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

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    $\begingroup$ Interesting question. Have you tried anything yet? $\endgroup$ – Mohammad Zuhair Khan Oct 24 '18 at 8:30
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    $\begingroup$ The case $a=2$ is popular, see here and the related links. $\endgroup$ – Dietrich Burde Oct 24 '18 at 8:48
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    $\begingroup$ $n$ is not a prime number. It also must be odd. $\endgroup$ – Oldboy Oct 24 '18 at 18:16
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    $\begingroup$ My original problem: Let $a\in\mathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)\mid \left(a^n+1\right).$$ $\endgroup$ – Drona Oct 26 '18 at 19:54
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    $\begingroup$ @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. $\endgroup$ – Oldboy Oct 27 '18 at 8:39


$a^{p_1} ≡ a \mod p_1$

$a^{p_2 } ≡a \mod p_2$

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

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

$a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$

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.


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