Exercise 5.6.5 (Rational points on the curve $y^2 = x^3 + 2x^2$)

Question

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

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
Put $f(x,y) = y^2 - x^3 - 2x^2$. Then
$$\frac{\partial f}{\partial x}(x,y) = 3x^2 + 4x,\qquad \frac{\partial f}{\partial x}(x,y)  = 2y,$$
thus $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x}$ vanish simultaneously if and only if $(x,y) = (0,0)$. The curve has a double point at $(0,0)$.

The parametrization $y = xt$ starting from this double point gives 
\begin{align*}
(x,y) \in \mathscr{C}_f(\Q) & \iff (x,y) = (0,0) 	ext{ or } ( x 
e 0 	ext{ and }  \exists t \in \Q, \ y = tx 	ext{ and } t^2 x^2 - x^3 - 2 x^2 = 0)\
&\iff (x,y) = (0,0) 	ext{ or }  \exists t \in \Q, \ 
\left\{
\begin{array}{l}
x = t^2 - 2,\
y = t^3 - 2 t.
\end{array}
ight.
\end{align*}
The rational points on the curve $y^2 = x^3 + 2x^2$ are $(0,0)$ with the points $(t^2 - 2, t^3 - 2t)$, where $t$ takes any rational value.
\end{proof}

Exercise 5.6.5 (Rational points on the curve $y^2 = x^3 + 2x^2$)

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Exercise 5.6.5 (Rational points on the curve $y^2 = x^3 + 2x^2$)

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