For what values of $m$ is the polynomial

$$
x^{m}+y^{m}+z^{m}-(x+y+z)^{m}
$$

divisible by the polynomial $(y+z)(z+x)(x+y)$?
I. solution. Let $a=x+y, b=z+x, c=y+z$. Then $x+y+z=\frac{a+b+c}{2}, x=\frac{-a+b+c}{2}, y=\frac{a-b+c}{2}$, $z=\frac{a+b-c}{2}$, and the problem's question modifies as follows: for what values of $m$ is the polynomial

$$
p(a, b, c)=\left(\frac{-a+b+c}{2}\right)^{m}+\left(\frac{a-b+c}{2}\right)^{m}+\left(\frac{a+b-c}{2}\right)^{m}-\left(\frac{a+b+c}{2}\right)^{m}
$$

divisible by $a b c$.

By expanding the parentheses, according to the polynomial theorem, we get

$$
\begin{aligned}
p(a, b, c)=\frac{1}{2^{m}}\left[\sum_{p+q+r=m} \frac{m!}{p!q!r!}(-a)^{p} b^{q} c^{r}+\right. & \sum_{p+q+r=m} \frac{m!}{p!q!r!} a^{p}(-b)^{q} c^{r}+ \\
& \left.+\sum_{p+q+r=m} \frac{m!}{p!q!r!} a^{p} b^{q}(-c)^{r}-\sum_{p+q+r=m p!q!r!} \frac{m!}{p!q!r!} a^{p} b^{q} c^{r}\right]
\end{aligned}
$$

In this polynomial, for divisibility by $abc$, it is sufficient to consider the terms where one of the exponents $p, q, r$ is 0. Let's look at the sum of the terms where $r=0$:

$$
\begin{gathered}
\frac{1}{2^{m}}\left[\sum_{p+q=m} \frac{m!}{p!q!}(-a)^{p} b^{q}+\sum_{p+q=m} \frac{m!}{p!q!} a^{p}(-b)^{q}+\sum_{p+q=m} \frac{m!}{p!q!} a^{p} b^{q}-\sum_{p+q=m} \frac{m!}{p!q!} a^{p} b^{q}\right]= \\
=\frac{1}{2^{m}}\left[\sum_{p+q=m} \frac{m!}{p!q!}(-a)^{p} b^{q}+\sum_{p+q=m} \frac{m!}{p!q!} a^{p}(-b)^{q}\right]
\end{gathered}
$$

By the binomial theorem, this is exactly

$$
\frac{1}{2^{m}}\left[(b-a)^{m}+(a-b)^{m}\right]
$$

A similar statement can be made for the cases where $p=0$ and $q=0$. Thus, the sum of the terms where one of $p, q, r$ (at least one and at most two) is zero:

$$
\begin{array}{ll}
\frac{1}{2^{m}}\left[(b-a)^{m}+(a-b)^{m}+(c-a)^{m}+(a-c)^{m}+(c-b)^{m}+(b-c)^{m}\right]- & \\
& -\frac{1}{2^{m}}\left[\left(a^{m}+b^{m}+c^{m}\right)-\left(a^{m}+b^{m}+c^{m}\right)\right]
\end{array}
$$

In summary: the polynomial $x^{m}+y^{m}+z^{m}-(z+y+z)^{m}$ is divisible by $(x+y)(y+z)(z+x)$ if and only if the polynomial

$$
(b-a)^{m}+(a-b)^{m}+(b-c)^{m}+(c-b)^{m}+(a-c)^{m}+(c-a)^{m}
$$

is divisible by $a b c$.

This is ultimately true if and only if $m$ is odd.

Remark. Most solvers used symmetric polynomials.

II. solution. We use the fact that in the ring of polynomials in three variables, a version of the fundamental theorem of arithmetic holds. As a consequence, a polynomial is divisible by $(y+z)(z+x)(x+y)$ if and only if it is divisible by any of the factors $(y+z)$, $(z+x)$, $(x+y)$. These factors are pairwise relatively prime, meaning they have no common divisors other than non-zero constant polynomials. The expression is symmetric in $x, y, z$, so it suffices to check divisibility by $(x+y)$. Since

$$
x^{m}+y^{m}+z^{m}-(x+y+z)^{m}=x^{m}+y^{m}-\sum_{i=1}^{m}\binom{m}{i}(x+y)^{i} z^{m-i}
$$

the question simplifies to the divisibility of $x^{m}+y^{m}$ by $(x+y)$. Here,

$x^{m}+y^{m}=$

$$
=(x+y)\left(x^{m-1}-x^{m-2} y+x^{m-3} y^{2}-+\cdots+(-1)^{m-1} y^{m-1}\right)+\left(y^{m}-(-1)^{m-1} y^{m}\right)
$$

shows that divisibility follows if $y^{m}-(-1)^{m-1} y^{m}=\left(1+(-1)^{m}\right) y^{m}$ is divisible by $(x+y)$, which is true if $m$ is odd.