1
$\begingroup$

Let $\phi(n)$ be the Euler phi-function. If $a>1$ is an integer, then what is the remainder when $\phi(a^n - 1)$ is divided by $n$ in accordance with the Euclidean algorithm?

$\endgroup$
6
$\begingroup$

The group $(\mathbb{Z}/(a^n-1)\mathbb{Z})^\times$ has an element of order $n$, namely $\overline{a}$, because obviously $$a^n\equiv 1\bmod (a^n-1),$$ and $a^k\not\equiv 1\bmod (a^n-1)$ for any $0<k<n$ because $1<a^k-1<a^n-1$ for any $0<k<n$.

Because the group has an element of order $n$, the order of the group is divisible by $n$.

Since $\varphi(N)=|(\mathbb{Z}/N\mathbb{Z})^\times|$, we have that $\varphi(a^n-1)$ is divisible by $n$. Thus the remainder is $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.