Prime divisors of $a^{p - 1} + \ldots + a + 1$ for $p$ odd prime Suppose $p$ prime and odd, $a \in \Bbb Z$, $n=a^{p-1}+...+a+1$. How to prove:
Prime divisors of 
$n =\begin{cases}
       p,     & \text{or} \\
       2px+1, & \text{$x\in \Bbb N$}
    \end{cases}$
 A: Hints: If $q$ is a prime dividing $n$, then since $a^p-1$ is a multiple of $n$ (why?) we also have $q \mid a^p - 1$.  If $a \not\equiv 1 \pmod q$, then this determines the exact multiplicative order of $a$ modulo $q$.  What do you know about $\text{ord}_q(a)$?
On the other hand, if $a \equiv 1 \pmod q$, this tells you a lot about the original sum $n = a^{p-1} + \cdots + 1$ modulo $q$.  What can you then deduce from $q \mid n$?
A: $(1)$ If odd prime $q\mid a, n\equiv1\pmod q\implies q\not\mid n$
$(2)$ Else if $q\mid(a-1)\iff a\equiv1\pmod q\implies a^r\equiv1\pmod q$ for integer $r\ge0$
So, $n\equiv\sum_{0\le r\le p-1}1\pmod q\equiv p\pmod q$
So, $q$ will divide $n$ iff $q\mid p\implies q=p$ as $p,q$ are primes.
$(3)$ Else $q\not\mid a\implies (a,q)=1$ and 
$q\not\mid (a-1)\implies (a-1,q)=1$
$\implies (a(a-1),q)=1$
Now, $n=\frac{a^p-1}{a-1}\implies q\mid \frac{a^p-1}{a-1}\implies q(a-1)\mid (a^p-1) \implies q\mid (a^p-1)$ as $(a-1,q)=1$
Using Fremat's Little Theorem, $a^{q-1}\equiv1\pmod q\implies ord_qa\mid (p,q-1)$
But $(p,q-1)=1$ or $p$ as $p$ is prime
If $(p,q-1)=1, ord_qa\mid1\implies q\mid (a-1)$ which is impossible as $(a-1,q)=1$
So, $(p,q-1)=p\implies p\mid(q-1)$
But $q-1$ is even as $q$ is odd $\implies 2\mid(q-1)$
So, lcm$(2,p)\mid(q-1)\implies 2p\mid(q-1) $
