Recursively differentiable quasigroups of orders 14 and 18: do they exist? (last open cases of the Couselo-González-Markov-Nechaev conjecture)

A *quasigroup* $(Q,\cdot)$ of order $q$ is a set of $q$ elements with a binary operation whose multiplication table is a Latin square (each element occurs exactly once in every row and every column). Its *recursive derivative* is the operation defined by $x*y = y\cdot(x\cdot y)$. Call $(Q,\cdot)$ *recursively differentiable* if $*$ is again a quasigroup operation (its table is also a Latin square). Couselo, González, Markov and Nechaev (1998) conjectured that a recursively differentiable quasigroup of order $q$ exists for every $q\notin\{2,6\}$ (equivalently, that a full recursive $[4,2,3]_q$ MDS code exists), and verified all orders except $q\in\{14,18,26,42\}$. Since then $q=42$ (2008) and $q=26$ (2026) have been settled affirmatively, and $q\in\{2,6\}$ are known to be impossible. Only $q=14$ and $q=18$ remain: does there exist a recursively differentiable quasigroup of order 14 (respectively 18)?

Acceptance. FULLY RESOLVES (per order, independently): EITHER an explicit Cayley table of a quasigroup of order 14 (resp. 18) with a verification script confirming (i) the table is a Latin square and (ii) its recursive derivative x*y = y·(x·y) is also a Latin square — a finite, machine-checkable certificate; OR a rigorous impossibility proof that no recursively differentiable quasigroup of that order exists. ADVANCES: a certified exhaustive search over a large autotopism/symmetry class at order 14 or 18 that provably rules out a substantial family (e.g. all quasigroups admitting a prescribed nontrivial automorphism), with re-runnable code. Numerical near-misses without an exact Latin-square certificate do NOT qualify.