completely multiplicative and periodic function

Question

Let $f$ be a completely multiplicative and periodic arithmetic function. Then $f$ are either zero or roots of unity $(f(n)^m = 1 \ \mbox{forsome}\ m, f \ \mbox{is complex value})$.

Proof. Let $k$ be the period. Then $$f(n+k) = f(k)$$ for any $n$. Then $$f(k) = f(2k) = f(3k) = ...$$ or precisely, $$f(i)f(k) = f(j)f(k)$$ for any $i,j \in \mathbb{N}$.

Case $1$ : $f(k) \neq 0$

Then $f(i) = f(j)$ for any $i,j$. Particularly, $f(n)^m = 1$, for any $n$, that is, $f$ is a root of unity.

Case $2$ : $f(k) = 0$ If $f = 0$, it is done. Assume that $f \neq 0$.

I am not sure what I should expect in this case. I guess that at any point that $f$ is not zero, it is a root of unity. However, I cannot produce the proof.

Any help please ?

PS. http://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Binder.pdf

I find some similar problems to this problem, but the consequence is a bot different. Moreoevr, the process of proving is different form mine.

Answer 1

For any periodic arithmetic completely multiplicative function with period $k$ note the following :-

1. $f(1.1)=f(1)^2$ $\Rightarrow$ $f(1)=1$
2. $f(n)^{\phi(k)}=f(n^{\phi(k)})=f(1)$ for any $(n,k)=1$. Therefore $f(n)$ is a $\phi(k)$-th root of unity.

3. For $n$ where $(n,k)>1$, $f(n)=0$. Because of the following:-

   Let $(n,k)=d >1$ and $f(n)=f(r)f(d)\neq0$ where $n=rd$ and $k=sd$. Then for any $a$ $f(a)f(d)=f(ad+sd)=f(d)f(a+s)$

$\Rightarrow$ $f(a)=f(a+ s)$ for all $a$. But $k$ is the smallest period.