Four, (50 points) Find all real-coefficient polynomials $f(x)$ such that for all real numbers $a, b, c$ satisfying $a b+b c+c a=0$, we have
$$
f(a-b)+f(b-c)+f(c-a)=2 f(a+b+c) .
$$
Let $a=b=c=0$ in equation (1).
Then $3 f(0)=2 f(0) \Rightarrow f(0)=0$.
Let $b=c=0$.
Then $f(a)+f(-a)=2 f(a) \Rightarrow f(-a)=f(a)$.
Therefore, $f(x)$ is an even function.
Assume $f(x)=a_{n} x^{2 n}+a_{n-1} x^{2 n-2}+\cdots+a_{1} x^{2}$,
where $a_{n} \neq 0$.
Let $a=x, b=2 x, c=-\frac{2}{3} x$. Then
$$
\begin{array}{l}
f(-x)+f\left(\frac{8}{3} x\right)+f\left(-\frac{5}{3} x\right)=2 f\left(\frac{7}{3} x\right) \\
\Rightarrow \sum_{k=1}^{n} a_{k}\left[1+\left(\frac{8}{3}\right)^{2 k}+\left(\frac{5}{3}\right)^{2 k}-2\left(\frac{7}{3}\right)^{2 k}\right] x^{2 k}=0 .
\end{array}
$$

Therefore, the coefficients of this polynomial are all 0.
But when $n \geqslant 3$,
$$
\frac{\left(\frac{8}{3}\right)^{2 n}}{\left(\frac{7}{3}\right)^{2 n}}=\left(\frac{8}{7}\right)^{2 n} \geqslant\left(\frac{8}{7}\right)^{6}>2 \text {, }
$$

Hence $1+\left(\frac{8}{3}\right)^{2 n}+\left(\frac{5}{3}\right)^{2 n}-2\left(\frac{7}{3}\right)^{2 n}>0$.
So, $n \leqslant 2$.
Therefore, $f(x)=a x^{4}+\beta x^{2}(\alpha, \beta \in \mathbf{R})$.
Thus $\sum(a-b)^{4}-2(a+b+c)^{4}$
$$
=\sum(a-b)^{4}-2\left(a^{2}+b^{2}+c^{2}\right)^{2}
$$
$=\sum\left(a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}\right)-$
$2\left(\sum a^{4}+2 \sum a^{2} b^{2}\right)$
$=-4 \sum a^{2}(a b+a c)+2 \sum a^{2} b^{2}$
$=4 \sum a^{2} b c+2 \sum a^{2} b^{2}$
$=2(a b+b c+c a)^{2}=0$,
where “$\sum$” denotes the cyclic symmetric sum, and
$$
\begin{array}{l}
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}-2(a+b+c)^{2} \\
=-4(a b+b c+c a)=0 .
\end{array}
$$

Therefore, $f(x)=\alpha x^{4}+\beta x^{2}(\alpha, \beta \in \mathbf{R})$ satisfies the conditions.
(Chen Xianqing, Anqing No.1 High School, Anhui, 246000 Huang Quanfu, Jiangzhen High School, Huaining County, Anhui, 246142)