One Fox coloring prime does not classify knots

By [F1], the trefoil system over ${\mathbb{F}}_5$ consists of the crossing equations in the Counterexample. Subtract the second equation $2b-c-a=0$ from the first $2a-b-c=0$. This gives $3(a-b)=0$; since $3^{-1}=2$ in ${\mathbb{F}}_5$, $a=b$. Substitution into the first equation then gives $c=a$. Conversely, every constant triple solves all three equations, so there are exactly five solutions, all constant.

The unknot diagram has one arc and no crossings, so its single arc may receive any of the five residues; it has exactly five colorings, all constant. By [F2] these are the algorithmic Fox counts.

Over ${\mathbb{F}}_3$, the same trefoil equations reduce to the single condition $a+b+c=0$, giving nine solutions, while the unknot has three. By [F3] and [A1] these counts are ambient-isotopy invariants; because they differ, the two knots are not ambient-isotopic. Thus they disprove the refuted statement at $p=5$.

For completeness compute the groups in the same diagrams. The crossing-free diagram gives $G_U=⟨x\mid ⟩\cong \mathbb{Z}$. Label the trefoil meridians so its three crossing relations are $c=ab{a}^{-1}$, $a=bc{b}^{-1}$ and $b=ca{c}^{-1}$ (the last is the first crossing in the cyclic order used for its Fox system). Eliminating $c$ from the first two gives $aba=bab$; under that relation the third follows since $ca{c}^{-1}=aba{b}^{-1}{a}^{-1}=b$. Thus [F4] gives $G_T\cong ⟨a,b\mid aba=bab⟩$. The assignments $a\mapsto (12)$ and $b\mapsto (23)$ satisfy this relation, because both triple products equal $(13)$, and generate $S_3$. This defines a surjection onto a nonabelian group, so $G_T$ is nonabelian and cannot be isomorphic to $G_U$. This group computation supplements, and agrees with, the Fox-count distinction.