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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

3 Answers 3

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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 $p-1.$

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    Was the Q altered after your answer? Because my A is a proof. Note that the $a$ in the Q is prime. Was that always there? – 2017-01-02
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    @user254665 No, it was not. Interesting. – 2017-01-02
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(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.

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    Could you explain how $$x^a-1=(1+py)^a-1=\left(1+\binom {a}{1}py+...\right)-1$$ is equal to $$apy +(py)^2z$$ where $z$ is an integer – 2017-01-02
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    $(1+py)^a-1=apy+\sum_{j=2}^a\binom {a}{j}(py)^j.$ Now each $\binom {a}{j}$ is an integer. And $(py)^j$ is divisible by $(py)^2$ when $j\geq 2.$ So each term $\binom {a}{j}(py)^j,$ for $j\geq 2$ , is divisible by $(py)^2 .$ – 2017-01-03
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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 $p$.

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    How did you come to this : And then, $a$ being prime, $x$ has order either $1$ or $a$ modulo $p$. Of course order $1$ means $x \equiv 1 \pmod{p}$. Is there a rule for that part ? – 2017-01-02
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    I am using the fact that if an element $x$ of a finite group has order $m$, then $x^a = 1$ iff $m \mid a$. – 2017-01-02