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. 
 A: 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 $x\le y\le z$.
A: Assume, without loss of generality, that $x\leq y\leq z$. We see from the original equation that $x>1$ (since $x = 1$ means $\frac1x + \cdots > 1$). At the same time, we must have $x<4$, as otherwise the sum is clearly less than $1$.
So, where does $x = 2$ actually take us? We insert and get
$$
\frac12 + \frac1y + \frac1z + \frac1{2y} + \frac1{2z} + \frac1{yz} + \frac1{2yz} = 1\\
\frac3{2y} + \frac3{2z} + \frac3{2yz} = \frac12\\
3z + 3y + 3 = yz\\
12 = yz - 3y - 3z + 9 = (y-3)(z-3)
$$
Since $y$ and $z$ are integers, this is an easy solve.
What about $x = 3$? We get
$$
\frac13 + \frac1y + \frac1z + \frac1{3y} + \frac1{3z} + \frac1{yz} + \frac1{3yz} = 1\\
\frac4{3y} + \frac4{3z} + \frac4{3yz} = \frac23\\
4z + 4y + 4 = 2yz\\
2z + 2y + 2 = yz\\
6 = yz - 2y-2z + 4 = (y-2)(z-2)
$$
which, again, is an easy solve using the fact that $y, z$ are integers.
A: This thing factors nicely after adding 1 to both sides. Not that it helps us much, though.
$$\left(1+{1\over x}\right)\cdot\left(1+{1\over y}\right)\cdot\left(1+{1\over z}\right)=2$$
Basically, it all boils down to brute force search which can easily be done by hand.
Let's assume WLOG $x\leqslant y\leqslant z$. Then just run through the possible values:


*

*$x=1$: LHS is too big regardless of $y$ and $z$.

*$x=2$:


*

*$y=2$: LHS is too big regardless of $z$.

*$y=3$: same as above.

*$y=4$: need to check.

*$y=5$: need to check.

*$y=6$: need to check.

*$y\geqslant7$: LHS is too small regardless of $z$.


*$x=3$:


*

*$y=3$: need to check.

*$y=4$: need to check.

*$y\geqslant5$: LHS is too small regardless of $z$.


*$x\geqslant4$: LHS is too small regardless of $y$ and $z$.


After checking each of the cases marked "need to check", you'll end up with quite a bunch of nice positive solutions.
A: Rewrite the equation as
$$
z( - xy + x + y + 1) + (x+1)(y+1)=0.
$$
Suppose first that $xy=x+y+1$, i.e., $y(1-x)=x+1$. Then $x\neq 1$, so that $y=\frac{x+1}{1-x}$. Substituting this into the original equation gives
$$
x(x+1)=0,
$$
a contradiction. Now assume that $- xy + x + y + 1\neq 0$. Then
$$
z=\frac{(x+1)(y+1)}{xy-x-y-1}.
$$
When can this be a positive integer? For $x=1$ we have $z=-y-1<0$, a contradiction. For $x=2$ we have 
$$
z=\frac{3y+3}{y-3},
$$
which has solutions for $y=4,5, 6,7,9,15$. For $x=3$ we obtain
$$
z=\frac{2y+2}{y-2}.
$$
This is an integer for $y=3,4,5,8.$ For $x\ge 4$ it is easy to see that we have no solution in positive integers.
A: Let $x\geq y\geq z$.
Thus, $$1\leq\frac{1}{z^3}+\frac{3}{z^2}+\frac{3}{z}$$ or
$$z^3\leq1+3z+3z^2$$ and we got some values of $z$: 
$$1\leq z\leq3.$$
Now, for all value of $z$ we can make the similar thing with $y$.
