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I'm trying to use Sage on-line,but I meet some trouble with the code of it. I want to find the rational points on an ellipse curve,such as $$1 + 18 x + 81 x^2 + 44 x^3 = y^2,\tag1$$ I know that $(x,y)=(0,1)(1,\pm12)(-\frac{1}{11},0)$ are on the curve,but I don't know how to find more,so I try to solve it with the math-software Sage.

What should I input?Thanks in advance!

Edit: Multiply by $44^2$,we can difine the ellipse curve by this way,

E = EllipticCurve([0,81,0,44*18,44^2])

Then how to find the rational points on it?

lsr314
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  • Since this question has been linked to elliptic curves, should it be better to tag it as such? Regards. – awllower May 30 '13 at 13:57
  • You could always try asking here. – 77474 May 30 '13 at 14:40
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    http://math.stackexchange.com/questions/203484/how-to-compute-rational-or-integer-points-on-elliptic-curves has links to some examples . – zyx May 30 '13 at 17:59
  • BTW, E.rational_points(bound=1000) took several days to return the output [(-4 : 0 : 1), (0 : -44 : 1), (0 : 1 : 0), (0 : 44 : 1), (44 : -528 : 1), (44 : 528 : 1)] -- so knowing the mathematics in the answers below certainly helps. :-) – ShreevatsaR Jun 10 '13 at 06:41

2 Answers2

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By the Mordell-Weil theorem, the group of rational points $E(\mathbb{Q})$ is finitely generated and therefore isomorphic to a group of the form $T\oplus \mathbb{Z}^R$, where $T$ is finite abelian formed by those points of finite order, and $R\geq 0$ is the rank of the elliptic curve, and it counts how many $\mathbb{Z}$-linearly independent points of infinite order there are.

With Sage, define your curve by

E=EllipticCurve([0,81,0,44*18,44^2]);

Now you can ask Sage to calculate the rank $R$ with

E.rank();

and the answer is $0$. So there are no points of infinite order, only finite order points. Let us calculate the torsion subgroup as follows

T=E.torsion_subgroup();

if you type ``T;'' and evaluate, Sage will tell you that $T\cong \mathbb{Z}/6\mathbb{Z}$. Finally, if you type

T.0;

Sage returns a generator of the torsion subgroup, namely $P=(44,528)$, which is a point of exact order $6$. Hence, $E(\mathbb{Q})=\langle P \rangle$ or $$E(\mathbb{Q})=\langle P \rangle = \{ \mathcal{O}, (44,528),(0,44), (-4,0), (0,-44), (44,-528)\}.$$

Álvaro Lozano-Robledo
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In case a curve has positive rank, you'd like to know its generators by E.gens().

In the case 

E = EllipticCurve([0,121,0,22*52,52^2]);

the rank is 1

sage: E.rank()
1

and its generator is 

sage: E.gens()
[(-104 : 260 : 1)]

In this case you can construct infinitely many points on E, by adding a torsion point and some multiple of the generator, for instance

sage: T = E(0,52)
sage: P = E(-104,260) 
sage: T + 2*P
(480/49 : 55796/343 : 1)
sage: T + 3*P
(-12662312/485809 : -65467251044/338608873 : 1)
sage: T - 5*P
(-70265016926482573/11385999920450401 : 5045808098009295785013222/1214945410097644049235601 : 1)

and so on.

Jesper Petersen
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