Let $\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\mathcal{B}$ denote the subset of $\mathcal{A}$ formed by all polynomials which can be expressed as 
$$ (x+y+z) P(x, y, z)+(x y+y z+z x) Q(x, y, z)+x y z R(x, y, z) $$
with $P, Q, R \in \mathcal{A}$. Find the smallest non-negative integer $n$ such that $x^{i} y^{j} z^{k} \in \mathcal{B}$ for all nonnegative integers $i, j, k$ satisfying $i+j+k \geqslant n$. (Venezuela) Answer: $n=4$.
We start by showing that $n \leqslant 4$, i.e., any monomial $f=x^{i} y^{j} z^{k}$ with $i+j+k \geqslant 4$ belongs to $\mathcal{B}$. Assume that $i \geqslant j \geqslant k$, the other cases are analogous. Let $x+y+z=p, x y+y z+z x=q$ and $x y z=r$. Then 
$$
0=(x-x)(x-y)(x-z)=x^{3}-p x^{2}+q x-r
$$
therefore $x^{3} \in \mathcal{B}$. Next, $x^{2} y^{2}=x y q-(x+y) r \in \mathcal{B}$. If $k \geqslant 1$, then $r$ divides $f$, thus $f \in \mathcal{B}$. If $k=0$ and $j \geqslant 2$, then $x^{2} y^{2}$ divides $f$, thus we have $f \in \mathcal{B}$ again. Finally, if $k=0, j \leqslant 1$, then $x^{3}$ divides $f$ and $f \in \mathcal{B}$ in this case also. In order to prove that $n \geqslant 4$, we show that the monomial $x^{2} y$ does not belong to $\mathcal{B}$. Assume the contrary:
$$
x^{2} y=p P+q Q+r R
$$
for some polynomials $P, Q, R$. If polynomial $P$ contains the monomial $x^{2}$ (with nonzero coefficient), then $p P+q Q+r R$ contains the monomial $x^{3}$ with the same nonzero coefficient. So $P$ does not contain $x^{2}, y^{2}, z^{2}$ and we may write
$$
x^{2} y=(x+y+z)(a x y+b y z+c z x)+(x y+y z+z x)(d x+e y+f z)+g x y z
$$
where $a, b, c ; d, e, f ; g$ are the coefficients of $x y, y z, z x ; x, y, z ; x y z$ in the polynomials $P$; $Q ; R$, respectively (the remaining coefficients do not affect the monomials of degree 3 in $p P+q Q+r R)$. By considering the coefficients of $x y^{2}$ we get $e=-a$, analogously $e=-b$, $f=-b, f=-c, d=-c$, thus $a=b=c$ and $f=e=d=-a$, but then the coefficient of $x^{2} y$ in the right hand side equals $a+d=0 \neq 1$.

Comment 1. The general question is the following. Call a polynomial $f\left(x_{1}, \ldots, x_{n}\right)$ with integer coefficients nice, if $f(0,0, \ldots, 0)=0$ and $f\left(x_{\pi_{1}}, \ldots, x_{\pi_{n}}\right)=f\left(x_{1}, \ldots, x_{n}\right)$ for any permutation $\pi$ of $1, \ldots, n$ (in other words, $f$ is symmetric and its constant term is zero.) Denote by $\mathcal{I}$ the set of polynomials of the form
$$
p_{1} q_{1}+p_{2} q_{2}+\ldots+p_{m} q_{m}
$$
where $m$ is an integer, $q_{1}, \ldots, q_{m}$ are polynomials with integer coefficients, and $p_{1}, \ldots, p_{m}$ are nice polynomials. Find the least $N$ for which any monomial of degree at least $N$ belongs to $\mathcal{I}$.

