A22 (17-6, UK) Determine the polynomial $P$ that satisfies the following conditions:
(a) $P$ is a homogeneous polynomial of degree $n$ in $x, y$, i.e., for all real numbers $t, x, y$, we have $P(t x, t y)=t^{n} P(x, y)$, where $n$ is a positive integer;
(b) For all real numbers $a, b, c$, we have $P(a+b, c)+P(b+c, a)+P(c+a, b) = 0$;
(c) $P(1,0)=1$.
Proof 1: From (b) and (c), we know that $P(x, y)$ is not a constant, i.e., $n \geqslant 1$.
In (b), let $a=b, c=-2a$, we get
$$
\begin{aligned}
0 & =P(2a, -2a) + P(-a, a) + P(-a, a) \\
& =\left[(-2)^{n} + 2\right] P(-a, a).
\end{aligned}
$$

When $n>1$, we have $P(-a, a)=0$, or equivalently, for any $y=-x$, $P(x, y)=0$. This means that $P(x, y)$ is divisible by $x+y$, so we can set
$$
P(x, y) = (x+y) P_{1}(x, y).
$$

It is easy to prove that $P_{1}(x, y)$ is a homogeneous function of degree $n-1$ and satisfies conditions (a) to (c). In fact:
(a)
$$
\begin{aligned}
P_{1}(tx, ty) & = \frac{P(tx, ty)}{t(x+y)} \\
& = \frac{t^{n-1} P(x, y)}{x+y} \\
& = t^{n-1} P_{1}(x, y);
\end{aligned}
$$
(b)
$$
\text{b) } \begin{aligned}
& P_{1}(a+b, c) + P_{1}(b+c, a) + P_{1}(c+a, b) \\
= & \frac{1}{a+b+c}[P(a+b, c) + P(b+c, a) + P(c+a, b)] \\
= & 0.
\end{aligned}
$$

Note: When $a+b+c=0$, the continuity of the polynomial ensures this.
(c) $P_{1}(1,0) = P(1,0) = 1$.

Continuing this process, we have
$$
\begin{aligned}
P(x, y) & = (x+y) P_{1}(x, y) \\
& = (x+y)^{2} P_{2}(x, y) \\
& = \cdots \cdots \\
& = (x+y)^{n-1} P_{n-1}(x, y).
\end{aligned}
$$

Here, $P_{n-1}(x, y) = Ax + By$ is a linear homogeneous polynomial, still satisfying (a) to (c).
Let $a=b=c=x$, we get $3 P_{n-1}(2x, x) = 0$ or $2Ax + Bx = 0$. By the arbitrariness of $x$, we know $2A + B = 0$. Also, by $P_{n-1}(1,0) = 1$, we know $A = 1$, so $B = -2$, thus $P_{n-1}(x, y) = x - 2y$.
Finally, we get $P(x, y) = (x+y)^{n-1}(x-2y)$. Including the case $n=1$.
Proof 2: Let $a=b=c=x$, then we have
$P(2x, x) = 0$. That is, when $x = 2y$, the value of the polynomial is zero. Hence
$$
P(x, y) = (x-2y) Q_{n-1}(x, y),
$$

where $Q_{n-1}(x, y)$ is a homogeneous polynomial of degree $n-1$.
Let $a=b=x, c=2y$, then we get
$$
P(2x, 2y) = -2 P(x+2y, x).
$$

By homogeneity, we get
$$
2^{n-1} P(x, y) = -P(x+2y, x).
$$

Substituting (1) into (2), we get
$$
\begin{aligned}
& 2^{n-1}(x-2y) Q_{n-1}(x, y) \\
= & -(2y-x) Q_{n-1}(x+2y, x),
\end{aligned}
$$

i.e.,
$$
2^{n-1} Q_{n-1}(x, y) = Q_{n-1}(x+2y, x).
$$

Using condition (3), we get
$$
Q_{n-1}(1,0) = 1.
$$

Substituting $x=1, y=0$ into (3), we have
$$
Q_{n-1}(1,1) = 2^{n-1};
$$

Substituting $x=1, y=1$ into (3), we have
$$
Q_{n-1}(3,1) = 4^{n-1}.
$$

Similarly, we get:
$$
\begin{array}{c}
Q_{n-1}(5,3) = 8^{n-1}, \\
Q_{n-1}(11,5) = 16^{n-1}, \\
Q_{n-1}(21,11) = 32^{n-1}, \\
\cdots \cdots
\end{array}
$$

Therefore, there are infinitely many pairs of $x, y$ that satisfy
$$
Q_{n-1}(x, y) = (x+y)^{n-1}.
$$

Since $Q_{n-1}(x, y)$ is a polynomial, (4) is an identity.
So $P(x) = (x-2y)(x+y)^{n-1}$.