Find all positive integer satisfying $5^n+1 \vdots n$
Here is my attempt: Introducing $p$ as the smallest prime divisor of $n$. Let $ord_p(5)=h \in \mathbb{N^*}$ Therefore $h \mid 2n; h \mid p-1$ on account of FLT. Then $h \mid \text{GCD} (p-1;2n)$. Since $p$ is the smallest prime divisor of $n$, $\text{GCD} (p-1;n)=1$. Hence, $h \mid 2 \implies h \in \{ 1;2\} \implies p \in \{ 2;3 \}$.
- $p=2$. Let $n=2^k.t; k;t$ are positive integer and $t$ is odd. We have $2^k \mid 5^{2^k.t}+1=(5^{2^k}+1)[5^{2^k(t-1)}-5^{2^k(t-2)}+...+1] $. Noting that $t$ is odd, the formula in the bracket is odd. Therefore $2^k \mid 5^{2^k}+1 \equiv 2 \pmod 4 \implies k=1$. Therefore $t \mid 5^{2t}+1$.
It is noticeable that $t=1;n=2$ satisfies the conditions.
When $t \ge 3$ then there exist the smallest prime divisor $q$ of $t$. Let $ord_q(5)=g$. We have $g \mid q-1; g\mid 4t$; this implies $g \mid 4$ since $\text{GCD} (q-1;t)=1$. Hence, $q \in \{ 3;13 \}$. Since $5^{2t}+1 \equiv 2 \pmod 3$, we have $q=13 \implies t=13^s.u;u;s \in \mathbb{N^*}; (u;(2;3;5;7;11;13))=1$.
+, It is easy to see that $n=2.13^s; \forall s$ satisfies the conditions as according to LTE lemma, we have $v_{13}(5^n+1)=v_{13}(25^t+1)=v_{13}(25+1)+v_{13}(t)=s+1 > s=v_{13}(13^s); 5^n+1 \vdots 2$. +, However, I can't solve for $u > 1$, $p=3$.
** Despite what I've illustrated, I found out that if $5^w+1 \vdots r$; for some $r$ prime then $n \vdots r$. I'm looking forward to the solution to this problems :)
It can be proved that for any number $a \in \mathbb N$ such that $a\neq 2^k; k\in \mathbb N$ there are infinitely many number like $n\in\mathbb N$such that $n|a^n+1$.For proving this theorem following Lemma is used:
Lemma: If for integer $k\geq0$, $a^{p^k}+1$ is divisible by $p^{k+1}$ (a is natural number and p is odd prime), then:
$p^{k+2}|a^{p^{k+1}}$
For when n is odd and $a=5$ we may write:
$(6-1)^n+1=6k\Rightarrow 6|5^n+1 $
Also :
$5^p\equiv 5\bmod p$
$5^3+1\equiv 6\equiv 0\bmod 3\bmod \Rightarrow 3|5^3+1$
Let $p=3$, Using the lemma we have:
$3^{k+2}|5^{3^{k+1}+1}$
For $k=0$:
$ 3^{0+2}|5^{3^{0+1}}+1\Rightarrow 3^2|5^3+1 $
So we have infinite numbers like $n=3^k$; $k\in \mathbb N$ which suffices the condition.
$9|5^9+1$, $27|5^{27}+1$, $81|5^{81}+1$. . . .