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

If $n > 1$, then $P(-a, a) = 0$, hence $P(x, y)$ is divisible by $x + y$.
Let $P(x, y) = (x + y) P_1(x, y)$, then from (2) we have
$$
P_1(a + b, c)(a + b + c) + P_1(b + c, a)(a + b + c) + P_1(c + a, b)(a + b + c) = 0,
$$

thus $P_1(a + b, c) + P_1(b + c, a) + P_1(c + a, b) = 0$. $P_1$ is still homogeneous, and $P_1(1, 0) = P_1(1, 0)(1 + 0) = P(1, 0) = 1$, meaning $P_1$ also satisfies conditions (1) $\sim$ (3), and its degree is $n-1$.
If $n-1 > 1$, we can similarly construct another homogeneous polynomial $P_2$ of degree $n-2$ that satisfies conditions (1) $\sim$ (3), and
$$
P_1(x, y) = (x + y) P_2(x, y),
$$

and so on. Eventually, we will obtain a linear polynomial $P_{n-1}(x, y) = mx + ny$.
By condition (2) with $a = 1, b = c = 0$, we get $2m + n = 0$, and from (3) we get $m = 1$, hence $n = -2$, and
$$
P(x, y) = (x + y)^{n-1} P_{n-1}(x, y) = (x + y)^{n-1}(x - 2y),
$$

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