<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] 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.
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