2 Find all real-coefficient polynomials $P(x)$ such that for all real numbers $a, b, c$ satisfying $ab + bc + ca = 0$, we have
$$
P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c).
$$
Given $a \cdot b, c \in \mathbf{R}, a b+b c) c a= c 0$. We have
$$
P(a-b)+P(b-c)+P(c-a)=2 P(a+b+c) .
$$

Let $a=b=c=0$ in (1). Then $P(0)=0$.
Let $b=c=0$ in (1). Then $P(a)=P(a)$ for any real number $a$.
Thus, all the coefficients of the odd terms of $P(x)$ are 0. Without loss of generality, assume
$$
P(x)=a_{n} x^{2 n}+\cdots+a_{1} \cdot x^{2}, a_{n} \neq 0 .
$$

Let $b=2 a, c=-\frac{2}{3} a$ in (1). Then
$$
P(-a)+P\left(\frac{8}{3} a\right)+P\left(-\frac{5}{3} a\right)=2 P\left(\frac{7}{3} a\right) .
$$

That is,
$$
\begin{array}{c}
a_{n}\left[1+\left(\frac{8}{3}\right)^{2 n}+\left(\frac{5}{3}\right)^{2 n}-2\left(\frac{7}{3}\right)^{2 n}\right] a^{2 n}+\cdots+ \\
a_{1}\left[1+\left(\frac{8}{3}\right)^{2}+\left(\frac{5}{3}\right)^{2}-2\left(\frac{7}{3}\right)^{2}\right] a^{2}=0
\end{array}
$$

holds for all $a \in \mathbf{R}$, so all the coefficients of the polynomial in $a$ are 0.
When $n \geqslant 3$, by $8^{6}=262144>235298=2 \times 7^{6}$, we have
$$
\left(\frac{8}{7}\right)^{2 n} \geq\left(\frac{8}{7}\right)^{n}>2 \text {. }
$$

Thus,
$$
1+\left(\frac{8}{3}\right)^{2 n}+\left(\frac{5}{3}\right)^{2 n}-2\left(\frac{7}{3}\right)^{2 n}>0 .
$$

Therefore, $n \leqslant 2$. Assume $P(x)=\alpha x^{4}+\beta x^{2}, \alpha, \beta \in \mathbf{R}$.
Now verify that $P(x)=\alpha x^{4}+\beta x^{2}$ satisfies the requirement.
Let $a, b, c \in \mathbf{R}, a b+b c+c a=0$, then
$$
\begin{aligned}
& (a-b)^{4}+(b-c)^{4}+(c-a)^{4}-2(a+b+c)^{4} \\
= & \sum\left(a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}\right)-2\left(a^{2}+b^{2}+c^{2}\right)^{2} \\
= & \sum\left(a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}\right)- \\
& 2 a^{4}-2 b^{4}-2 c^{4}-4 a^{2} b^{2}-4 a^{2} c^{2}-4 b^{2} c^{2} \\
= & \sum\left(-4 a^{3} b+2 a^{2} b^{2}-4 a b^{3}\right) \\
= & -4 a^{2}(a b+c a)-4 b^{2}(b c+a b)-4 c^{2}(c a+b c)+ \\
= & 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \\
= & 2\left(a b+4 b^{2} c a+4 c^{2} a b+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}\right. \\
& (a-b)^{2}+(b-c)^{2}=(0, \\
= & \sum\left(a^{2}-2 a b+b^{2}\right)-2 \sum a^{2}-4 \sum a b \\
= & 0 .
\end{aligned}
$$

Therefore, $P(x)=\alpha x^{4}+\beta x^{2}$ satisfies the requirement.