# Rational Points on the genus $2$ curve $y^2 = x^6 - 4x^2+4$.

I would like to find a rational solution(s) to elliptic curve $$y^2 = x^3 - 4x+4$$ where $$x$$ is a square (other than $$x,y=±1$$).

This problem is equivalent to finding a rational point on the curve:

$$y^2 = x^6 - 4x^2 + 4$$

There are lots of points on the elliptic but not necessarily with $$x$$ also a square:

Can someone help me with how I could find a solution where $$x$$ is a square as well?

• $y^2 = x^6 - 4x^2 + 4$ is a curve of genus $2$, there are many results on the Web about rational points on such curves. Jan 3, 2022 at 18:01
• You do have $x,y=\pm1$ Jan 3, 2022 at 18:17
• Yes, thanks, looking for solution other than 1. Jan 3, 2022 at 18:39
• See, e.g., Flynn, Wetherell, "Finding Rational Points on Bielliptic Genus 2 Curves": citeseerx.ist.psu.edu/viewdoc/… Jan 3, 2022 at 18:54
• To those who voted to close as a duplicate. This question is misrepresented by its title - it is about computing points on the genus 2 curve covering a rank 1 elliptic curve, which is a nontrivial extension Jan 6, 2022 at 19:52

The first question is simply answered by asking a computer for the generators of the Mordell-Weil group of the elliptic curve $$E : y^2 = x^3 - 4x + 4$$. The group of rational points $$E(\mathbb{Q})$$ is isomorphic to $$\mathbb{Z}$$ and is generated by the point $$(2, -2)$$. This can be achieved via, e.g., Generators in Magma (and probably some analogy in Sage will work since this is rather an easy example as the $$2$$-torsion in sha is trivial).

The second question is rather more interesting. Let $$C$$ be the genus $$2$$ curve $$C: y^2 = x^6 - 4x^2 + 4$$ It is a theorem of Faltings that a curve of genus $$\geq 2$$ will have finitely many rational points. A quick search for points using a computer yeilds the rational points $$(-1 : -1), (-1 : 1), (0 : -2), (0 , 2), (1 , -1), (1 , 1)$$

I claim these are all the rational points on (the affine part of) $$C$$. Magma checks that the rank of $$J = \operatorname{Jac}(C)$$ is $$1$$. Indeed the point $$[\infty_+] - [(-1,-1)] \in J(\mathbb{Q})$$ has infinite order.

Since that rank of $$J$$ is less than its dimension Chabauty's method may be applied to determine the rational points on $$C$$. In this case we are especially lucky since it is implemented in Magma for genus $$2$$ curves.

Actually there is an even easier way. Since we have a covering $$C \to E : [x,y] \mapsto [x^2, y]$$ it is a hint that $$C$$ must have split Jacboian - i.e., $$J$$ is isogenous to a product of elliptic curves. One of which is $$E$$, which has rank $$1$$, hence the other must have rank $$0$$ (since $$J$$ has rank $$1$$).

In fact $$C$$ admits a morphism to the elliptic curve $$E : y^2 = 4x^3 - 4x^2 + 1$$ via the map $$(x,y) \mapsto (1/x^2, y/x^3)$$. This curve has rank $$0$$ and its only rational torsion points are the $$5$$-torsion points $$(0, 1), (0, -1), (1, -1), (1, 1)$$ (experts may spot that this is one of the curves with minimal conductor $$11$$, Cremona label $$11a3$$). The result then follows from computing the preimages of these points on $$C$$.

The following code then verifies our Chabauty claim.

_<x> := PolynomialRing(Rationals());
C := HyperellipticCurve(x^6 - 4*x^2 + 4);
pts := Points(C : Bound:=10);
J := Jacobian(C);
RankBound(J);

P := J![C![1,1,0], C![-1,-1,1]];
Order(P);
ptsChb := Chabauty(P);
ptsChb eq pts;