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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 $(a^{p-1}+a^{p-2}+...+1)$ ??

The Short One
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a1bcdef
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2 Answers2

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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.

André Nicolas
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  • how can you prove that the gcd of $a-1$ and $f(a)$ is the gcd between $a-1$ and $p$? – Elaqqad Apr 06 '15 at 20:00
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    It is a general result you have probably already seen before, since it is at the basis of the Euclidean Algorithm. If $b=qa+r$, then $\gcd(b,a)=\gcd(a,r)$. I imagine you know a proof. Briefly if $k$ divides $a$ and $b$, it divides $r$ (and $a$), and if $k$ divides $r$ and $a$ then $k$ divides $b$ (and $a$). So $a$ and $b$ and $a$ and $r$ have the same set of common divisors, and in particular the same greatest common divisor. – André Nicolas Apr 06 '15 at 20:05
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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 $d=p$.

Elaqqad
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