<s>[INST] Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$ I need to determine all positive integer solutions for the equation: 
$$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$
This is how I have tried to do it: 
Mulitiplied both sides by $xyz$ to get
$$yz+xz+xy+z+x+y+1=xyz.$$
Factor it:
\begin{align}
x(y+z+1-yz)+yz+y+z &= -1   \\
x(y(1-z)+z+1)+y(1+z)+z &= -1
\end{align}
If $z=0$, we get
\begin{align}
x(y+1)+y=-1 &\iff xy+x+y=-1   \\
&\iff (x+1)(y+1)=0,
\end{align}
which gives us $x=y=-1$.
Is this all positive integer solutions? Or have I missed something?
EDIT: I am stupid. 
New attempt. 
If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.
Then there is no solutions of positive integers for both $x$ and $y$ at the same time.  
If I try for $z=2$, I get
\begin{align}
x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1   \\
&\iff 3x+3y+3-xy=0
\end{align}
and I won't get a solution where all the variables are positive integers. 
 [/INST] Hint: We can write the equation as
$$
\left(1+\frac1 x\right)\left(1+\frac1 y\right)\left(1+\frac1 z\right)=2.
$$ Assume without loss of generality $x\le y\le z$ by rearranging. Observe that
$
1+\frac 1 x<2\le \left(1+\frac1 x\right)^3 \implies x\in \{2,3\}
$ and using this information, investigate each case $x=2,3$ and so on.


Addendum, Solution.


*

*For $x=2$, we have that $\frac{y+1}{y}\cdot\frac{z+1}{z}=\frac 43$. Then $y>3$ and only $(y,z)=(4,15), (5,9), (6,7)$ work. (Check this using $\frac{z+1}z =\frac{4}{3}\frac{y}{y+1}$.)

*For $x=3$, we have that $\frac{y+1}{y}\cdot\frac{z+1}{z}=\frac 32$. Using $y\ge x=3$ and $\frac{z+1}z =\frac{3}{2}\frac{y}{y+1}$, check that only $(y,z)=(3,8), (4,5)$ work. 
By rearranging, every solution is a permutation of such solutions with $x\le y\le z$.
</s>