4. For the set of non-negative real numbers, solve the system of equations:

$$
\begin{aligned}
& a \sqrt{b}-c=a \\
& b \sqrt{c}-a=b \\
& c \sqrt{a}-b=c
\end{aligned}
$$
4. In the case that one of $a, b, c$ is equal to 0, it is easy to see that the remaining two must also be 0, which gives one solution. Now assume that $a, b, c$ are positive. From the first equation, it follows that $\sqrt{b}>1$ (otherwise the left side would be less than $a$), and similarly from the second and third equations, $\sqrt{c}>1$ and $\sqrt{a}>1$. Without loss of generality, let $a=\max \{a, b, c\}$. From the second equation, we have $a=b(\sqrt{c}-1) \leqslant a(\sqrt{c}-1)$, from which it follows that $c \geqslant 4$ (so $a \geqslant c \geqslant 4$). From the first equation, we have $c=a(\sqrt{b}-1) \geqslant c(\sqrt{b}-1)$, from which it follows that $b \leqslant 4$. However, now from $b=c(\sqrt{a}-1) \geqslant 4(\sqrt{4}-1)=4$ it follows that equality must be achieved in the last inequality, i.e., $a=b=c=4$.

Therefore, the system has two solutions: $a=b=c=0$ and $a=b=c=4$.