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 $$(a-2{)}^2+(b-1{)}^2=5(a,b\ge 0).$$ 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 $\{0\}\cup [4,20]$.