Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$. Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of case work?
Let $x$, $y$ and $z$ be natural numbers satisfying
$$x+2y+2z=xyz.$$
If $xyz=0$ then $x=y=z=0$, so suppose $x,y,z&gt;0$. A bit of algebra shows that
$$(xy-2)(xz-2)=x^2+4.\tag{1}$$
Of course without loss of generality $y\leq z$. If $x=1$ then the above simplifies to
$$(y-2)(z-2)=5,$$
and so $y=3$ and $z=7$. Otherwise $x\geq2$ and then from $y\leq z$ it follows that
$$(x+1)^2&gt;x^2+4=(xy-2)(xz-2)\geq(xy-2)^2,$$
and hence that $x\geq xy-2$, or equivalently $2\geq x(y-1)$. This means that either $x=y=2$, in which case $z=3$, or $y=1$, in which case $(1)$ simplifies to
$$(x-2)(xz-2)=x^2+4.\tag{2}$$
Clearly $x\neq2$, and so from the identity
$$xz-2=\frac{x^2+4}{x-2}=x+2+\frac{8}{x-2},$$
we see that $x-2$ divides $8$, so $x-2\in\{1,2,4,8\}$. Plugging these values into $(2)$ yields
\begin{eqnarray*}
1\cdot(3z-2)&amp;=&amp;13,\\
2\cdot(4z-2)&amp;=&amp;20,\\
4\cdot(6z-2)&amp;=&amp;40,\\
8\cdot(10z-2)&amp;=&amp;104.
\end{eqnarray*}
The first three have the solutions $z=5$, $z=3$ and $z=2$, respectively, and the last has no integral solution.
In summary, up to permutation of $y$ and $z$, the solutions $(x,y,z)$ are
$$(0,0,0),\quad(1,3,7),\quad(2,2,3),\quad(3,1,5),\quad(4,1,3),\quad(6,1,2).$$
&quot;OP&quot; equation is:
$x+2y+2z=xyz$  ---(1)
Eqn. (1) has parameteric solution's shown below.
$(x,y,z)=(3k^2+1)/(k),(3k),(1/k)$
For, $k=1$, we get: $(x,y,z)=(4,3,1)$
&amp; for, $k=1/3$, we get: $(x,y,z)=(4,1,3)$
Another parametrization is:
$(x,y,z)=(4k^2+2)/(k),(2k),(1/k)$
For, $k=1$, we get: $(x,y,z)=(6,2,1)$
&amp; for, $k=1/2$, we get: $(x,y,z)=(6,1,2)$