# Rational $3$-torsion points of an elliptic curve.

Let $$E$$ be an elliptic curve defined over $$\mathbb{Q}$$ with equation $$y^2=x^3+(ax+b)^2$$, where $$a,b \in \mathbb{Q}$$.
I have to prove that $$0$$ and $$(0, \pm b)$$ are the rational $$3$$-torsion points of $$E$$. I know that if $$Q$$ is a $$3$$-torsion point then $$[2]Q=-Q$$ and that the $$x$$-coordinate of a $$3$$-torsion point satisfies $$x(3x^3+4a^2x^2+12abx+12b^2)=0$$
Now, clearly $$0$$ is a $$3$$-torsion point and if I choose $$x=0$$ and I plug it into the equation of $$E$$ I get $$y^2=b^2$$, thus $$y= \pm b$$.
But now I don't know to continue and prove that there are no other rational $$3$$-torsion points (for example if $$a,b \in \mathbb{Z}$$ I would have used the rational root theorem, but I don't know in this case..)

• Over the algebraic closure $\overline{\Bbb{Q}}$ the curve has the full 3-torsion of eight points ($E[3]\simeq C_3\times C_3$). If you are familiar with the Weil pairing then the claim follows easily. For if $E[3]$ is rational over a field $K$, then $K$ must contain the values of the Weil pairing, that is, the third roots of unity. This rules out the possibility that $E(\Bbb{Q})$ could ever contain the full 3-torsion subgroup. – Jyrki Lahtonen Feb 23 at 20:37
• I know that on $\overline{\mathbb{Q}}$ there are nine $3$-torsion points (the roots of $\psi_3^2$ and $0$), but I don't know Weil pairing, and I haven't proved anything yet about the group structure of $E[m]$ – user289143 Feb 23 at 20:39
• Mind you, that curve has a singularity at $(-4,0)$ if $a=3,b=4$. In that case (probably also for other pairs $(a,b)$) you don¨t have an elliptic curve, This may be needed to rule out eventual other zeros of $\psi_3$. – Jyrki Lahtonen Feb 23 at 20:46
• That's maybe helpful.. since $3x^3+4a^2x^2+12abx+12b^2=3(x^3+a^2x^2+2abx+b^2)+a^2x^2+6axb+9b^2=3y^2+(ax+3b)^2$ and this is $0$ iff $y=0$ and $x=-\frac{3b}{a}$ – user289143 Feb 23 at 20:50

Now $$\Delta_E=16b(4a^3b^2-27b^3)$$.
We want to find the solution for $$3x^3+4a^2x^2+12abx+12b^2=0$$ and we can rewrite it as
$$3(x^3+a^2x^2+2abx+b^2)+a^2x^2+6abx+9b^2=3y^2+(ax+b)^2$$ and this is equal to $$0$$ iff $$y=0$$ and $$x=-\frac{3b}{a}$$.
Thus, if we plug those values to the equation of $$E$$ we get $$0=-\frac{27b^3}{a^3}+4b^2 \Leftrightarrow 4a^3b^2-27b^3=0 \Leftrightarrow \Delta_E=0$$ but then we get a singular curve, hence it cannot be.