Show that a periodic, completely multiplicative arithmetic function is a Dirichlet character to some module $q$ 
Show that if $f$ is a periodic, completely multiplicative arithmetic function, then $f$ is a Dirichlet character to some modulus $q$.

A Dirichlet character modulo $q$ is an arithmetic function $\chi$ which is periodic module $q$(i.e., $\chi(n+q)=\chi(n)$), completely multiplicative (in particular $\chi(1)=1$) and $\chi(n)\neq 0$ iff $\gcd(n,q)=1$.
Suppose that $q$ is the minimal period of $f$. Since $f$ is periodic module $q$, we can think $f$ as a function on $\mathbf{Z}/q\mathbf{Z}$. Then how to show that $\chi(n)\neq 0$ iff $\gcd(n,q)=1$?
 A: Note that if $f\equiv1$ we have obviously that $f$ is the principal character mod $1$. So assume that $f\not\equiv1$, assume that $f\not\equiv0$ and also assume that $\left(n,q\right)=1$. Since $f$ is completely multiplicative and $q$ periodic (and we consider $q$ as the minimum period), we have (remember we can work in $\left(\mathbb{Z}/q\mathbb{Z}\right)^{*}$) $$f^{\phi\left(q\right)}\left(n\right)\equiv f\left(n^{\phi\left(q\right)}\right)\equiv f\left(1\right)\mod q
  $$ from the Euler's theorem. So if we assume for absurd that $f\left(n\right)=0
 $ hence we have $f\left(1\right)=0
 $, and this is absurd, then $f\left(n\right)\neq0
 $. Now assume that exists an $n
 $ such that $\left(n,q\right)=d>1
 $ and $f\left(n\right)\neq0
 $. Then exists $a,b\in\mathbb{N}
 $ such that $q=da
 $ and $n=db
 $. Since $$f\left(n\right)=f\left(d\right)f\left(b\right)
 $$ we have $f\left(d\right)\neq0
 $. Now for all $m$ we have $$f\left(m\right)f\left(d\right)=f\left(dm\right)=f\left(dm+q\right)=f\left(d\right)f\left(m+\frac{q}{d}\right)
 $$ and so $$f\left(m\right)=f\left(m+\frac{q}{d}\right)
 $$ and it is absurd because $q$ is the minimum period.
