# Number of Rational Points on $C : ax^2 + bxy + cy^2 = dz^2$ over finite field

Let $$p \neq 2$$ be a prime, let $$a,b,c,d \in \mathbb{F}_p$$ satisfy $$acd \neq 0$$, and let $$C$$ be the conic given by the homogeneous equation $$C : ax^2 + bxy + cy^2 = dz^2.$$

a) If $$b^2 \neq 4ac$$, prove that $$\#C(\mathbb{F}_p)=p+1$$.

b) If $$b^2 = 4ac$$, prove either $$\#C(\mathbb{F}_p) = 1$$ or $$2p+1$$. Give examples for $$p = 3$$ to show that both possibilities can occur. More generally, show that both possibilities occur for all odd primes.

I don't really know where to begin with a question like this. I have tried a few examples where $$p =3$$ and $$a=b=c=d = 1$$ but I don't really have any intuition. I'm using Silverman and Tate's Rational Points on Elliptic Curves.

• Hint $b^2=4ac$ means that $ax^2+bxy+cy^2=A(Bx+Cy)^2$, If you need another hint, do ping! Mar 14 '19 at 20:42

General idea is that if you take a point on your conic then there are $$p+1$$ lines through that point. Each of them will intersect the conic in one well-defined other point, except for one line, the tangent, where that second point is in fact the same one you started from. So the $$p+1$$ lines through the point correspond $$1:1$$ to the $$p+1$$ points on the conic.
For the degenerate case over the reals, you might want to distinguish three cases. Either your conic is a pair of real projective lines, or a single line with algebraic multiplicity $$2$$, or a pair of complex conjugate lines which intersect in a single real point. Transported to finite fields, the first of these cases would yield $$2(p+1)-1=2p+1$$ points since the point of intersection is only counted once. The third case yields $$1$$ point. The second case again yields $$p+1$$ points, so at first glance I would expect that to be an option as well. But since your formula is not fully general due to the absence of mixed terms for $$xz$$ and $$yz$$, it can't express duplicate lines while maintaining $$d\neq 0$$.