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 
 A: The statement is not correct. If $x \equiv 1 \mod p,$ and $a\equiv 0 \mod p,$ then $p$ divides your expression.If $x$ is not $1$ modulo $p,$ the denominator is not divisible by $p,$ so the expression is divisible by $p$ if and only if the numerator is, or, if and only if $x^a \equiv 1 \mod p.$ This last has a solution whenever $a$ divides $p-1,$ or, more generally, $a$ is not relatively prime to $p-1.$
A: (1). If $1\ne x\equiv 1 \pmod  p,$ let $x=1+py.$ Then by the Binomial Theorem, $$x^a-1=(1+py)^a-1=\left(1+\binom {a}{1}py+...\right)-1=apy +(py)^2z$$ where $z$ is an integer . So $$\frac {1}{p} \frac {x^a-1}{x-1}=\frac {1}{p} \frac {apy +(py)^2z}{py}=\frac {a}{p}+yz.$$ This is an integer iff $p|a.$  If $a$ and  $p$ are prime, $p|a\implies p=a.$
(2A).  For prime $p$ and $p\not | x,$ let $ord_p(x)$ be the least $n>0$ such that $x^n\equiv 1 \pmod p.$ Then for $m>0$ we have $$x^m\equiv 1 \pmod p\implies ord_p (x)|m.$$ Proof:  There are positive integers $A, B$ with $$A\cdot ord_p(x)-Bm=\gcd (ord_p(x),m).$$ Therefore $$1\equiv x^{A\cdot ord_p(x)}\equiv x^{Bm}x^{\gcd (ord_p(x),m)}\equiv x^{\gcd (ord_p(x),m)} \pmod p.$$ So by the def'n of $ord_p(x)$ we must have $\gcd (ord_p(x),m)\geq ord_p (x).$ But $\gcd (ord_p (x),m)$ cannot be greater than $ ord_p (x).$  So $ord_p (x)=\gcd (ord_p (x),m),$ which is equivalent to $ord_p(x)|m.$ 
(2B). From (2A) and Fermat's Little Theorem , we have: If $p,a$ are primes and $x\not \equiv 1 \pmod p$  then   $$x^a\equiv 1\pmod p\implies (a=ord_p(x)\land a|(p-1).)$$ Because  $ord_p(x)>1,$ and (2A) implies $ord_p(x)|a,$ so $ord_p(x)=a.$ Then by  Fermat we have $a=ord_p(x)|(p-1).$
(3). We  use (2B) to handle your Q for the case $x\not \equiv 1 \mod p.$
(i). If $p|x$ then $p\not | \;(x^a-1)$ so this case is moot.
(ii). If $0\not \equiv x\not \equiv 1 \pmod p$ then $$p|(x^a-1)/(x-1)\implies p|(x^a-1)$$ with prime $a,$ so by (2B), $a|(p-1).$ That is, $p\equiv 1 \pmod a. $
Remark: The results of (2A) and (2B) are commonly used tools.
A: 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 $p$.
