4-258 $p, q, r$ are pairwise distinct real numbers, satisfying the system of equations
$$
\left\{\begin{array}{l}
q=p(4-p), \\
r=q(4-q), \\
p=r(4-r) .
\end{array}\right.
$$

Find all possible values of $p+q+r$.
[Solution] From the given,
$$
p=r(4-r)=-(r-2)^{2}+4 \leqslant 4,
$$

Similarly, we can get
$$
q \leqslant 4, r \leqslant 4 .
$$

Next, we prove $p \geqslant 0$ using proof by contradiction.
If $p<0$, then from $4-p \geqslant 0$ we get
$$
q \leqslant 0 .
$$

Then from $4-q \geqslant 0$ we get
$$
r \leqslant 0 .
$$

Thus, we have
$$
p+q+r<0 .
$$

Adding the three equations in the problem, we get
$$
p^{2}+q^{2}+r^{2}=3(p+q+r)
$$

From the above two equations, we get
$$
p^{2}+q^{2}+r^{2}<0
$$

which is a contradiction.
Therefore, $4 \geqslant p \geqslant 0$.

Similarly, we can get
$$
4 \geqslant q \geqslant 0,4 \geqslant r \geqslant 0 .
$$

Let
$$
p=4 \sin ^{2} \theta, \theta \in\left[0, \frac{\pi}{2}\right] .
$$

Substituting the above into the first equation of the original system, we get
$$
q=4 \sin ^{2} 2 \theta,
$$

Substituting the above into the second equation of the original system, we get
$$
r=4 \sin ^{2} 4 \theta .
$$

Substituting the above into the third equation of the original system, we get
$$
p=4 \sin ^{2} 8 \theta .
$$

Thus, we have
$$
\sin 8 \theta= \pm \sin \theta .
$$

Since $0 \leqslant 8 \theta \leqslant 4 \pi$, there are 8 possible cases:
$$
\begin{array}{l}
8 \theta=\theta, 8 \theta=\pi-\theta, 8 \theta=\pi+\theta, 8 \theta=2 \pi-\theta, \\
8 \theta=2 \pi+\theta, 8 \theta=3 \pi-\theta, 8 \theta=3 \pi+\theta, 8 \theta=4 \pi-\theta .
\end{array}
$$

That is, $\theta \in\left\{0, \frac{\pi}{9}, \frac{\pi}{7}, \frac{2 \pi}{9}, \frac{2 \pi}{7}, \frac{\pi}{3}, \frac{3 \pi}{7}, \frac{4 \pi}{9}\right\}$
When $\theta=0$ or $\frac{\pi}{3}$, $p=q=r=0$ or 3, which are discarded.
When $\theta=\frac{\pi}{9}, \frac{2 \pi}{9}, \frac{4 \pi}{9}$,
$$
\begin{aligned}
p+q & +r=4\left(\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{2 \pi}{9}+\sin ^{2} \frac{4 \pi}{9}\right) \\
& =6-2\left(\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cos \frac{8 \pi}{9}\right) \\
& =6-2\left(2 \cos \frac{\pi}{3} \cdot \cos \frac{\pi}{9}-\cos \frac{\pi}{9}\right) \\
& =6,
\end{aligned}
$$

When $\theta=\frac{\pi}{7}, \frac{2 \pi}{7}, \frac{3 \pi}{7}$,
$$
\begin{aligned}
p+q+r & =4\left(\sin ^{2} \frac{\pi}{7}+\sin ^{2} \frac{2 \pi}{7}+\sin ^{2} \frac{4 \pi}{7}\right) \\
& =6-2\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7}\right) \\
& =6-2\left(2 \cos \frac{3 \pi}{7} \cdot \cos \frac{\pi}{7}-\cos \frac{\pi}{7}\right) \\
& =6-2 \cos \frac{\pi}{7}\left(2 \cos \frac{3 \pi}{7}-1\right) \\
& =6-\frac{\sin \frac{2 \pi}{7}\left(2 \cos \frac{3 \pi}{7}-1\right)}{\sin \frac{\pi}{7}} \\
& =6-\frac{1}{\sin \frac{\pi}{7}} \cdot\left(\sin \frac{5 \pi}{7}-\sin \frac{\pi}{7}-\sin \frac{2 \pi}{7}\right) \\
& =6+1=7 .
\end{aligned}
$$

Therefore, the only possible values of $p+q+r$ are 6 and 7.