For $a,b \in \mathbb{Z}$ we define the binary quartic form

$$\displaystyle F_{a,b}(u,v) = a(u^2 - v^2)^2 + 4bu^2 v^2.$$

We shall assume throughout that the discriminant $$\Delta(F_{a,b}) = 4096a^2 b^2 (a-b)^2$$ of $F_{a,b}$ is non-zero; that is, the form $F_{a,b}$ is non-singular. Consider the twist family of genus one curves given by

$$\displaystyle d z^2 = F_{a,b}(u,v), d \in \mathbb{Z}.$$

Is it known that for infinitely many $d$, the curve given above has a rational point *and* has jacobian whose rank is at least one (so that the curve is isomorphic to its jacobian, which has positive rank)?

The Jacobian of the curve $z^2 = F_{a,b}(u,v)$ (i.e., when $d = 1$) is given by

$$y^2 = x^3 - \frac{16(a^2 - ab + b^2)}{3} - \frac{64(a+b)(2a-b)(a-2b}{27} $$ $$= \left( x - \frac{4b - 8a}{3}\right)\left( x - \frac{4a - 8b}{3}\right)\left( x - \frac{4b + 4a}{3}\right),$$

which is an elliptic curve with full rational 2-torsion. Therefore it is amenable to the techniques introduced by A. Smith (https://arxiv.org/abs/1702.02325), so we know that for 100% of quadratic twists of this curve the $2^\infty$-Selmer rank is at most one. In particular, ordering the twists by $|d| \leq X$, that only $o(X)$ of such twists will have rank at least two.

This is a stronger conclusion that can be reached for any large family of elliptic curves, since in that case we only have control of the sizes of $p$-Selmer groups for $p = 2,3,5$ on average due to work of Bhargava and Shankar. However in this setting there is work due to Bhargava and Skinner, and subsequently Bhargava-Skinner-Zhang, which shows that in fact a positive proportion of curves will have positive rank in any large family.