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$$\small \begin{array}{rl} F=&x^8+2x^6z^2-2x^5z^3-4x^5zy^2+3x^4z^4+9x^4z^2y^2+x^4y^4-2x^3z^5-16x^3z^3y^2\\\\ &-6x^3zy^4+3x^2z^6+x^2z^4y^2+20x^2z^2y^4+x^2y^6-2xz^7-4xz^5y^2-8xzy^6+z^8\\\\ &+z^6y^2+z^4y^4+y^8\\\\ \sim&(x^4+0.4271x^2z^2-1.4271xy^2z+xz^3+0.3568y^4-0.5y^2z^2-0.7454z^4)^2 +\\\\ &(-1.0704x^3z+0.5352x^2y^2+1.8684x^2z^2-3.7368xy^2z-0.3819xz^3+0.9342y^4+0.1910y^2z^2+0.6666z^4)^2 \end{array}$$

We have the following classical result \begin{theorem} $K$ is a totally real number field iff its trace form is positive definite \end{theorem} Hillar's result can easily be proven by $LDL^\top$ decomposition of the trace form in $K$ \dots \begin{theorem} If $g\in K[x,y]$ then $\tau(g,g)\in \sum\Q[x,y]^2$ \end{theorem} \begin{proof} \small{ Let $v_1,\dots, v_n$ be a basis of $K$ over $\Q$ and $B\in\Q^{n\times n}$ be the associated matrix of $\tau$. We can write $g(x,y)=\sum_i a_i(x,y)v_i$ where $a_i\in \Q[x,y]$. Denote $\vec a:=(a_1,\dots,a_n)^\top$ then $$\tau(g,g) =\vec a^\top B\vec a = (L^\top\vec a)^\top D(L^\top\vec a)=\sum d_ig_i^2$$ where $g_i(x,y)$ is the $i$-th entry of $L^\top \vec a$ and $d_i$ is the $i$-th diagonal entry of $D$. } \end{proof}