if three integer such diophantine equation How find $x+y+z$

Question

following Diophantine equation $$xy^2+yz^2+zx^2=x^2y+y^2z+z^2x+x+y+z$$ ie:$(x-y)(y-z)(z-x)=x+y+z$

where $x,y,z$ are integers.

can find $x+y+z$

I tried some values and got some near equalities when $x,y,z$ at least two are equal, $x+y+z=0$,for other case, who have an idea to discuss...plz. Thanks in advance.

Answer 1

Suppose $x\le y\le z$ and put $y=x+a=z-b$ where $a,b\ge 0$

Then you will find $$-a\cdot b\cdot-(a+b)=ab(a+b)=3y+a-b$$ so that $$3y= ab(a+b)+b-a$$

Then you can search for values of $a,b$ which give an integer multiple of $3$ on the right-hand side. So if $a=b=3$ you get $3y=54$ so that $y=18$.

Note that if $a$ and $b$ are not both multiples of $3$ then one of $ab(a+b)$ and $a-b$ is a multiple of $3$ and the other is not.

So put $a=3c, b=3d$ and obtain the general solution $$y=9cd(c+d)+d-c, x=y-3c, z=y+3d$$

Answer 2

We just need to figure this fact: given $x,y,z\in\mathbb{Z}$, if we replace them with $x+n,y+n,z+n$ the function $$ P(x,y,z)=(x-y)(y-z)(z-x) $$ stays unchanged, while the sum $x+y+z$ is increased by $3n$. So, in order to have a solution, we just need that $x,y,z$ belong to the right residue classes $\!\!\pmod{3}$.