6 - 130 Let the function
$$
f_{k}(x, y)=\frac{x^{k}+y^{k}+(-1)^{k}(x+y)^{k}}{k}, k \in Z, k \neq 0 .
$$

Find all non-zero integer pairs $(m, n)$ such that $m \leqslant n, m+n \neq 0$, and
$$
f_{m}(x, y) f_{n}(x, y) \equiv f_{m+n}(x, y),
$$

where $x, y \in R, x y(x+y) \neq 0$.
Hint: The pairs $m=2, n=3$ and $m=2, n=5$ satisfy the given conditions.
$\begin{array}{l}\text { [Solution] Let the pair }(n, m) \text { satisfy } m \leqslant n, n m(m+n) \neq 0 \text {, and } \\ f_{m}(x, y) f_{n}(x, y) \equiv f_{m+n}(x, y), x, y \in R, x y(x+y) \neq 0 \\ \text { where } f_{k}(x, y)=\frac{x^{k}+y^{k}+(-1)^{k}(x+y)^{k}}{k}, k \in Z, k \neq 0 \text {. (1) } \\ \text { First, when } k=1 \text {, we have } \\ f_{1}(x, y) \equiv 0, x, y \in R \\ \text { Next, for other values of } k \text {, provide three limit formulas. If } k1 \text {, it is impossible, so } q>1 \text {, thus from equations (3) } \\ \text { and (4) we get } \\ \lim _{x \rightarrow \infty} \frac{f_{p}\left(x, y_{0}\right) f_{q}\left(x, y_{0}\right)}{x^{\rho+q-1}}=\lim _{x \rightarrow \infty} \frac{f_{p}\left(x, y_{0}\right)}{x^{p}} \lim _{x \rightarrow \infty} \frac{f q\left(x, y_{0}\right)}{x^{q-1}} \\ =-\frac{2 y_{0}}{p}, \\ \text { since } p+q \text { is odd, so from equation (4) } \\ \lim _{x \rightarrow \infty} \frac{f_{p+q}\left(x, y_{0}\right)}{x^{p+q-1}}=-y_{0}, \\ \text { thus } p=2 \text {. From the identity, we get } \\ \begin{array}{l}\frac{3\left(2-2^{q}\right)}{q}=f_{2}(1,1) f_{q}(1,1)=f_{2+q}(1,1)=\frac{2-2^{2+q}}{2+q}, \\ 3(2+q)\left(2-2^{q}\right)=q\left(2-2^{2+q}\right),\end{array} \\ \text { thus } \begin{array}{c}q \\ 3(2+q)\left(2-2^{q}\right)\end{array}=q\left(2-2^{2+q}\right) \text {, } \\ \text { i.e., } 3+q=(6-q) 2^{q-2} \\ \text { therefore, } q1 \text {, so } q \in\{3,5\} \text {, which is the case mentioned in the hint. It is easy to verify that the pairs } (2,3) \text { and } (2,5) \text { indeed satisfy the requirements of the problem } \\ \text { (4) } mm+n \text {, so regardless of the } \\ \text { sign and parity of } m+n \text {, we have } \\ \lim _{x \rightarrow \infty} \frac{f_{m+n}\left(x, y_{0}\right)}{x^{n}}=0 \\ \text { On the other hand, from equations (2) and (3), we have } \\ \lim _{n \rightarrow \infty} \frac{f_{m}\left(x, y_{0}\right) f_{n}\left(x, y_{0}\right)}{x^{n}}=\lim _{x \rightarrow \infty} f_{m}\left(x, y_{0}\right) \lim _{x \rightarrow \infty} \frac{f_{n}\left(x, y_{0}\right)}{x^{n}} \\ =\frac{2 y_{0}^{\prime \prime}}{m n} \neq 0 . \\ \text { (5) } m>0 \text {, and } n0, m+n>0 \text {, so } \\ \text { since } m n>0, m+n>0 \text {, we have } \\ \frac{y_{0}^{m+n}}{m n} \neq \frac{y_{0}^{m+n}}{m+n}, \\ \text { thus } f_{m}(x, y) f_{n}(x, y) \neq f_{m+n}(x, y) \text {, a contradiction. } \\\end{array}$