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Does $(x^2 - 1)(y^2 - 1) = c z^4$ have a rational point, with z non-zero, for any given rational c?
I need this result for something else. It seems fairly hard, but I may be missing something obvious.
Just one non-trivial solution for any given $c$ would be fine (for my application).
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Is the set of points on an abelian surface which project to rational points on the Kummer surface a subgroup?
Let $C$ be a hyperelliptic curve of genus 2 defined over $\mathbb{Q}$, let $J$ be its Jacobian, and let $X$ be the Kummer surface associated to $J$ (i. e. $X$ is the singular Kummer surface which ...
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