11. (German Mathematical Competition 2002-2003) Find the integer solutions of the system of equations
$$
\left\{\begin{array}{l}
x^{3}-4 x^{2}-16 x+60=y \\
y^{3}-4 y^{2}-16 y+60=z \\
z^{3}-4 z^{2}-16 z+60=x
\end{array}\right.
$$
11. From the three equations of the known system, they can be transformed into $(x-4)^{2}(x+4)=y+4$, (1) $\quad(y-4)^{2}(y+4)=z+4$, (2) $(z-4)^{2}(z+4)=x+4$, (3) Therefore, $(x, y, z)=(-4,-4,-4)$ is a solution to the system of equations.
If $x \neq-4$, from (3) we know $z \neq-4$, hence $y \neq-4$.
By (1) - (2) - (3) and simplifying, we get $(x-4)^{2}(y-4)^{2}(z-4)^{2}=1$, whose integer solutions can only be $x=3$ or $5, y=3$ or $5, z=3$ or $5$. Combining with the known equations, we obtain three solutions to the system of equations: $(x, y, z)=(3,3,3),(5,5,5),(-4,-4,-4)$.