Sorry if the title is not very explicit, but I didn't find a good title for the problem. In fact, this is the problem :
Let $P \in \mathbb{R}[X]$ a polynomial which verify the following condition : for all $a,b,c$ such that $ab+ac+bc=0$, we have : $P(a-b)+ P(b-c) + P(c-a) = 2P(a+b+c)$. Considering $ax$, $bx$, $cx$ for all $x \in \mathbb{R}$, show that P is the sum of a monomial of degree $2$ and a monomial of degree $4$.
I didn't find how to find the solution. With the indication, I really don't know what to do actually, so I've tried several others method, including the one which consists of writing :
$p_0 + p_1((a-b)+(b-c)+(c-a))+...+ p_n((a-b)^n + (b-c)^n + (c-a)^n) = p_0 + p_1(a+b+c)+...+p_n(a+b+c)^n$
and try to reasoning on it to find a condition of the $p_i$ (the coefficients of $P$), but it's seems very complicated. I thought about using symmetric polynomial, but I don't see how.
Maybe we should prove that for all $p_i$, except $i=2,4$, $\sigma_2(a,b,c) / p_i$ for a good choice of $a,b,c$ which verify $ab+ac+bc=0$ but I don't succeed to do it...
Someone could help me, please (using the idea suggered by the exercise if it's possible) ? :)