Exercise 5.7.11 (Rational points on the elliptic curve $y^2 = x^3 + 4x$)

Proof. Let $(x,y)\in {\mathbb{Q}}^2$ be a solution of $y^2=x^3+4x$. If $x=0$, then $y=0$. We suppose now that $x\ne 0$. Then $m=y∕x\in \mathbb{Q}$, and $m^2{x}^2=x^3+4x.$ Since $x\ne 0$,

By completing the square, we obtain ${\left(x-\frac{m^2}{2}\right)}^2=\frac{m^4}{4}-4$, and by multiplying by $4$, this gives

Put $n=2x-m^2$. Then $n\in \mathbb{Q}$, and the discriminant

is the square of a rational number.

Since $(m,n)\in {\mathbb{Q}}^2$, there are integers $X,Y,Z$ such that $Z\ne 0$ and $m=X∕Z,n=Y∕Z$. Then

If we put $a=2Z,b=YZ,c=X$, we obtain an integral solution of

This is the second equation of Problem 5.4.14. We proved in the solution of this problem that this Diophantine equation has no positive solution.

It remains to examine the solutions where $a,b$ or $c$ is zero.

Here $a=2Z\ne 0$, therefore $c^4>0$, so $c\ne 0$. If $b=0$, then $a^4=c^4$, so $a=\pm c$. This gives $Y=0$ and $X=\pm 2Z$. Thus $n=0$ and $m=\pm 2$, so $x=2$ and $y=\pm 4$.

This shows that the only rational points of the elliptic curve associate to $f(x,y)=y^2-x^3-4x$ are $(0,0),(2,4),(2,-4)$.

So the group $E_f(\mathbb{Q})$ has four elements, the points $T=(0,0),A=(2,4),-A=(2,-4)$ and the point at infinity $O=(0:1:0)$.

The tangent to the curve at point $A=(2,4)$ is given by the equation

so the equation of the tangent is $y=2x$. This gives $A+A=T$. Since $2T=O$ and $T\ne O$, we obtain that the order of $A$ is $4$. Therefore

□