Finding integer pairs having same functional values

Question

We have a curve $$z=cxy-x^2y-xy^2,$$ where $c$ is some positive integer constant. How to find the set of integer pairs $(x,y)$ such that they give the same nonzero integral value of $z$.

Answer 1

A partial answer.

Since $cxy-x^2y-xy^2 = x y (c - x - y)$, every pair $(x,y)$ with $x+y=c$ gives $z=0$.

Answer 2

Set $u:=x, v:=y, w:=(c-x-y)$. This means you want to find 3 integers whose sum is $c$ and whose produdct is $z$:

$$u+v+w=c$$

$$uvw = z$$

Since this is totally symmetric in $u,v,w$, you can get additional solutions by exchanging $u,v$ and $w$: $$u:=x, v:=c-x-y$$ and $$u:=c-x-y, v:=y.$$

This is another way to get what Nicolas FRANCOIS wrote above.

It may or may not be easier to tackle the question if there are more solutions to this equaion than simple permutation (it may even be the point from which you are coming). If we change the view to complex numbers, each $u$ that is chosen will result in a quadratic equation for $v,w$ with a single solution (up to exchanging $v$ and $w$). I don't see why, for some $c$ and $z$, chosing another integer as $u$ might not result in integer $v,w$.

Added: An example would be $z=36, c=13$, which allows the essentially different solution (1,6,6) and (2,2,9). This also means that the "only if" part of Nicolas FRANCOIS' comment is not correct.

Further addition: I can't find a better method at the moment to find all solutions as to have $x=u$ run through all the divsiors of $z$, and solve the quadratic equation in each case, checking if an integer $v,w$ is produced.