Sum of Jacobi's symbol is $0$

Question

Let $m \in \mathbb{N}$ be an odd number such that $m = p_1 \cdots p_n$ (where $p_1 , \cdots, p_n$ are different primes, no repetitions), prove that$$\sum_{a \in \mathbb{Z}_m^*}{\left(\frac{a}{m}\right)} = 0.$$

Can anyone help?

Answer 1

The Jacobi symbol is a character on $\Bbb Z_m^*$, and is non-trivial, not all its values are $1$. In general $$\sum_{a\in G}\chi(a)=0$$ when $G$ is a finite group, and $\chi$ is a non-trivial character on $G$ ($\chi:G\to\Bbb C^*$ is a homomorphism). The standard proof is to take $b\in G$ with $\chi(b)\ne1$ and observe $$\sum_{a\in G}\chi(a)=\sum_{a\in G}\chi(ba)= \sum_{a\in G}\chi(b)\chi(a)=\chi(b)\sum_{a\in G}\chi(a).$$