# Are there infinitely many integers $m$ such that $a+m$ divides $a^m+1$?

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)

• 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? – YCor Oct 21 '18 at 8:03
• @YCor In fact I haven't found any solutions for $a=2$ yet. I have edited my question. – apple Oct 21 '18 at 10:04
• $m=43\cdot 281\cdot 331\cdot 5419-2$ 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. – 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 $$7$$.
• Thanks for your answer. Can you find any $m$ for other $a$'s ? – apple Oct 21 '18 at 10:05