elementary number theory - Ordered pair solutions of $x^2-y^2 equiv a pmod p.$

This is the full question:

show that if $p$ is an odd prime then the number of ordered pair solutions of the congruence$x^2-y^2 \equiv a \pmod p,$ is $p-1$ unless $a \equiv 0 \pmod p,$ in which case number of solutions is $2p-1$.

Considering $x,y \,\in \mathbb{Z}$. In the second case, since $a \equiv 0 \pmod p,$ it follows that $p\mid (x-y)(x+y)$, but then there will be infinitely many solutions of this congruence relation because there are no bounds mentioned in the question on $x$ and $y$.

So is the question incomplete? or is it implicitly stated that $0\leq x,y .For this bound, do we get $2p-1$ solutions?

Answer

ETA: For whatever reason I thought the question here was to prove that the statement in the OP's thread in the shaded yellow box, which is what I did below.

You already answered your own question for the case where $a$ is zero so we now consider the case where $a$ is nonzero i.e., $a \in (\mathbb{F}_p)^{\times}$.

1. Let $a \in (\mathbb{F}_p)^{\times}$. For every such nonzero $b \in (\mathbb{F}_p)^{\times}$, there is exactly one $c \in (\mathbb{F}_p)^{\times}$ satisfying $a=bc$.




2. The above equation factors to $(x-y)(x+y) = a; x,y \in (\mathbb{F}_p)^{\times}$. By the above, for each nonzero $b=(x-y)\in (\mathbb{F}_p)^{\times}$, there is exactly one $c=(x+y)\in (\mathbb{F}_p)^{\times}$ such that $(x-y)(x+y)=a$.




3. So for each nonzero $b\in (\mathbb{F}_p)^{\times}$, there is exactly one pair $(x,y); x,y \in (\mathbb{F}_p)^{\times}$ such that both $(x-y)=b$ and $(x-y)(x+y)=a$.

Can you finish from here.