Let $p$ and $q$ be two distinct odd primes. Let $\omega$ be a primitive $q$-th root of unity. Consider the sum
$$S = \sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^x.$$ Prove that
$$S^2 = q \left (\frac {-1} {q} \right ).$$
I have proved the above as follows $:$
$$\begin{align} S^2 & = \sum_{x,y \in \Bbb {F_q}^*} \left ( \frac {xy} {q} \right ) {\omega}^{x+y}. \end{align}$$
Putting $x = yz$ we have
$$\begin{align} S^2 & = \sum_{y,z \in \Bbb {F_q}^*} \left (\frac {y^2 z} {q} \right ) {\omega}^{y(z+1)} \\ & = \sum_{y,z \in \Bbb {F_q}^*} \left (\frac z q \right ){\omega}^{y(z+1)} \\ & = \sum_{y \in \Bbb {F_q}^*} \left ( \frac {-1} {q} \right ) + \sum_{z \neq -1} \left ( \frac z q \right ) \sum_{y \in \Bbb {F_q}^*} {\omega}^{y(z+1)} \\ & = \left ( \frac {-1} {q} \right )(q-1) + (-1) \sum_{z\neq -1} \left (\frac z q \right ) \\ & = q \left ( \frac {-1} {q} \right ) + (-1) \sum_{z \in \Bbb {F_q}^*} \left (\frac z q \right ) \\ & = q \left (\frac {-1} {q} \right ). \end{align}$$
But I don't find any proper reason as to why should I take $\omega$ as a primitive $q$-th root of unity instead of taking any ordinary $q$-th root of unity. Would anybody please point out that where have I used this fact implicitly?
Thank you very much.
The step where you simplify $\sum_{y \in \Bbb {F_q}^*} {\omega}^{y(z+1)}$ to $-1$ is only valid if $\omega^{z+1}\neq 1$. It seems you mean to write $z\neq -1$ in the outer summation instead of $z\neq 1$, so that $\omega^{z+1}\neq 1$ for all such $z$ as long as $\omega$ is a primitive $q$th root of unity. On the other hand, if $\omega=1$, then $\omega^{z+1}=1$ for all $z$ and we get $\sum_{y \in \Bbb {F_q}^*} {\omega}^{y(z+1)}=q-1$.