How to prove equality of sum of Legendre symbols I have to prove that the next equality holds:
$$\sum_{k=0}^{p-1} \left( \frac{k(k+a)}{p} \right)=\sum_{k=0}^{p-1} \left( \frac{k(k+1)}{p} \right)$$
with $a \in \mathbb{Z}$ and $a$ not divisible by $p$ and p prime. I am supposed to use a substitution for this, but I have no idea which one. 
Afterwards I have to use this equality, together with this one (which I already proved) 
$$\sum_{k,l=1}^{p-1} \left( \frac{kl}{p} \right)=0$$
to prove that
$$\sum_{k=1}^{p-2} \left( \frac{k(k+1)}{p} \right)=-1.$$
Any help would be appreciated!
 A: Hint
If I understand you correctly, you cannot prove the first equality, nor use this to prove the second and thrid?
For the third note that for $1 \leq k \leq p-2$, $k$ has a unique inverse between $1$ and $p-2$. 
\begin{align}
\sum_{k=1}^{p-2}{\left(\frac{k(k+1)}{p}\right)}&=\sum_{k=1}^{p-2}{\left(\frac{(\frac{k+1}{k})}{p}\right)}\\
&=\sum_{k=1}^{p-2}{\left(\frac{1+k^{-1}}{p}\right)}\\
&=\sum_{k=1}^{p-2}{\left(\frac{1+k}{p}\right)}\\
&=\sum_{k=2}^{p-1}{\left(\frac{k}{p}\right)}\\
&=-1
\end{align}
The first equality holds since 
\begin{align}
\left(\frac{a(a+1)}{p}\right)&=\left(\frac{(\frac{a+1}{a})}{p}\right)\left(\frac{a^2}{p}\right)\\
&=\left(\frac{(\frac{a+1}{a})}{p}\right)
\end{align}
The last one also holds because $\sum\limits_{a=1}^{p-1}{\left(\frac{a}{p}\right)}=0 $ and $\left(\frac{1}{p}\right)=1$
A: $$\sum_{k\in\mathbb{F}_p}\left(\frac{k(k+a)}{p}\right)=\sum_{k\in a\mathbb{F}_p}\left(\frac{k(k+a)}{p}\right)=\sum_{k\in\mathbb{F}_p}\left(\frac{ak(ak+a)}{p}\right)=\sum_{k\in\mathbb{F}_p}\left(\frac{a^2}{p}\right)\left(\frac{k(k+1)}{p}\right)=\sum_{k\in\mathbb{F}_p}\left(\frac{k(k+1)}{p}\right).$$
