Cox proves the following statement using quadratic reciprocity:
If $D\equiv 0, 1\mod 4$ is a nonzero integer, then there is a unique homomorphism $\chi:(\mathbb{Z}/D\mathbb{Z})^*\to\{\pm 1\}$ such that $\chi([p]) = (D/p)$ for odd primes $p$ not dividing $D$.
Exercise 1.13 then asks to show that the above statement implies quadratic reciprocity, showing that the two are equivalent. The hint given is as follows:
If $p$, $q$ are distinct odd primes, let $q^* = (-1)^{(q-1)/2}q$; then $q^*\equiv 1\mod{4}$, so that $(q^*/\cdot)$ gives a homomorphism from $(\mathbb{Z}/q\mathbb{Z})^*\to\{\pm 1\}$. $(\cdot/q)$ is also a homomorphism between these groups. Show that these two homomorphisms are the same.
The second homomorphism, $(./q)$, is obviously nontrivial. If we can show that the first is nontrivial, then the two must be equal, since they must both equal $-1$ on some generator of $(\mathbb{Z}/q\mathbb{Z})^*$. But I can't see why (without using quadratic reciprocity or heavier machinery) the first map must be nontrivial.
So why is that true?