\documentclass[12pt]{article} \usepackage{e-jc} % Please remove all other commands that change parameters such as % margins or pagesizes. % only use standard LaTeX packages % only include packages that you actually need % we recommend these ams packages \usepackage{amsthm,amsmath,amssymb} % we recommend the graphicx package for importing figures \usepackage{graphicx} % use this command to create hyperlinks (optional and recommended) \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} % use these commands for typesetting doi and arXiv references in the bibliography \newcommand{\doi}[1]{\href{http://dx.doi.org/#1}{\texttt{doi:#1}}} \newcommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{\texttt{arXiv:#1}}} % all overfull boxes must be fixed; % i.e. there must be no text protruding into the margins % declare theorem-like environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{fact}[theorem]{Fact} \newtheorem{observation}[theorem]{Observation} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{open}[theorem]{Open Problem} \newtheorem{problem}[theorem]{Problem} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{note}[theorem]{Note} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % if needed include a line break (\\) at an appropriate place in the title \title{\bf The Orthomorphism Graph $\mathcal{L}_3(q)$} % input author, affilliation, address and support information as follows; % the address should include the country, and does not have to include % the street address \author{Anthony~B. Evans\\ \small Department of Mathematics and Statistics\\[-0.8ex] \small Wright State University\\[-0.8ex] \small Dayton, Ohio, U.S.A.\\ \small\tt anthony.evans@wright.edu } % \date{\dateline{submission date}{acceptance date}\\ % \small Mathematics Subject Classifications: comma separated list of % MSC codes available from http://www.ams.org/mathscinet/freeTools.html} \date{\dateline{Mar 10, 2017}{Feb 3, 2018}\\ \small Mathematics Subject Classifications: 05B15} \begin{document} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{abstract} Orthomorphisms of groups and pairwise orthogonal orthomorphisms have been used in several constructions of combinatorial designs, in particular in the construction of mutually orthogonal sets of latin squares based on groups. In this paper we will use difference equations to construct orthomorphisms in $\mathcal{L}_3(q)$, an orthomorphism graph of $GF(q)^{+}\times GF(3)^+$, and to establish lower bounds for the number of pairwise orthogonal orthomorphisms in $\mathcal{L}_3(q)$. % keywords are optional \bigskip\noindent \textbf{Keywords:} latin squares; orthomorphisms; MOLS \end{abstract} \section{Introduction}\label{Section:Intro} For $G$ a finite abelian group, a mapping $\theta\colon G\to G$ is an \emph{orthomorphism} of $G$ if the mappings $x\mapsto \theta (x)$ and $x\mapsto \theta (x)-x$ are both bijections. Two mappings $\theta,\phi\colon G\to G$ are said to be \emph{orthogonal} if the mapping $x\mapsto \theta(x)-\phi(x)$ is a bijection. An orthomorphism $\theta$ of $G$ is \emph{normalized} if $\theta(0)=0$. If $\theta$ is an orthomorphism of $G$, then the mapping $x\mapsto \theta(x)-\theta(0)$ is a normalized orthomorphism of $G$: normalization does not affect orthogonality. We will use $Orth(G)$ to denote the set of normalized orthomorphisms of $G$ as well as the graph in which adjacency is orthogonality. Some automorphisms of $Orth(G)$, described in~\cite{Evans:1992}, are $H_{\alpha}$, $\alpha\in Aut(G)$, defined by $H_{\alpha}[\theta]=\alpha\theta\alpha^{-1}$; $T_{g}$, $g\in G$, defined by $T_{g}[\theta](x)=\theta(x+g)-\theta(g)$; and $R$ defined by $R[\theta](x)=x+\theta(-x)$. These automorphisms will prove useful in counting orthomorphisms in Section~\ref{Section:Number}. An \emph{orthomorphism graph} of $G$ is a subgraph of $Orth(G)$. For an orthomorphism graph $\mathcal{H}$ of a group $G$ one parameter is of particular interest: the \emph{clique number} of $\mathcal{H}$, written $\omega(\mathcal{H})$, is the largest possible number of pairwise adjacent vertices of $\mathcal{H}$. The reason for interest in this parameter is that from $r$ pairwise orthogonal orthomorphisms of $G$ we can construct $r+1$ mutually orthogonal latin squares (MOLS) of order $|G|$. Several constructions of large sets of $MOLS$ arise in this way: see~\cite{Abel/Colbourn/Dinitz:2007} and~\cite{Evans:2007a} in the Handbook of Combinatorial Designs~\cite{ColbournEtal:2007}. For more on orthomorphism graphs see~\cite{Evans:1992}, and for more on latin squares and MOLS see~\cite{DenesEtal:2015} and~\cite{Laywine/Mullen:1998}. Closely related to orthomorphisms is the concept of a complete mapping. For $G$ a finite abelian group, a mapping $\theta\colon G\to G$ is a \emph{complete mapping} of $G$ if the mappings $x\mapsto \theta (x)$ and $x\mapsto \theta (x)+x$ are both bijections. Note that a bijection $\theta\colon G\to G$ is a complete mapping of $G$ if and only if the mapping $x\mapsto \theta (x)+x$ is an orthomorphism of $G$, and an orthomorphism of $G$ if and only if the mapping $x\mapsto \theta (x)-x$ is a complete mapping of $G$. There is an extensive literature of constructions of complete mapping polynomials, i.e., polynomials that represent complete mappings of elementary abelian groups: see the survey~\cite{Winterhof:2014}. Complete mappings and orthomorphisms have a number of applications: see the papers in \cite{Hsu:1985} and \cite{Hsu:1987} for instance. In \cite{SchellenbergEtal:1978} a set of four MOLS of order 15 was constructed via a computer search. This construction implicitly used orthomorphisms of $\mathbb{Z}_{15}$, which turned out to have a very striking property: each orthomorphism had the following form. \[ \theta(x)= \begin{cases} ax \pmod{5} &\text{if $x\equiv 0\pmod{3}$}\\ ax+3 \pmod{5} &\text{if $x\equiv 1\pmod{3}$}\\ ax-3 \pmod{5} &\text{if $x\equiv 2\pmod{3}$} \end{cases} \] This suggests trying to construct large sets of MOLS of order $3p$, $p$ a prime, $p\ne 2,3$, using orthomorphisms of $\mathbb{Z}_{3p}$ that have a similar form. \[ \theta(x)= \begin{cases} ax \pmod{p} &\text{if $x\equiv 0\pmod{3}$}\\ ax+b \pmod{p} &\text{if $x\equiv 1\pmod{3}$}\\ ax-b \pmod{p} &\text{if $x\equiv 2\pmod{3}$} \end{cases} \] Here $b$ is fixed. In \cite{Evans:1999} all orthomorphisms of this type were determined, and it was shown that no orthogonal pair of orthomorphisms can be constructed from this class of orthomorphisms if $b=0$, and at most three if $b\ne 0$. In~\cite{Evans:2002} this class of orthomorphisms was generalized to the group $\mathbb{Z}_{3n}$, where $n>3$ is not divisible by either $2$ or $3$, constructing orthogonal pairs of orthomorphisms whenever $b\ne 0$: this was subsequently improved in~\cite{Evans:2007} to three pairwise orthogonal orthomorphisms when $n\ne 7,17$, $n$ not divisible by $2$ or $3$. A number of special cases of this latter result had already been proved: see~\cite{Abeletal:2006} for a summary of these special cases. In this paper we will generalize these results by considering the orthomorphism graph $\mathcal{L}_3(q)$ of $GF(q)^{+}\times GF(3)^{+}$. In Section~\ref{Section:L3q} we will introduce the orthomorphism graph $\mathcal{L}_3(q)$ and show that orthomorphisms and orthogonalities in this graph can be obtained by solving systems of difference equations. In Section~\ref{Section:Constr} we will use these difference equations to construct orthomorphisms in $\mathcal{L}_3(q)$, in Section~\ref{Section:Number} we will derive lower bounds for $|\mathcal{L}_3(q)|$, and in Section~\ref{Section:Bounds} we will give lower bounds for $\omega(\mathcal{L}_3(q))$. %%% \section{The orthomorphism graph $\mathcal{L}_3(q)$}\label{Section:L3q} In this section we will define the orthomorphism graph $\mathcal{L}_3(q)$, $q>3$ a prime power. We will see that the elements of $\mathcal{L}_3(q)$ and orthogonalities in $\mathcal{L}_3(q)$ can be determined by solving systems of difference equations. First note that any mapping $GF(q)\times GF(3)\to GF(q)\times GF(3)$, and hence any orthomorphism of $GF(q)^{+}\times GF(3)^{+}$, can be written as $\theta (x,y)=(\theta _{1,y}(x),\theta_{2,x}(y))$, where for each $y\in GF(3)$, $\theta _{1,y}$ is a mapping $GF(q)\to GF(q)$, and for each $x\in GF(q)$, $\theta _{2,x}$ is a mapping $GF(3)\to GF(3)$. We will call $\theta_{1,y}$, $y\in GF(3)$, the \emph{first component mappings} of $\theta$; and $\theta_{2,x}$, $x\in GF(q)$, the \emph{second component mappings} of $\theta$. The orthomorphism graph $\mathcal{L}_3(q)$ is the induced subgraph of $Orth(GF(q)^+\times GF(3)^+)$ whose orthomorphisms have first component mappings $\theta_{1,y}(x)=ax+b_y$, $a\ne 0,1$ and $b_0=0$. From the first component mappings of a bijection $GF(q)\times GF(3)\to GF(q)\times GF(3)$ the form of the corresponding second component mappings can be determined. \begin{lemma}\label{lem:2ndCompMap} Let the mapping $\theta\colon GF(q)\times GF(3)\to GF(q)\times GF(3)$ have first component mappings $\theta_{1,y}(x)=ax+b_y$, $a\ne 0$ and $b_0=0$. Then $\theta$ is a bijection if and only if each second component mapping can be written as \begin{equation}\label{eqn:2ndCompMap} \theta_{2,x}(y)=y \mathrm{def}_{\theta}(x+b_y/a)+ \mathrm{sum}_{\theta}(x+b_y/a) \end{equation}\label{eqn:2ndCompMaps} for some $\mathrm{def}_{\theta},\mathrm{sum}_{\theta}\colon GF(q)\to GF(3)$, $\mathrm{def}_{\theta}(x)\ne 0$ for all $x\in GF(q)$. \end{lemma} \begin{proof} Assume that $\theta$ is a bijection. For any given $z\in GF(q)$ the mapping $y\to \theta_{2,(z-b_y)/a}(y)$ is a bijection $GF(3)\to GF(3)$, as $\theta$ is a bijection and $\theta((z-b_y)/a,y)=(z,c)=\theta((z-b_w)/a,w)$ whenever $\theta_{2,(z-b_y)/a}(y)=\theta_{2,(z-b_w)/a}(w)=c$. Thus, using the fact that any permutation of the elements of $GF(3)$ can be represented as a nonconstant linear polynomial over $GF(3)$, \[ \theta_{2,(z-b_y)/a}(y)=yk(z)+h(z) \] for some maps $k,h\colon GF(q)\to GF(3)$, $k(z)\ne 0$ for all $z\in GF(q)$. This equation holds for all $z\in GF(q)$ and $y\in GF(3)$. Setting $x=(z-b_y)/a$, $\mathrm{def}_{\theta}(x)=k(ax)$, and $\mathrm{sum}_{\theta}(x)=h(ax)$ yields Equation~(\ref{eqn:2ndCompMap}). The converse is routine. \end{proof} We will call the function $\mathrm{def}_{\theta}$ in Lemma~\ref{lem:2ndCompMap} the \emph{defining sequence} for $\theta$ and the function $\mathrm{sum}_{\theta}$ the \emph{sum function} for $\theta$. Any element $\theta\in\mathcal{L}_3(q)$ can be specified by stating its first component mappings, its sum function $\mathrm{sum}_{\theta}$, and its defining sequence $\mathrm{def}_{\theta}$. These functions must satisfy a system of difference equations. \begin{theorem}\label{thm:IsOrtho} Let $\theta\colon GF(q)\times GF(3) \to GF(q)\times GF(3)$ be a bijection with first component mappings $\theta_{1,y}(x)=ax+b_y$, $a\ne 0,1$ and $b_0=0$, defining sequence $\mathrm{def}_{\theta}$, and sum function $\mathrm{sum}_{\theta}$, $\mathrm{sum}_{\theta}(0)=0$. Then $\theta\in\mathcal{L}_3(q)$ if and only if $\mathrm{def}_{\theta}$ and $\mathrm{sum}_{\theta}$ satisfy the following system of difference equations. \begin{equation}\label{eqn:IsOrtho} \mathrm{sum}_{\theta}(x+C_{\theta,y})-\mathrm{sum}_{\theta}(x)= y(\mathrm{discr}_{\theta}(x)-\mathrm{def}_{\theta}(x+C_{\theta,y})+1) \end{equation} for all $y\in GF(3)$ and some $\mathrm{discr}_{\theta}\colon GF(q)\to GF(3)$, $\mathrm{discr}_{\theta}(x)\ne 0$ for all $x\in GF(q)$, where \[ C_{\theta,y}=\frac{-b_y}{a(a-1)}. \] \end{theorem} \begin{proof} Assume that $\theta$ is an orthomorphism. For the mapping $(x,y)\mapsto\theta(x,y)-(x,y)$ the three images with first component $(a-1)x$ have second components $\mathrm{sum}_{\theta}(x+C_{\theta,y})+ y\mathrm{def}_{\theta}(x+C_{\theta,y})-y$, and so, as these must all be different and any permutation of the elements of $GF(3)$ can be represented as a nonconstant linear polynomial over $GF(3)$, \[ \mathrm{sum}_{\theta}(x+C_{\theta,y})+ y\mathrm{def}_{\theta}(x+C_{\theta,y})-y= y\mathrm{discr}_{\theta}(x)+c_{\theta}(x) \] for some $\mathrm{discr}_{\theta},c_{\theta}\colon GF(q)\to GF(3)$, $\mathrm{discr}_{\theta}(x)\ne 0$ for all $x\in GF(q)$. Setting $y=0$ we see that $c_{\theta}(x)=\mathrm{sum}_{\theta}(x)$. This yields the system of equations~(\ref{eqn:IsOrtho}). The converse is routine. \end{proof} We will call the function $\mathrm{discr}_{\theta}$ in Theorem~\ref{thm:IsOrtho} the \emph{discriminant