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

Again, 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
$$p^{2}+q^{2}+r^{2}<0$$

This 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:
$$\begin{array}{l} 
8 \theta=\theta, 8 \theta=\pi-\theta, 8 \theta=\pi+\theta, 8 \theta=2 \pi-\theta \\
8 \theta=2 \pi+\theta, 8 \theta=3 \pi-\theta, 8 \theta=3 \pi+\theta, 8 \theta=4 \pi-\theta . \\
\text { That is } \quad \theta \in\left\{0, \frac{\pi}{9}, \frac{\pi}{7}, \frac{2 \pi}{9}, \frac{2 \pi}{7}, \frac{\pi}{3}, \frac{3 \pi}{7}, \frac{4 \pi}{9}\right\}
\end{array}$$

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.