IMO 2019 Shortlisted Problems

Solution 1. In the first two steps, we deal with any polynomial $P(x,y,z)$ satisfying $P(x,y,z)=P(x,y,xy-z)$. Call such a polynomial weakly symmetric, and call a polynomial satisfying the full conditions in the problem symmetric.

Step 1. We start with the description of weakly symmetric polynomials. We claim that they are exactly the polynomials in $x,y$, and $z(xy-z)$. Clearly, all such polynomials are weakly symmetric. For the converse statement, consider $P_1(x,y,z):=P\left(x,y,z+\frac{1}{2}xy\right)$, which satisfies $P_1(x,y,z)=P_1(x,y,-z)$ and is therefore a polynomial in $x,y$, and $z^2$. This means that $P$ is a polynomial in $x,y$, and ${\left(z-\frac{1}{2}xy\right)}^2=-z(xy-z)+\frac{1}{4}{x}^2{y}^2$, and therefore a polynomial in $x,y$, and $z(xy-z)$.

Step 2. Suppose that $P$ is weakly symmetric. Consider the monomials in $P(x,y,z)$ of highest total degree. Our aim is to show that in each such monomial $\mu x^a{y}^b{z}^c$ we have $a,b⩾c$. Consider the expansion $$P(x,y,z)=\sum _{i,j,k}{\mu }_{ijk}{x}^i{y}^j(z(xy-z){)}^k\text{\hspace{1em}(1.1)}$$ The maximal total degree of a summand in (1.1) is $m={\max }_{i,j,k:{\mu }_{ijk}\ne 0}(i+j+3k)$. Now, for any $i,j,k$ satisfying $i+j+3k=m$ the summand ${\mu }_{i,j,k}{x}^i{y}^j(z(xy-z){)}^k$ has leading term of the form $\mu x^{i+k}{y}^{j+k}{z}^k$. No other nonzero summand in (1.1) may have a term of this form in its expansion, hence this term does not cancel in the whole sum. Therefore, $\mathrm{deg}P=m$, and the leading component of $P$ is exactly $$\sum _{i+j+3k=m}{\mu }_{i,j,k}{x}^{i+k}{y}^{j+k}{z}^k$$ and each summand in this sum satisfies the condition claimed above.

Step 3. We now prove the problem statement by induction on $m=\mathrm{deg}P$. For $m=0$ the claim is trivial. Consider now a symmetric polynomial $P$ with $\mathrm{deg}P>0$. By Step 2, each of its monomials $\mu x^a{y}^b{z}^c$ of the highest total degree satisfies $a,b⩾c$. Applying other weak symmetries, we obtain $a,c⩾b$ and $b,c⩾a$; therefore, $P$ has a unique leading monomial of the form $\mu (xyz{)}^c$. The polynomial $P_0(x,y,z)=P(x,y,z)-\mu {\left(xyz-x^2-y^2-z^2\right)}^c$ has smaller total degree. Since $P_0$ is symmetric, it is representable as a polynomial function of $xyz-x^2-y^2-z^2$. Then $P$ is also of this form, completing the inductive step.

Solution 2. We will rely on the well-known identity $${\cos }^{2}u+{\cos }^{2}v+{\cos }^{2}w-2\cos u\cos v\cos w-1=0\text{whenever}u+v+w=0.\text{\hspace{1em}(2.1)}$$ Claim 1. The polynomial $P(x,y,z)$ is constant on the surface $$\mathfrak{S}=\{(2\cos u,2\cos v,2\cos w):u+v+w=0\}$$ Proof. Notice that for $x=2\cos u,y=2\cos v,z=2\cos w$, the Vieta jumps $x\mapsto yz-x$, $y\mapsto zx-y,z\mapsto xy-z$ in ( $\ast$ ) replace ( $u,v,w$ ) by ( $v-w,-v,w$ ), ( $u,w-u,-w$ ) and ( $-u,v,u-v$ ), respectively. For example, for the first type of jump we have $$yz-x=4\cos v\cos w-2\cos u=2\cos (v+w)+2\cos (v-w)-2\cos u=2\cos (v-w).$$ Define $G(u,v,w)=P(2\cos u,2\cos v,2\cos w)$. For $u+v+w=0$, the jumps give $$\begin{array}{rl}G(u,v,w)& =G(v-w,-v,w)=G(w-v,-v,(v-w)-(-v))=G(-u-2v,-v,2v-w)\\ & =G(u+2v,v,w-2v)\end{array}$$ By induction, $$G(u,v,w)=G(u+2kv,v,w-2kv)(k\in \mathbb{Z}).\text{\hspace{1em}(2.2)}$$ Similarly, $$G(u,v,w)=G(u,v-2\ell u,w+2\ell u)(\ell \in \mathbb{Z}).\text{\hspace{1em}(2.3)}$$ And, of course, we have $$G(u,v,w)=G(u+2p\pi ,v+2q\pi ,w-2(p+q)\pi )(p,q\in \mathbb{Z})\text{\hspace{1em}(2.4)}$$ Take two nonzero real numbers $u,v$ such that $u,v$ and $\pi$ are linearly independent over $\mathbb{Q}$. By combining (2.2-2.4), we can see that $G$ is constant on a dense subset of the plane $u+v+w=0$. By continuity, $G$ is constant on the entire plane and therefore $P$ is constant on $\mathfrak{S}$.

