I am trying to solve

Let $1 \neq \psi$ be a charachter of $\mathbb{F}_p$ and define $$G(\psi) = \sum_{x\in \mathbb{F}_p} \psi(x^2) $$ Proof that $|G(\psi)|^2 = p$.

What I tried so far:

$$|G(\psi)|^2 = G(\psi)\cdot \overline{G(\psi)} = \sum_{x\in \mathbb{F}_p} \psi(x^2) \cdot \sum_{y\in \mathbb{F}_p} \overline{\psi(y^2)} $$

And then $$ \sum_{x\in \mathbb{F}_p} \psi(x^2) \cdot \sum_{y\in \mathbb{F}_p} \overline{\psi(y^2)} = \sum_{x\in \mathbb{F}_p} \sum_{y\in \mathbb{F}_p} \psi(x^2) \psi( (y^{-1})^2) = \sum_{x\in \mathbb{F}_p} \sum_{y\in \mathbb{F}_p} \psi((xy^{-1})^2) $$

Now I believe we can say that per $x \neq 0$, we get that $$\sum_{y\in \mathbb{F}_p} \psi((xy^{-1})^2) = \sum_{u\in \mathbb{F}_p} \psi(u^2) = G(\psi)$$

Then $$\sum_{x\in \mathbb{F}_p} \sum_{y\in \mathbb{F}_p} \psi((xy^{-1})^2 = \sum_{y\in \mathbb{F}_p} \psi(0) + (p-1)G(\psi) $$ This seems like a weird result to me.

I have also not used the facts that $$ \sum_{x \in \mathbb{F}_p} \psi(x) = 0 \text{ and } \sum_{x \in \mathbb{F}_p} \psi(x)\overline{\psi(x)} = p $$


For the proof, which is a bit longer, see Proposition $8.2.2$ in Ireland and Rosen's book, "A Classical Introduction to Modern Number Theory". The main idea is to evaluate $$ \sum_x \psi(x)\overline{\psi(x)} $$ in two different ways.


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