Will Sawin and Michael Stoll have noted that, as a consequence of Faltings's "Big Theorem," a hyperelliptic equation $y^2 = f(x)$ with $\deg{f} > 6$ (genus $> 2$) and not admitting a degree $2$ non-constant map to an elliptic curve, has all but finitely many of its quadratic solutions $x, y \in \mathbb{Q}(\sqrt{d}), \, d \in \mathbb{Z}$, satisfy $x \in \mathbb{Q}$. We may add to this an argument due to Granville in *Rational and integral points of quadratic twists of a given hyperelliptic curve* to show:

**Claim.** *The ABC conjecture implies, for a fixed $f$ having $\deg{f} > 6$ and no repeated roots, that the number of squarefree $d$ in $|d| \leq D$ for which the equation $dy^2 = f(x)$ has a rational solution with $y \neq 0$, is $O(D^{2/3+o(1)})$.*

This answers Will's question under ABC. Hence ABC takes care of the problem for most $y^2 = f(x)$ - save for the ones of genus one or two or those doubly covering an elliptic curve. (For rational curves $C/\mathbb{Q}$ the problem is easy: Hasse's theorem shows that $C(\mathbb{Q}) = \emptyset$ is only possible when $C(\mathbb{Q}_p) = \emptyset$ for some prime $p$, but then $C$ will not have points in any quadratic field $\mathbb{Q}(\sqrt{d})$ split by $p$.) It seems to me that the case of hyperelliptic curves of genus $> 2$ doubly covering an elliptic curve can be settled under ABC by similar methods (Faltings's theorem and a modification of Granville's argument), whereas the genus one case should be solved unconditionally by Kolyvagin's theorem and non-vanishing results, cf. Chris Wurthrich's comment in the linked question. I am not sure about the genus two case though - it is barely missed by the argument below.

*Proof of the claim.* (Granville). M. Langevin has noted (cf. Thm. 12.2.12 in *Heights in Diophantine Geometry* by Bombieri and Gubler) that Elkies's construction for "ABC $\Rightarrow$ Roth" via Belyi maps yields in fact much more than Roth's theorem:

**Lemma.** *Let $\varepsilon > 0$ and let $F \in \mathbb{Z}[X,Y]$ be a homogeneous polynomial with distinct linear factors over $\mathbb{C}$. Then for all co-prime $m,n$ with $F(m,n) \neq 0$, the (strong) ABC conjecture implies $\mathrm{rad}(F(m,n)) \gg_{\varepsilon,F} \max(|m|,|n|)^{\deg{F}-2-\varepsilon}$.*

The ABC conjecture is recovered as the special case $F(X,Y) = XY(X+Y)$, whereas Roth's theorem is just the weakening of this statement dropping the radical.

We apply this is follows. Consider the $\asymp T^2$ rational values $x = m/n \in \mathbb{Q}$ with $T \leq \max(|m|, |n|) < 2T$, $(m,n)=1$, $n > 0$, and $f(x) \neq 0$. Writing $du^2 = F(m,n)$ with $F$ the homogenization of $f$, the above quoted form of ABC yields
$$
(DT^{\deg{f}})^{1/2} \gg_f |dF(m,n)|^{1/2} = |du| \geq \mathrm{rad}(F(m,n)) \gg_{\varepsilon,F} T^{\deg{f} - 2 - \varepsilon},
$$
or $D > T^{\deg{f} - 4 - o(1)}$, in $|d| \leq D$. As the irreducible fraction $x$ uniquely determines the squarefree part $d$, we get what we want by splitting the range $T < D^{1/(\deg{f}-4-o(1))}$ into dyadic intervals.

singlequadratic extension of $\mathbb{Q}$, then it already has a point over $\mathbb{Q}$. $\endgroup$ – Daniel Loughran Mar 6 '15 at 9:221more comments