TITLE: Multivariate polynomial functional equation
QUESTION [6 upvotes]: I’m having some difficulties solving the following functional equation:

Find all polynomials $P(x,y)\in\mathbb{R}[X,Y]$ for which:

$P(x,y)$ is homogeneous (so $\exists n\in\mathbb{N}, \forall x,y,t\in\mathbb{R}: P(tx,ty)=t^n\cdot P(x,y)$).
$\forall a,b,c\in\mathbb{R}: P(a+b,c)+P(b+c,a)+P(c+a,b)=0$
$P(1,0)=1$



My (useful) observations:

$P(x,0)=x^n$
$P(0,x)=-2x^n$
$P(2x,x)=0$
$P(x,x)=\frac{-1}{2}\cdot2^nx^n$
$P(-x,-x)=-P(x,x)= \frac{1}{2}\cdot2^nx^n$
$P(y,x)=-P(x,y)-(x+y)^n$


When we write
$$P(x,y) = \sum_{i=0}^na_i\cdot x^iy^{n-i}$$
These observations imply that:

$a_n=1$
$a_0=-2$
$\sum_{i=0}^na_i=\frac{-1}{2}\cdot2^n$
$\sum_{i=0}^n2^ia_i=0$

Also, the fifth observation implies that $n$ is odd.

I’ve noticed that $P(x,y)=x-2y$ satisfies the conditions, but I don’t know how to prove it’s the only solution.
Can someone please give me a hint how to proceed?

REPLY [3 votes]: From $P_n(t x, t y) = t^nP_n(x,y)$ we conclude that $P_n(x,y) = \sum_{k=0}^n a_k x^k y^{n-k}$
now considering
$$
P_n(2x,y) + 2P_n(x+y,x) = 0\Rightarrow \sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(x+y\right)^k x^{n-k}
$$
or
$$
\sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(\sum_{j=0}^k C_j^k x^j y^{k-j}\right) x^{n-k} = \sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(\sum_{j=0}^k C_j^k x^{n-(k-j)} y^{k-j}\right)=0
$$
or making $\nu=k-j$
$$
\sum_{k=0}^n 2^k a_k x^k y^{n-k}+2\sum_{k=0}^na_k\left(\sum_{\nu=k}^0 C_{k-\nu}^k x^{n-\nu} y^{\nu}\right)=0
$$
so to guarantee the polynomial identity we conclude
$$
2^{n-k}a_{n-k}+2 \sum_{j=0}^{n-k}a_{j+k}C_j^{j+k}= 0
$$
with $a_n = 1$ due to $P_n(1,0) = 1$
This is a almost triangular linear system. For $n = 3$ we have
$$
\begin{cases}
2^3a_3+2\left(a_0C_0^0+a_1C_1^1+a_2C_2^2+a_3C_3^3\right)=0\\
2^2a_2+2\left(a_1C_0^1+a_2C_1^2+a_3C_2^3\right)=0\\
2^1a_1+2\left(a_2C_0^2+a_3C_1^3\right)=0\\
2^0a_0+2a_3C_0^3 = 0
\end{cases}
$$
or
$$
\left\{
\begin{array}{rcl}
 2 a_0+2 a_1+2 a_2+10 a_3& = &0 \\
 2 a_1+8 a_2+6 a_3&=&0 \\
 2 a_1+2 a_2+6 a_3&=&0 \\
 a_0+2 a_3&=&0 \\
\end{array}
\right.
$$
and solving we have
$$
a_0 = -2, a_1 = -3, a_2 = 0, a_3 = 1
$$
or
$$
P_3(x,y) = (x-2y)(x+y)^2
$$
and for $n$
$$
P_n(x,y) = (x-2y)(x+y)^{n-1}
$$
as can be easily verified.