Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$ 
Find $x,y,z \in \mathbb Q$ such that: $$x + \frac 1y, y + \frac 1z, z+
 \frac 1x \in \mathbb Z$$

Here is my thinking:
$$x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z\\ \implies \left ( x + \frac 1y\right ) \left ( y + \frac 1z\right )\left ( z + \frac 1x\right )\in \mathbb Z \\ \iff \frac 1 {xyz} + xyz + \left( x+\frac 1y \right) + \left( y+\frac 1z \right) + \left( z+\frac 1x \right) \in \mathbb Z \\ \iff xyz + \frac 1 {xyz} \in \mathbb Z \\ \iff |xyz| = 1 \text{ (I proved it, easily)}$$
Case 1:
$xyz=1 \implies \exists a,b,c \in \mathbb Z$ such that: $x = \frac ab,y=\frac bc,z=\frac ca$
$$\implies \frac {a+c} b = x+\frac 1y \in \mathbb Z$$
and so on.
Now, I am stuck. Do you thing I am going a right way ? 

It is most interesting to consider $x>0,y>0,z>0$, because there is a lots of solutions in other case.
A.
If $(x,y,z)$ is a solution, then one can construct other solutions:
$$\begin{array}{ccc}
(x,y,z) & (y,z,x) & (z,x,y) \\
(-x,-y,-z) & (-y,-z-,x) & (-z,-x,-y) \\
(\frac{1}{z},\frac{1}{y},\frac{1}{x}) & (\frac{1}{x},\frac{1}{z},\frac{1}{y}) & (\frac{1}{y},\frac{1}{x},\frac{1}{z}) \\
(-\frac{1}{z},-\frac{1}{y},-\frac{1}{x}) & (-\frac{1}{x},-\frac{1}{z},-\frac{1}{y}) & (-\frac{1}{y},-\frac{1}{x},-\frac{1}{z}). \\
\end{array}\tag{1}
$$
B.
For each pair $a,b\in \mathbb{N}$, such that $GCD(a,b)=1$, triple 
$$
\Bigl(x = \frac{a}{b}, \quad y = -\frac{b}{a+b}, \quad z = -\frac{a+b}{a} \Bigr)
\tag{2}
$$
is a solution: $x+\frac{1}{y} = -1$, $y+\frac{1}{z} = -1$, $z+\frac{1}{x} = -1$. 
Other derived triples (see A.) are solutions too.
C.
For each $n \in \mathbb{N}$ triple 
$$
\Bigl(x = n, \quad y=-1, \quad z = -\frac{1}{n} \Bigr)
\tag{3}
$$
is a solution: $x+\frac{1}{y} = n-1$, $y+\frac{1}{z} = -n-1$, $z+\frac{1}{x} = 0$. 
Other derived triples (see A.) are solutions too.
D. There are other (most interesting for me) solutions: 
$\diamond \qquad (1,1,1)\quad$ (and derived modification $(-1,-1,-1)$);  
$\diamond \qquad (1, \frac{1}{2}, 2)\quad$ (and modifications); 
$\diamond \qquad (\frac{1}{2}, \frac{2}{3}, 3) \quad$ (and modfications).

They have $x,y,z$ all positive (or all negative).
Maybe this list has continuation (?).
