4. A4 (KOR) ${ }^{\mathrm{IMO} 2}$ Find all polynomials $P(x)$ with real coefficients that satisfy the equality  
$$ P(a-b)+P(b-c)+P(c-a)=2 P(a+b+c) $$  
for all triples $a, b, c$ of real numbers such that $a b+b c+c a=0$.
4. Let $P(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$. For every $x \in \mathbb{R}$ the triple $(a, b, c)=$ $(6 x, 3 x,-2 x)$ satisfies the condition $a b+b c+c a=0$. Then the condition on $P$ gives us $P(3 x)+P(5 x)+P(-8 x)=2 P(7 x)$ for all $x$, implying that for all $i=0,1,2, \ldots, n$ the following equality holds:  $$ \left(3^{i}+5^{i}+(-8)^{i}-2 \cdot 7^{i}\right) a_{i}=0 $$  Suppose that $a_{i} \neq 0$. Then $K(i)=3^{i}+5^{i}+(-8)^{i}-2 \cdot 7^{i}=0$. But $K(i)$ is negative for $i$ odd and positive for $i=0$ or $i \geq 6$ even. Only for $i=2$ and $i=4$ do we have $K(i)=0$. It follows that $P(x)=a_{2} x^{2}+a_{4} x^{4}$ for some real numbers $a_{2}, a_{4}$. It is easily verified that all such $P(x)$ satisfy the required condition.