Balti Tee 1995 · Ülesanne 1

Solution: From the first equation it follows that $x$ is even. The second equation implies $x>y$ and $x>z$. Hence $4x > 2(y+z) = x^{2}$, and therefore $x=2$ and $y+z=2$, so $y=z=1$. It is easy to check that the triple $(2,1,1)$ satisfies the given system of equations.