function} for $\theta$ and we will call the system of equations~(\ref{eqn:IsOrtho}) in Theorem~\ref{thm:IsOrtho} the \emph{difference equations} for $\theta$. Orthogonality in $\mathcal{L}_3(q)$ depends similarly on solutions to a system of equations. \begin{theorem}\label{thm:Orthog} Let $\theta,\phi\in\mathcal{L}_3(q)$ have first component mappings $\theta_{1,y}(x)=ax+b_y$, $a\ne 0,1$ and $b_0=0$, and $\phi_{1,y}(x)=Ax+B_y$, $A\ne 0,1, a$ and $B_0=0$, respectively, sum functions $\mathrm{sum}_\theta$ and $\mathrm{sum}_\phi$, respectively, and defining sequences $\mathrm{def}_\theta$ and $\mathrm{def}_\phi$, respectively. Then $\theta$ is orthogonal to $\phi$ if and only if \begin{equation}\label{eqn:Orthog} \left(\mathrm{sum}_\theta(x+C_{\theta,\phi,y})-\mathrm{sum}_\theta(x)\right)- \left(\mathrm{sum}_\phi(x+C_{\phi,\theta,y})-\mathrm{sum}_\phi(x)\right)= \end{equation} \[ y(\mathrm{discr}_{\theta,\phi}(x)-\mathrm{def}_{\theta}(x+C_{\theta,\phi,y})+ \mathrm{def}_{\phi}(x+C_{\phi,\theta,y})) \] for all $y\in GF(3)$, and some $\mathrm{discr}_{\theta,\phi}\colon GF(q)\to GF(3)$, $\mathrm{discr}_{\theta,\phi}(x)\ne 0$ for all $x\in GF(q)$, where \[ C_{\theta,\phi,y}=\frac{aB_y-Ab_y}{a(a-A)} \quad\text{and}\quad C_{\phi,\theta,y}=\frac{Ab_y-aB_y}{A(A-a)}. \] \end{theorem} \begin{proof} Similar to the proof of Theorem~\ref{thm:IsOrtho}. \end{proof} We will call the function $\mathrm{discr}_{\theta,\phi}$ in Theorem~\ref{thm:Orthog} the \emph{discriminant function} for $\theta$ by $\phi$ and we will call the system of equations~(\ref{eqn:Orthog}) in Theorem~\ref{thm:Orthog} the \emph{adjacency difference equations} for $\theta$ and $\phi$. Note that, in Theorem~\ref{thm:Orthog}, a necessary condition for $\theta$ to be orthogonal to $\phi$ is $A\ne a$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Constructions of orthomorphisms in $\mathcal{L}_3(q)$}\label{Section:Constr} In this section we will give some constructions of orthomorphisms in $\mathcal{L}_3(q)$. Throughout this paper, for a bijection $\theta\colon GF(q)^+\times GF(3)^+\to GF(q)^+\times GF(3)^+$; $\mathrm{def}_{\theta}$ will denote the defining sequence for $\theta$, $\mathrm{def}_{\theta}(x)\ne 0$ for all $x\in GF(q)$; $\mathrm{sum}_{\theta}$ will denote the sum function for $\theta$; and, if the first component mappings for $\theta$ are $ax+b_y$, $y\in GF(3)$, then, as in Theorem~\ref{thm:IsOrtho}, $C_{\theta,y}=-b_y/(a(a-1))$. To construct orthomorphisms in $\mathcal{L}_3(q)$ we first specify first component mappings $ax+b_y$, $a\ne 0,1$ and $b_0=0$, and then solve the difference equations~(\ref{eqn:IsOrtho}) for $\mathrm{def}_{\theta}$ and $\mathrm{sum}_{\theta}$. As the difference equation~(\ref{eqn:IsOrtho}) with $y=0$ is $0=0$, we need only solve the difference equations~(\ref{eqn:IsOrtho}) for $y=1,2$. If we regard $GF(q)$ as a vector space over its prime subfield then $\{C_{\theta,y}\mid y=1,2\}$, equivalently $\{b_{y}\mid y=1,2\}$, spans a subspace of $GF(q)$ of dimension at most two: the dimension of this subspace will be called the \emph{dimension} of $\theta$. The simplest