Proving the Formula: $\sum_{a=1}^{p-2} \left(\frac{a(a+1)}{p}\right) =\ -1$

[Vespero]

Homework Statement

Prove the formula
$$\sum_{a=1}^{p-2} \left(\frac{a(a+1)}{p}\right) =\ -1$$

Homework Equations

$\left(\frac{xy}{p}\right) = \left(\frac{x}{p}\right) \left(\frac{y}{p}\right)$

$\left(\frac{x}{p}\right) \equiv x^{\frac{p-1}{2}}\ (mod\ p)$ (Euler's Criterion)

and other basic properties of modular arithmetic and the Legendre symbol.

The Attempt at a Solution

For the first few p's, I calculated the values of the terms as

p=5 => 2,1,2
p=7 => 2,6,5,6,2
p=11 => 2,6,1,9,8,9,1,6,2
p=13 => 2,6,12,7,4,3,4,7,12,6,2

There is symmetry resulting from the fact that $P = \{1, 2, ..., (p-2), (p-1)\}$ can be rewritten as $P = \{1, 2, ..., \frac{p-1}{2}, -\frac {p-1}{2}, ..., -2, -1\}$.

Could I somehow show that the number of consecutive quadratic residues and consecutive quadratic nonresidues (which would each produce a quadratic residue and thus a Legendre symbol of 1 when multiplied together) is one less than the number of pairs containing a quadratic residue and a quadratic nonresidue together (which would produce a quadratic nonresidue and thus a Legendre symbol of -1 when multiplied together?

 

[Nino MMP]

In other words, is there a way to show that the sum of all the Legendre symbols is -1? I think this would work since each consecutive pair of quadratic residues and quadratic nonresidues would have a Legendre symbol of 1, except for the two pairs on the end, which would have a Legendre symbol of -1. Any help is appreciated.