exist an integer $m>1$ such that: $$m \ |\ a^m+1 \ $$ I have tried to use the Lifting the Exponent Lemma and I found that if $m|a+1$, $m$ is odd and $a+1$ is not a power of $2$, then $m|a^m+1$. But for the question $a+m \ |\ a^m+1 \ $, I cannot make any progress, especially the 2nd question. Are there any ways to solve the questions ?

(Sorry, English is my second language)

  • 1
    $\begingroup$ Do you have a particular reason to exclude $a=2$? for $a=3$ one has the solution $m=2$. Have you obtained other solutions? Do you have some particular motivation? $\endgroup$
    – YCor
    Oct 21, 2018 at 8:03
  • $\begingroup$ @YCor In fact I haven't found any solutions for $a=2$ yet. I have edited my question. $\endgroup$
    – apple
    Oct 21, 2018 at 10:04
  • $\begingroup$ $m=43\cdot 281\cdot 331\cdot 5419-2$ is an odd multiple of $105=3\cdot 5\cdot 7$, so $2^{m}+1$ is divisible by $2^{105}+1$, which is a multiple of $m+2$ ($43$, etc. are prime factors of $2^{105}+1$, thanks to alpertron.com.ar/ECM.HTM ). This makes me suspect that the answer is "yes" more often than not, but I don't have a proof of anything yet. $\endgroup$
    – fedja
    Oct 23, 2018 at 1:32

1 Answer 1


If $2a-1$ is a prime number and $a^{a-1}\equiv -1 \pmod{2a-1}$, then we have a solution with $m=a-1$. For example, take $a=6$ or $7$.

  • $\begingroup$ Thanks for your answer. Can you find any $m$ for other $a$'s ? $\endgroup$
    – apple
    Oct 21, 2018 at 10:05

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