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Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $ divisible by $n$?

Klangen
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Drona
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    Interesting question. Have you tried anything yet? – For the love of maths Oct 24 '18 at 08:30
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    The case $a=2$ is popular, see here and the related links. – Dietrich Burde Oct 24 '18 at 08:48
  • I doubt that anyone can fully answer this problem. – Piquito Oct 24 '18 at 11:19
  • OEIS contains the list of the least $n$ for each $a\geq 3$. – Wojowu Oct 24 '18 at 11:31
  • @Drona I think this question is false. We need some information like that $n\mid a$. or just $n$ divisible by $a^n+1$. Check, please. – 1ENİGMA1 Oct 24 '18 at 11:39
  • It is fact that $n\mid a^n+1$. Why did this necessitate $n\mid a^n-a+1$ – 1ENİGMA1 Oct 24 '18 at 12:25
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    $n$ is not a prime number. It also must be odd. – Saša Oct 24 '18 at 18:16
  • OK Oldboy. I have tried to solve the problem in the case of $n = pq$, where $p,, q$ are prime numbers. Then the problem becomes: prove that there exist prime numbers $p$ and $q$ satisfying $a ^ p-a + 1$ divisible by $q$ and $a ^ q-a + 1$ divisible by $p.$

    But, that is also very difficult!

     – Drona Oct 24 '18 at 19:34
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    I doubt that we can solve the problem in the general form. Looking at the OEIS sequence mentioned by @Wojowu it seems that itćs extremely difficult to find a single solution, not infinitely many. For example, for $a=6$ the smallest solution is $n= 4021227877$. And for some relatively small values of $a$ the smallest value of $n$ is still unknown. Why don't you post a simpler problem? Like: "Prove that for a=4 there is at least one solution". Maybe we learn something from it. You also mentioned that your original problem leads to the problem described here. Can you post the original problem? – Saša Oct 26 '18 at 19:47
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    My original problem: Let $a\in\mathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)\mid \left(a^n+1\right).$$ – Drona Oct 26 '18 at 19:54
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    @Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it. – Saša Oct 27 '18 at 08:39
  • @Drona: The original problem is from what source? – quasi Oct 28 '18 at 21:42
  • @Drona: Oldboy has posted your original problem as a new question: https://math.stackexchange.com/questions/2974395 – quasi Oct 28 '18 at 22:27
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    @Oldboy: There is no mistake here! I did not say the original problem was equivalent to the one I mentioned here. The problem I have raised, is a strong expansion from the original problem. – Drona Oct 29 '18 at 08:18
  • @Oldboy, but the number you gave for a=6 ie. 4021227877 is prime and does not satisfy the relation! it satisfy $a^n-a≡0\mod n$. – sirous Nov 03 '18 at 08:29
  • @sirous According to OEIS sequence A128149: $6^ {4021227877} \equiv 5 \bmod 4021227877$ which means that $4021227877 \mid 6^ {4021227877} - 6 +1$. Check for yourself. – Saša Nov 03 '18 at 08:51
  • $(a, n)=1$ ⇒ $a^{n-1} ≡1\mod n$ ⇒ $a^n ≡a\mod n$ ⇒ $a^n -a≡0\mod n$ ⇒ $6^{4021227877}-6 ≡0\mod (4021227877)$⇒ $6^{4021227877}-7+1 ≡0\mod (4021227877)$. There must be a mistake! – sirous Nov 03 '18 at 10:03

1 Answers1

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$N=a^n-a+1$

$a^{p_1} ≡ a \mod p_1$

$a^{p_2 } ≡a \mod p_2$

$(a^{p_1})^{p_2} ≡ a^{p_2} \mod p_1 ≡(a\ mod p_2) \mod p_1= k_1 p_1 + k_2 p_2 + a$

$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$

⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$

If $n=p_1p_2 | a^n-a+1$ then we must have:

$p_1p_2 | k_1p_1 + k_2p_2+1$

So we have following linear equation:

$k_1p_1 + k_2 p_2 = m p_1p_2-1$

For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:

with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:

$k_1= 7 t + 11$ and $k_2= -5 t+7$.

Now in first step problem reduces to:

Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.

For example for $a=3$, $p_1=5$ and $p_2=7$ we have:

$3^{35}-3+1=105$ and $(35, 105)=5$

In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.

Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1\mod a$ and $p_2≡1\mod a$, then:

$k_1p_1+k_2p_2= M(a)$

Or: $a | k_1p_1+k_2p_2$

We can see this in solution $n=409\times 9831853$ for $a=6$; $409=48\times 6 +1$ and $9831853=1638642 \times 6 +1$

This can help us in choosing a and primes $p_1$ and $p_2$

I see no reason for the lack of more solutions.

sirous
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