The answer is $n(n-1) / 2+1$. The lower bound follows from the following claim: the polynomial
$$
F\left(x_{1}, \ldots, x_{n}\right)=x_{2} x_{3}^{2} x_{4}^{3} \cdot \ldots \cdot x_{n}^{n-1}
$$
does not belong to $\mathcal{I}$. Assume that $F=\sum p_{i} q_{i}$, according to (2). By taking only the monomials of degree $n(n-1) / 2$, we can additionally assume that every $p_{i}$ and every $q_{i}$ is homogeneous, $\operatorname{deg} p_{i}>0$, and $\operatorname{deg} p_{i}+\operatorname{deg} q_{i}=$ $\operatorname{deg} F=n(n-1) / 2$ for all $i$.

Consider the alternating sum
$$
\sum_{\pi} \operatorname{sign}(\pi) F\left(x_{\pi_{1}}, \ldots, x_{\pi_{n}}\right)=\sum_{i=1}^{m} p_{i} \sum_{\pi} \operatorname{sign}(\pi) q_{i}\left(x_{\pi_{1}}, \ldots, x_{\pi_{n}}\right):=S
$$
where the summation is done over all permutations $\pi$ of $1, \ldots n$, and $\operatorname{sign}(\pi)$ denotes the sign of the permutation $\pi$. Since $\operatorname{deg} q_{i}=n(n-1) / 2-\operatorname{deg} p_{i}1$ and that the proposition is proved for smaller values of $n$.

We proceed by an internal induction on $S:=\left|\left\{i: c_{i}=0\right\}\right|$. In the base case $S=0$ the monomial $h$ is divisible by the nice polynomial $x_{1} \cdot \ldots x_{n}$, therefore $h \in \mathcal{I}$. Now assume that $S>0$ and that the claim holds for smaller values of $S$. Let $T=n-S$. We may assume that $c_{T+1}=\ldots=c_{n}=0$ and $h=x_{1} \cdot \ldots \cdot x_{T} g\left(x_{1}, \ldots, x_{n-1}\right)$, where $\operatorname{deg} g=n(n-1) / 2-T+1 \geqslant(n-1)(n-2) / 2+1$. Using the outer induction hypothesis we represent $g$ as $p_{1} q_{1}+\ldots+p_{m} q_{m}$, where $p_{i}\left(x_{1}, \ldots, x_{n-1}\right)$ are nice polynomials in $n-1$ variables. There exist nice homogeneous polynomials $P_{i}\left(x_{1}, \ldots, x_{n}\right)$ such that $P_{i}\left(x_{1}, \ldots, x_{n-1}, 0\right)=p_{i}\left(x_{1}, \ldots, x_{n-1}\right)$. In other words, $\Delta_{i}:=p_{i}\left(x_{1}, \ldots, x_{n-1}\right)-P_{i}\left(x_{1}, \ldots, x_{n-1}, x_{n}\right)$ is divisible by $x_{n}$, let $\Delta_{i}=x_{n} g_{i}$. We get
$$
h=x_{1} \cdot \ldots \cdot x_{T} \sum p_{i} q_{i}=x_{1} \cdot \ldots \cdot x_{T} \sum\left(P_{i}+x_{n} g_{i}\right) q_{i}=\left(x_{1} \cdot \ldots \cdot x_{T} x_{n}\right) \sum g_{i} q_{i}+\sum P_{i} q_{i} \in \mathcal{I}
$$
The first term belongs to $\mathcal{I}$ by the inner induction hypothesis. This completes both inductions.

Comment 2. The solutions above work smoothly for the versions of the original problem and its extensions to the case of $n$ variables, where all polynomials are assumed to have real coefficients. In the version with integer coefficients, the argument showing that $x^{2} y \notin \mathcal{B}$ can be simplified: it is not hard to show that in every polynomial $f \in \mathcal{B}$, the sum of the coefficients of $x^{2} y, x^{2} z, y^{2} x, y^{2} z, z^{2} x$ and $z^{2} y$ is even. A similar fact holds for any number of variables and also implies that $N \geqslant n(n-1) / 2+1$ in terms of the previous comment.