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.

  • 1
    $\begingroup$ Hint $b^2=4ac$ means that $ax^2+bxy+cy^2=A(Bx+Cy)^2$, If you need another hint, do ping! $\endgroup$ Mar 14 '19 at 20:42

I would try to aim for the following intuition: Your conic is either non-degenerate or degenerate, which is what your discriminant decides. If it is non-degenerate, then over the reals you'd get something that is topologically equivalent to a projective line. You can define a projective basis on it and describe every point in a unique fashion with respect to that basis. All of this should carry over to finite fields just fine.

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 $d\neq 0$.


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