We fix a binary form $F \in \mathbb{Z}[x,y]$ with non-zero discriminant and degree $d = 2g+2$, and consider the hyperelliptic curve


$$C\_F: \displaystyle z^2 = F(x,y).$$


We say that a point $(x,y,z)$ is a *quadratic point* if there exists a quadratic extension $K/\mathbb{Q}$ over which $x,y,z$ are defined.


Then $C\_F$ is easily seen to have infinitely many quadratic points: indeed, for any *rational integers* $x,y$ we obtain the quadratic point $(x,y, \sqrt{F(x,y)})$. These are the so-called obvious quadratic points. If we take the height of the point $(x,y,z) \in C\_F$ to simply be the Weil height of the point $[x:y]$ on $\mathbb{P}^1$, then counting the obvious quadratic points of $C\_F$ having bounded height $X$ is tantamount to counting the number of co-prime integers $x,y$ such $\max\{|x|, |y|\} \leq X$.


Do we expect this count to be dominant? That is, do the *non-obvious* quadratic points on $C\_F$, i.e., those where $x,y,z$ are all in some quadratic extension and at least two of them not in $\mathbb{Q}$, more/less/equally dense than the obvious points?



Density of quadratic points on a hyperelliptic curve
Yes, this is (provably!) dominant.


As Jackson Morrow pointed out, quadratic points on $C$ are the same as rational points on the symmetric square of $C$. The obvious points are the rational points on a $\mathbb P^1$. Contracting this $\mathbb P^1$, we get a surface in an abelian variety, hence by Falting's theorem (as Jackson Morrow also pointed out), contained in finitely many abelian subvarieties plus finitely many points.


For each abelian variety, the number of points of height at most $X$ will be $O ((\log X)^{r/2})$ where $r$ is the rank. On the other hand, $\mathbb P^1$ has $\approx X^2$ points of height $<X$.


So we can see that the obvious points are dense, as soon as we check that the heights in $\operatorname{Sym}^2 C$ and $C$ are comparable. But this is straightforward - we can map $\operatorname{Sym}^2 C \to \operatorname{Sym}^2 \mathbb P^1 \to \mathbb P^2$ by $( (x\_1:y\_1),(x\_2:y\_2)) \to ( x\_1x\_2: x\_1 y\_2+ x\_2 y\_1 : y\_1 y\_2)$ which makes it clear that if $(x\_1:y\_1)$ and $(x\_2:y\_2)$ are conjugate and have the same Weil height, their heights are at least the square root of the height of $( x\_1x\_2: x\_1 y\_2+ x\_2 y\_1 : y\_1 y\_2)$.


