Cycles of quadratic Latin squares and anti-perfect $1$-factorisations

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