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?
 A: Add $xyz+1$ to both sides to obtain $$2xyz+1=xyz+xy+yz+zx+x+y+z+1=(x+1)(y+1)(z+1)$$
The left-hand side is odd, so the right-hand side must also be odd, whence $x,y,z$ are all even.
Now divide by $xyz$ to obtain $$\left(1+\frac 1x\right)\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{xyz}\gt 2$$
Now note that $\left(\frac 54\right)^3=\frac {125}{64}\lt 2$, so that one of $x,y,z$ at least must be equal to $2$ - say $x$.
We then have $$\frac 32\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{2yz}$$ which gives $$\left(1+\frac 1y\right)\left(1+\frac 1z\right)\gt \frac 43$$ from which it follows that one of $x$ and $y$ must be less than $8$. Note that it can't be $6$, because that would not work in the top equation of this answer taken mod $3$. Once $y$ is fixed, there is at most one possibility for $z$ (the equation is linear in $z$) and any answer must be a permutation of one with $x\le y\le z$.
Using this method, you should be able to find the solutions.
A: If $xyz = xy + xz + yz + x + y + z$  is a natural solution.
$(xy - x - y - 1)z = xy + x + y > 0$
$z = \frac {xy + x + y}{xy - x - y - 1}$
$= 1 + \frac {2x + 2y + 1}{xy -  x-y -1}\in \mathbb N$ so 
$xy - x - y -1\le 2x + 2y + 1$
$xy \le 3x + 3y + 2$
Likewise $xz \le 3x + 3z + 2$
And $yz \le 3y + 3z + 2$.
If we relabel $c = \max(x,y,z); a = \min(x,y,z)$ and $b$ is the median of $x,y,z$.
then
$ab \le 3a + 3b + 2 \le 6b + 2$ so $a \le 6 + \frac 2b$ has only finitely many possible values.
And $bc \le 3b + 3c +2 \le 6c + 2$ so $b \le 6 + \frac 2c$ has only finitely many possible values and
$c =\frac {ab + a + b}{ab - a - b - 1}$ has only finitely many possible values.
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We can work our way through these: $a,b < 6$ and
If $a = 1$ then $c = \frac {2b + 1}{-2}$ which is not possible.
$a = 2$  then $c = \frac {3b + 2}{b-3}$ if $b = 4$ then $c = 14$
If $b = 5$ we have $\frac {odd}{even}$ which is not an integer .
If $a=3$ then $c = {4b +3}{2b -4}$ which is an impossible $\frac {odd}{even}$
If $a = 4$ then $c = {5b + 4}{3b -5}$.  If $b = 4;c={24}{7}$. No good. $b=5$ yields $\frac {odd}{even}$.
If $a =5; b=5$ then $c = \frac {29}{10}$. No go.
So, unless I made an error $\{x,y,z\} = \{2,4,14\}$ are the only solutions.
A: 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 $z$.
