Real points of order 3 on an elliptic curve.

Question

This comes from Silverman's Rational Points on Elliptic Curves:

Consider the elliptic curve (non singular) $y^2=x^3+ax^2+bx+c=f(x)$ after some computations we can see that points of order 3 in this curve satisfy $\psi_3(x))=0$ $\psi_3(x)=2f(x)f''(x)-f'(x)^2=3x^4+4ax^3+6bx^2+12cx+(4ac-b^2)$

I want two show that there are only two real roots of $\psi_3$ if a,b,c $\in \mathbb{R}$:

First $\psi_3$ goes to infinity if $x \rightarrow \pm \infty$ and $f''(\frac{-a}{3})=0 \implies \psi_3(\frac{-a}{3})<0$ by continuity we get $\alpha_1$, $\alpha_2$ $\in \mathbb{R}$ roots of $\psi_3$. What do I do to show that these are the only real roots?

Answer 1

Let $E: y^2=f(x)$ and $\psi_3(x) = 2f(x)f''(x)-(f'(x))^2$, where $f(x)$ is a monic cubic polynomial with three distinct roots (because $E$ is non-singular!), as above. This can be shown using the following hint:

• A quartic polynomial $p(x)=a_4x^4+\cdots+a_0$, with $a_4>0$, has exactly two real roots if $p(x)$ takes negative values at all the zeros of the derivative $p'(x)$.

Thus, you just need to show that $\psi_3(x)$ is negative at the zeros of $\psi_3'(x)$, using the fact that $\psi_3'(x)$ has a simple expression in terms of $f(x)$. You also need to use the fact that $f(\alpha)=f'(\alpha)=0$ is not possible under the above hypotheses.

Answer 2

I think I got a solution so I will share it so you can give me your opinion:

We know that this quartic polynomial will have at least two real roots , suppose we have 4 roots $\{ \alpha_1,\beta, \gamma,\alpha_2 \}$ conveniently orderded $\alpha_1< \beta <\gamma< \alpha_2$.

Note that $f''(x)=6x+2a$ and $\psi_3'=12f(x)$

Since $\psi_3(\alpha_1)=0 \implies f(x)<0$ ( $\alpha_1< \frac{-a}{3}$).

Now to reach again $0$, $\psi_3$ must increase its value so $f(\beta)>0$ and this implies $f''(\beta)>0$ $\iff$ $\beta > \frac{-a}{3}$.

The same argument show us $f(\gamma)<0 \iff f''(\gamma)<0 \iff \gamma <\frac{-a}{3}$ and $\gamma < \beta$ hence contradiction.