12.43 Find all integer pairs $(x, y)$ that satisfy the equation $x^{2}=y^{2}+2 y+13$.
(46th Moscow Mathematical Olympiad, 1983)
［Solution］ Given the equations can be transformed into
$$
\begin{array}{c}
x^{2}-(y+1)^{2}=12, \\
(x+y+1)(x-y-1)=12 .
\end{array}
$$

Since $x+y+1$ and $x-y-1$ have the same parity, and 12 is an even number, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x + y + 1 = 2 , } \\
{ x - y - 1 = 6 , }
\end{array} \quad \left\{\begin{array}{l}
x+y+1=6, \\
x-y-1=2,
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x + y + 1 = - 2 , } \\
{ x - y - 1 = - 6 , }
\end{array} \quad \left\{\begin{array}{l}
x+y+1=-6, \\
x-y-1=-2 .
\end{array}\right.\right. \\
\end{array}
$$

From the above, we can obtain four sets of integer solutions:
$$
(x, y)=(4,-3),(4,1),(-4,1),(-4,-3) \text {. }
$$