orthomorphisms in $\mathcal{L}_3(q)$ to construct are those of dimension $0$. \begin{theorem}\label{thm:dim0Orthos} If $\theta\colon GF(q)\times GF(3)\to GF(q)\times GF(3)$ is a bijection with first component mappings $ax$ for all $y\in GF(3)$, $a\ne 0,1$, then $\theta\in\mathcal{L}_3(q)$ if and only if $\mathrm{def}_{\theta}(x)=-1$ for all $x\in GF(q)$ and $\mathrm{sum}_{\theta}(0)=0$. \end{theorem} \begin{proof} The difference equation~(\ref{eqn:IsOrtho}) for $\theta$ with $y=1$ and also $y=2$ is \[ \mathrm{discr}_{\theta}(x)-\mathrm{def}_{\theta}(x)+1=0. \] As $\mathrm{discr}_{\theta}(x),\mathrm{def}_{\theta}(x)\ne 0$ for all $x\in GF(q)$, it follows that $\mathrm{discr}_{\theta}(x)=1$ for all $x\in GF(q)$ and $\mathrm{def}_{\theta}(x)=-1$ for all $x\in GF(q)$. Hence \begin{equation}\label{eqn:dim0Orthos} \theta(x,y)=(ax,-y+\mathrm{sum}_{\theta}(x)), \end{equation} $a\ne 0,1$, $\mathrm{sum}_{\theta}(0)=0$. It is routine to show that, if $\theta$ satisfies Equation~(\ref{eqn:dim0Orthos}), then $\theta$ satisfies the difference equations~(\ref{eqn:IsOrtho}). \end{proof} In the following theorems we will describe more orthomorphisms in $\mathcal{L}_3(q)$ of dimension one. \begin{theorem}\label{thm:dim1Orthos} If $\theta\colon GF(q)\times GF(3)\to GF(q)\times GF(3)$ is a bijection with first component mappings $ax+b_y$, $a\ne 0,1$, $b_0=b_1=0$, $b_2\ne 0$, then $\theta\in\mathcal{L}_3(q)$ if and only if $\mathrm{def}_{\theta}(x)=-1$ for all $x\in GF(q)$, $\mathrm{sum}_{\theta}$ is constant on orbits of $x\mapsto x+C_{\theta,2}$, and $\mathrm{sum}_{\theta}(0)=0$. \end{theorem} \begin{proof} The difference equation~(\ref{eqn:IsOrtho}) for $\theta$ with $y=1$ is \[ \mathrm{discr}_{\theta}(x)-\mathrm{def}_{\theta}(x)+1=0. \] As $\mathrm{discr}_{\theta}(x),\mathrm{def}_{\theta}(x)\ne 0$ for all $x\in GF(q)$, it follows that $\mathrm{discr}_{\theta}(x)=1$ for all $x\in GF(q)$ and $\mathrm{def}_{\theta}(x)=-1$ for all $x\in GF(q)$. It follows that the difference equation~(\ref{eqn:IsOrtho}) for $\theta$ with $y=2$ is \[ \mathrm{sum}_{\theta}(x+C_{\theta,2})-\mathrm{sum}_{\theta}(x)=0. \] Hence, $\mathrm{sum}_{\theta}$ is constant on orbits of $x\mapsto x+C_{\theta,2}$. We leave it to the reader to verify that the defining sequence, sum function, and discriminant function we have constructed satisfy the conditions of Theorem~\ref{thm:IsOrtho}. \end{proof} \begin{theorem}\label{thm:SVVb} Let $p$ be the characteristic of $GF(q)$, $p$ odd. If $\theta\colon GF(q)\times GF(3)\to GF(q)\times GF(3)$ is a bijection with first component mappings $ax+b_y$, $a\ne 0,1$, $b_0=0$, $b_2=2b_1\ne 0$, then $\theta\in\mathcal{L}_3(q)$ if and only if $\mathrm{sum}_{\theta}(0)=0$ and, for each orbit $O$ of $x\mapsto x+C_{\theta,1}$, one of the following holds. \begin{description} \item[i]\label{thmpt:SVVbi} $\sum_{x\in O}\mathrm{def}_{\theta}(x)=-p$ and \[ \mathrm{sum}_{\theta}(w+nC_{\theta,1})=\mathrm{sum}_{\theta}(w)+n+\sum_{i=1}^n\mathrm{def}_{\theta}(w+iC_{\theta,1}), \] where $w$ is a given element of $O$. \item[ii]\label{thmpt:SVVbii} $p=3$, $\mathrm{def}_{\theta}(x)=1$ for all $x\in O$, and \[ \mathrm{sum}_{\theta}(w+nC_{\theta,1})=\mathrm{sum}_{\theta}(w)+n, \] where $w$ is a given element of $O$. \item[iii]\label{thmpt:SVVbiii} $p=3$, $\mathrm{def}_{\theta}(x)=-1$ for all $x\in O$, and \[ \mathrm{sum}_{\theta}(w+nC_{\theta,1})=\mathrm{sum}_{\theta}(w)+n, \] where $w$ is a given element of $O$. \end{description} \end{theorem} \begin{proof} Adding the difference equation~(\ref{eqn:IsOrtho}) for $\theta$ with $y=1$ to the difference equation~(\ref{eqn:IsOrtho}) for $\theta$ with $y=1$ and $x$ replaced by $x+C_{\theta,1}$ yields \begin{align*} \mathrm{sum}_{\theta}(x+2C_{\theta,1})-\mathrm{sum}_{\theta}(x)&= \mathrm{discr}_{\theta}(x+C_{\theta,1})+\mathrm{discr}_{\theta}(x)\\ &-\mathrm{def}_{\theta}(x+C_{\theta,1})-\mathrm{def}_{\theta}(x+2C_{\theta,1})-1. \end{align*} Comparing this with the difference equation~(\ref{eqn:IsOrtho}) for $\theta$ with $y=2$ yields \[ \mathrm{discr}_{\theta}(x+C_{\theta,1})+\mathrm{def}_{\theta}(x+2C_{\theta,1})=\mathrm{discr}_{\theta}(x)+\mathrm{def}_{\theta}(x+C_{\theta,1}). \] Thus $\mathrm{discr}_{\theta}(x)+\mathrm{def}_{\theta}(x+C_{\theta,1})$ is constant on orbits of $x\mapsto x+C_{\theta,1}$. Let $O$ be an orbit of $x\mapsto x+C_{\theta,1}$. If $\mathrm{discr}_{\theta}(x)+\mathrm{def}_{\theta}(x+C_{\theta,1})=0$ for all $x\in O$, then \[ \mathrm{sum}_{\theta}(x+C_{\theta})-\mathrm{sum}_{\theta}(x)=\mathrm{def}_{\theta}(x+C_{\theta,1})+1, \] from which (\textbf{i}) follows. If $\mathrm{discr}_{\theta}(x)+\mathrm{def}_{\theta}(x+C_{\theta,1})=1$ for all $x\in O$, then $\mathrm{discr}_{\theta}(x)=\mathrm{def}_{\theta}(x+C_{\theta,1})=-1$ for all $x\in O$. If $\mathrm{discr}_{\theta}(x)+\mathrm{def}_{\theta}(x+C_{\theta,1})=-1$ for all $x\in O$, then $\mathrm{discr}_{\theta}(x)=\mathrm{def}_{\theta}(x+C_{\theta,1})=1$ for all $x\in O$. In either case $\mathrm{sum}_{\theta}(x+C_{\theta,1})-\mathrm{sum}_{\theta}(x)=1$, which can only occur if $p=3$. (\textbf{ii}) and (\textbf{iii}) follow. We leave it to the reader to verify that the defining sequence, sum function, and discriminant function we have constructed satisfy the conditions of Theorem~\ref{thm:IsOrtho}. \end{proof} Let us consider more generally, for $GF(q)$ of odd characteristic, the case of orthomorphisms in $\mathcal{L}_3(q)$ of dimension $1$ that have first component mappings of the form $ax+b_y$, where $b_1,b_2\ne 0$. For such an orthomorphism $b_2=m.b_1\ne 0=b_0$ for some positive integer $m$, $1\le m3$}; \] and \[ M_{(b,3b)}=M_{(b,5b)}=(q-2)3^{(q/p)-1}8^{q/7},\text{ if $p=7$}; \] If $q$ is not prime, let $b,c\in GF(q)$ span a $2$-dimensional subspace of $GF(q)$, viewed as a vector space over its prime subfield. Then the following hold. \[ M_{(b,c)}=(q-2)3^{(q/4)-1}5^{q/4},\text{ if $p=2$}; \] \[ M_{(b,c)}\ge (q-2)3^{(q/9)-1}4^{q/9},\text{ if $p=3$}; \] and \[ M_{(b,c)}\ge (q-2)3^{(q/p^2)-1}\left(\sum_{i\ge 0}\binom{p}{3i}\right)^{q/p^2},\text{ if $p>3$}. \] \end{lemma} \begin{proof} The calculation of $M_{(0,0)}$; $M_{(0,b)}$; $M_{(b,2b)}$; $M_{(b,3b)}$ for $p=7$; and $M_{(b,c)}$ is straightforward from the characterizations of