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Integral points on a cubic curve which is not in Weierstrass form [closed]

asked 2021-12-11 09:43:10 +0100

benwh1 gravatar image

How can I compute the integral points on a cubic curve which is not in Weierstrass form? For example something like x^3 + y^3 + z^3 = 6xyz. I have tried using EllipticCurve_from_cubic, but this automatically transforms it into an equivalent (over Q) curve in Weierstrass form, which doesn't preserve the integral points.

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Closed for the following reason the question is answered, right answer was accepted by benwh1
close date 2021-12-12 01:44:37.247093


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slelievre gravatar imageslelievre ( 2021-12-11 14:25:07 +0100 )edit

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answered 2021-12-12 01:43:58 +0100

benwh1 gravatar image

Wait, never mind. I just remembered that this is a homogeneous curve in 3 variables rather than a projective curve in 2 variables, so the integral points of this curve are the same as the rational points of the projective curve. So it doesn't matter that EllipticCurve_from_cubic rewrites it in Weierstrass form.

A.<x,y,z> = PolynomialRing(QQ, 3)
phi = EllipticCurve_from_cubic(x^3+y^3+z^3-6*x*y*z)
C = phi.codomain()
P = C.gens()[0]
P2 = phi.inverse()(5*P)
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slelievre gravatar imageslelievre ( 2021-12-12 04:27:06 +0100 )edit

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Asked: 2021-12-11 09:43:10 +0100

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Last updated: Dec 12 '21