[proofplan] We first prove the basic orthogonality identity for additive characters modulo $p$. Expanding $G_p\overline{G_p}$ and using the invertible change of variables $u=x-y$ and $v=x+y$ gives $|G_p|^2=p$. To determine the sign of $G_p^2$, we rewrite $G_p$ as the quadratic-character Gauss sum $\tau(\chi)$, square this sum, and evaluate it by separating the zero terms and applying the same additive orthogonality. [/proofplan] [step:Establish additive orthogonality modulo $p$] Let $R:=\mathbb Z/p\mathbb Z$. Define the additive-character sum as a map $A:R\to\mathbb C$ by \begin{align*} A:R\to\mathbb C,\qquad c\mapsto \sum_{v\in R}e\left(\frac{cv}{p}\right) \end{align*} where $cv$ is computed using integer representatives modulo $p$. We claim that $A(0)=p$ and $A(c)=0$ for every $c\in R^\times$. If $c=0$ in $R$, every summand is $1$, so $A(0)=p$. If $c\ne 0$ in $R$, choose the representatives $v=0,1,\dots,p-1$ and set $r:=e(c/p)$. Since $c\not\equiv 0\pmod p$, one has $r\ne 1$, while $r^p=e(c)=1$. Therefore the finite geometric-sum identity gives \begin{align*} A(c)=\sum_{v=0}^{p-1}r^v=\frac{1-r^p}{1-r}=0. \end{align*} [guided] The only cancellation principle used in the proof is the orthogonality of the nontrivial additive characters modulo $p$, so we prove it explicitly. Define the additive-character sum $A:R\to\mathbb C$ by \begin{align*} A(c):=\sum_{v\in R}e\left(\frac{cv}{p}\right). \end{align*} This is well-defined because replacing $cv$ by another integer representative changes $cv/p$ by an integer, and $e(t+ m)=e(t)$ for every $m\in\mathbb Z$. If $c=0$ in $R$, then $cv=0$ in $R$ for every $v\in R$, and every term in the sum equals $e(0)=1$. Hence \begin{align*} A(0)=p. \end{align*} Now suppose $c\ne 0$ in $R$. Use the complete residue system $v=0,1,\dots,p-1$ and put \begin{align*} r:=e\left(\frac{c}{p}\right). \end{align*} Because $c\not\equiv 0\pmod p$, the number $c/p$ is not an integer, so $r\ne 1$. On the other hand, \begin{align*} r^p=e(c)=1. \end{align*} Thus the geometric-sum formula applies with nonunit denominator $1-r$, and gives \begin{align*} A(c)=\sum_{v=0}^{p-1}r^v=\frac{1-r^p}{1-r}=0. \end{align*} This is the exact form of additive orthogonality needed later: summing $e(cv/p)$ over a full residue system in $v$ kills every nonzero frequency $c$ and leaves $p$ only at frequency $0$. [/guided] [/step] [step:Compute $|G_p|^2$ by the change of variables $u=x-y$ and $v=x+y$] Since $\overline{e(t)}=e(-t)$ for every $t\in\mathbb R$, we have \begin{align*} |G_p|^2=G_p\overline{G_p}=\sum_{x\in R}\sum_{y\in R}e\left(\frac{x^2-y^2}{p}\right). \end{align*} Define the map $\Phi:R^2\to R^2$ by \begin{align*} \Phi(x,y)=(x-y,x+y). \end{align*} Since $p$ is odd, the element $2\in R$ is invertible, and $\Phi$ has inverse \begin{align*} (u,v)\mapsto \left(\frac{v+u}{2},\frac{v-u}{2}\right). \end{align*} Thus $\Phi$ is a bijection. Under the notation $u=x-y$ and $v=x+y$, one has $x^2-y^2=uv$ in $R$. Therefore \begin{align*} |G_p|^2=\sum_{u\in R}\sum_{v\in R}e\left(\frac{uv}{p}\right)=\sum_{u\in R}A(u)=A(0)=p. \end{align*} [/step] [step:Rewrite $G_p$ as a quadratic character Gauss sum] Define $\chi:R\to\{-1,0,1\}$ by $\chi(t):=\chi_p(t)$, where $\chi_p$ is the Legendre-symbol map from the statement. For each $t\in R$, the number of solutions $x\in R$ of $x^2=t$ is $1+\chi(t)$: it is $1$ when $t=0$, it is $2$ when $t$ is a nonzero square, and it is $0$ when $t$ is a nonsquare. Hence \begin{align*} G_p=\sum_{t\in R}(1+\chi(t))e\left(\frac{t}{p}\right). \end{align*} Define the quadratic Gauss sum $\tau(\chi)\in\mathbb C$ by \begin{align*} \tau(\chi):=\sum_{t\in R}\chi(t)e\left(\frac{t}{p}\right). \end{align*} Using the additive orthogonality identity with $c=1$, we have \begin{align*} \sum_{t\in R}e\left(\frac{t}{p}\right)=A(1)=0. \end{align*} Consequently, \begin{align*} G_p=\tau(\chi). \end{align*} [/step] [step:Evaluate the square of the quadratic character Gauss sum] We use the elementary quadratic-character facts that $\chi(ab)=\chi(a)\chi(b)$ for all $a,b\in R$ and that \begin{align*} \sum_{u\in R}\chi(u)=0. \end{align*} Indeed, by the standard theorem that the multiplicative group of a finite field is cyclic, choose a generator $g$ of $R^\times$. Then the nonzero squares are exactly