Show that the following sum of legendre symbols is -1. Let. $p$ be an odd prime. Consider the following sum of Legendre Symbols:
$(\frac{1}{p})(\frac{2}{p}) + (\frac{2}{p})(\frac{3}{p}) + \cdots + (\frac{p-2}{p})(\frac{p-1}{p})$.
Show that this sum is equal to $-1$. 
Using the algebra of the Legendre symbol i can show that this sum is the same as 
$\sum_{i = 2}^{\frac{p-1}{2}}(\frac{i-1}{p})(\frac{i}{p}) + (\frac{\frac{p-1}{2}}{p})(\frac{\frac{p+1}{2}}{p})$. 
I can also show that this last term is equal to 1 if $p \equiv 1 \mod 4$ and $-1$ if $p \equiv 3 \mod 4$ via Guass' Lemma. I'm really more interested in a hint than a full solution but any help would be greatly appreciated. 

With the assumptions $\left(\frac{0}{p}\right)=0$, $p\equiv 1\pmod{2}$, by exploiting the multiplicativity of the Legendre symbol we have
$$ \sum_{k=1}^{p-2}\left(\frac{k}{p}\right)\left(\frac{k+1}{p}\right)=\sum_{k=1}^{p-2}\left(\frac{k^2+k}{p}\right)=\sum_{k=1}^{p-2}\left(\frac{1+k^{-1}}{p}\right)$$
where $k^{-1}$ stands for the inverse of $k$ in $\mathbb{F}_p^*$. Now it is enough to consider how that map $k\mapsto 1+k^{-1}$ acts on $\{1,2,\ldots,p-1\}$ and recall that
$$ \sum_{k=1}^{p-1}\left(\frac{k}{p}\right)=0, $$
i.e. that in $\mathbb{F}_p^*$ there are as many quadratic residues as quadratic non-residues.
