Prove that $\sum_{X=0}^{p-1} \left(\frac{X^{2}+A}{p}\right)=-1$ 
Let $\left(\dfrac{a}{p}\right)$ denote the Legendre symbol and $p \geq 3$ a prime number and $A$ an integer such that $p \nmid A$. Prove that $$\sum_{X=0}^{p-1} \left(\dfrac{X^{2}+A}{p}\right)=-1 .$$

I wasn't sure how to manipulate this sum. Maybe if we show that $\left(\dfrac{X^{2}+A}{p}\right)$ can't be a quadratic residue for more than some amount of values it will help.
 A: Fix an odd prime natural number $p$.  Let
$$S_p(A):=\sum_{x\in\mathbb{F}_p}\,\left(\frac{x^2+A}{p}\right)$$
for all $A\in\mathbb{F}_p^\times$, where $\mathbb{F}_p^\times$ is the group $\mathbb{F}_p\setminus\{0\}$ of units modulo $p$.  Note that
$$
\begin{align}
S_p(A)
&=\left(\frac{A}{p}\right)+\sum_{x\in\mathbb{F}_p^\times}\,\left(\frac{x^2+A}{p}\right)=\left(\frac{A}{p}\right)+\sum_{x\in\mathbb{F}_p^{\times}}\,\left(\frac{1+Ax^2}{p}\right)\left(\frac{x^{-2}}{p}\right)
\\
&=\left(\frac{A}{p}\right)+\sum_{x\in\mathbb{F}_p^{\times}}\,\left(\frac{1+Ax^2}{p}\right)=\left(\frac{A}{p}\right)\,\left(1+\sum_{x\in\mathbb{F}_p^{\times}}\,\left(\frac{x^2+A^{-1}}{p}\right)\right)
\\
&=\left(\frac{A}{p}\right)\left(1-\left(\frac{A^{-1}}{p}\right)+\sum_{x\in\mathbb{F}_p}\,\left(\frac{x^2+A^{-1}}{p}\right)\right)=\left(\frac{A}{p}\right)-1+\left(\frac{A}{p}\right)\,S_p\left(A^{-1}\right)\,.
\end{align}$$
If $\left(\dfrac{A}{p}\right)=-1$, then we have
$$S_p(A)+S_p\left(A^{-1}\right)=-2\,.\tag{1}$$
From the claim in this answer, we have
$$-1=-\left(\frac{1}{p}\right)=\sum_{y\in\mathbb{F}_p}\,\left(\frac{y(y+A)}{p}\right)=\frac{1}{2}\,\sum_{x\in\mathbb{F}_p^{\times}}\,\left(\frac{x^2+A}{p}\right)+\frac{1}{2}\,\sum_{x\in\mathbb{F}_p^{\times}}\,\left(\frac{x^2+Au}{p}\right)\,,$$
where $u$ is an arbitrary quadratic nonresidue modulo $p$.  Hence,
$$S_p(A)+S_p(Au)=-2\,.\tag{2}$$
If $\left(\dfrac{A}{p}\right)=-1$, then $(2)$ with $u:=A^{-1}$ yields
$$S_p(A)+S_p(1)=-2\,.\tag{3}$$
Comparing $(3)$ with $(1)$, we have $S_p\left(A^{-1}\right)=S_p(1)$.  However, this is true for all $A$ with $\left(\dfrac{A}{p}\right)=-1$, whence $S_p(A)=S_p(1)$.  Consequently, $S_p(A)=-1$.
Now, if $\left(\dfrac{A}{p}\right)=+1$, then we know from the paragraph above that $S_p(Au)=-1$.  Ergo, $S_p(A)=-1$ as well.

In general, let $p$ be an odd prime natural number and $a,b,c\in\mathbb{F}_p$.  We have
$$\sum_{x\in\mathbb{F}_p}\,\left(\dfrac{ax^2+bx+c}{p}\right)=\begin{cases}
p\,\left(\dfrac{c}{p}\right)&\mbox{if }a=0\mbox{ and }b=0\,,\\
0&\mbox{if }a=0\mbox{ and }b\neq0\,,\\
(p-1)\,\left(\dfrac{a}{p}\right)&\mbox{if }a\neq 0\mbox{ and }b^2-4ac=0\,,\\
-\left(\dfrac{a}{p}\right)&\mbox{if }a\neq 0\mbox{ and }b^2-4ac\neq 0\,.\\
\end{cases}$$

