We denote by $\mathbb{Z}[x, y, z]$ the set of polynomials with integer coefficients in the three variables $x, y$, and $z$. We then say that a polynomial $P \in \mathbb{Z}[x, y, z]$ is Olympic if $\mathbb{Z}[x, y, z]$ contains polynomials $A, B$, and $C$ such that

$$
P(x, y, z)=(x+y+z) A(x, y, z)+(x y+y z+z x) B(x, y, z)+x y z C(x, y, z).
$$

Find the largest integer $n$ for which there exist natural numbers $i, j$, and $k$ with sum $i+j+k=n$ and such that the polynomial $x^{i} y^{j} z^{k}$ is not Olympic.
Note: A polynomial with integer coefficients in the variables $x, y$, and $z$ is a function that can be written as a sum of terms of the form $\lambda x^{i} y^{j} z^{k}$, where $\lambda$ is an integer and $i, j$, and $k$ are natural numbers. For example, $x-y+1$ and $x y+y z+z x$ are such polynomials, but $\pi x y z, \exp (x), x /\left(y^{2}+1\right)$, and $\sqrt{x y+z}$ are not.
Let $s_{1}=x+y+z, s_{2}=x y+y z+z x$ and $s_{3}=x y z$. We observe that

$$
\begin{cases}x^{3} \equiv-x^{2}(y+z) \equiv-x(x y+x z) \equiv x y z \equiv 0 & \left(\bmod s_{1}, s_{2}, s_{3}\right) \\ x^{2} y^{2} \equiv x y(-x z-y z) \equiv-x y z(x+y) \equiv 0 & \left(\bmod s_{1}, s_{2}, s_{3}\right)\end{cases}
$$

which means that $x^{3}$ and $x^{2} y^{2}$ are Olympic. Moreover, the set of Olympic polynomials is stable under multiplication by any polynomial and permutation of the variables.
Therefore, consider a polynomial $x^{i} y^{j} z^{k}$, which is not Olympic. Without loss of generality, we assume that $i \geqslant j \geqslant k$. If $i \geqslant 3, j \geqslant 2$ or $k \geqslant 1$, our polynomial is a multiple of $x^{3}$, $x^{2} y^{2}$, or $x y z$, so it is Olympic. We deduce that $i \leqslant 2, j \leqslant 1$ and $k \leqslant 0$, so that $i+j+k \leqslant 3$. This already ensures that $n \leqslant 3$.
Conversely, let us consider the set $\mathcal{E}$ of Olympic polynomials whose monomials all have a total degree of 3, i.e., polynomials $x^{i} y^{j} z^{k}$ such that $i+j+k=3$. By restricting $A, B$ and $C$ to their monomials of total degrees 2, 1, and 0 respectively, each polynomial in $\mathcal{E}$ can be written in the form

$$
\begin{aligned}
P= & (x+y+z)\left(a x^{2}+b y^{2}+c z^{2}+d x y+e x z+f y z\right)+(x y+y z+z x)(g x+h y+i z)+j x y z \\
= & a x^{3}+b y^{2}+c z^{3}+(a+d+g) x^{2} y+(a+e+g) x^{2} z+(b+d+h) y^{2} x+(b+f+h) y^{2} z \\
& \quad+(c+e+i) z^{2} x+(c+f+i) z^{2} y+(d+e+f+g+h+i+j) x y z .
\end{aligned}
$$

If $x^{2} y$ is Olympic, we then have

$$
1=(a+d+g)+(b+f+h)+(c+e+i)-(a+e+g)-(b+d+h)-(c+f+i)=0,
$$

which is absurd. Thus, the polynomial $x^{2} y$ is not Olympic, and the largest integer sought is $n=3$.

**Comment from the graders** The problem was difficult, but many students made significant progress by showing that several polynomials were Olympic. However, be careful of calculation errors!
Some tried to show that $x^{2} y$ was not Olympic by doing arithmetic on polynomials in several variables. This approach was risky and, like any complicated path to the solution, led to many errors.
Here, the key was to look at the terms of degree 3 but in two variables, i.e., $x^{2} y, x^{2} z, z^{2} x, z^{2} y, y^{2} x$ and $y^{2} z$. However, be careful: some students proceeded by contradiction and, from polynomials $(A, B, C)$ that allowed obtaining $x^{2} y$, exhibited new $\left(A^{\prime}, B^{\prime}, C^{\prime}\right)$ that allowed obtaining $\pm y^{2} x$, and then concluded directly that since they reduced to the same case (up to symmetry) it was impossible. There is no a priori reason for this to be the case: there could be many triplets $(A, B, C)$ that allow obtaining that $x^{2} y$ is Olympic!
Finally, after restricting to terms of degree 3, some forgot the terms in $x^{2}, y^{2}$ and $z^{2}$ in $A$, which is a pity.