China-TST-2025A

Denote the given polynomial by $F(x,y,z)$. By contradiction, assume there exist real-coefficient polynomials $f_1(x,y,z)$, $f_2(x,y,z)$, ..., $f_m(x,y,z)$ satisfying $$F(x,y,z)=\sum _{i=1}^m{f}_i(x,y,z{)}^2.(1)$$ First, all $f_i$ must have degree at most 3. If some $f_i$ had degree $d>3$, then the sum of squares of their degree $d$ homogeneous parts would be 0, implying all these parts vanish, which is a contradiction! Thus all $\mathrm{deg}(f_i)\le 3$. By replacing each $f_i(x,y,z)$ with its degree 3 homogeneous part, we may assume each $f_i(x,y,z)$ is a homogeneous real-coefficient polynomial of degree 3. Setting $x=y$ in (1) yields $$z^4(z-y{)}^2=\sum _{i=1}^m{f}_i(y,y,z{)}^2.(2)$$ Let $g_i(y,z)=f_i(y,y,z)$. Setting $y=z$ in (2) gives $0={\sum }_{i=1}^m{g}_i(y,y{)}^2$, so $g_i(y,y)=0$ for all $i$. Thus $z-y\mid g_i(y,z)$; similarly, setting $z=0$ shows $z\mid g_i(y,z)$. Therefore we can write $g_i(y,z)=z(z-y)h_i(y,z)$, where $h_i(y,z)$ is linear in $y$ and $z$. Substituting into (2) and canceling $z^2(z-y{)}^2$ gives $$z^2=\sum _{i=1}^m{h}_i(y,z{)}^2.$$ Setting $z=0$ shows $h_i(y,z)={\alpha }_{i}z$, so $f_i(y,y,z)={\alpha }_i{z}^2(z-y)$, where ${\alpha }_i\in \mathbb{R}$. In particular, the coefficients of $y^3$ and $y^{2}z$ in $f_i(y,y,z)$ are 0, and the sum of coefficients of $z^3$ and $z^{2}y$ is 0. Now fix one $f_i$ and write $f_i={\sum }_{\begin{array}{c}u,v,w\ge 0\\ u+v+w=3\end{array}}{a}_{u,v,w}{x}^u{y}^v{z}^w$. Then: $$\{\begin{array}{l}{a}_{3,0,0}+a_{2,1,0}+a_{1,2,0}+a_{0,3,0}=0;\\ a_{2,0,1}+a_{1,1,1}+a_{0,2,1}=0;\\ a_{1,0,2}+a_{0,1,2}+a_{0,0,3}=0.\end{array}(3)$$ These relations can be visualized in Figure 1, where dashed boxes enclose elements summing to 0.

Figure 1

By symmetry, we also have: $$\{\begin{array}{l}{a}_{3,0,0}+a_{2,0,1}+a_{1,0,2}+a_{0,0,3}=0;\\ a_{0,3,0}+a_{0,2,1}+a_{0,1,2}+a_{0,0,3}=0;\\ a_{1,2,0}+a_{1,1,1}+a_{1,0,2}=0;\\ a_{2,1,0}+a_{1,1,1}+a_{0,1,2}=0.\end{array}(4)$$ In the figure, this corresponds to sums of elements along edges and diagonals being 0. From (3), the sum of all $a_{u,v,w}$ is 0. Removing the corner triangles shows $a_{1,1,1}=0$. Then (4) implies: $$a_{2,1,0}+a_{0,1,2}=0,a_{2,0,1}+a_{0,2,1}=0,a_{1,2,0}+a_{1,0,2}=0.$$ Let $a=a_{2,1,0}$, $b=a_{0,2,1}$, $c=a_{1,0,2}$, then $a_{0,1,2}=-a$, $a_{2,0,1}=-b$, $a_{1,2,0}=-c$. Now, the coefficient of $x^2{y}^2{z}^2$ in $f_i(x,y,z{)}^2$ is: $$a_{1,1,1}^2+2a_{2,1,0}\cdot a_{0,1,2}+2a_{2,0,1}\cdot a_{0,2,1}+2a_{1,2,0}\cdot a_{1,0,2}=-2a^2-2b^2-2c^2\le 0,$$ while $F(x,y,z)$ has coefficient 0 for $x^2{y}^2{z}^2$. From (1), we must have $a=b=c=0$ for each $f_i$. Substituting back into (3) and (4) shows $f_i=0$, contradicting (1). Therefore, $F(x,y,z)$ cannot be expressed as a finite sum of squares. $\square$

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Alternative solution.

