Prove $p\mid\frac{x^{a}-1}{x-1}$ using Fermat's little theorem Fermat's little theorem states that if $p$ is a prime number, then for any integer $a$, the number $a^{p} − a $ is an integer multiple of $p$. In the notation of modular arithmetic, this is expressed as $a^{p} \equiv a \pmod p.$
Using this theorem prove:
Given a prime number $p$, show that if there are a positive integer $x$ and a prime number $a$ such that  $p$ divides 
$\frac{x^{a}-1}{x-1}$, then either $a = p$ or $p \equiv 1 \pmod a$.
$$p\mid\frac{x^{a}-1}{x-1}$$
So, I'm thinking:
$$\frac{x^{a}-1}{x-1} = x^{a-1}+x^{a-2}+...+1$$ 
I tried the telescoping technique but that doesn't work, assuming $a = p$, shows that $x^{p-1}\equiv 1 \pmod p$.
So, what else can I do?
 A: You have:
$p|\frac{x^a-1}{x-1}\Rightarrow p|(\frac{x^a-1}{x-1})(x-1) \Rightarrow p|x^a-1 \Rightarrow x^a \equiv 1 (\mod p)$
Hence if $m$ is the order of $x$ with respect to $p$, then $m|a \Rightarrow m=1$ or $m=a$, since $a$ is prime.
If $m=1$, then by the definition of the order, we have that
$x\equiv 1(\mod p)$ and so by the hypothesis we get:
$0\equiv \frac{x^a-1}{x-1}\equiv x^{(a-1)}+...+1\equiv 1^{(a-1)}+...+1\equiv a(\mod p)$
Hence in this case $p|a \Rightarrow p=1\  \text{or}\  p=a$, since $a$ is a prime. The first equality is rejected, because $p>1$ as a prime number.
If $m=a$, then $a|p-1 \Rightarrow p\equiv 1 (\mod a)$, because it is known that for the order $m$ we have $m|φ(p)$. $φ$ is the Euler totient function.
The proof of the property that $m$ holds uses the Fermat's little Theorem.
Links:
1.Orders
2.Euler's Totient Function
A: If $p$ divides $\frac{x^a-1}{x-1}$, it must also divide $x^a-1$, so we can conclude $$x^a\equiv 1\mod p$$
Let $o$ be the smallest non-zero number $o$ with $x^o\equiv 1\mod p$, also called the order of $x$ modulo $p$
Then $a$ must be a mutiple of $o$, but $a$ is prime. So, either we have $o=1$, in which case we have $p|x-1$ , in other words $x\equiv 1\mod p$ or we must have $a=o$, in which case we can conclude $a|p-1$ , in other words $p\equiv 1\mod a$
A: Since $p\mid \frac{x^a-1}{x-1}$ can be expressed as $\frac{x^a-1}{x-1} = pk + 0$ thus operating we have $x^a = (x-1)kp + 1$ and in modular form $x^a \equiv 1 \pmod p$.
Then through the last one we know that $Ord_p(x)=a$ and that $a|\varphi(p)$ that is $p-1 \equiv 0 \pmod a$ so $p \equiv 1 \pmod a$ holds.
