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If $p$ is prime and $ a \ge 2$, prove that

$$

d = (a – 1, \frac{a^p – 1}{a – 1}) =

\begin{cases}

p & \text{if } p \mid (a – 1)\\

1 & \text{if } p \nmid (a – 1) \, .

\end{cases}

$$

I was thinking that since $(a^p-1)/(a-1) = (a^{p-1}+a^{p-2}+…+1)$ if p divides a-1 then p should divide $(a^{p-1}+a^{p-2}+…+1)$ ??

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We find the remainder when the **polynomial** $f(x)=x^{p-1}+x^{p-2}+\cdots+1$ is divided by $x-1$. This is $f(1)$, which is $p$.

So the gcd of $a-1$ and $f(a)$ is the same as the gcd of $a-1$ and $p$, and we are finished.

Let $d=\gcd(a-1,\frac{a^p-1}{a-1})$, now let $q$ be a prime dividing $a-1$ and let $v_q(a-1)$ be the greatest power of $q$ dividng $a-1$,using the lifting’s exponent lemma we have:

$$v_q\left(\frac{a^p-1}{a-1}\right)=v_q(a^p-1)-v_q(a-1)=v_q(p)$$.

- If $q\neq p$ then $v_q\left(\frac{a^p-1}{a-1}\right)=0$ then $q$ does not divide $d$.
- If $q=p$ then $v_q\left(\frac{a^p-1}{a-1}\right)=1$ so the greatest power of $p$ dividing $\frac{a^p-1}{a-1}$ is $p$ hence the power greatest power of $p$ dividing $d$ is $p$.

Finally $d=p$.

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