I am reading Cox's Primes of the Form $x^2+ny^2$ and solving Exercise 1.13, which depends on Lemma 1.14.
Lemma 1.14. If $D\equiv 0, 1\pmod{4}$ 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 $D$. Furthermore,
$$
\chi([-1])=\left\{
\begin{array}{ll}
1 & \text{when }D>0, \\
-1 & \text{when }D>0 \\
\end{array}
\right.
$$
Exercise 1.13. We will assume that Lemma 1.14 holds for all nonzero integers $D\equiv 0,1 \pmod{4}$, and we will prove quadratic reciprocity and the supplementary laws.
(a) Let $p$ and $q$ be distinct odd primes, and let $q^*=(-1)^{(q-1)/2}q$. By applying the lemma with $D=q^*$, show that $(q^*/\cdot)$ induces a homomorphism from $(\mathbb{Z}/q\mathbb{Z})^*$ to $\{\pm 1\}$. Since $(\cdot /q)$ can be regarded as a homomorphism between the same two groups and $(\mathbb{Z}/q\mathbb{Z})^*$ is cyclic, conclude that the two are equal.
My goal is proving that $(\cdot /q)$ and $(q^*/\cdot)$ are not trivial. That is, there exists $a, b\in (\mathbb{Z}/q\mathbb{Z})^*$ such that $(a/q)=(q^*/b)=-1$.
I have no problem with $(\cdot /q)$.
Suppose that $(\mathbb{Z}/q\mathbb{Z})^*=\langle g\rangle$. I am trying to show that $(q^*/g)=-1$.
I have proven the case $q\equiv 1\pmod{4}$, but my argument uses the Generalized Quadratic Reciprocity Law, which I am not supposed to use it because this exercise is a proof of that law. Here is my argument: since $(q/g)(g/q)=(-1)^{(q-1)/2 \cdot (g-1)/2}=1$, $(q/g)$ and $(g/q)$ have the same sign. Since $(g/q)=g^{(q-1)/2}=-1 \pmod{q}$, we have $(g/q)=-1$.
I am stuck here.
Best Answer
When $q\equiv 3 \pmod 4, $ the lemma says $(q^*/\cdot)$ is not trivial; hence, $(q^*/\cdot)=(\cdot/q).$
Now we prove $(q/\cdot)$ is nontrivial when $q\equiv 1 \pmod 4.$ Note that it is enough to check on odd primes ( it is a homomorphism and we can write $-1\equiv 2q-1, 2\equiv q+2$ which are product of odd primes).
Suppose $(q/\cdot)$ is trivial. We know $(\cdot/q)$ is not trivial. Therefore we can find an odd prime $p$ such that $(q/p)=1,$ but $(p/q)=-1.$
If $p\equiv 3 \pmod 4,$ we have $(p^*/\cdot)=(\cdot/p)$ by the discussion above. Therefore, $$ (q/p)=(p^*/q)=(-1/q)(p/q)=(-1)^{(p-1)/2}(p/q)=-1. $$
If $p\equiv 1 \pmod 4$, then $(p/\cdot)$ is nontrivial by $(p/q)=-1.$ Therefore $(p/\cdot)=(\cdot /p).$ So $$ (q/p)=(p/q)=-1. $$
In both cases, we prove $(q/p)=-1$ which contradicts to the assumption. So we conclude $(q/\cdot)$ is nontrivial.