the even powers of $g$, the nonsquares are exactly the odd powers of $g$, and there are equally many of each. Since $\chi(0)=0$, the zero terms do not contribute to $\tau(\chi)^2$. Thus \begin{align*} \tau(\chi)^2=\sum_{a\in R^\times}\sum_{b\in R^\times}\chi(a)\chi(b)e\left(\frac{a+b}{p}\right). \end{align*} For each fixed $a\in R^\times$, make the bijective change of variables $b=au$ with $u\in R^\times$. Multiplicativity gives $\chi(a)\chi(au)=\chi(a)^2\chi(u)=\chi(u)$, since $\chi(a)^2=1$ for $a\in R^\times$. Hence \begin{align*} \tau(\chi)^2=\sum_{u\in R^\times}\chi(u)\sum_{a\in R^\times}e\left(\frac{a(1+u)}{p}\right). \end{align*} If $u=-1$, the inner sum equals $p-1$. If $u\ne -1$, then $1+u\ne 0$, and additive orthogonality gives \begin{align*} \sum_{a\in R^\times}e\left(\frac{a(1+u)}{p}\right)=\sum_{a\in R}e\left(\frac{a(1+u)}{p}\right)-1=-1. \end{align*} Therefore \begin{align*} \tau(\chi)^2=(p-1)\chi(-1)-\sum_{\substack{u\in R^\times, u\ne -1}}\chi(u). \end{align*} Since $\sum_{u\in R}\chi(u)=0$ and $\chi(0)=0$, we have \begin{align*} \sum_{\substack{u\in R^\times, u\ne -1}}\chi(u)=-\chi(-1). \end{align*} Substitution gives \begin{align*} \tau(\chi)^2=(p-1)\chi(-1)+\chi(-1)=p\chi(-1). \end{align*} [guided] The goal is to determine the sign of $G_p^2$, and the previous step reduced this to computing $\tau(\chi)^2$. We first record the two elementary facts about the quadratic character that will be used. The multiplicative group $R^\times$ is cyclic; choose a generator $g\in R^\times$. Then every nonzero residue has the form $g^m$, the nonzero squares are precisely the even powers $g^{2j}$, and the nonsquares are precisely the odd powers $g^{2j+1}$. It follows that $\chi(ab)=\chi(a)\chi(b)$ for all $a,b\in R$ and that the sum of $\chi$ over $R$ is zero, because there are exactly $(p-1)/2$ nonzero squares and $(p-1)/2$ nonsquares: \begin{align*} \sum_{u\in R}\chi(u)=0. \end{align*} Now expand the square: \begin{align*} \tau(\chi)^2=\sum_{a\in R}\sum_{b\in R}\chi(a)\chi(b)e\left(\frac{a+b}{p}\right). \end{align*} The terms with $a=0$ or $b=0$ vanish because $\chi(0)=0$, so this becomes \begin{align*} \tau(\chi)^2=\sum_{a\in R^\times}\sum_{b\in R^\times}\chi(a)\chi(b)e\left(\frac{a+b}{p}\right). \end{align*} For each fixed $a\in R^\times$, the substitution $b=au$ is a bijection from $R^\times$ to $R^\times$. This is the useful substitution because it separates the multiplicative character from the additive exponential. Under this change of variables, \begin{align*} \chi(a)\chi(b)=\chi(a)\chi(au)=\chi(a)^2\chi(u)=\chi(u), \end{align*} and \begin{align*} a+b=a+au=a(1+u). \end{align*} Hence \begin{align*} \tau(\chi)^2=\sum_{u\in R^\times}\chi(u)\sum_{a\in R^\times}e\left(\frac{a(1+u)}{p}\right). \end{align*} We now evaluate the inner additive sum. If $u=-1$, then $1+u=0$ in $R$, so every summand is $1$ and \begin{align*} \sum_{a\in R^\times}e\left(\frac{a(1+u)}{p}\right)=p-1. \end{align*} If $u\ne -1$, then $1+u\ne 0$ in $R$. Additive orthogonality gives the full sum over $R$ as zero: \begin{align*} \sum_{a\in R}e\left(\frac{a(1+u)}{p}\right)=0. \end{align*} Removing the omitted $a=0$ term, which equals $1$, gives \begin{align*} \sum_{a\in R^\times}e\left(\frac{a(1+u)}{p}\right)=-1. \end{align*} Therefore the $u=-1$ term contributes $(p-1)\chi(-1)$, and every other nonzero $u$ contributes $-\chi(u)$. Thus \begin{align*} \tau(\chi)^2=(p-1)\chi(-1)-\sum_{\substack{u\in R^\times, u\ne -1}}\chi(u). \end{align*} Since the total character sum is zero and $\chi(0)=0$, \begin{align*} \sum_{\substack{u\in R^\times, u\ne -1}}\chi(u)=-\chi(-1). \end{align*} Substituting this identity yields \begin{align*} \tau(\chi)^2=(p-1)\chi(-1)+\chi(-1)=p\chi(-1). \end{align*} This is exactly the sign refinement: the square of the quadratic Gauss sum is $p$ times the value of the quadratic character at $-1$. [/guided] [/step] [step:Conclude the Gauss sum evaluation] From the previous two steps, $G_p=\tau(\chi)$ and \begin{align*} \tau(\chi)^2=p\chi(-1)=\left(\frac{-1}{p}\right)p. \end{align*} Thus \begin{align*} G_p^2=\left(\frac{-1}{p}\right)p. \end{align*} Taking absolute values in the already proved identity $|G_p|^2=p$ gives \begin{align*} |G_p|=p^{1/2}. \end{align*} This proves both assertions. [/step]