Subfield: Algebra


Question: A polynomial $P(x, y, z)$ in three variables with real coefficients satisfies the identities

$$
P(x, y, z)=P(x, y, x y-z)=P(x, z x-y, z)=P(y z-x, y, z)
\tag{*}
$$

Prove that there exists a polynomial $F(t)$ in one variable such that

$$
P(x, y, z)=F\left(x^{2}+y^{2}+z^{2}-x y z\right)
$$

Solution:
In the first two steps, we deal with any polynomial $P(x, y, z)$ satisfying $P(x, y, z)=$ $P(x, y, x y-z)$. 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 $z(x y-z)$. Clearly, all such polynomials are weakly symmetric. For the converse statement, consider $P_{1}(x, y, z):=P\left(x, y, z+\frac{1}{2} x y\right)$, 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 $\left(z-\frac{1}{2} x y\right)^{2}=-z(x y-z)+\frac{1}{4} x^{2} y^{2}$, and therefore a polynomial in $x, y$, and $z(x y-z)$.

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 $a, b \geqslant c$. Consider the expansion

$$
P(x, y, z)=\sum_{i, j, k} \mu_{i j k} x^{i} y^{j}(z(x y-z))^{k}
\tag{1.1}
$$

The maximal total degree of a summand in (1.1) is $m=\max _{i, j, k: \mu_{i j k} \neq 0}(i+j+3 k)$. Now, for any $i, j, k$ satisfying $i+j+3 k=m$ the summand $\mu_{i, j, k} x^{i} y^{j}(z(x y-z))^{k}$ 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, $\operatorname{deg} P=m$, and the leading component of $P$ is exactly

$$
\sum_{i+j+3 k=m} \mu_{i, j, k} x^{i+k} y^{j+k} z^{k}
$$

and each summand in this sum satisfies the condition claimed above.

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

$$
\cos ^{2} u+\cos ^{2} v+\cos ^{2} w-2 \cos u \cos v \cos w-1=0 \quad \text { whenever } u+v+w=0 \text {. }
\tag{2.1}
$$

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 $x \mapsto y z-x$, $y \mapsto z x-y, z \mapsto x y-z$ in $(*)$ 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

$$
y z-x=4 \cos v \cos w-2 \cos u=2 \cos (v+w)+2 \cos (v-w)-2 \cos u=2 \cos (v-w) \text {. }
$$

Define $G(u, v, w)=P(2 \cos u, 2 \cos v, 2 \cos w)$. For $u+v+w=0$, the jumps give

$$
\begin{aligned}
G(u, v, w) & =G(v-w,-v, w)=G(w-v,-v,(v-w)-(-v))=G(-u-2 v,-v, 2 v-w) \\
& =G(u+2 v, v, w-2 v)
\end{aligned}
$$

By induction,

$$
G(u, v, w)=G(u+2 k v, v, w-2 k v) \quad(k \in \mathbb{Z}) .
\tag{2.2}
$$

Similarly,

$$
G(u, v, w)=G(u, v-2 \ell u, w+2 \ell u) \quad(\ell \in \mathbb{Z})
\tag{2.3}
$$

And, of course, we have

$$
G(u, v, w)=G(u+2 p \pi, v+2 q \pi, w-2(p+q) \pi) \quad(p, q \in \mathbb{Z}) .
\tag{2.4}
$$

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 $T(x, y, z)=x^{2}+y^{2}+z^{2}-x y z-4$ 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

$$
P(x, y, z)-P(2,2,2)=T(x, y, z) \cdot R(x, y, z)+A(y, z) x+B(y, z)
\tag{2.5}
$$

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 $P$.
Let $Q=x^{2}+y^{2}+z^{2}-x y z$ 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 $x y z=8$ 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

$$
\iota_{1}:(x, y, z) \mapsto(x, y, x y-z), \quad \iota_{2}:(x, y, z) \mapsto(x, x z-y, z), \quad \iota_{3}:(x, y, z) \mapsto(y z-x, y, z)
$$

$\Gamma$ 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 $M_{z}^{-1}=M_{z}^{a d j}$. Note $\operatorname{det} M_{z}=1$ and $\operatorname{tr} M_{z}=z^{2}-2$. 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 $\mathbb{C}^{2} \backslash\{(0,0)\}$ is unbounded. For example, fixing $z=3$ we find $\left(3 F_{2 k+1}, 3 F_{2 k-1}, 3\right) \in U$ for every $k \in \mathbb{Z}$, where $\left(F_{n}\right)_{n \in \mathbb{Z}}$ is the Fibonacci sequence with $F_{0}=0, F_{1}=1$.

Now we may start at any point $\left(3 F_{2 k+1}, 3 F_{2 k-1}, 3\right)$ 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 $x-3 F_{2 k+1}=0$. (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.

### 分析
The polynomial $x^{2}+y^{2}+z^{2}-x y z$ satisfies the condition $(*)$, so every polynomial of the form $F\left(x^{2}+y^{2}+z^{2}-x y z\right)$ does satisfy $(*)$. We will use without comment the fact that two polynomials have the same coefficients if and only if they are equal as functions.