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 '18 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 '18 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 '18 at 1:32

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 $7$.

| cite | improve this answer | |
  • $\begingroup$ Thanks for your answer. Can you find any $m$ for other $a$'s ? $\endgroup$ – apple Oct 21 '18 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.