<s>[INST] 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. 
 [/INST] 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 $z$.
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