Ask Your Question
1

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.

edit retag flag offensive reopen merge delete

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

Comments

Welcome to Ask Sage! Thank you for your question.

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

1 Answer

Sort by ยป oldest newest most voted
1

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)
P2.clear_denominators()
edit flag offensive delete link more

Comments

Click the accept button (with a check mark) near the top left of the answer to mark the question as answered.

slelievre gravatar imageslelievre ( 2021-12-12 04:27:06 +0100 )edit

Question Tools

1 follower

Stats

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

Seen: 201 times

Last updated: Dec 12 '21