# Show that $S^p = \sum\limits_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^{xp}.$

Let $$a \in \Bbb Z$$ and $$p,q$$ be primes. Define $$\left (\frac a p \right )$$ as follows $$:$$

$$\left(\frac{a}{p}\right) = \begin{cases}\;\;\,0&\text{ if }p \text { divides } a\\+1&\text{ if } a \operatorname{R} p \text{ and }p \text { does not divide } a\\-1&\text{ if }a \operatorname{N} p \text{ and }p \text{ does not divide } a\end{cases}$$

Let $$\omega$$ be a primitive $$q$$-th root of unity. Let $$S = \sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^x.$$ Show that $$S^p = \sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^{xp}.$$

That is not true: when $$p \neq q$$, one can show that $$S^p$$ has modulus $$q^{p/2}$$, while the RHS has modulus $$q^{1/2}$$.
You must have misunderstood something. What is true, is that $$S^p$$ is congruent to $$\sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^{xp}$$ modulo the ideal $$(p)$$, inside $$\mathbb Z[\omega]$$.