Claim 2. The polynomial $T(x,y,z)=x^2+y^2+z^2-xyz-4$ divides $P(x,y,z)-P(2,2,2)$. Proof. By dividing $P$ by $T$ with remainders, there exist some polynomials $R(x,y,z),A(y,z)$ and $B(y,z)$ such that $$P(x,y,z)-P(2,2,2)=T(x,y,z)\cdot R(x,y,z)+A(y,z)x+B(y,z).\text{\hspace{1em}(2.5)}$$ On the surface $\mathfrak{S}$ the LHS of (2.5) is zero by Claim 1 (since $(2,2,2)\in \mathfrak{S}$ ) and $T=0$ by (2.1). Hence, $A(y,z)x+B(y,z)$ vanishes on $\mathfrak{S}$. Notice that for every $y=2\cos v$ and $z=2\cos w$ with $\frac{\pi }{3}<v,w<\frac{2\pi }{3}$, there are two distinct values of $x$ such that $(x,y,z)\in \mathfrak{S}$, namely $x_1=2\cos (v+w)$ (which is negative), and $x_2=2\cos (v-w)$ (which is positive). This can happen only if $A(y,z)=B(y,z)=0$. Hence, $A(y,z)=B(y,z)=0$ for $|y|<1,|z|<1$. The polynomials $A$ and $B$ vanish on an open set, so $A$ and $B$ are both the zero polynomial. The quotient $(P(x,y,z)-P(2,2,2))/T(x,y,z)$ is a polynomial of lower degree than $P$ and it also satisfies (). The problem statement can now be proven by induction on the degree of $P$.

Solution 3 (using algebraic geometry, just for interest). Let $Q=x^2+y^2+z^2-xyz$ and let $t\in \mathbb{C}$. Checking where $Q-t,\frac{\partial Q}{\partial x},\frac{\partial Q}{\partial y}$ and $\frac{\partial Q}{\partial z}$ vanish simultaneously, we find that the surface $Q=t$ is smooth except for the cases $t=0$, when the only singular point is $(0,0,0)$, and $t=4$, when the four points $(\pm 2,\pm 2,\pm 2)$ that satisfy $xyz=8$ are the only singular points. The singular points are the fixed points of the group $\Gamma$ of polynomial automorphisms of ${\mathbb{C}}^3$ generated by the three Vieta involutions $${\iota }_1:(x,y,z)\mapsto (x,y,xy-z),{\iota }_2:(x,y,z)\mapsto (x,xz-y,z),{\iota }_3:(x,y,z)\mapsto (yz-x,y,z).$$ $\Gamma$ acts on each surface ${\mathcal{V}}_t:Q-t=0$. If $Q-t$ were reducible then the surface $Q=t$ would contain a curve of singular points. Therefore $Q-t$ is irreducible in $\mathbb{C}[x,y,z]$. (One can also prove algebraically that $Q-t$ is irreducible, for example by checking that its discriminant as a quadratic polynomial in $x$ is not a square in $\mathbb{C}[y,z]$, and likewise for the other two variables.) In the following solution we will only use the algebraic surface ${\mathcal{V}}_0$. Let $U$ be the $\Gamma$-orbit of $(3,3,3)$. Consider ${\iota }_3\circ {\iota }_2$, which leaves $z$ invariant. For each fixed value of $z$, ${\iota }_3\circ {\iota }_2$ acts linearly on $(x,y)$ by the matrix $$M_z:=\left(\begin{array}{cc}{z}^2-1& -z\\ z& -1\end{array}\right)$$ The reverse composition ${\iota }_2\circ {\iota }_3$ acts by $M_z^{-1}=M_z^{\text{adj}}$. Note $\det M_z=1$ and $\mathrm{tr}{M}_z=z^2-2$. When $z$ does not lie in the real interval $[-2,2]$, the eigenvalues of $M_z$ do not have absolute value 1, so every orbit of the group generated by $M_z$ on ${\mathbb{C}}^2\setminus \{(0,0)\}$ is unbounded. For example, fixing $z=3$ we find $(3F_{2k+1},3F_{2k-1},3)\in U$ for every $k\in \mathbb{Z}$, where $(F_n{)}_{n\in \mathbb{Z}}$ is the Fibonacci sequence with $F_0=0,F_1=1$. Now we may start at any point $(3F_{2k+1},3F_{2k-1},3)$ and iteratively apply ${\iota }_1\circ {\iota }_2$ to generate another infinite sequence of distinct points of $U$, Zariski dense in the hyperbola cut out of ${\mathcal{V}}_0$ by the plane $x-3F_{2k+1}=0$. (The plane $x=a$ cuts out an irreducible conic when $a\notin \{-2,0,2\}$.) Thus the Zariski closure $\bar{U}$ of $U$ contains infinitely many distinct algebraic curves in ${\mathcal{V}}_0$. Since ${\mathcal{V}}_0$ is an irreducible surface this implies that $\bar{U}={\mathcal{V}}_0$. For any polynomial $P$ satisfying (), we have $P-P(3,3,3)=0$ at each point of $U$. Since $\bar{U}={\mathcal{V}}_0$, $P-P(3,3,3)$ vanishes on ${\mathcal{V}}_0$. Then Hilbert's Nullstellensatz and the irreducibility of $Q$ imply that $P-P(3,3,3)$ is divisible by $Q$. Now $(P-P(3,3,3))/Q$ is a polynomial also satisfying (*), so we may complete the proof by an induction on the total degree, as in the other solutions.