816. Solve in integers $x, y, z$ and $t$ the system of equations

$$
\left\{\begin{array}{l}
x y+z t=1 \\
x z+y t=1 \\
x t+y z=1
\end{array}\right.
$$
$\triangle$ Subtract, for example, the second equation from the first in the system:

$$
x(y-z)+t(z-y)=0, \quad(y-z)(x-t)=0
$$

Then $y-z=0$ or $x-t=0$. Let's consider both cases.

1) Let $y-z=0$, i.e., $y=z$.

The system of equations takes the following form:

$$
\left\{\begin{array}{l}
x z+z t=1 \\
x z+z t=1 \\
x t+z^{2}=1
\end{array}\right.
$$

Here, the first and second equations coincide.

We can represent the first equation in the following form:

$$
z(x+t)=1 \text{. }
$$

From this, either $z=1, x+t=1$, or $z=-1, x+t=-1$.

If $z=1, x+t=1$, i.e., $x=1-t$, then the third equation of the last system becomes:

$$
(1-t) t+1=1, \quad(1-t) t=0
$$

Then $t=1$ or $t=0$. We obtain two solutions to the system: $(0 ; 1 ; 1 ; 1)$ and $(1 ; 1 ; 1 ; 0)$.

If $z=-1, x+t=-1$, i.e., $x=-1-t$, then

$$
(-1-t) t+1=1, \quad-(1+t) t=0
$$

From this, $t=-1$ or $t=0$. We have found two more solutions: $(0 ;-1 ;-1 ;-1)$ and $(-1 ;-1 ;-1 ; 0)$.

2) Let $x-t=0$, i.e., $x=t$.

The initial system of equations takes the following form:

$$
\left\{\begin{array}{l}
t y+z t=1 \\
t z+y t=1 \\
t^{2}+y z=1
\end{array}\right.
$$

This system can be solved in the same way, but it is simpler to proceed differently. It differs from the system obtained in the first case in that $t$ is replaced by $z$, $z$ by $t$, and $x$ by $y$ (this is especially evident when comparing the last equations of both systems). Taking into account the specified substitutions, we can use the four solutions already found for the initial system to obtain four more:

$$
(1 ; 0 ; 1 ; 1),(1 ; 1 ; 0 ; 1),(-1 ; 0 ;-1 ;-1),(-1 ;-1 ; 0 ;-1)
$$

Answer: $(0 ; 1 ; 1 ; 1),(1 ; 0 ; 1 ; 1),(1 ; 1 ; 0 ; 1),(1 ; 1 ; 1 ; 0),(0 ;-1 ;-1 ;-1),(-1 ; 0 ;-1 ;-1)$, $(-1 ;-1 ; 0 ;-1),(-1 ;-1 ;-1 ; 0)$.