* For each natural number $m$, find all integer solutions $x, y, z$ of the equation $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=m$, where $x, y, z$ are pairwise coprime.
The original equation is equivalent to the equation $x^{2} z+y^{2} x+z^{2} y=m x y z$, where $x, y, z$ are non-zero integers, and pairwise coprime. Thus, $y\left|x^{2} z, z\right| y^{2} x, x \mid z^{2} y$. Also, $(x, y)=1,(z, y)=1$, so $\left(x^{2} z\right.$, $y)=1$. From $y \mid x^{2} z$ we get $y= \pm 1$, similarly we get $z= \pm 1, x= \pm 1$.

If $x, y, z$ are all positive or all negative, then from the original equation we get $m=3$. If two of $x, y, z$ are positive and one is negative, or two are negative and one is positive, then from the original equation we get $m$ as a negative number, which contradicts the problem statement.

Therefore, the original equation, when $m=3$, has integer solutions $x, y, z$ that are pairwise coprime, with only two sets of solutions: $x=y=z=1, x=y=z=-1$. When $m \neq 3$, the original equation has no solutions that satisfy the problem conditions.