Counting the solutions of a quadratic equation

Question

I have read that a non-singular conic will contain $p+1$ points on the finite field $\mathbb{F}_{p}$, but there is a exercise on Silverman's Rational Points on Elliptic Curves, p.142, 4.8 that tells us that if $p\equiv3\pmod 4$ is a prime and $b\in\mathbb{F}_{p}^{*}$, the equation $v^2=u^4–4d$ has $p-1$ points in $\mathbb{F}_{p}\times\mathbb{F}_{p}$. But $u^2\mapsto u^4$ is an automorphism in the group of quadratic residues. So it should contain exactly same number of points with the conic case. So where are the missing two points?

Answer 1

The missing points are on the line at infinity.

The statement that the number of $\Bbb{F}_p$-rational points on a non-singular conic is $p+1$ applies to projective plane curves. This exercise is about the number of points on the affine plane $(u,v)\in\Bbb{F}_p^2$.

Notice that the corresponding quadratic curve $$ v^2=u^2-4d $$ has two points on the line at infinity. Namely the ideal points of the two lines $v=\pm u$. Or, if you prefer projective coordinates of the curve $$ V^2=U^2-4dW^2, $$ then those two points have homogeneous coordinates $$[U:V:W]=[1:\pm1:0].$$