$4 \cdot 104$ Find the integer solutions of the system of equations
$$
\left\{\begin{array}{l}
x z-2 y t=3, \\
x t+y z=1 .
\end{array}\right.
$$
［Solution］From the system of equations, we can obtain
$$
(x z-2 y t)^{2}+2(x t+y z)^{2}=\left(x^{2}+2 y^{2}\right)\left(z^{2}+2 t^{2}\right)=11 .
$$

Therefore, we have $x^{2}+2 y^{2}=1$ or $z^{2}+2 t^{2}=1$.
If $x^{2}+2 y^{2}=1$, then we get $y=0, x= \pm 1$, and from the second equation, we get $t= \pm 1$, and from the first equation, we get $z= \pm 3$.
If $z^{2}+2 t^{2}=1$, then we get $t=0, z= \pm 1$, and then $y= \pm 1, x= \pm 3$.

Direct verification shows that the 4 sets of solutions $(x, y, z, t)$ satisfy the original system of equations. They are $(1,0,3,1),(-1,0,-3,-1),(3,1,1,0),(-3,-1,-1,0)$.