RagvaloD wrote: Solve in positives $$x^y=z,y^z=x,z^x=y$$ If $x>1$, then $z=x^y>1$ and $y=z^x>1$ and so : $z=x^y>x$ and $x=y^z>y$ and $y=z^x>z$ impossible If $x<1$, then $z=x^y<1$ and $y=z^x<1$ and so : $z=x^y>x$ and $x=y^z>y$ and $y=z^x>z$ impossible So $\boxed{x=y=z=1}$ which indeed is a solution

We claim that the only solutions are $\boxed{x=y=z=1}$, $\boxed{x=y=z=-1}$ First take, the equation $ z^{x} =y $. The multiply both sides by $ x$ so that we get \begin{align*} z^{x} \cdot y^{z} &= xy \\ &= z^{\frac{1}{y}} \cdot y \end{align*}$$ \implies x = \frac{1}{y} , z = 1$$Which also implies that $\boxed{ xy = 1} $ and so, in that sense we can infer that $ (x,y) = \{ (1,1) , (-1,-1) \} $ But now notice that if $(x,y) = (-1,-1) $ then $z = -1$ which is not true so, the solutions are $\boxed{x=y=z=1}$. Nevertheless, if we consider the solution $\boxed{x=y=z=-1}$ the equations will also hold true. Finally there are no more solutions because if we follow the process made above we'll have equations of the form $ a_{j} \cdot a_{i} =1 $ $( i \neq j)$ from where the $a's$ can be $ x, y $ or $z$ and thus is imposible for other solutions, not equal to ones, to exist.