Rational curve

[Papapetros Vaggelis]

Consider the algebraic curve \(\displaystyle{V(u^3+v^3-3\,u\,v)\subseteq \mathbb{R}^2}\) .

Prove that this curve is a rational curve and use this result in order to find five triplets \(\displaystyle{\left(x,y,z\right)\in\mathbb{N}^3}\)

such that \(\displaystyle{{\gcd}(x,y,z)=1}\) and \(\displaystyle{x^3+y^3=3\,x\,y\,z}\) .

[Papapetros Vaggelis]

Let \(\displaystyle{\left(x,y,z\right)\in\mathbb{N}\times\mathbb{N}\times\mathbb{N}}\) be a triplet with \(\displaystyle{\rm{gcd}(x,y,z)=1}\)

and \(\displaystyle{x^3+y^3=3\,x\,y\,z}\) . Then,

\(\displaystyle{\left(\dfrac{x}{z}\right)^3+\left(\dfrac{y}{z}\right)^3=3\,\dfrac{x}{z}\,\dfrac{y}{z}}\), which means that the rational

point \(\displaystyle{\left(\dfrac{x}{z},\dfrac{y}{z}\right)}\) is a point of the plane algebraic curve \(\displaystyle{V\left(u^3+v^3-3\,u\,v\right)\subseteq \mathbb{R}^2}\) .

This curve is a rational curve. Indeed, consider the lines \(\displaystyle{\left(\varepsilon_{t}\right): v=t\,u\,,t\in\mathbb{R}}\) .

If \(\displaystyle{u^3+v^3-3\,u\,v=0}\) and \(\displaystyle{v=t\,u}\), then :

\(\displaystyle{u^3+t^3\,u^3-3\,t\,u^2=0\iff u^2\,\left(\left(1+t^3\right)\,u-3\,t\right)=0}\), so :

\(\displaystyle{u=0}\) (double) and \(\displaystyle{u=\dfrac{3\,t}{1+t^3}}\) for \(\displaystyle{t\neq -1}\) .

The rational functions \(\displaystyle{u(t)=\dfrac{3\,t}{1+t^3}\,,v(t)=t\,u(t)=\dfrac{3\,t^2}{1+t^3}\,,t\neq -1}\) satisfy the equation

\(\displaystyle{u^3(t)+v^3(t)-3\,u(t)\,v(t)=0}\) and we observe that

\(\displaystyle{t\in\mathbb{Q}-\left\{-1\right\}\implies u(t)\,,v(t)\in\mathbb{Q}}\) . Therefore,

\(\displaystyle{u(1)=\dfrac{3}{2}\,,v(1)=\dfrac{3}{2}\,,\rm{gcd}(3,3,2)=1}\)

\(\displaystyle{u(2)=\dfrac{6}{9}=\dfrac{2}{3}\,,v(2)=\dfrac{12}{9}=\dfrac{4}{3}\,,\rm{gcd}(2,4,3)=1}\)

\(\displaystyle{u(4)=\dfrac{12}{65}\,,v(4)=\dfrac{48}{65}\,\,,\rm{gcd}(12,48,65)=1}\)

\(\displaystyle{u\,\left(\dfrac{1}{3}\right)=\dfrac{27}{28}\,,v\,\left(\dfrac{1}{3}\right)=\dfrac{9}{28}\,\,,\rm{gcd}(27,9,28)=1}\)

and thus, five triplets \(\displaystyle{\left(x,y,z\right)\in\mathbb{N}\times\mathbb{N}\times\mathbb{N}}\) with \(\displaystyle{\rm{gcd}(x,y,z)=1}\)

and \(\displaystyle{x^3+y^3=3\,x\,y\,z}\) are the following:

\(\displaystyle{\left(x,y,z\right)\in\left\{\left(3,3,2\right)\,,\left(2,4,3\right)\,,\left(4,2,3\right)\,,\left(12,48,65\right)\,,\left(27,9,28\right)\right\}}\) .