2. Find all real-coefficient polynomials $P(x)$ such that for all real numbers $a, b, c$ satisfying $a b+b c+c a=0$, we have
$$
P(a-b)+P(b-c)+P(c-a)=2 P(a+b+c) .
$$
Solution: (Based on the solution by Huang Zhiyi) For any given $a, b, c \in \mathbf{R}$, satisfying $ab + bc + ca = 0$, we have
$$
P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c).
$$

In equation (1), let $a = b = c = 0$, then $P(0) = 0$.
Let $b = c = 0$, for any real number $a_1$ we have
$$
P(-a) = P(a).
$$

Therefore, all the coefficients of the odd powers of $P(x)$ are 0. Without loss of generality, assume
$$
P(x) = a_n x^{2n} + \cdots + a_1 x^2 + a_0, \quad a_n \neq 0.
$$

In equation (1), let $b = 2a, c = -\frac{2}{3}a$, then
$$
P(-a) + P\left(\frac{8}{3}a\right) + P\left(-\frac{5}{3}a\right) = 2P\left(\frac{7}{3}a\right).
$$

Thus, $a_n \left[1 + \left(\frac{8}{3}\right)^{2n} + \left(\frac{5}{3}\right)^{2n} - 2\left(\frac{7}{3}\right)^{2n}\right] a^{2n} +$
$$
\cdots + a_1 \left[1 + \left(\frac{8}{3}\right)^2 + \left(\frac{5}{3}\right)^2 - 2\left(\frac{7}{3}\right)^2\right] a^2 + a_0 = 0.
$$

The above equation holds for all $a \in \mathbf{R}$, so all the coefficients of the polynomial in $a$ are 0.
When $n \geq 3$, by $8^6 = 262144 > 235298 = 2 \times 7^6$ we know $\left(\frac{8}{7}\right)^{2n} \geq \left(\frac{8}{7}\right)^6 > 2$.
Thus, $1 + \left(\frac{8}{3}\right)^{2n} + \left(\frac{5}{3}\right)^{2n} - 2\left(\frac{7}{3}\right)^{2n} > 0$.
Therefore, $n \leq 2$.
Assume $P(x) = \alpha x^4 + \beta x^2, \alpha, \beta \in \mathbf{R}$.
We now verify that $P(x) = \alpha x^4 + \beta x^2$ satisfies the requirement.
Let $a, b, c \in \mathbf{R}, ab + bc + ca = 0$, then
$$
\begin{aligned}
& (a-b)^4 + (b-c)^4 + (c-a)^4 - 2(a+b+c)^4 \\
= & \sum \left(a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4\right) - \\
& 2\left(a^2 + b^2 + c^2\right)^2 \\
= & \sum \left(a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4\right) - \\
& 2a^4 - 2b^4 - 2c^4 - 4a^2b^2 - 4a^2c^2 - 4b^2c^2 \\
= & \sum \left(-4a^3b + 2a^2b^2 - 4ab^3\right) \\
= & -4a^2(ab + ca) - 4b^2(bc + ab) - \\
& 4c^2(ca + bc) + 2\left(a^2b^2 + b^2c^2 + c^2a^2\right) \\
= & 4a^2bc + 4b^2aa + 4c^2ab + 2a^2b^2 + 2b^2c^2 + 2c^2a^2 \\
= & 2(ab + bc + aa)^2 = 0; \\
& (a-b)^2 + (b-c)^2 + (c-a)^2 - 2(a+b+c)^2 \\
= & \sum \left(a^2 - 2ab + b^2\right) - 2 \sum a^2 - 4 \sum ab = 0.
\end{aligned}
$$

Here, $\sum$ denotes the cyclic sum.
Therefore, $P(x) = \alpha x^4 + \beta x^2$ satisfies the requirement.