Find all possible values of $x+y+z$. Let $x, y, z$ be non-zero integers such that ${x\over y}+{y\over z}+{z\over x}$ is an integer. Find all possible values of $x+y+z$. Please provide a proof with all solutions. 

Note: this assumes that the question is really about non-zero integers, and not about positive integers. I think the answer is more difficult when having to deal with positive integers.
Now, we can get any non-zero integer $t$ by choosing $(x,y,z)=(t,t,-t)$. 
This gives $\frac{t}{t} + \frac{t}{-t} + \frac{-t}{t} = -1$ and $x+y+z=t$. 

Now suppose $x+y+z=0$, so $z=-x-y$. 
This gives $\frac{x}{y} - \frac{y}{x+y} - \frac{x+y}{x} \in \mathbb Z$, hence $\frac{x}{y} - \frac{y}{x+y} - \frac{y}{x} \in \mathbb Z$.
Hence $\dfrac{x^2(x+y)-y^2(x+y)-y^2x}{xy(x+y)}  \in \mathbb Z$.
Note that numerator and denominator are both homogenous of degree 3, so we can divide out any common prime factor. So we can assume that $\gcd(x,y)=1$. Suppose there is a prime $p \mid x+y$. We get $p \mid x+y \mid y^2x$. So $p \mid x$ or $p \mid y$, and $p \mid x+y$ gives that $p \mid x$ and $p \mid y$. Contradiction with $\gcd(x,y)=1$. Therefore, we have $x+y=1$ or $x+y=-1$. 
The first case gives $\dfrac{x^2-y^2-y^2x}{xy}  \in \mathbb Z$, hence $\dfrac{x^2-y^2}{xy}  \in \mathbb Z$. 
The second case gives $\dfrac{-x^2+y^2-y^2x}{xy}  \in \mathbb Z$, hence $\dfrac{-x^2+y^2}{xy}  \in \mathbb Z$, so $\dfrac{x^2-y^2}{xy}  \in \mathbb Z$. 
Now, if $p \mid x$, then $p \mid y$, contradiction with $\gcd(x,y)=1$. So $x=\pm1$, and analogous $y=\pm1$. But then $x+y$ can't be $1$ or $-1$. So there are no solutions with $x+y+z=0$. 
