# Proof that a Gauss sum associated with a quadratic character mod p is the same as a quadratic Gauss sum

In this note I want to quickly record a solution I have found to the following problem: If $p$ is an odd prime such that $p \nmid n$ and $\chi(r) = (r|p)$, prove that

$G(n, \chi) = \sum_{r \text{mod } p} (r|p) e^{2 \pi i n r/p} = \sum_{r = 1}^p e^{2 \pi i n r^2/p} = G(n ; p)$

My solution is as follows. We have

$G(n, \chi) = \sum_{r \text{mod } p} (r|p) e^{2 \pi i n r/p}$

$= \sum_{\substack{\text{quadratic}\\\text{residues }r}} e^{2 \pi i n r/p} - \sum_{\substack{\text{quadratic}\\\text{non-residues }s}} e^{2 \pi i n s/p}$

whereas

$G(n ; p) = \sum_{r = 1}^p e^{2 \pi i n r^2/p}$

$= \sum_{r = 1}^{p-1} e^{2 \pi i n r^2/p} + 1$

$= 2\sum_{\substack{\text{quadratic}\\\text{residues }r}} e^{2 \pi i n r/p} + 1$

(since $r$ mod $p$ will range over both quadratic residues and non-residues, so $r^2$ will range over the quadratic residues twice)

$= \sum_{\substack{\text{quadratic}\\\text{residues }r}} e^{2 \pi i n r/p}$

$+ \big \{ \sum_{\substack{\text{quadratic}\\ \text{residues }r}} e^{2 \pi i n r/p} + 1 \big \}$

Therefore $G(n, \chi) = G(n ; p)$ if and only if

$- \sum_{\substack{\text{quadratic}\\\text{non-residues }s}} e^{2 \pi i n s/p}$

$= \sum_{\substack{\text{quadratic}\\ \text{residues }r}} e^{2 \pi i n r/p} + 1$

$\iff$

$\sum_{\substack{\text{quadratic}\\ \text{residues }r}} e^{2 \pi i n r/p} + \sum_{\substack{\text{quadratic}\\ \text{non-residues }s}} e^{2 \pi i n s/p} = -1$

But this is true because

$\sum_{\substack{\text{quadratic}\\ \text{residues }r}} e^{2 \pi i n r/p} + \sum_{\substack{\text{quadratic}\\ \text{non-residues }s}} e^{2 \pi i n s/p} = \sum_{r=1}^{p-1} e^{2 \pi inr/p}$

This is a geometric sum of the form

$S = \sum_{r=1}^{p-1} x^r = x + x^2 + \cdots + x^{p-1}$

where

$x = e^{2 \pi in/p}$

Therefore

$xS = x^2 + x^3 + \cdots + x^p$

Subtracting the expression for $S$ from this we get

$(x - 1)S = x^p - x = 1 - x$

(the last equality is true because $x$ is a $p$-th root of unity)

$\iff$

$S = -1$

Therefore $G(n, \chi) = G(n ; p)$.