4A. In the set of real numbers, solve the system

$$
\left\{\begin{array}{l}
x y z=x+y+z \\
y z t=y+z+t \\
z t u=z+t+u \\
t u x=t+u+x
\end{array}\right.
$$
Solution. From the first equation, we subtract the second and get $y z(x-t)=x-t$, or $(y z-1)(x-t)=0$. Possible cases are $y z-1=0$ or $x-t=0$. If $y z-1=0$, by substitution into the first equation we get $0=y+z$. Then, from $y z-1=0$ and $y+z=0$ it follows that $y^{2}=-1$, so in this case there is no solution. Therefore, $x-t=0$, i.e., $x=t$. If we repeat the procedure with the second and third equations, as well as with the third and fourth, we get $x=t=y=z$. Then, $x^{3}=3 x, x\left(x^{2}-3\right)=0$ and from this the set of solutions to the system is $\{(0,0,0,0),(\sqrt{3}, \sqrt{3}, \sqrt{3}, \sqrt{3}),(-\sqrt{3},-\sqrt{3},-\sqrt{3},-\sqrt{3})\}$.