# Show that the curve $y^2 = x^3 + 2x^2$ has a double point, and find all rational points

Show that the curve $y^2 = x^3 + 2x^2$ has a double point. Find all rational points on this curve.

By implicit differentiation of $x$, $-3x^2 - 4x$ vanishes iff $x = -4/3$ and $0$. By implicit differentiation of $y$, $2y$ vanishes iff $y = 0$.

Taking the second derivative, I got $-6x-4$ and then using the point on the curve $(0,0)$ I got $-4$. Is this my double point?

Thanks for any help!

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I apologize I am new to this. I have been trying to figure out how to get started on this problem for a while, curves and geometry have always been confusing to me. The book I am using uses more words to teach than examples and I am a visual learner. Any help on how I would go about finding double and rational points on a curve would be greatly appreciated. – Alyssa Jul 19 '12 at 5:54
For the first question, you can perform implicit differentiation on your Cartesian equation; you should find that there is a value of $x$ and $y$ that would make your derivative indeterminate. That is your double point. – J. M. Jul 19 '12 at 5:57

Since you are a visual person, you should try drawing it. It seems fairly obvious from the real picture that there is a double point at zero, and then you just use whatever methods your book usually employs to prove that it actually is a double point.

On the other hand, finding all rational solutions means finding all rational points such that

$$y = \pm\sqrt{x^3+2x^2} = \pm~ x\sqrt{x+2}.$$

I will omit the sign for the rest of this post. Note that $x+2$ must have a rational root. In other words, $x+2=\frac{p^2}{q^2}$ for integers $p$ and $q$, so $x=\frac{p^2-2q^2}{q^2}$. This yields

$$y = \frac{p^2-2q^2}{q^2} \cdot \frac{p}{q}$$

and multiplying by $q^3$, we obtain that

$$q^3y = p^3 - 2pq^2$$

over the integers. Hence, any rational solution of your equation is parametrized by $p,q\in\mathbb{Z}$ and then,

\begin{align*} y &= \frac{p^3-2pq^2}{q^3} \\ x &= \frac{p^2-2q^2}{q^2}. \end{align*}

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As you have already found by using the partial derivatives, the double point of the curve $\mathcal{C}: y^2 = x^3 + 2x^2$ is at the origin $(0, 0)$. Now let me try to explain you a way to make use of this double point to find all the rational points in the curve. Basically the process will give you a rational parametrization of the coordinates of any point on the curve.

The idea is to "project" from the double point onto some rational line. The picture below shows the curve $\mathcal{C}$ in red, in which first of all, you can see the double point at the origin where the curve actually intersects itself.

Also, I've plotted the rational line $x = 1$, which is the vertical line shown in blue. And to explain the idea of the projection I've plotted the general line shown in green in the picture.

The basic idea of the projection from the double point is the following. Since a line in the plane has an equation of the form $y = mx + b$ and the polynomial defining your curve is the cubic polynomial $f(x, y) := y^2 - x^3 - 2x^2$, then when you look at the intersection points of the line $y = mx + b$ and the curve, you will have to substitute this $y$ in $f(x, y)$ to get the polynomial $f(x, mx + b) = (mx + b)^2 - x^3 - 2x^2$, which is a cubic in $x$. So in general it has three roots, which will correspond to the $x$ coordinates of the intersection points of the line with the curve $\mathcal{C}$.

But now comes into play the importance of the double point at the origin. When you have a double point, that means that this point is counted twice in the intersection, that means geometrically that there is only one different point in the intersection of the line and the curve, as you can see in the picture.

Then the idea of the projection is to match up this point on the curve, with the corresponding point in the line, which in this case I choose to be the vertical line shown in blue, but you could pick any line you like with rational coefficients. Then from this you can find a rational parametrization of the curve, which in turn gives you all the rational points on the curve.

To be more explicit, any point on the blue line has the form $(1, t)$ for $t \in \mathbb{R}$. Then the equation of the line passing through the double point at the origin, and the point $(1, t)$ is $y = tx$. To find the intersection point different from the origin, we solve the system

$$\begin{cases} y^2 = x^3 + 2x^2\\ y = tx \end{cases} \implies t^2 x^2 = x^3 + 2x^2 \implies x^3 + 2x^2 - t^2 x^2 = 0$$

$$\implies x^2 (x + 2 - t^2 ) = 0 \implies x = 0 \ \ \text{or} \ \ x = t^2 - 2$$

The solution $x = 0$, which has multiplicity 2, corresponds to the double point at the origin. The other solution $x = t^2 - 2$ corresponds to the other intersection point on the curve. Then since $y = tx$, this implies that $y = t(t^2 - 2)$.

And this is basically it, we have the correspondence between the points on the curve, and the points on the line $x = 1$ given by

$$\begin{array} \{ \text{points on the curve y^2 = x^3 + 2x^2} \} &\longleftrightarrow \{ \text{points on the line x = 1} \}\\ \quad \quad \quad \quad \quad \quad (t^2 - 2, t(t^2 - 2))& \longleftrightarrow \quad (1, t) \end{array}$$

So all the rational points on the curve have the form $(t^2 - 2, t(t^2 - 2))$ for $t \in \mathbb{Q}$.

You should read at least the first chapter of Silverman and Tate's book Rational Points on Elliptic Curves, where they explain this idea beautifully.

I hope that this helps you a little.

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Having parametric equations for your curve is a wonderful thing... :) – J. M. Jul 19 '12 at 7:29
@J.M. It sure is, math is a wonderful thing... ;-) – Adrián Barquero Jul 19 '12 at 7:33

You may find the following fact useful: suppose that a plane curve is described by a polynomial equation $F(x,y)=0$. Write $$F(x,y)=\sum_{d\geq0}F_d(x,y)\qquad\qquad(*)$$ where in $F_d(x,y)$ you collect all the monomials in $F$ of degree exactly $d$. Then:

• $(0,0)$ belongs to the curve if and only if $F_0\equiv0$ (by which, I mean that is $0$ identically).

• $(0,0)$ is a singular point of the curve if and only if $F_0\equiv0$ and $F_1\equiv0$.

Moreover if $d_0\geq2$ is the smallest degree appearing in $(*)$, then $F_d=0$ is the tangent cone at $(0,0)$. Note that this answers immediately and trivially to your first question.

Finally, the method applies to just any point $(x_0,y_0)$ after a simple change of variables $X=x-x_0$, $Y=y-y_0$.

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