6. 59 Let $S_{r}=x^{r}+y^{r}+z^{r}$, where $x, y, z$ are real numbers. It is known that when $S_{1}=$ 0, for
$$
(m, n)=(2,3),(3,2),(2,5) \text { or }(5,2) \text {, }
$$

we have
$$
\frac{S_{m+n}}{m+n}=\frac{S_{m}}{m} \cdot \frac{S_{n}}{n}
$$

Determine all other positive integer pairs $(m, n)$ that satisfy (1).
【Sol】Obviously, when $x=k+1, y=-k, z=-1$, it can make $S_{1}=x+y+z=0$.
If $\frac{S_{m+n}}{m+n}=\frac{S_{m}}{m} \cdot \frac{S_{n}}{n}$

then $\frac{x^{m+n}+y^{m+n}+z^{m+n}}{m+n}=\frac{x^{m}+y^{m}+z^{m}}{m} \cdot \frac{x^{n}+y^{n}+z^{n}}{n}$
Substituting $x=k+1, y=-k, z=-1$ into the above equation, we get
$$
\begin{aligned}
& \frac{(k+1)^{m+n}+(-k)^{m+n}+(-1)^{m+n}}{m+n} \\
= & \frac{(k+1)^{m}+(-k)^{m}+(-1)^{m}}{m} \cdot \frac{(k+1)^{n}+(-k)^{n}+(-1)^{n}}{n}
\end{aligned}
$$

Obviously, both sides of (2) can be seen as polynomials in $k$.
Case 1: If $m$ and $n$ are both odd, then $m+n$ is even. At this point, the left side of (2) is a polynomial of degree $m+n$ in $k$, and the right side of (2) is a polynomial of degree $m+n-2$ in $k$, which is a contradiction. This means that $m$ and $n$ cannot both be odd.

Case 2: If $m$ and $n$ are both even, then $m+n$ is also even. At this point, comparing the coefficients of the highest degree terms on both sides of (2), we get
$$
\frac{2}{m+n}=\frac{2}{m} \cdot \frac{2}{n}
$$

which is $\frac{1}{\frac{m}{2}}+\frac{1}{\frac{n}{2}}=1$.
Since $\frac{m}{2}, \frac{n}{2}$ are positive integers, therefore $\frac{m}{2}=\frac{n}{2}=2$, thus $m=n=4$.
On the other hand, when $m=n=4$, we take $k=1$, i.e., $x=2, y=-1$, $z=-1$, then we get
$$
\begin{array}{l}
S_{4}=2^{4}+(-1)^{4}+(-1)^{4}=18, \\
S_{8}=2^{8}+(-1)^{8}+(-1)^{8}=258,
\end{array}
$$

but $\quad \frac{S_{8}}{8}=\frac{258}{8} \neq \frac{18}{4} \cdot \frac{18}{4}=\frac{S_{4}}{4} \cdot \frac{S_{4}}{4}$.
This shows that $m$ and $n$ cannot both be even.
Case 3: If one of $m$ and $n$ is odd and the other is even, then $m+n$ is odd. At this point, comparing the coefficients of the highest degree terms of $k$ on both sides of (2), we get
$$
\begin{array}{c}
1=1 \cdot \frac{2}{n}, \\
n=2 .
\end{array}
$$

When $m=1$, by $S_{1}=0$, the right side of (2) is zero, so (2) cannot hold at this time.
When $m=3$, (2) holds by the given condition.
When $m>3$, the left side of (2) is
$$
\begin{aligned}
\frac{S_{m+2}}{m+2} & =\frac{(k+1)^{m+2}-k^{m+2}-1}{m+2} \\
& =k^{m+1}+\frac{m+1}{2} k^{m}+\frac{(m+1) m}{6} k^{m-1}+\cdots,
\end{aligned}
$$

while the right side of (2) is
$$
\begin{aligned}
\frac{S_{m}}{m} \cdot \frac{S_{2}}{2}= & \frac{(k+1)^{m}-k^{m}-1}{m} \cdot \frac{(k+1)^{2}+k^{2}+1}{2} \\
= & {\left[k^{m-1}+\frac{m-1}{2} k^{m-2}+\frac{(m-1)(m-2)}{6} k^{m-3}+\cdots\right] \cdot\left(k^{2}\right.} \\
& +k+1) \\
= & k^{m+1}+\frac{m+1}{2} k^{m}+\left[\frac{(m-1)(m-2)}{6}+\frac{m-1}{2}+1\right] k^{m-1} \\
& +\cdots
\end{aligned}
$$

Comparing the coefficients of $k^{m-1}$ on both sides, we get
$$
\frac{(m+1) m}{6}=\frac{(m-1)(m-2)}{6}+\frac{m-1}{2}+1 .
$$

Solving this, we get $m=5$. And $(m, n)=(5,2)$ is precisely the solution given in the problem.
Therefore, $(m, n)=(2,3),(3,2),(2,5),(5,2)$ are all the positive integer solutions that satisfy (1).

Note: The problem has the condition "When $S_{1}=0$, $(m, n)=(2,3),(3,2),(2,5)$ or $(5,2)$ satisfy (1)". If this condition is removed, then only the following supplement to the solution is needed.
$$
\text { Let } x y+y z+z x=b \text {, }
$$
$$
x y z=c \text {. }
$$
$$
\begin{array}{l}
\text { Also, from } S_{1}=x+y+z=0 \text {, we know that } x, y, z \text { are the three roots of the equation } \\
u^{3}+b u-c=0 \\
\text { Multiplying (3) by } u^{m-3} \text { gives } \\
u^{m \prime}+b u^{m-2}-c u^{m \prime-3}=0 .
\end{array}
$$

Substituting $u=x, y, z$ into the above equation and adding them up, we get
$$
\text { Since } \begin{aligned}
S_{m} & +b S_{m-2}-C S_{m-3}=0 \quad(m=3,4, \cdots) \\
S_{0} & =x^{0}+y^{0}+z^{0}=3, \\
S_{1} & =0, \\
S_{2} & =x^{2}+y^{2}+z^{2} \\
& =(x+y+z)^{2}-2(x y+y z+z x) \\
& =-2 b .
\end{aligned}
$$

Therefore,
$$
\begin{array}{l}
S_{3}=3 c, \\
S_{4}=2 b^{2}, \\
S_{5}=-5 b c, \\
S_{7}=7 b^{2} c .
\end{array}
$$

Thus, $\quad \frac{S_{5}}{5}=\frac{-5 b c}{5}=\frac{3 c}{3} \cdot \frac{-2 b}{2}=\frac{S_{3}}{3} \cdot \frac{S_{2}}{2}$,
$$
\frac{S_{7}}{7}=\frac{7 b^{2} c}{7}=\frac{-5 b c}{5} \cdot \frac{-2 b}{2}=\frac{S_{5}}{5} \cdot \frac{S_{2}}{2} .
$$