A problem with showing that $g_a^2=(-1)^{\tfrac{p-1}{2}}p$ for quadratic Gauss sums $g_a$

Question

Let $p>2$ be prime, and let $a\in\Bbb Z$. Also, let $\left(\frac{n}{p}\right)$ denote the Legendre symbol.

Then let the quadratic Gauss sum be $$g_a=\sum_{x=1}^{p-1}\left(\frac{x}{p}\right)\zeta_p^{ax},$$ where $\zeta_p=\exp(2i\pi/p).$ My professor makes the claim that $g_a^2=(-1)^{\tfrac{p-1}{2}}p.$ His proof is as follows.

Proof: We will compute the sum $$\sum_{a\in(\Bbb Z/p\Bbb Z)^\times}g_a\cdot g_{-a}$$ in two ways. First, $$\begin{align} \sum_{a\in(\Bbb Z/p\Bbb Z)^\times}g_a\cdot g_{-a}&=\sum_{a\in(\Bbb Z/p\Bbb Z)^\times}\left(\frac{a}{p}\right)\left(\frac{-a}{p}\right)g_1^2\\ &=\sum_{a\in(\Bbb Z/p\Bbb Z)^\times}\left(\frac{a}{p}\right)^2\left(\frac{-1}{p}\right)g_1^2\\ &=\sum_{a\in(\Bbb Z/p\Bbb Z)^\times}(-1)^{\tfrac{p-1}{2}}g_1^2\\ &=(p-1)(-1)^{\tfrac{p-1}{2}}g_1^2 \end{align}$$ I'm fine with this, it makes sense.

The second way is $$\begin{align} \sum_{a\in(\Bbb Z/p\Bbb Z)^\times}g_a\cdot g_{-a}&=\sum_{a}\left(\sum_{x}\left(\frac{x}{p}\right)\zeta_p^{ax}\right)\left(\sum_{y}\left(\frac{y}{p}\right)\zeta_p^{-ay}\right)\\ &=\sum_{a}\sum_{x,y}\left(\frac{x}{p}\right)\left(\frac{y}{p}\right)\zeta_p^{a(x-y)}\\ &=\sum_{x,y}\left(\frac{x}{p}\right)\left(\frac{y}{p}\right)\sum_{a}\zeta_p^{a(x-y)}. \end{align}$$ Then my professor claims that the sum $\sum_{a}\zeta_p^{a(x-y)}$ is equal to $p$ when $x\equiv y\mod p$ and equal to $0$ otherwise. I cannot justify this. Specifically, when $x\equiv y\mod p$, $p|x-y$ and thus $\zeta_p^{a(x-y)}=1$ for all $a\in(\Bbb Z/p\Bbb Z)^\times$, so that

$$\sum_{a}\zeta_p^{a(x-y)}=\sum_{a=1}^{p-1}1=\color{red}{p-1}.$$ When $x\not\equiv y\mod p$, the map $q\mapsto (x-y)q$ is a bijection $(\Bbb Z/p\Bbb Z)^\times\rightarrow (\Bbb Z/p\Bbb Z)^\times$ and thus $$\sum_{a}\zeta_p^{a(x-y)}=\sum_{a}\zeta_p^a=\sum_{a=1}^{p-1}\zeta_p^{a}=\frac{\zeta_p}{\zeta_p-1}(\zeta_p^{p-1}-1)=\frac{\zeta_p}{\zeta_p-1}(\zeta_p^{-1}-1)=\frac{1-\zeta_p}{\zeta_p-1}=-1.$$ So in both cases, my computations suggest that we should be summing over $\Bbb Z/p\Bbb Z$ instead of $(\Bbb Z/p\Bbb Z)^\times$, but I don't know if this will affect the proof.

Did I mess up somewhere? Why am I unable to justify his claim? Thanks.

Answer 1

No, you are very correct, the sum $\sum_{a \neq 0} \zeta_p^{a(x - y)}$ is indeed $p - 1$ and $-1$ respectively, as you calculated, and what your professor stated is incorrect. However, the rest of the calculation is correct. you can follow this up, since we can manipulate the sum as follows:

\begin{align*} \sum_{x, y} \left(\frac{x}{p}\right)\left(\frac{y}{p}\right) \sum_a \zeta_p^{a(x - y)} &= \underbrace{\sum_x \left(\frac{x}{p}\right)^2 (p - 1)}_{x = y} + \underbrace{\sum_{x \neq y} \left(\frac{x}{p}\right)\left(\frac{y}{p}\right)} (-1) \\ &= \underbrace{(p - 1)}_{\left(\frac{x}{p}\right)^2 = 1 \iff x \neq 0}(p - 1) - (1 - p) \\ &= p(p - 1) \end{align*}

And the rest follows! Hope this helps :)

P.S. This proof is presented in Ireland and Rosen's book "A Classical Introduction to Modern Number Theory" which has a clear presentation as well, in chapter 8 p.92. I have copied it below:

Answer 2

You are correct, it is in error. We can fix it, however.

Note, $g_0=0$ follows from the definition, when $p$ is odd, so adding $g_0g_{-0}=0$ in your sum on the left changes nothing.

This makes all your sums over $a\in F=\mathbb Z/p\mathbb Z.$ Then you see the results the professor states.

When $p\not\mid (x-y),$ the argument that $a\mapsto (x-y)a$ is a bijection applies to all of $F,$ not just on $F^{\times},$ so the same argument follows.