Non-Existence of Linear-Quartic 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 $p\neq q>0$. Writing $Q_{p,q}(t)$ as a quintic in $t^{2}$ produces an associated monic quintic polynomial. After the weighted normalization $r=p/q$ and $s=r^{2}$ we obtain a one-parameter family $P_s(x)\in\mathbb{Q}[x]$ such that $$Q_{p,q}(t)=q^{20}\,P_s\!\left(\frac{t^{2}}{q^{4}}\right)\qquad\text{with}\qquad s=\left(\frac{p}{q}\right)^{2}.$$ We show that for every rational $s>0$ with $s\neq 1$ the equation $P_s(x)=0$ has no rational solutions. Equivalently, $P_s$ admits no $1+4$ factorization over $\mathbb{Q}$. The proof uses an explicit quotient by the inversion involution $(s,y)\mapsto(1/s,1/y)$ and reduces the rational-root problem for $P_s$ to rational points on the fixed genus-$2$ hyperelliptic curve $$C:\quad w^2=t^5+21t^4+26t^3+10t^2+5t+1=(t+1)(t^4+20t^3+6t^2+4t+1).$$ Using Magma and Chabauty's method on the Jacobian of $C$, we compute $C(\mathbb{Q})$ exactly and conclude that the only parameter value producing a rational root is the excluded case $s=1$ (equivalently $p=q$). As a consequence, for coprime $p\neq q>0$ the polynomial $Q_{p,q}(t)$ has no rational roots (hence no linear factor over $\mathbb{Q}$, and in particular no linear factor over $\mathbb{Z}$).