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Let $a\in\mathbb Z$ and $a\gt3$. Prove that there exist infinitely many positive integers $n$ satisfying $(n+a)\mid(a^n+1)$.

This problem was mentioned for the first time in this post, so all the credits should go to Drona. The author (wrongly, I think) thought that the two problems were equivalent. I made a comment about that but it went unnoticed because it was the last one in a pretty long chain. I asked Drona to post the original question but did not hear from him since then. I believe that this problem is too interesting to be left buried in some hidden comment, so I decided to post it here.

It's relatively easy to prove that $a$ and $n$ must be coprime. But apart from that simple fact I did not get much further.

Saša
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  • For $a=12$, can you find any qualifying positive integer value of $n$, other than $n=1$? – quasi Oct 28 '18 at 10:01
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    Unless you know the statement is true, to be fair to would-be solvers, you should state the problem as either a question (e.g., must there be infinitely many . . . ?), or a conjecture, or a prove/disprove statement, rather than a claim. – quasi Oct 28 '18 at 10:13
  • Also, why did you skip the cases $a=2$, and $a=3$? – quasi Oct 28 '18 at 10:40
  • In the referenced post, Drona never claimed the problems were equivalent. What's true (as Drona surely knew) is that the two problems are related in the following way: If $a$ is a positive integer such that $k{,\mid,}(a^k - a + 1)$ for infinitely many positive integers $k$, then $(n+a){,\mid,}(a^n+1)$ for infinitely many positive integers $n$. – quasi Oct 28 '18 at 22:07
  • @quasi He started his problem with: "I have a problem, and it leads to the following problem..." – Saša Oct 28 '18 at 22:12
  • What he mean't was that if he could solve that one with a "yes" answer, then he could solve his original one with a "yes" answer. – quasi Oct 28 '18 at 22:14
  • Isn't this the same problem? – Yong Hao Ng Oct 29 '18 at 02:49
  • I'm interested in the problem at https://math.stackexchange.com/questions/2968838/an-a-1-divisible-by-n, not this problem. I did not say that the two problems were equal. This problem is only a consequence of the problem I mentioned. Do you understand? – Drona Oct 29 '18 at 08:12
  • @YongHaoNg It definitely is. Somehow MSE did not suggest an obvious duplicate. – Saša Oct 29 '18 at 08:49
  • @quasi $12+1,022,924,329\mid12^{1,022,924,329}+1$, found by Julián Aguirre. – Saša Oct 29 '18 at 08:51
  • So $a=12$ finally succumbed, but it held out for quite a while! – quasi Oct 29 '18 at 08:56
  • My prior point is still valid. Unless you actually know that a statement is true, it's misleading to state the problem as "Prove . . ." – quasi Oct 29 '18 at 08:59

2 Answers2

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Ok, so this may not be a proper solution or a solution at all but I'll try my best. Sorry in advance.

If $ (a+n)|(a^n+1)$ then by long division we get the remainder $(-1)^kn^ka^{n-k} +1$ for each k times that we divide. When we keep on dividing then k finally becomes $n$. Then the remainder becomes $(-1)^nn^n+1$. Since $(a+n)|(a^n+1) \rightarrow (a+n)|((-1)^nn^n+1)$ in order for the remainder to vanish. The term $(-1)^nn^n+1$ is dependent only on n, and we can find its factors($x$) greater than $3+n$ where they exist. Then we have the desired values of $n$ for each $a = x-n$.

nnut
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  • Basically, you have replaced the original problem with the following one: Prove that there exists $n$ such that $(-1)^nn^n+1$ has a factor greater than 3+n. It could be a way to proceed but it's still not a complete solution. – Saša Oct 30 '18 at 10:07
  • Yes, I couldn't make any argument except that "hey that's obvious, if you don't find such factors for the term, then choose another n". – nnut Oct 30 '18 at 10:25
  • Many terms definitely has such factors when n is odd. But I'm all out of arguments. – nnut Oct 30 '18 at 10:27
  • True, most numbers like $(-1)^nn^n+1$ have factors and for each factor you will get some value for $a$. But you cannot guarantee that every possible value of $a$ will be hit. Actually, you are proving that there is a solution for some values of $a$ but not for all of them. Don't get me wrong, it's still something :) – Saša Oct 30 '18 at 10:38
  • Yes, I realize the mistake. Should I delete the post? – nnut Oct 30 '18 at 10:40
  • No, no... It's not a full solution but it's a legitimate idea. It is possible that your equivalent statement is easier to prove than the original one. Someone could still find that inspiring. – Saša Oct 30 '18 at 10:46
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This is also not a proper solution but might be a good direction to follow. Suppose $a-1$ is not a power of 2 and is not a prime number. Let $s\geq 3$ be an odd prime factor of $a-1$ and $r=(a-1)/s \geq 2$. Suppose that $A=a^{r}+1$ has a prime factor $p$ greater than $\sqrt{A}$. Then $p>\sqrt{A}>a$. We claim that $n+a \mid a^n+1$, where $n=p-a>0$. One has, by Fermat's little theorem, that $$a^n+1 \equiv a^{p-a}+1 \equiv a^{1-a}(a^{a-1}+1) \equiv a^{1-a}(a^{r}+1)B \equiv 0 \pmod p,$$ for some integer $B$.

The probability of a number $A$ having a prime factor greater than $\sqrt{A}$ is $\log 2$. So this argument proves the claim for at least 30 percent of numbers.

Marco
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