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Jan 10, 2022 at 20:20 vote accept Adithya Chakravarthy
Jan 10, 2022 at 15:24 comment added Wojowu In 2, do you mean to ask for $E'$ to be not isomorphic/isogenous to $E$? Because I don't see why you couldn't take $E'=E$.
Jan 10, 2022 at 14:50 answer added Olivier timeline score: 5
Jan 10, 2022 at 10:29 comment added Alex B. See also mathoverflow.net/questions/158702/…
Jan 10, 2022 at 9:38 comment added Chris Wuthrich Tom Fisher has a paper "Explicit moduli spaces for congruences of elliptic curves" for $p<12$ and one for $p=13$ and one for $17$, the last containing a strong conjecture as when elliptic curves could be congruent modulo $p\geq 17$ strengthening the Frey-Mazur conjecture.
Jan 10, 2022 at 2:21 history edited Adithya Chakravarthy CC BY-SA 4.0
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Jan 10, 2022 at 2:13 comment added Noam D. Elkies Problem 2 is surely too hard, because it asks for a second rational point on the twist of ${\rm X}(p)$ by $E[p]$; once $p \geq 7$ that's a curve of genus $ > 1$ that already has one rational point, so there's no local obstruction to the existence of another rational point, but typically no such point exists. (For $p=2,3,5$ there are infinitely many such $E'$: for these small $p$, the curve ${\rm X}(p)$ has genus zero, so any twist with a rational point is a rational curve has infinitely many rational points.)
Jan 10, 2022 at 1:15 history edited Adithya Chakravarthy CC BY-SA 4.0
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Jan 10, 2022 at 1:06 history asked Adithya Chakravarthy CC BY-SA 4.0