Let $f$ be a totally multiplicative function, $f(n+N)=f(n)$ for every $n \in \mathbb{N}$. Prove that $|f(n)|=1$ for every $n \in \mathbb{N}$, where $gcd(n,N)=1$
My approach is to prime factorise $n+N=\prod_{i} p_i^{a_i}$, then $f(n+N)=f(p_{1}^{a_{1}})...f(p_{n}^{a_{n}})$
We also prime factorise $n=\prod_i p_i^{b_i}$, then $f(n)=f(q_1^{b_{1}})...f(q_n^{b_{n}})$
$f(p_1^{a_{1}})...f(p_n^{a_{n}})=f(q_1^{b_{1}})...f(q_n^{b_{n}})$ Since $gcd(n,N)=1$, implies that $gcd(n,n+N)=1$. However as the primes on left are different than the right, the only possible case where the two are equal is when all the exponents $a_i,b_i=0$. Thus $f(n)=1$.
However, not sure where does the absolute value comes in. Any help would be appreciated, thank you