Showing that $\sum_{x \in \mathbb{F}_p} \left(\frac{x^2-1}{p}\right) = -1$, where $\left(\frac{x}{p}\right)$ is the Legendre symbol The question is, 

Show that $$\sum_{x \in \mathbb{F}_p} \left(\frac{x^2-1}{p}\right) = -1$$ where the operation $\left(\frac{x}{p}\right) = \pm 1$ if $x$ is a quadratic residue/non-residue and $0$ otherwise.

I was able to follow the argument given here Prove that $\sum_{X=0}^{p-1} \left(\frac{X^{2}+A}{p}\right)=-1$ for the $\left(\frac{-1}{p}\right) = -1$ case, but I'm not sure how to make progress on the other case (i.e $p\equiv 1$ mod $4$).
One thought I had that I couldn't quite make work was to try and use the formula $\left(\frac{x}{p}\right) \equiv x^r$ mod $p$, where $r = (p-1)/2$ and then expanding using the binomial theorem.
Any thoughts or hints would be greatly appreciated!

This is my attempt at what I believe is a full answer following the hints given in the comments.
We first show that there are exactly $p-1$ solutions to the equation $$y^2 = x^2 - 1$$ for pairs $(x,y)\in \mathbb{F}_p^2$. If $y^2 = x^2 -1$ then $(x-y)(x+y) = 1$. Letting $$u = x-y$$ $$v = x+y$$ we see that for any choice of $(u,v) \in \mathbb{F}_p^2$ we can uniquely solve the system of equations for $x$ and $y$. Thus the number of solutions to $1 = (x-y)(x+y)$ is just the number of solutions to $1 = uv$. There are $p-1$ choices for $u$, and once $u$ is chosen, there is only one $v$ for which $1 = uv$. We conclude that there are exactly $p-1$ solutions to $y^2 = x^2 -1$ 
We will use this information to count the number of distinct $x$ values for which $(\frac{x^2-1}{p}) = 1$. We have the two solutions $(\pm1,0)$, and then for each other value of $x$ for which $y^2 = x^2 -1$ has a solution, we get two solutions, $(x,\pm y)$. Thus we conclude that there should be $\frac{p-3}{2}$ distinct values of $x$ for which $(\frac{x^2-1}{p}) = 1$. Since there are two values of $x$ for which $(\frac{x^2-1}{p}) = 0$, we conclude that the remaining $p-\frac{p-3}{2} - 2$ values of $x$ evaluate $(\frac{x^2-1}{p}) $ as $-1$. Thus our sum is just $\frac{p-3}{2} - (p-\frac{p-3}{2}-2) = p-3+2-p = -1$. 
