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. 
 A: If $x = \frac{z}{4}$, and $y= \frac{z}{2}$, then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} = \frac{1}{2}+\frac{1}{2} + 4 = 5$. Thus, whenever $z = 4t, t \in \mathbb{N}$, the solution set $(x,y,z) = (t,2t,4t)$ satisfies, and thus are infinitely many forms of $x+y+z$ with the given constraints - notably of the form $7t$.
These aren't all the solutions, but it's at least one class.
EDIT: Doing some further work, we can utilize this method to greatly expand the solution class. Say that $x | z$ and that $\frac{x}{y} < 1, \frac{y}{z} < 1$. Thus, in order for the original equation to be an integer, $\frac{x}{y} + \frac{y}{z} = 1$. Then for some $m,n \in \mathbb{N}$ where $m< n$, $\frac{x}{y} = \frac{m}{n}, \frac{y}{z} = \frac{n-m}{n}$.
Thus $nx = my$ and $ny = (n-m)z$, implying that $y = (1-\frac{m}{n})z$ and that $x = (\frac{m}{n} - \frac{m^{2}}{n^{2}})z$. Therefore $x+y+z = (\frac{m}{n} - \frac{m^{2}}{n^{2}})z + (1-\frac{m}{n})z + z = (2-\frac{m^{2}}{n^{2}})z$. Set $z = n^{2}t$ and the sum becomes $(2n^{2}-m^{2})t$, where $m,n,t \in \mathbb{N}$.
Thus the solution set $(mnt - m^{2}t,n^{2}t-mnt,n^{2}t)$ suffices to greatly expand the number of possible sums of $x+y+z$. In short, any multiple of any integer of the form $2n^{2} - m^{2}$ where $m < n$ can be expressed as a sum of $x,y,$ and $z$.
A: 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 $x+y+z=0$. 
A: First there are infintely many solutions $x=y=z$.
Let $x,y,z$ be such that $x/y+y/z+z/x$ is an integer.
Then $k= \frac{x^2 z + y^2 x + z^2 y}{xyz} = x/y+y/z+z/x$ is an integer.
Consider the polynomial:
$p(t) = (t-x)(t-y)(t-z) = t^3-a_2 t^2+a_1 t -a_0$
Then since $k a_0 = kxyz = x^2 z + y^2 x + z^2 y$, this polynomial satisfies:
1) $x,y,z \in \mathbb{N}$
2) $x^2 z + y^2 x + z^2 y \equiv 0 \mod(a_0)$
Conversely it is easy seen, that every such polynomial satisfying 1 & 2 gives rise to a solution.
Hence there is a bijection between all solutions and the polynomials with the property 1 & 2.
