This question comes from a previous question of mine.
I am trying to prove that the equation $$xy(x+y+1)=1$$ has no rational solutions. (Such a solution would give three numbers, namely $-x, -y, 1-x-y$, whose sum and product are equal to $1$).
This can be thought as a homogeneous equation over the integers $$XY(X+Y+Z)=Z^3$$ or equivalently $$X^2Y+XY^2+XYZ-Z^3=0$$ where $x= X/Z$ and $y=Y/Z$.
Clearly there are some trivial solutions for $Z=0$, but those are not interesting for my purposes.
So the question is
Does $$X^2Y+XY^2+XYZ-Z^3=0$$ have non-trivial integer solutions ($Z \neq 0$)?
Depends on how elementary you wish the proof to be.
Using elliptic curves it is straightforward. $xy(x+y+1)=1$ give the quadratic in $x$, $y x^2+(y^2+y)x-1=0$. For rational $x$, the discriminant must be a rational square, so there must exist rational $D,y$ such that \begin{equation*} D^2=y^4+2y^3+y^2+4y \end{equation*}
This can be transformed to the elliptic curve \begin{equation*} v^2=u^3+(u+4)^2 \end{equation*} with \begin{equation*} y=\frac{v-u-4}{2u} \end{equation*}
The elliptic curve has the points of order $3$ at $(0,\pm 4)$, and no others. It has rank $0$ according to Simon's ellrank code.
There are thus NO rational points on the curve which give a point on the quadratic.
I cannot believe I am the first person to do this !!