Let $m = p_1\cdots p_k$, where $p_i \ne p_j$ for all $i\ne j$.
Prove that: $$\sum_{a\in\mathbb{Z}_m^\star} (a/m)= 0$$
I've noticed that $$(m-a/m) =(-a/m) = (-1/m)(a/m)$$
Now, $(-1/m) = (-1)^{\frac{m-1}{2}}$. If I knew that $(-1/m) = -1$ then the proposition holds.
$m-1$ is even so it can be of the form $4k$ or $4k+2$. Now, if it has the form $4k+2$ then $\frac{m-1}{2}$ must be odd and we're done.
I couldn't prove the last assumption though. Is my idea correct? Is there another approach?
If $\mathbb{Z}/(m\mathbb{Z})^*$ is a cyclic group we have as many quadratic residues as quadratic non-residues, since the set of quadratic residues is a subgroup of index $2$, given by the even powers of some generator.
If you prove the statement in the case $m=2^k$ and invoke the Kronecker decomposition theorem, the statement in its general case follows from the multiplicativity of the Jacobi symbol.
A simpler approach is to notice that once we have some $b\in\mathbb{Z}/(m\mathbb{Z})^*$ such that $\left(\frac{b}{m}\right)=-1$, by reindexing through $a=ba'$ we get that the original sum equals it opposite, hence it equals zero.