4・161 Solve the equation $2 \sqrt{x}+\sqrt{2 y}=\sqrt{32}, x, y$ are integers.
［Solution］It is easy to know that $x \geqslant 0, y \geqslant 0$.
From the original equation, we get
$$
\sqrt{2 y}=\sqrt{32}-2 \sqrt{x},
$$

Squaring both sides and rearranging, we obtain
$$
y=16+2 x-8 \sqrt{2 x} \text {, }
$$

Thus, $8 \sqrt{2 x}$ should be an integer. Let it be $a$, then we have
$$
2 x=\left(\frac{a}{8}\right)^{2},
$$

Hence, $a$ is a multiple of 8. Let $a=8 b$, substituting gives $2 x=b^{2}$, so $b^{2}$ is even. Let $b=2 c$, substituting again gives $x=2 c^{2}$.

From the original equation, if $y=0$, then $x=8$; if $y \neq 0$, then $x<8$. Therefore, $0 \leqslant x \leqslant 8$, and thus $0 \leqslant c \leqslant 2$. It is easy to calculate the values of $c, x, y$ as shown in the table below:
\begin{tabular}{c|c|c|c}
\hline$c$ & 0 & 1 & 2 \\
\hline$x$ & 0 & 2 & 8 \\
\hline$y$ & 16 & 4 & 0 \\
\hline
\end{tabular}