5-147 Find all polynomials in two variables that satisfy the following conditions:
(1) $P$ is homogeneous of degree $n$. That is, for all real numbers $t, x, y$, we have $P(t x, t y)=t^{n} P(x, y)$;
(2) For all real numbers $a, b, c$,
$$
P(a+b, c)+P(b+c, a)+P(c+a, b)=0 ;
$$
(3) $P(1,0)=1$.
[Solution] Let $P$ be an $n$-degree polynomial satisfying conditions (1) and (3). Since a constant polynomial cannot satisfy both (2) and (3) simultaneously, we have $n \geqslant 1$.
Substituting $a=b, c=-2a$ in (2), and using the homogeneity of $P$, we get
$$
\begin{aligned}
0 & =P(2a, -2a) + P(-a, a) + P(-a, a) \\
& =\left[(-2)^{n} + 2\right] P(-a, a),
\end{aligned}
$$

If $n>1$, then $P(-a, a)=0$, so $P(x, y)$ is divisible by $x+y$.
$$
\text{Let } P(x, y) = (x+y) P_{1}(x, y),
$$

Then from (2) we have
$$
\begin{array}{l}
P_{1}(a+b, c)(a+b+c) + P_{1}(b+c, a)(a+b+c) + \\
P_{1}(c+a, b)(a+b+c) = 0,
\end{array}
$$

Thus, $P_{1}(a+b, c) + P_{1}(b+c, a) + P_{1}(c+a, b) = 0$, and $P_{1}$ is still homogeneous, with $P_{1}(1,0) = P_{1}(1,0)(1+0) = P(1,0) = 1$,
i.e., $P_{1}$ also satisfies conditions (1) to (3), and its degree is $n-1$.
If $n-1 > 1$, we can similarly construct an $(n-2)$-degree homogeneous polynomial $P_{2}$ that satisfies conditions (1) to (3), and
$$
P_{1}(x, y) = (x+y) P_{2}(x, y),
$$

Continuing this process, we will eventually obtain a linear polynomial
$$
P_{n-1}(x, y) = mx + ny.
$$

By condition (2), substituting $a=1, b=c=0$, we get $2m + n = 0$,
and from (3) we get $m = 1$, so $n = -2$, and
$$
\begin{aligned}
P(x, y) & = (x+y)^{n-1} P_{n-1}(x, y) \\
& = (x+y)^{n-1}(x-2y),
\end{aligned}
$$

It is not difficult to verify that polynomials of this form indeed satisfy conditions (1) to (3). Thus, we have found all homogeneous polynomials that satisfy conditions (1) to (3).