 Solution

Use $R={\mathbb{F}}_2[t]$. At $t=1$ the specialization is $X^5+X^2+1$, already proved irreducible and separable. Because the original polynomial is monic, this proves irreducibility over ${\mathbb{F}}_2(t)$ and gives a transitive group on five roots.

At $t=0$ the specialization factors asNeither factor has a root in ${\mathbb{F}}_2$, so they are irreducible; the factors are coprime and the specialization is separable. For example its derivative is $(X+1{)}^4$ and its value at one is one. The unramified reduction test gives a $(2)(3)$ permutation whose cube is a transposition. The same primitive prime-degree graph argument now givesNo characteristic-zero discriminant-sign argument is being transferred to characteristic two: good specialization and the explicit cycle test are sufficient.