How to prove that $m \mid \phi (a^m - 1)$ if $gcd(a,m) = 1$
$\phi$ is euler function
How to prove that $m \mid \phi (a^m - 1)$ if $gcd(a,m) = 1$
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elementary-number-theory
totient-function
1 Answers
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Euler's theorem says that $$b^{\phi(a^m-1)}\equiv 1\pmod{a^m-1}$$ when $\gcd(b,a^m-1)=1$.
Since $\gcd(a,a^m-1)=1$, $$a^{\phi(a^m-1)}\equiv 1\pmod{a^m-1}$$
We have also that $$a^m\equiv 1\pmod{a^m-1}$$
Note that $m$ is the order of $a$ in the multiplicative group $\Bbb Z_{a^m-1}^\times$, because $1 Then $m\mid\phi(a^m-1)$. I have not used that $a$ and $m$ are coprime. Are you sure that this is necessary?
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0Thank you! But why we know that $gcd(a,a^m-1) =1$? – 2017-01-30
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0If you do the Euclidean division $(a^m-1)/a$ you get $a-1$ as remainder. Just like $9999$ and $10$ are coprime. – 2017-01-30
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0@EugeneKorotkov If a prime $p$ divides $a$, then $p$ divides $a^m$, so $p$ doesn't divide $a^m-1$. Because otherwise $p$ divides $a^m-\left(a^m-1\right)=1$, contradiction. – 2017-01-31