3

I do have a few books that discuss elliptic curves, however...

What are the rational points on $$ y^2 = 4 x^3 - 4 x = 4 x(x-1)(x+1)? $$ I think it ought to be $(-1,0), (0,0), (1,0).$ Maybe it's just me. I also think this must be a famous first case, written in some popular book I don't have. I do not think there is any important difference if you prefer $$ w^2 = v^3 - v. $$ The curve in the real plane has two connected components.

If this works the way I want it will finish There is no Pythagorean triple in which the hypotenuse and one leg are the legs of another Pythagorean triple.

Community
  • 1
Will Jagy
  • 139,541
  • These are indeed the only rational points of this curve, but I’m unfortunately not in a position to give you a reference. – Lubin Mar 03 '14 at 01:28

1 Answers1

7

Let $A,B$ be integers, and let $E_{A,B}: y^2=x^3+Ax^2+Bx$ be an elliptic curve. Then, the rank of the Mordell-Weil group $E_{A,B}(\mathbb{Q})$ is less or equal to $\nu(A^2-4B)+\nu(B)-1$, where $\nu(N)$ is the number of positive prime divisors of $N$. This is shown in Milne's "Elliptic Curves", Proposition 5.6, or here (Proposition 1.1).

In particular, $E_{0,-1}:y^2=x^3-x$ has rank less or equal to $1+0-1=0$, and so the rank is $0$. The torsion subgroup can be shown to be $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, so $E_{0,-1}(\mathbb{Q})=\{\infty, (0,0),(1,0),(-1,0)\}.$

By the way, notice that the natural number $n$ is a congruent number if and only if the elliptic curve $E: y^2=x^3-n^2x$ is a curve of rank $0$. The proof that $n=1$ is not a congruent number is due to Fermat, and it is `elementary'. The proof can be found in these notes, Theorem 2.1, by Keith Conrad.

Álvaro Lozano-Robledo
  • 15,390
  • @WillJagy, in the new edited version, I added a note about an "elementary" proof of the fact that the rank of $y^2=x^3-x$ is zero. – Álvaro Lozano-Robledo Mar 03 '14 at 02:37
  • Oooh, paragraph just before Example 4.3: "For example, since $1$ is not congruent (Theorem 2.1), the only rational solutions to $y^2 = x^3 - x$ have $y=0.$" – Will Jagy Mar 03 '14 at 02:52