What is the number of solution for $(x-1)(y-1)+(xy-1)=0$ over the finite field $\mathbb{F}_{p}$?

Question

I know how to find the number solutions of polynomials of single variable over finite field. I curious about equations with multiple variable.

Answer 1

This one is simple, as it defines $y$ as a homographic function of $x$ (or vice-versa).

Rewrite this equation as $$(2x-1)y=x.$$ You have two cases

1. If $p=2$, it reduces to $y=-x$, so there are $2$ solutions (and $2^n$ if the field is $\mathbf F_{2^n}$).
2. If $p\ne 2$, $2$ has an inverse in $\mathbf F_p$, so if $x\ne 2^{-1}$, we have $$y=x(2x-1)^{-1}=x(2x-1)^{p-2}.$$

Answer 2

Over $\mathbb{F}_2$, this equation becomes $x + y = 0$, which has 2 solutions.

Over $\mathbb{F}_p$ for $p > 2$, we can rewrite the equation as $$(2x - 1)(2y - 1) = 1$$ but since $x \mapsto 2x - 1$ is an invertible map $\mathbb{F}_p \to \mathbb{F}_p$, the number of solutions is the same as the number of solutions of $ab = 1$ over $\mathbb{F}_p$, which is clearly $p-1$.