China Mathematical Competition

Let $\sqrt{y} = a$, $\sqrt{x-y} = b$ ($a, b \ge 0$). Then $x = y + (x-y) = a^2 + b^2$. The equation in the question becomes $a^2 + b^2 - 4a = 2b$, which is equivalent to
$$(a - 2)^2 + (b - 1)^2 = 5 \quad (a, b \ge 0).$$
As seen in Fig. 7.1, the trace of point $(a, b)$ in plane $aOb$ is the part of the circle with center $(2, 1)$ and radius $\sqrt{5}$ satisfying $a, b \ge 0$, i.e., the union of point $O$ and arc $\widehat{ACB}$. Then

Fig. 7.1
$$\sqrt{a^2 + b^2} \in \{0\} \cup [2, 2\sqrt{5}].$$
Therefore, $x = a^2 + b^2 \in \{0\} \cup [4, 20]$.
The answer is $\{0\} \cup [4, 20]$.