[Algebra]
Problem: 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 $i+j+k \geqslant n$.

Solution: 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:

$$
x^{2} y=p P+q Q+r R
\tag{1}
$$

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 $a+d=0 \neq 1$.

Final answer: 4