Cycles of quadratic Latin squares and antiperfect 1‐factorisations

A Latin square of order n $n$ is an × $n\times n$ matrix symbols, such that each symbol occurs exactly once in row and column. For odd prime power q $q$ let F ${{\mathbb{F}}}_{q}$ denote the finite field . quadratic a L [ , b ] ${\rm{ {\mathcal L} }}[a,b]$ defined by ( ) i j = + − if residue otherwise ${({\mathscr{L}}[a,b])}_{i,j}=\left\{\begin{array}{cc}i+a(j-i) & \,\text{if}\,\,j-i\,\,\text{is in}\,\,{{\mathbb{F}}}_{q},\\ i+b(j-i) \,\text{otherwise},\end{array}\right.$ for some { } ⊆ $\{a,b\}\subseteq {{\mathbb{F}}}_{q}$ $ab$ 1 $(a-1)(b-1)$ are residues Quadratic squares have previously been used to construct perfect 1-factorisations, mutually orthogonal atomic squares. We first characterise which devoid 2 $2\times 2$ subsquares. Let G $G$ be graph F} }}$ 1-factorisation If union every pair 1-factors induces Hamiltonian cycle then called perfect, there no induce antiperfect. use new examples antiperfect 1-factorisations complete graphs bipartite graphs. also demonstrate p $p$ only finitely many orders powers could or