Evaluate the sum of the following Legendre symbols: $\sum\limits_{a\in\mathbb{F}_p \ \text{and} \ a \neq 0,1} \left(\frac{a-a^2}{p} \right)$ 
Evaluate the sum of the following Legendre symbols: 
   $$\sum\limits_{a\in\mathbb{F}_p \ \text{and} \  a \neq 0,1} \left(\frac{a-a^2}{p} \right).$$

I am stuck in this step:  $\sum\limits_{a\in\mathbb{F}_p \ \text{and} \  a \neq 0,1} \left(\frac{-1}{p} \right)*(\frac{a^2 -a}{p})$
could anyone help me please?
 A: Use the claim below and your problem is trivial.

Claim: For an odd prime natural number $p$ and $a,b\in\mathbb{F}_p$ with $b\neq 0$, we have $$\sum_{x\in\mathbb{F}_p}\,\left(\frac{x(ax+b)}{p}\right)=-\left(\frac{a}{p}\right)\,.$$

Proof:  Let $x^{-1}$ be the inverse of $x$ modulo $p$.  We have $$\begin{align}
\sum_{x\in\mathbb{F}_p}\left(\frac{x(ax+b)}{p}\right)&= \sum_{x\in\mathbb{F}_p^\times} \left(\frac{x}{p}\right)\left(\frac{ax+b}{p}\right) = \sum_{x\in\mathbb{F}_p^{\times}} \left(\frac{x^{-1}}{p}\right)\left(\frac{ax+b}{p}\right)
\\&= \sum_{x\in\mathbb{F}_p^{\times}}\left(\frac{x^{-1}(ax+b)}{p}\right)=\sum_{x\in\mathbb{F}_p^{\times}}\left(\frac{a+bx^{-1}}{p}\right)\,,
\end{align}$$
where $\mathbb{F}_p^\times:=\mathbb{F}_p\setminus\{0\}$. Therefore,
$$ \sum_{x\in\mathbb{F}_p}\left(\frac{x(ax+b)}{p}\right)= \sum_{y\in\mathbb{F}_p^{\times}}\left(\frac{a+by}{p}\right) = \sum_{y\in\mathbb{F}_p}\left(\frac{a+by}{p}\right)-\left(\frac{a}{p}\right)\,.$$ Because $p\nmid b$, we have $\left\{a+by \,\big|\, y\in\mathbb{F}_p\right\}$ is a complete residue system modulo $p$, and thus, $\sum_{y\in\mathbb{F}_p}\left(\frac{a+by}{p}\right) = 0$.  This proves the desired equality.
A: You can seperate the cases when $p\equiv 3\pmod{4}$ and $p\equiv 1\pmod{4}$. Note that $$\sum_{a\in \mathbf{F}_{p}} \left(\frac{a^{2}-a}{p}\right) = \sum_{a \in \mathbf{F}_{p}} \left(\frac{a}{p}\right)\cdot \left(\frac{1-a}{p}\right) = \sum_{a \in \mathbf{F}_{p}} \left(\frac{a^2}{p}\right)\left(\frac{a^{-1}-1}{p}\right)$$
Now think what happens when to $a^{-1}-1$ as $a$ varies over $\mathbf{F}_{p}$.
