Can't come even near to the solution.. Can somebody take a look? 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 $$a^p \equiv a \mod p.$$
Use Fermat's little theorem to prove that:
Given prime number $p$, show that if there are positive integer $x$ and prime number $a$ such that  $p$ divides 
 $\frac{x^a – 1}{x – 1}$, then either $а= p$ or $p \equiv 1 \mod a$.
I tried to connect  $\frac{x^a – 1}{x – 1}$ to the theorem, but without any success.. 
Anything will help..
Thanks in advance
 Picture in addition same as the text 

We have that $p \mid x^a - 1$, or $x^a \equiv 1 \pmod{p}$. So first of all $x \not\equiv 0 \pmod{p}$, so that $x^ p - x \equiv 0 \pmod{p}$ implies $x^{p-1} \equiv 1 \pmod{p}$. (As the prime $p$ divides $x^ p - x = x (x^{p-1} - 1)$, and $p \nmid x$.)
And then, $a$ being prime, $x$ has order either $1$ or $a$ modulo $p$. Of course order $1$ means $x \equiv 1 \pmod{p}$. But then
$$
\frac{x^a - 1}{x - 1} = x^{a-1} + \dots + x + 1 \equiv a \equiv 0 \pmod{p},
$$
so that $p \mid a$ and thus $p = a$.
If $x$ has order $a$ modulo $p$, then $a$ must divide the order $p-1$ of the group of invertible integers modulo $p$. I guess this is where Fermat' s Little Theorem might come in. First of all,
$$
x^a \equiv 1 \pmod{p}, \quad x^{p-1} \equiv 1 \pmod{p}
$$ 
imply that if $p-1 = a q + r, 0 \le r < a$ (so that $r$ is the remainder of the division of $p-1$ by $a$) then
$$
x^{p-1} \equiv x^{a q + r} \equiv x^r \equiv 1 \pmod{p},
$$
so that $r = 0$, as $r$ is less than the order $a$ of $x$ modulo $p$.
