<s>[INST] Prove that the equation $ xyz=xy+xz+yz+x+y+z$ has finite numbers of natural solutions. Prove that the equation 
$$\begin{equation}
xyz=xy+xz+yz+x+y+z\tag{*}
\end{equation}$$
has finitely many natural solutions.
My attempt.  Suppose that $x \leq y \leq z$  then 
$$
xyz\leq xz+xz+yz+3z \implies xy \leq 2x+y+3 \implies x \leq \frac{y+3}{y-2}
$$ 
Since $$x \leq \frac{y+3}{y-2}=\frac{y-2+5}{y-2}=1+\frac{5}{y-2} \implies (x-1)(y-2) \leq 5,$$
we can conclude that the last inequality has finite number of natural solutions, so there are only finite numbers for $x,y.$
For any fixed $x,y$ the equation (*) reduces to linear equation for $z$ with finite numbers of solutions.
Is it ok?
Is there any other solutions?
 [/INST] Suppose 
$$xyz = xy + xz + yz + x + y + z   $$ 
is a natural solution with $$x \leq y \leq z$$  then 
$z = \frac {xy + x + y}{xy - x - y - 1}= 1 + \frac {2x + 2y + 1}{xy -  x-y -1}$
So it must be $$xy - x - y -1\le 2x + 2y + 1$$ or 
$$xy \le 3x + 3y + 2$$ which actually means at least $x$ is smaller than 8.
Likewise we can conclude that $y$ is smaller than 8. Now the $$z = \frac {xy + x + y}{xy - x - y - 1}= 1 + \frac {2x + 2y + 1}{xy -  x-y -1}$$ gives us finitely many possibilities for $z$.
</s>