Non-Existence of Quintic Factorization for the Second Cuboid Polynomial $Q_{p,q}(t)$

We consider the even monic degree-$10$ second cuboid polynomial $Q_{p,q}(t)\in \mathbb{Z}[t]$ depending on coprime integers $p\mathrm{\ne }q>0$. We exclude the existence of a splitting of type $5+5$ over $\mathbb{Q}$, i.e., a factorization of $Q_{p,q}(t)$ into two irreducible quintic polynomials. Since $Q_{p,q}(t)$ is even and satisfies $Q_{p,q}(0)\mathrm{\ne }0$, any such $5+5$ splitting is necessarily symmetric, meaning that it can be written in the normal form $Q_{p,q}(t)=R_{p,q}(t)\cdot (-R_{p,q}(-t))$. After a weighted normalization reducing to a one-parameter polynomial $Q_r(u)$ with $r=p\mathrm{/}q\in {\mathbb{Q}}_{>0}$, coefficient comparison and elimination via resultants show that a $5+5$ splitting forces the existence of a rational point on an explicitly defined plane curve $F(r,a)=0$. Passing to the quotient parameters $a=ry$ and $s=r^2$ yields an affine curve $f(s,y)=0$ such that, for each fixed $s>0$, the polynomial $f(s,\cdot )$ is of degree $16$. We compute and factor the discriminant ${\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{c}}_y(f)$ and then use Sturm root counts to certify that $f(s,\cdot )$ has no real roots for every rational $s>0$ with $s\mathrm{\ne }1$. Hence $f(s,y)=0$ admits no rational solutions with $s>0$, $s\mathrm{\ne }1$, and consequently no quintic $5+5$ factorization occurs for $Q_{p,q}(t)$ when $p\mathrm{\ne }q$.