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If $E_1$ and $E_2$ are two elliptic curves over $\mathbb{Q}$ such that $|E_1(\mathbb{F}_p)|=|E_2(\mathbb{F}_p)|$ for all primes $p$, what does this tell us about the relationship between $E_1$ and $E_2$?

E.g. the identity holds (at least for all primes $p<500$) when $E_1$ and $E_2$ have the equations $y^2+y=x^3+x^2+2x+4$ and $y^2+y=x^3+x^2-208x-1256$. In what way are these curves similar?

They aren't isomorphic because their reduced forms ($y^2=x^3+ax+b$) are distinct. One of them has a torsion subgroup over $\mathbb{Q}$ of size 5, and the other has trivial torsion subgroup. The difference between the RHS's is $210x+1260=(2)(3)(5)(7)(x+6)$; is it relevant that this has a primorial factor? (Obviously it explains why $|E_1(\mathbb{F}_p)|=|E_2(\mathbb{F}_p)|$ for $p=2,3,5,7$, but why should this still hold for larger $p$ ??)

Many thanks for any help with this!

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A little bird once told me "if and only if they are isogenous," but don't take my word for it. –  Qiaochu Yuan Sep 2 '12 at 19:35
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Ah, never mind; some Googling reveals that what I said is true at a fixed prime $p$ (apparently it is a theorem of Tate) but I don't know if it's true globally. –  Qiaochu Yuan Sep 2 '12 at 19:40
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The global theorem is apparently a theorem of Faltings (at least if I've got the hypotheses straight; I'm a little concerned about the primes of bad reduction). –  Qiaochu Yuan Sep 2 '12 at 19:44
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Your elliptic curve $E$ is a group, and it has a finite subgroup $S$. Then $E/S$ is also an elliptic curve. The primitive way of finding $E/S$ is to get the subfield of its function field that’s fixed under the action of $S$, acting as translations of the underlying space. But Qiaochu may be able to tell you better methods. –  Lubin Sep 2 '12 at 19:56
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You probably knew this, but these are the two curves c1 and c2 of level 75 in Cremona's tables. So one way to see that they are isogenous is just to look them up. –  Jeremy Teitelbaum Sep 2 '12 at 20:12

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