20 Find all integer pairs $(x, y)$, such that $x^{2}+3 y^{2}=1998 x$.
20. Let $(x, y)$ be the integer pairs that satisfy the equation, it is known that $3 \mid x^{2}$, hence $3 \mid x$, set

then
$$\begin{array}{l}
x=3 x_{1} \\
3 x_{1}^{2}+y^{2}=1998 x_{1}
\end{array}$$

Thus,
$$3 \mid y^{2},$$

hence
$$3 \mid y \text {. }$$

Set again
$$y=3 y_{1},$$

then
$$x_{1}^{2}+3 y_{1}^{2}=666 x_{1}$$

Proceeding similarly, we can set
$$x=27 m, y=27 n$$

we get
$$m^{2}+3 n^{2}=74 m$$

Rearranging and completing the square, we get
$$(m-37)^{2}+3 n^{2}=37^{2}$$

Analyzing the parity of both sides of this equation, we find that $m$ and $n$ are both even, thus
$$3 n^{2}=37^{2}-(m-37)^{2} \equiv 0(\bmod 8)$$

Hence
set
then
thus
hence
$$\begin{array}{c}
n=4 r \\
48 r^{2} \leqslant 37^{2} \\
r^{2} \leqslant 28 \\
|r| \leqslant 5
\end{array}$$

For each
$$|r|=0,1, \cdots, 5$$

calculate the value of $37^{2}-48 r^{2}$, we find that only when $|r|=0,5$, $37^{2}-48 r^{2}$ is a perfect square, so $(x, y)=(0,0),(1998,0),(1350, \pm 540),(648, \pm 540)$