Conics, rational points and probability

Question

Given a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ with integers and random coefficients, what is more probable? To find a rational point on the conic or not?

Answer 1

For any given prime number $p > 2$ the probability that there is no $p$-adic point is $\ge c/p$ for some constant $c > 0$ indpendent of $p$. Since the sum over $1/p$ diverges, this implies that the `probability' (or rather, density) of conics with a rational point is zero.

EDIT: I assume that the `probability' is meant in the sense of density, i.e. $$ \lim_{N \to \infty} \frac{\#\{(A,B,C,D,E,F) \in ({\mathbb Z} \cap [-N,N])^6 : \exists \text{ rational point}\}}{(2N+1)^6}. $$ The probability of $p$-adic solubility (with respect to the standard measure on ${\mathbb Z}_p^6$) comes down to the $p$-adic density of pairs of $p$-adic integers $a, b$ such that the Hilbert symbol $(a,b)_p = -1$. This is the case (e.g.) when $v_p(a) = 1$, $v_p(b) = 0$ and $b$ is a nonsquare mod $p$; the probability for this is $\frac{p-1}{p^2} \cdot \frac{p-1}{2p} \ge \frac{2}{9p}$.

EDIT 2: I realized that the map that produces $a,b$ from $A,B,C,D,E,F$ very likely does not preserve $p$-adic measure. So here is an alternative and more direct argument. We work with projective coordinates, as suggested by Joe Silverman. By Hensel's Lemma, the non-existence of a smooth ${\mathbb F}_p$-point on the reduction mod $p$ of the conic is a necessary condition for the non-existence of $p$-adic points, and this is also sufficient when the conic is regular over ${\mathbb Z}_p$. So whenever the reduction mod $p$ is the product of two conjugate and distinct linear forms with coefficients in ${\mathbb F}_{p^2}$ and the intersection point of the two corresponding lines is regular, then there is no $p$-adic point on the conic. The $p$-adic measure (as a subset of ${\mathbb P}^5({\mathbb Q}_p)$) of the corresponding set is $$\frac{1}{2} \frac{\#{\mathbb P}^2({\mathbb F}_{p^2}) - \#{\mathbb P}^2({\mathbb F}_p)}{\#{\mathbb P}^5({\mathbb F}_p)} \frac{p-1}{p} = \frac{(p-1)^2}{2(p^3+1)} \ge \frac{3}{14p} .$$ (The factor $(p-1)/p$ is the probability that the intersection point is regular.) The paper by Bhargava, Cremona, Fisher, Jones and Keating linked in post.as.a.guest's answer gives the precise value $p/(2(p+1)^2)$.

Answer 2

Problems of this type are considered by Serre in the paper:

Serre - Spécialisation des éléments de $\mathrm{Br}_2(\mathbb{Q}(T_1,\ldots, T_n))$

The case relevant to you is Exemple 4. Here Serre shows that

\begin{align*} &\#\{|A|,|B|,|C|,|D|,|E|,|F| \leq N : \\ & Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \text{ has a rational point} \} \ll \frac{N^6}{(\log N)^{1/2}}. \end{align*}

This gives a more precise quantitative version of "probability $0$" mentioned by Michael Stoll.

Hooley has in fact shown that Serre's bound is sharp. This result is the object of the paper:

Hooley - On ternary quadratic forms that represent zero II.

Answer 3

There is a paper, that answers this in more generality.

https://www.dpmms.cam.ac.uk/~taf1000/papers/isotropic.html

In your case, the probability is 0 (asymptotically) under conditions that the distribution is piecewise smooth and rapidly decaying, though likely can be generalized, in your case.