Diophantine equation and triplets

Find five (5) triplets $(x, y, z)\in \mathbb{N}^3$ such that $\gcd(x, y, z)=1$ and
$$x^3+y^3 = 3xyz$$

Solution



Let  $\left(x,y,z\right)\in\mathbb{N}\times\mathbb{N}\times\mathbb{N}$ be a triplet with $\gcd(x,y,z)=1$ and $x^3+y^3=3xyz$ . Then,

$$\left(\frac{x}{z}\right)^3+\left(\frac{y}{z}\right)^3=\frac{3xy}{z^2}$$

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

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

If  $u^3+v^3-3\,u\,v=0$  and $v=t\,u$, then :

$$u^3+t^3\,u^3-3\,t\,u^2=0 \Leftrightarrow u^2\,\left(\left(1+t^3\right)\,u-3\,t\right)=0$$

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

The rational functions $u(t)=\dfrac{3t}{1+t^3}, \; v(t)=tu(t)=\dfrac{3t^2}{1+t^3}, \;t\neq -1$ satisfy the equation
$$u^3(t)+v^3(t)-3\,u(t)\,v(t)=0$$
and we observe that $t\in\mathbb{Q}-\left\{-1\right\}\Rightarrow u(t), \; v(t)\in\mathbb{Q}$ . Therefore,

${\color{gray} \blacksquare} \;\; u(1)=\frac{3}{2}\,,v(1)=\frac{3}{2}\,,\rm{gcd}(3,3,2)=1$
${\color{gray} \blacksquare} \;\; u(2)=\frac{6}{9}=\frac{2}{3}\,,v(2)=\frac{12}{9}=\dfrac{4}{3}\,,\rm{gcd}(2,4,3)=1$
${\color{gray} \blacksquare} \;\; u(4)=\frac{12}{65}\,,v(4)=\frac{48}{65}\,\,,\rm{gcd}(12,48,65)=1$
${\color{gray} \blacksquare} \;\; u\,\left(\frac{1}{3}\right)=\frac{27}{28}\,,v\,\left(\frac{1}{3}\right)=\frac{9}{28}\,\,,\rm{gcd}(27,9,28)=1$

and thus, five triplets $\left(x,y,z\right)\in\mathbb{N}\times\mathbb{N}\times\mathbb{N}$ with $\rm{gcd}(x,y,z)=1$ and $x^3+y^3=3\,x\,y\,z$ are the following:

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