Computational Evidence Against Quadratic-Cubic Factorization for the Second Cuboid Quintic

Let $Q_{p,q}(t)\in \mathbb{Z}[t]$ be Sharipov’s even monic degree-$10$ second cuboid polynomial depending on coprime integers $p\mathrm{\ne }q>0$. Writing $Q_{p,q}(t)$ as a quintic in $t^2$ produces an associated monic quintic polynomial. After the weighted normalization $r=p\mathrm{/}q$ and $s=r^2$ we obtain a one-parameter family $P_s(x)\in \mathbb{Q}[x]$ such that $$Q_{p,q}(t)=q^{20}{P}_s\text{\hspace{0.17em}⁣}\left(\frac{t^2}{q^4}\right)\text{with}s={\left(\frac{p}{q}\right)}^2\mathrm{.}$$ Assuming a quadratic divisor $x^2+ax+b$ with $a,b\in \mathbb{Q}$, we reduce divisibility of $P_s(x)$ to the vanishing of an explicit remainder $$R(x)=R_1(s,a,b)x+R_0(s,a,b)\mathrm{.}$$ A key structural observation is that $R_1$ and $R_0$ are quadratic in $b$ and that, on the equation $R_1=0$, the second condition becomes linear in $b$. This yields a one-direction elimination to a plane obstruction curve $F(s,a)=0$ with $F\in \mathbb{Z}[s,a]$, without any lifting-back issues: when the linear coefficient is nonzero, the parameter $b$ is forced to be the rational value $b=C\mathrm{/}L$. We isolate the degenerate locus $L=C=0$ and show it produces only $s=\pm 1$ (hence only $s=1$ in the cuboid domain $s>0$). Let $\bar{C}\subset {\mathbb{P}}^2$ be the projective closure of $F(s,a)=0$. Using Magma we perform a height-bounded search for rational points on $\bar{C}$. With bound $H={10}^9$, the search returns $8$ rational points, whose affine part has $s\in \{-1,0,1\}$. In particular, no affine rational point with $s>0$ and $s\mathrm{\ne }1$ is found up to this bound. This provides strong computational evidence that for rational $s>0$, $s\mathrm{\ne }1$, the quintic $P_s(x)$ admits no quadratic factor over $\mathbb{Q}$ (equivalently, no $2+3$ (quadratic-cubic) factorization over $\mathbb{Q}$), and yields a conditional exclusion assuming completeness of the rational-point enumeration on $\bar{C}$.