Find all integer solutions of the system $\left\{\begin{array}{l}x z-2 y t=3 \\ x t+y z=1 .\end{array}\right.$
Solution

A good strategy will be to apply some algebraic manipulation, such as adding the equations, multiplying them, adding a correction factor, among others, to obtain some factorization involving these numbers. Squaring both equations, we have:

$$
\left\{\begin{aligned}
x^{2} z^{2}-4 x y z t+4 y^{2} t^{2} & =9 \\
x^{2} t^{2}+2 x y t z+y^{2} z^{2} & =1
\end{aligned}\right.
$$

Multiplying the second equation by two and adding it to the first, we get:

$$
\begin{aligned}
x^{2}\left(z^{2}+2 t^{2}\right)+2 y^{2}\left(z^{2}+2 t^{2}\right) & =11 \\
\left(x^{2}+2 y^{2}\right)\left(z^{2}+2 t^{2}\right) & =11
\end{aligned}
$$

Since each of the previous terms is a non-negative integer, we have two cases:

$$
\begin{gathered}
\left\{\begin{array}{l}
x^{2}+2 y^{2}=11 \\
z^{2}+2 t^{2}=1
\end{array} \Rightarrow(x, y, z, t)=( \pm 3, \pm 1, \pm 1,0)\right. \\
\left\{\begin{array}{l}
x^{2}+2 y^{2}=1 \\
z^{2}+2 t^{2}=11
\end{array} \Rightarrow(x, y, z, t)=( \pm 1,0, \pm 3, \pm 1)\right.
\end{gathered}
$$

Therefore, the only possible solutions are the quadruples $( \pm 1,0, \pm 3, \pm 1)$ and $( \pm 3, \pm 1, \pm 1,0)$.