Find all polynomials $P$ with real coefficients such that for all $a, b, c \in \mathbb{R}$ satisfying $a b+b c+c a=0$, we have

$$
P(a-b)+P(b-c)+P(c-a)=2 P(a+b+c)
$$
If we take $b=c=0$, any $a$ satisfies the condition $a b+b c+c a=0$, and we have that for any real $a, P(a)+P(-a)+P(0)=2 P(a)$, thus for $a=0$ we have $P(0)=0$ and this gives that $P$ is even, and contains only terms of even degree.

Noting that $6 \cdot 3+3 \cdot(-2)+6 \cdot(-2)=0$, we have that for any real $x$, the reals $a=6 x, b=3 x$ and $c=-2 x$ satisfy the condition of the statement, and therefore for any real $x$,

$$
P(3 x)+P(5 x)+P(-8 x)=2 P(7 x) .
$$

The study of the leading coefficients gives, denoting $n=\operatorname{deg}(P)$,

$$
3^{n}+5^{n}+(-8)^{n}=2 \cdot 7^{n}
$$

We can verify that this is only possible if $n \leq 4$, that is, $P$ is of the form $P=$ $r X^{4}+s X^{2}$. Conversely, for all $a, b, c$ such that $a b+b c+c a=0$ :

$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2\left(a^{2}+b^{2}+c^{2}\right)=2(a+b+c)^{2}
$$

and

$$
(a-b)^{4}+(b-c)^{4}+(c-a)^{4}=2\left(a^{4}+b^{4}+c^{4}\right)-4\left(a^{3} b+a^{3} c+b^{3} c+b^{3} a+c^{3} a+c^{3} b\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)
$$

Noting that

$$
a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=-2\left(a^{2} b c+b^{2} a c+c^{2} a b\right)=2\left(a^{3} b+a^{3} c+b^{3} c+b^{3} a+c^{3} a+c^{3} b\right)
$$

we find

$$
(a-b)^{4}+(b-c)^{4}+(c-a)^{4}=2(a+b+c)^{4}
$$

We therefore deduce

$$
P(a-b)+P(b-c)+P(c-a)=2 P(a+b+c)
$$

The polynomial solutions are therefore those of the form $r X^{4}+s X^{2}$.