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Let $E/\mathbb{Q}$ be an elliptic curve having a rational 3-torsion point. Then $E$ can be given an affine equation of the type $$y^{2} = x^{3} + (ax + b)^{2}$$ for $a, b, D \in \mathbb{Q}$. Has there been any work done about curves of this form having prime or almost prime conductors?

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Actually that's the form of the general elliptic curve with a rational $3$-isogeny. For $3$-torsion, you can take $D=1$, or equivalently write the curve as $y^2 + a_1 x y + a_3 y = x^3$ with a $3$-torsion point at $(x,y)=(0,0)$. Do you want a $3$-torsion point or just the rational $3$-isogeny? – Noam D. Elkies Jan 29 '13 at 19:20
I think it's known that the only $E / \mathbb Q$ of prime conductor with a $3$-torsion point, or even a rational $3$-isogeny, are the ones of conductor $19$ and $37$. What do you mean by "almost prime"? – Noam D. Elkies Jan 29 '13 at 19:23
@Noam Elkies: I mean a 3-torsion point, so I've edited the question to reflect this. By "almost prime" I had in mind that the conductor is of the form $pq$ or $pqr$; forcing $E$ to be semistable and somehow restricting the number of factors might do that (though that is also something I'm not clear about). Do you have a source for the prime conductor? – lk728 Jan 29 '13 at 19:43
Curves with prime-power conductor and $\ge 3$ torsion points were classified by Miyawaki: Osaka J. Math. 10 (1973) 309-323,… . (Also, if you've got square-free conductor and an isogeny $E\to E'$ of prime degree $p>2$, then either $E$ or $E'$ has an actual $p$-torsion point, so his result gives you a complete classification for prime conductor, even if you only have a 3-isogeny rather than a 3-torsion point.) – Tim Dokchitser Jan 29 '13 at 23:25
Miyawaki's work was generalized by Hadano in a paper in Manuscripta Math. in 1982. His Table I classifies (in some sense) curves with a $3$-torsion point and bad reduction at $2$ and a prime $p$. One can do something similar (and more precise) with $2$ replaced by $3$. In either case, the arguments extend easily to isogenies rather than torsion points. – Mike Bennett Jan 31 '13 at 3:46

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