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.



Rank of jacobians of twists of hyperelliptic curves of genus one
The answer is yes, and it's fairly elementary. By the usual 2-descent, the curve $C$ gives a class $c$ in $H^1(\mathbb{Q},E[2])$, where $E$ is the Jacobian you wrote down. As you vary $d$, the groups $H^1(\mathbb{Q},E\_d[2])$ are canonically isomorphic, and $c$ is also the class of $C\_d$. To answer your question, you should condition on the possibility that $c$ "comes from" the 2-torsion subgroup of $E$ via $E(\mathbb{Q})/2E(\mathbb{Q}) \to H^1(\mathbb{Q},E[2])$. I don't know what the conditions on $a,b$ for this to happen are, but it won't matter.


If $C$ doesn't come from 2-torsion, and if $C\_d$ has a rational point, then automatically $E\_d$ has rank at least one (by descent). And as Alex B. says, it's easy to see that there are infinitely many squarefree $d$ for which this happens (his argument doesn't quite show this, but it's not hard to fiddle with congruence conditions to produce infinitely many squareclasses).


If $C$ does come from 2-torsion, then $C\_d$ has a rational point for all $d$, but there is no guarantee that the rank is at least 1. So your question reduces to asking about the ranks of the twists of $E$. But now just apply the argument from the previous paragraph to some other class in $H^1(\mathbb{Q},E[2])$ represented by a binary quartic form (there are infinitely many of these).


