For any natural number $n$, determine the coefficients $a_{1}, a_{2}, \ldots, a_{n}$ such that for the bivariate polynomial

$$
P(x, y)=x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\ldots+a_{n-1} x y^{n-1}+a_{n} y^{n}
$$

the following holds:

$$
P(u+v, w)+P(v+w, u)+P(w+u, v)=0
$$

for all $u, v, w$.
For $n=1,(2) \mathrm{a}$

$$
2(u+v+w)+a_{1}(u+v+w)=0
$$

represents an identity, from which, for example, with $u=v=w=1$, it follows that $a_{1}=-2$, i.e., $P(x, y)=x-2 y$. For the rest, we assume that $n>1$.

Let $a$ be any real number, and substitute the values $u=a, v=a, w=1-2 a$ into (2):

$$
P(2 a, 1-2 a)+2 P(1-a, a)=0
$$

Accordingly, for the polynomial

$$
Q(z)=P(1-z, z)=b_{n} z^{n}+b_{n-1} z^{n-1}+\ldots+b_{1} z+b_{0}
$$

we have

$$
Q(1-2 a)+2 Q(a)=0
$$

which holds for any $a$, meaning that the polynomial

$$
R(z)=Q(1-2 z)+2 Q(z)
$$

has a value of 0 for any $z=a$. This can only be if all the coefficients of the polynomial $R(z)$ are 0. If the highest degree term in $Q$ with a non-zero coefficient is $b_{k} z^{k} (0 \leqq k \leqq n)$, then the highest degree term in $R$ is $\left[(-2)^{k}+2\right] b_{k} z^{k}$, which is only equal to 0 for $k=1$. Therefore, $Q(z)=b_{1} z+b_{0}$, and $R(z)=b_{1}+3 b_{0}$, so $b_{1}=-3 b_{0}$, and

$$
Q(z)=b_{0}(1-3 z)
$$

According to (1), for any pair of numbers $(a, b)$ such that $a+b \neq 0$,

$$
P(a, b)=(a+b)^{n} P\left(\frac{a}{a+b}, \frac{b}{a+b}\right)
$$

from which, based on (3) and (4),

$$
P(a, b)=(a+b)^{n} Q\left(\frac{b}{a+b}\right)=(a+b)^{n} b_{0}\left(1-\frac{3 b}{a+b}\right)=b_{0}(a+b)^{n-1}(a-2 b)
$$

follows. If $b=-a \neq 0$, substitute the values

$$
u=\frac{a}{2}, \quad v=\frac{a}{2}, \quad w=-a
$$

into (2), thus we get

$$
P(a,-a)+2 P\left(-\frac{a}{2}, \frac{a}{2}\right)=0
$$

Here, again based on (1), $P(a,-a)=a^{n} P(1,-1), P(-a / 2, a / 2)=(-a / 2)^{n} \cdot P(1,-1)$, so

$$
\left[1+2\left(-\frac{1}{2}\right)^{n}\right] a^{n} P(1,-1)=0
$$

from which, for $n>1, a \neq 0$, it follows that $P(1,-1)=0$. Therefore, (5) is also true for any pair of numbers $(a, b)$ where $b=-a \neq 0$, and it is obviously true for the pair $(a, b)=(0,0)$. This proves that (2) can only be satisfied by a polynomial $P(x, y)$ for which (5) holds. The converse of this is easily verifiable by substitution: if the polynomial $P(x, y)$ satisfies (5), then (2) is clearly satisfied by it. According to (5), $P(1,0)=b_{0}$, so for $P(1,0)=1$, the sought polynomial is

$$
P(x, y)=(x+y)^{n-1}(x-2 y)
$$

From this, using the binomial theorem, we can easily obtain the coefficients of $P$: $a_{k}=\binom{n-1}{k}-2\binom{n-1}{k-1}, k=1,2, \ldots, n$, which is also valid for the case $n=1$ discussed separately in the introduction.

Remark. At the beginning of the solution, we used the fact that the value of the polynomial $P(z)$ can only be 0 for every $z=a$ if all the coefficients of the polynomial are 0. This is due to the factorized form of polynomials: if the polynomial $p(x)$ has roots $x_{1}, x_{2}, \ldots, x_{k}$, then there is a polynomial $q(x)$ such that

$$
p(x)=\left(x-x_{1}\right) \ldots\left(x-x_{k}\right) q(x)
$$

which follows from the fact that $p(x)-p(a)$ is divisible by $(x-a)$ for any $a$. Thus, an $n$-degree polynomial can have at most $n$ roots, so if an at most $n$-degree polynomial has 0 value at $n+1$ different points, all its coefficients are 0.

We used the two-variable version of this statement at the end of the solution, when we concluded from the fact that (5) is true for every $(a, b)$ that (5) gives the form of the sought polynomial. Generally, if the values of the polynomials $P(x, y), Q(x, y)$ are equal for every $(a, b)$, then for any $b$, we can apply the above statement to the polynomials $P(x, b), Q(x, b)$, and since the coefficients of the polynomials are the values of some new polynomials at $b$, we get that the coefficients of $P$ and $Q$ are also equal.