I liked this MathOverflow question, which asks: are there two non-isogenous elliptic curves over Q, each one of which has a rational cyclic 13-isogeny, and such that the kernels of the two isogenies are isomorphic as Galois modules?

This is precisely to look for rational points on the modular surface S parametrizing pairs (E,E’,C,C’,φ), where E and E’ are elliptic curves, C and C’ are cyclic 13-subgroups, and φ is an isomorphism between C and C’.

S is a quotient of X_1(13) x X_1(13) by the diagonal in the natural (Z/13Z)^* x (Z/13Z)^* action.

Is S general type, rational, what?

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Suppose I look at a family of twists H_d : dy^2 = f(x) of a hyperelliptic curve, where f(x) has odd degree. Is it possible that for every d, H_d has at most 2 rational points?

(If the answer is negative we could presumably apply this to twists of X_1(13).)

Why require f to have odd degree (equivalently, H_d to have a rational Weierstrass point)?

X_1(13) has genus 2 and six rational points but none of them is Weierstrass.

Turns out the surface is honestly elliptic, and as it happens I already ran across a nontrivial example of one of its rational points in 2004! See my answer on mathoverflow.

NDE’s awesome answer:

http://mathoverflow.net/questions/129818/elliptic-curves-over-qq-with-identical-13-isogeny/130844#130844