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Let $a$ be a positive integer that satisfies $a>3$.

  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: Does there exist 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 2. question. Are there any ways to solve the questions ?

(Sorry, English is my second language)

apple
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  • For $a=2$ no such integer exists. – Servaes Oct 17 '18 at 13:24
  • @Servaes Sorry, it's actualy $a>3$. I have edited it. – apple Oct 17 '18 at 13:26
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    What is the source of this problem? Also, what techniques have you tried / progress have you made? Have you, for example, found any such pairs $(a,m)$, or proven that none exist for any specific $a>3$? – Carl Schildkraut Oct 17 '18 at 13:56
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    I did not find a solution for $\ \ a=12\ \ $ yet – Peter Oct 17 '18 at 16:48
  • @CarlSchildkraut : Sorry, I have written it wrong, in fact $a$ is given in the first place, not "For all $a$'s". I have tried to use LTE, but it may not work. (Sorry, English is my second language) – apple Oct 17 '18 at 18:53
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    Iterating to 10,000 on both a and m produces pairs: (4,9), (4,477), (4,1413), (4,6105), (5,9), (5,308), (6,5), (6,95), (6,1105), (7,6), (7,15), (7,18), (7,165), (7,174), (7,318), (7,375), (7,2346), (7,2431), (7,4195), (7,4518), (8,25), (8,35), (8,985), (8,5411), (8,8275), (8,9065), (9,49), (10,3), (10,9), (10,81), (10,1719), (11,50), (11,63), (11,6150), (11,6375), (13,405), (13,1785), (13,3825), (13,4620), (13,9093), (14,55), (14,65), (14,81), (14,291), (14,897), (14,5945), (14,7293), (15,14), (15,77), (15,98), (15,143), ... – Russ Oct 18 '18 at 01:58
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    @Servaes not true, when $a = 2$, $m = 49735$ works (but is the first such $m$), as does $m = 717025$ – Christopher Oct 18 '18 at 13:19
  • @Christopher That's a great find! I'll revisit my argument. How did you find this solution? – Servaes Oct 18 '18 at 19:09
  • @Servaes an unsubtle computer search – Christopher Oct 18 '18 at 19:10
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    After a few days improving my program, I found$$12+1,022,924,329\mid12^{1,022,924,329}+1$$ – Julián Aguirre Oct 20 '18 at 13:13

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