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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.

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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).

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  • $\begingroup$ Thank you!! This is immensely helpful. $\endgroup$ – Stanley Yao Xiao Jan 9 at 19:38
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If you are willing to assume finiteness of Sha, then the answer is "yes".

The first claim, that for infinitely many $d$ (square-free, presumably?) this has a rational point, is easy and does not need any conjectural input: start with any $u$, $v$ such that $F_{a,b}(u_0,v_0)$ is non-zero. Then by writing $F_{a,b}(u_0,v_0)$ as a square times the square-free part, you immediately see that there exists a unique square-free $d\in \mathbb{Z}$ such that $F_{a,b}(u_0,v_0)/d$ is a square, say $z_0^2$, so that $(u_0,v_0,z_0)$ is a rational point on $dz^2=F_{a,b}(u,v)$.

For the second part, observe, e.g. by using the explicit equation for the Jacobian that you have given, that the Jacobian of the twist by $d$ is the twist by $d$ of the Jacobian. Since these Jacobians have full $2$-torsion, it follows from work of Kane that infinitely many of these have $2$-Selmer of dimension $3$, and if Sha is finite, then $3$-dimensional $2$-Selmer implies rank $1$.

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