For which positive number quadruples $x, y, u, v$ does the following system of equations hold?

$$
x+y=u v x y=u+v x y u v=16
$$
It is visible that the system of equations is satisfied when $x=y=z=v=2$. We will show that there is no other solution. Let's write down the inequality between the arithmetic and geometric means for the numbers $x$ and $y$, and $u$ and $v$:

$$
u v=x+y \geq 2 \sqrt{x y}, \quad x y=u+v \geq 2 \sqrt{u v}
$$

and equality holds if and only if $x=y$, and $u=v$. Multiply the two inequalities:

$$
16=(u v)(x y) \geq 2 \sqrt{x y} \cdot 2 \sqrt{u v}=4 \sqrt{x y u v}=16
$$

Since equality must hold, according to the above, $x=y$ and $u=v$. Substituting these into the original equations:

$$
2 x=u^{2}, \quad 2 u=x^{2}
$$

from which $x^{4}=(2 u)^{2}=4 u^{2}=4 \cdot 2 x=8 x$, hence $x=2$. Similarly, we find that $u=2$.