One, (50 points) Try to find all real-coefficient polynomials $f(x)$, such that for all real numbers $a, b, c$ satisfying $a b + b c + c a = 0$, the following holds:
$$
f(a-b) + f(b-c) + f(c-a) = 2 f(a+b+c).
$$
Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$.
Since for all $a, b, c \in \mathbf{R}$ satisfying $a b+b c+c a=0$, we have
$$
\begin{array}{l}
f(a-b)+f(b-c)+f(c-a) \\
=2 f(a+b+c),
\end{array}
$$

Therefore, when $a=b=c=0$, we have
$$
3 f(0)=2 f(0), f(0)=a_{0}=0 \text {; }
$$

When $a=x, b=c=0$, we have
$$
f(x)+f(-x)=2 f(x) \text {. }
$$

Thus, $f(-x)=f(x)$ for all $x \in \mathbf{R}$.
Therefore, $a_{1}=a_{3}=\cdots=a_{2 k-1}=0$.
$$
\begin{array}{l} 
\text { Equation (1) } \Leftrightarrow \sum_{k=1}^{m} a_{2 k}\left[(a-b)^{2 k}+(b-c)^{2 k}+(c-a)^{2 k}\right] \\
= 2 \sum_{k=1}^{m} a_{2 k}(a+b+c)^{2 k} . \\
\text { Also, }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\
= 2\left(a^{2}+b^{2}+c^{2}\right)=2(a+b+c)^{2}, \\
(a-b)^{4}+(b-c)^{4}+(c-a)^{4} \\
= 2\left(a^{4}+b^{4}+c^{4}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)- \\
4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right) \\
= 2\left(a^{2}+b^{2}+c^{2}\right)^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+ \\
4\left(a^{2} b c+a b^{2} c+a b c^{2}\right) \\
= 2(a+b+c)^{4}+2(a b+b c+c a)^{2} \\
= 2(a+b+c)^{4} .
\end{array}
$$

Let $f(x)=g\left(x^{2}\right)$, and let $u=a-b, v=b-c$. Then
$$
u+v=a-c, a=u+v+c, b=c+v \text {. }
$$

Since $a b+b c+c a$
$$
\begin{array}{l}
=(u+v+c)(c+v)+c(c+v)+c(u+v+c) \\
=3 c^{2}+2(u+2 v) c+v(u+v)=0, \\
\Delta=4\left(u^{2}+u v+v^{2}\right) \geqslant 0,
\end{array}
$$

Therefore, for any $u, v$ there exists $c$.
Thus, $c, u, v$ determine $a, b, c$. Hence,
$$
\begin{array}{l}
f(a-b)+f(b-c)+f(c-a)=2 f(a+b+c) \\
\Leftrightarrow g\left(u^{2}\right)+g\left(v^{2}\right)+g\left(u^{2}+2 w+v^{2}\right) \\
\quad=2 g\left((a+b+c)^{2}\right)=2 g\left(a^{2}+b^{2}+c^{2}\right) \\
\quad=2 g\left(\frac{(a-b)^{2}}{2}+\frac{(b-c)^{2}}{2}+\frac{(c-a)^{2}}{2}\right) \\
\quad=2 g\left(u^{2}+w w+v^{2}\right) .
\end{array}
$$

Let $g(x)=\sum_{k=0}^{n} a_{k} x^{k}$, then
$$
\begin{array}{l}
g\left(u^{2}\right)+g\left(v^{2}\right)+g\left(u^{2}+2 w w+v^{2}\right) \\
=2 g\left(u^{2}+w w+v^{2}\right) \\
\Leftrightarrow \sum_{k=0}^{n} a_{k}\left[u^{2 k}+v^{2 k}+\left(u^{2}+2 w w+v^{2}\right)^{k}\right] \\
\quad=2 \sum_{k=0}^{n} a_{k}\left(u^{2}+u v+v^{2}\right)^{k} .
\end{array}
$$

Since $g(0)=0$, we have $a_{0}=0$.
Let $u=v$, we get $2 g\left(u^{2}\right)+g\left(4 u^{2}\right)=2 g\left(3 u^{2}\right)$.
Comparing the coefficients of $u^{2 n}$, we get $2+4^{n}=2 \times 3^{n}$.
Since when $n \geqslant 3$, $4^{n}>2 \times 3^{n}$, the integer solutions of the equation are $n=1$ or 2.
Thus, $g(x)=a_{2} x^{2}+a_{1} x$,
$$
f(x)=A x^{4}+B x^{2}(A, B \in \mathbf{R})
$$