$12 \cdot 120$ Solve the system of equations $\left\{\begin{array}{l}x z-2 y t=3, \\ x t+y z=1 .\end{array}\right.$ for integer solutions.
(25th All-Soviet Union Mathematical Olympiad, 1991)
[Solution] From the original system of equations, we get

i.e., $\square$
$$
\begin{array}{c}
(x z-2 y t)^{2}+2(x t+y z)^{2}=11, \\
\left(x^{2}+2 y^{2}\right)\left(z^{2}+2 t^{2}\right)=11 .
\end{array}
$$

Since 11 is a prime number, we have
I. $\left\{\begin{array}{l}x^{2}+2 y^{2}=1 \\ z^{2}+2 t^{2}=11\end{array}\right.$
II. $\left\{\begin{array}{l}x^{2}+2 y^{2}=11, \\ z^{2}+2 t^{2}=1\end{array}\right.$

From I and II, we can obtain four sets of integer solutions for the equations:
$$
\left\{\begin{array} { l } 
{ x = 1 , } \\
{ y = 0 , } \\
{ z = 3 , } \\
{ t = 1 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = - 1 , } \\
{ y = 0 , } \\
{ z = - 3 , } \\
{ t = - 1 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = 3 , } \\
{ y = 1 , } \\
{ z = 1 , } \\
{ t = 0 , }
\end{array} \quad \left\{\begin{array}{l}
x=-3, \\
y=-1, \\
z=-1, \\
t=0 .
\end{array}\right.\right.\right.\right.
$$