$12 \cdot 20$ Find all ordered integer pairs $(x, y)$ that satisfy the equation $x^{2}+2 x y+3 y^{2}-2 x+y+1=0$.
(China Beijing High School Mathematics Competition, 1988)
[Solution] Convert the given equation into a quadratic equation in terms of $x$
$$
x^{2}+2(y-1) x+\left(3 y^{2}+y+1\right)=0 \text {. }
$$

If the equation has integer solutions, then a necessary condition is
$$
\Delta=4\left[(y-1)^{2}-\left(3 y^{2}+y+1\right)\right] \geqslant 0,
$$

Solving this, we get
$$
-\frac{3}{2} \leqslant y \leqslant 0 \text {. }
$$

Since $y$ is an integer, then
$$
y=0,-1 \text {. }
$$

When $y=0$,
$$
\begin{array}{c}
x^{2}-2 x+1=0, \\
x=1 .
\end{array}
$$

When $y=-1$,
$$
\begin{array}{l}
x^{2}-4 x+3=0, \\
x=1, x=3 .
\end{array}
$$

Therefore, all the ordered integer pairs that satisfy the given equation are
$$
(x, y)=(1,0),(1,-1),(3,-1) .
$$