XXXI Brazilian Math Olympiad

Since $xyz=1$, $$\begin{array}{rl}x+\frac{1}{y}& =y+\frac{1}{z}⟺xyz+z=y^{2}z+y\\ ⟺1-y=-z(1-y^2)⟺y=1\text{ or }z=-\frac{1}{1+y}& \end{array}$$ If $y=1$, the system reduces to $xz=1⟺z=\frac{1}{x}$ and $$x+1=1+\frac{1}{z}=z+\frac{1}{x}⟺x+1=1+x=\frac{2}{x}⟺x=1\text{ or }x=-2$$ If $z=-\frac{1}{1+y}$, then $xyz=1⟺x=-\frac{1+y}{y}$ and $$-\frac{1+y}{y}+\frac{1}{y}=y-(1+y)=-\frac{1}{1+y}-\frac{y}{1+y}$$ (Solutions) --- which is true for every $y\ne 0,-1$ (all three expressions are equal to $-1$). So all solutions are $(1,1,1)$ and $(-\frac{1+t}{t},t,-\frac{1}{1+t})$ and its cyclic analogous triples. Notice that $t=1$ yields the solution $(-2,1,-\frac{1}{2})$.