Denote the polynomial by $F(x,y,z)$. By contradiction, assume $F$ can be written as a sum of squares of real-coefficient polynomials. Then there exist $f_i(x,y,z),g_i(x,y,z)\in \mathbb{R}[x,y,z]$ ($i=1,\dots ,m$) with: $$F=\sum _{i=1}^m(x{f}_i+g_i{)}^2=\sum _{i=1}^m{x}^2{f}_i^2+\sum _{i=1}^m{g}_i^2+2x\sum _{i=1}^m{f}_i{g}_i,(5)$$ where each monomial in $f_i(x,y,z)$, $g_i(x,y,z)$ has even degree in $x$. Since $\mathrm{deg}(F)=6$, as in Proof 1 we may assume $f_i,g_i$ are homogeneous of degrees 2 and 3 respectively, with ${\mathrm{deg}}_x{f}_i\le 2$, ${\mathrm{deg}}_x{g}_i\le 2$.

Setting $(x,y,z)=(0,1,1)$ in (5) gives: $$\sum _{i=1}^m{g}_i^2(0,1,1)=0,$$ so $g_i(0,1,1)=0$. Thus we can write $g_i=c_i{x}^2+(y-z)d_i$ for some $c_i,d_i\in \mathbb{R}[y,z]$. Setting $y=z=1$ in (5) and comparing coefficients of $x^2$ gives: $$\sum _{i=1}^m{f}_i^2(0,1,1)=0,$$ so $f_i(0,1,1)=0$. Thus $f_i=a_i{x}^2+(y-z)b_i$ for some $b_i\in \mathbb{R}[y,z]$ and $a_i\in \mathbb{R}$. Using the identity: $$(a{x}^3+b{)}^2+(c{x}^3+d{)}^2={\left(\sqrt{a^2+c^2}{x}^3+\frac{ab+cd}{\sqrt{a^2+c^2}}\right)}^2+{\left(\frac{ad-bc}{\sqrt{a^2+c^2}}\right)}^2,$$ we may assume $a_1=1$ and $a_i=0$ for $i\ge 2$. Comparing coefficients of $x^5$ gives: $$2\sum _{i=1}^m{a}_i{c}_i=-y-z,$$ so $c_1=-\frac{y+z}{2}$. Comparing coefficients of $x^4$ gives: $$2(y-z)b_1+\sum _{i=1}^m{c}_i^2=yz.$$ Thus: $$2(y-z)b_1+\frac{1}{4}(y-z{)}^2+\sum _{i=2}^m{c}_i^2=0.$$ Setting $y=z$ shows $y-z\mid c_i$ for $i\ge 2$.

Comparing coefficients of $x^2$ gives: $$(y-z)\sum _{i=1}^m{b}_i^2+2\sum _{i=1}^m{c}_i{d}_i=0.$$ Setting $y=z$ shows $y-z\mid d_1$. Let $d_1=(y-z)(sy+tz)$ for $s,t\in \mathbb{R}$. Comparing coefficients of $x^3$ gives: $$d_1+\sum _{i=1}^m{b}_i{c}_i=0.$$ Setting $y=z$ shows $y-z\mid b_1$. Let $b_1=k(y-z)$ for $k\in \mathbb{R}$. From $F(1,1,0)=0$, we get $1+k-\frac{1}{2}+s=0$, so $s=-\frac{1}{2}-k$. From $F(1,0,1)=0$, we get $1+k-\frac{1}{2}+t=0$, so $t=-\frac{1}{2}-k$. Thus: $$x{f}_1+g_1=x(x^2+k(y-z{)}^2)-\frac{y+z}{2}{x}^2-(\frac{1}{2}+k)(y+z)(y-z{)}^2.$$

Setting $x=1,y=1$ in $F-(x{f}_1+g_1{)}^2\ge 0$ gives: $$\begin{array}{rl}0& \le z^4(z-1{)}^2-{\left(\frac{1-z}{2}+k(1-z{)}^2-\left(\frac{1}{2}+k\right)(1-z{)}^2(1+z)\right)}^2\\ & =(z-1{)}^2\left[z^4-{\left(\frac{1}{2}+k(1-z)-\left(\frac{1}{2}+k\right)(1-z^2)\right)}^2\right].\end{array}$$ The inequality $z^2-((\frac{1}{2}+k)z-k{)}^2\ge 0$ holds for all real $z$ only when $k=0$. But then setting $z=0,x=7,y=2$ gives $F(x,y,z)-(x{f}_1+g_1{)}^2=-225<0$, a contradiction. Therefore, $F(x,y,z)$ cannot be expressed as a finite sum of squares of real-coefficient polynomials. $\square$