20.12*. (USA, 82). Find all pairs of non-zero integer values \( m \leqslant n \), satisfying the inequality \( m+n \neq 0 \) and the identity

\[ 
f_{m}(x, y) f_{n}(x, y)=f_{m+n}(x, y), \quad x, y \in \mathbf{R}, \quad x y(x+y) \neq 0, 
\]
where it is denoted

\[ 
f_{k}(x, y)=\left(x^{k}+y^{k}+(-1)^{k}(x+y)^{k}\right) / k 
\]

Hint. The pairs \( m=2, n=3 \) and \( m=2, n=5 \) satisfy the required conditions.
20.12. We will prove that no other pairs of the required values, except those given in the indication, satisfy the required conditions. Suppose that some integer pair $(m ; n)$, different from $(2 ; 3)$ and $(2 ; 5)$, satisfies the inequalities $m \leqslant n, m n(m+n) \neq 0$ and the identity

$$
f_{m}(x, y) f_{n}(x, y) \equiv f_{m+n}(x, y), \quad x, y \in \mathbb{R}, \quad x y(x+y) \neq 0
$$

where

$$
f_{k}(x, y)=\left(x^{k}+y^{k}+(-1)^{k}(x+y)^{k}\right) / k \quad \text { for } \quad k \in Z, \quad k \neq 0
$$

For each fixed value $y=y_{0} \neq 0$, the following statements are true. If $k1$, because if $q=1$; then $f_{q}(1,1)=0$, and by the identity we have

$$
f_{p+q}(1,1)=\frac{2-2^{p+q}}{p+q}=0
$$

which is impossible, since $p+q>1$. Therefore, we obtain

$$
\begin{gathered}
\lim _{x \rightarrow \infty} \frac{f_{p}\left(x, y_{0}\right) f_{\eta}\left(x, y_{0}\right)}{x^{p+q-1}}=\lim _{x \rightarrow \infty} \frac{f_{p}\left(x, y_{0}\right) f_{q}\left(x, y_{0}\right)}{x^{p}} \frac{2 y_{0}}{x^{q-1}}=-\frac{f_{p}}{p} \\
\lim _{x \rightarrow \infty} \frac{f_{p+q}\left(x, y_{0}\right)}{x^{p+q-1}}=-y_{0}
\end{gathered}
$$

from which $p=2$. From the identity we have

$$
3 \cdot\left(2-2^{q}\right) / q=f_{2}(1,1) f_{q}(1,1)=f_{2+q}(1,1)=\left(2-2^{2+q}\right) /(2+q)
$$

i.e.

$$
3(2+q)\left(1-2^{q-1}\right)=q\left(1-2^{q+1}\right), \quad \text { or } \quad 3+q=(6-q) 2^{q-2}
$$

Therefore, if $q1$, then $q \in\{3 ; 5\}$, which contradicts the made assumption.

g) Suppose $mm+n$, then regardless of the sign and parity of the number $m+n$ we have

$$
\lim _{x \rightarrow \infty} \frac{f_{m+n}\left(x, y_{0}\right)}{x^{\eta}}=0
$$

while

$$
\lim _{x \rightarrow \infty} \frac{f_{m}\left(x, y_{0}\right) f_{n}\left(x, y_{0}\right)}{x^{n}}=\lim _{x \rightarrow \infty} f_{m}\left(x, y_{0}\right) \lim _{x \rightarrow \infty} \frac{f_{2}\left(x, y_{0}\right)}{x^{n}}=\frac{2 y_{0}^{m}}{m n} \neq 0
$$

d) Suppose $mm+n$, and regardless of the sign and parity of the number $m+n$ we have

$$
\lim _{x \rightarrow \infty} \frac{f_{m+n}\left(x, y_{0}\right)}{x^{n-1}}=0
$$

Therefore, the number $m$ can only be equal to -1. Then the number $m+n=n-1$ is even and

$$
\lim _{x \rightarrow \infty} \frac{f_{m+n}\left(x, y_{0}\right)}{x^{n-1}}=\frac{2}{n-1}
$$

Thus, for each value $y_{0} \neq 0$ the equality $1 / y_{0}=$ $=2 /(n-1)$ holds, which is impossible.

e) Suppose $m, n0$, $m+n<0$.

Thus, the only pairs $(m ; n)$ that satisfy the conditions of the problem are $(2 ; 3)$ and $(2 ; 5)$.