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Take infinitely many 2-variable polynomials $p_k(X,Y)\in \mathbb{Q}[X]$ ($k\in \mathbb{N}$) and let $S_n$ be the surface given by $p_1(X,Y)=Z_1^2,\dots, p_n(X,Y)=Z_n^2$ Assume that no $p_k$ equals the product of any ``previous" $p_j$ up to squares, so that $S_n\to S_{n-1}$ is indeed a quadratic cover.

Is it possible for all $S_n$ ($n\in \mathbb{N}$) to have a Zariski dense set of rational points?

(Note that the answer to the same question for 1-variable polynomials is ``no" by Faltings' theorem, since taking the fiber product of quadratic covers of curves will quickly blow up the genus.)

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    $\begingroup$ Might it be better to formulate a geometric version first, i.e., is it possible that none of the $S_n$ are of general type? This is on the theory that surfaces that are not of general type may well have a Zariski dense set of rational points, but conjecturally general type surfaces do not. $\endgroup$
    – Joe Silverman
    Commented Apr 19, 2021 at 15:26
  • $\begingroup$ @Joe Silverman: That's basically what I had in mind, although there are cases which are trivially not of general type - e.g. if one of the variables doesn't feature at all or only in a very restricted way, one may get a product of a line and a high genus curve. In this case, the assertion reduces again to Faltings though, so the working hypothesis would be that excluding such extreme cases, one will eventually get to general type. $\endgroup$
    – Joachim König
    Commented Apr 19, 2021 at 15:57
  • $\begingroup$ If the irregularity grows with $n$ then you can deduce non-density from Faltings's theorem for closed subvarieties of abelian varieties most likely. (Actually, you just need the irregularity to be at least three for some n. Then you can apply Faltings.) $\endgroup$
    – Ariyan Javanpeykar
    Commented Apr 21, 2021 at 13:19

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