orthomorphisms given in Theorems~\ref{thm:dim0Orthos},~\ref{thm:dim1Orthos},~\ref{thm:SVVb},~\ref{thm:dim1char7},~\ref{thm:2dim2},~\ref{thm:dim3d-g}, and~\ref{thm:dim3d+g}. From the actions of automorphisms, described in Lemma~\ref{lem:Mbc}, we see that $M_{(0,b)}=M_{(b,0)}=M_{(b,b)}$, $M_{(b,2b)}=M_{(b,((p-1)/2)b)}=M_{(b,(p-1)b)}$ if $p\ne 2,3$, and $M_{(b,3b)}=M_{(b,5b)}$ if $p=7$. \end{proof} \begin{theorem}\label{thm:nochar2} Let $p$ be the characteristic of $GF(q)$. The following hold. \begin{description} \item[i] If $p=2$, then \begin{multline*} |\mathcal{L}_{3}(q)|=(q-2)3^{q-1}+(q-1)(q-2)3^{q/2}+ (q-2)^{2}(q-1)3^{(q/4)-1}5^{q/4}. \end{multline*} \item[ii] If $p=3$, then \begin{align*} |\mathcal{L}_{3}(q)|\ge (q-2)3^{q-1}+(q-1)(q-2)3^{q/3}+(q-1)(q-2)3^{(q/3)-1}4^{q/3}\\ +(q-1)(q-2)(q-3)3^{(q/9)-1}4^{q/9}. \end{align*} \item[iii] If $p=5$, then \begin{align*} |\mathcal{L}_{3}(q)|\ge (q-2)3^{q-1}+(q-1)(q-2)3^{q/5}+(q-1)(q-2)3^{q/5}11^{q/5}\\ +(q-1)(q-2)(q-5)3^{(q/25)-1}11^{(q/25)}. \end{align*} \item[iv] If $p=7$, then \begin{align*} |\mathcal{L}_{3}(q)|\ge (q-2)3^{q-1}+(q-1)(q-2)3^{q/7}+(q-1)(q-2)3^{q/7}43^{q/7}\\ +2(q-1)(q-2)8^{q/7}3^{(q/7)-1} + (q-1)(q-2)(q-7)3^{(q/49)-1}43^{q/49}. \end{align*} \item[v] If $p>7$, then \begin{align*} |\mathcal{L}_{3}(q)|\ge (q-2)3^{q-1}+(q-1)(q-2)3^{q/p}+(q-1)(q-2)3^{(q/p)-1}\left(\sum_{i\ge 0}\binom{p}{3i}\right)^{q/p}\\ +(q-1)(q-2)(q-p)3^{(q/p^{2})-1}\left(\sum_{i\ge 0}\binom{p}{3i}\right)^{q/p^{2}}, \end{align*} \end{description} \end{theorem} \begin{proof} If $q$ is not prime, let $b,c\in GF(q)$ span a $2$-dimensional subspace of $GF(q)$, viewed as a vector space over its prime subfield. If $q$ is prime, let $b\in GF(q)\setminus\{0\}$ and set $M_{(b,c)}=0$. If $p=2$, then \[ |\mathcal{L}_3(q)|=M_{(0,0)}+(q-1)(M_{(0,b)}+M_{(b,0)}+M_{(b,b)})+(q-1)(q-2)M_{(b,c)}. \] If $p=3$, then \begin{align*} |\mathcal{L}_3(q)|=M_{(0,0)}+(q-1)(M_{(0,b)}+M_{(b,0)}+M_{(b,b)})+(q-1)M_{(b,2b)}\\ +(q-1)(q-2)M_{(b,c)}. \end{align*} If $p=5$, then \begin{align*} |\mathcal{L}_3(q)|=M_{(0,0)}+(q-1)(M_{(0,b)}+M_{(b,0)}+M_{(b,b)})\\ +(q-1)(M_{(b,2b)} +M_{(b,3b)}+M_{(b,4b)}) +(q-1)(q-2)M_{(b,c)}. \end{align*} If $p=7$, then \begin{align*} |\mathcal{L}_3(q)|=M_{(0,0)}+(q-1)(M_{(0,b)}+M_{(b,0)}+M_{(b,b)})\\ +(q-1)(M_{(b,2b)} +M_{(b,4b)}+M_{(b,6b)})+ (q-1)(M_{(b,3b)}+M_{(b,5b)})\\ +(q-1)(q-2)M_{(b,c)}. \end{align*} If $p>7$, then \begin{align*} |\mathcal{L}_3(q)|\ge M_{(0,0)}+(q-1)(M_{(0,b)}+M_{(b,0)}+M_{(b,b)})\\ +(q-1)(M_{(b,2b)}+M_{(b,((p-1)/2)b)}+M_{(b,(p-1)b)})+ (q-1)(q-2)M_{(b,c)}. \end{align*} The result then follows from Lemma~\ref{lem:Mbc}. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Bounds on $\omega(\mathcal{L}_3(q))$}\label{Section:Bounds} In this section we will establish lower bounds on $\omega(\mathcal{L}_3(q))$ when $q$ is not a power of $3$. Throughout this section, for $\theta,\phi\in\mathcal{L}_3(q)$, $\mathrm{discr}_{\theta,\phi}$ will denote the discriminant function for $\theta$ by $\phi$, $\mathrm{discr}_{\phi,\theta}$ will denote the discriminant function for $\phi$ by $\theta$, and, if the first component mappings for $\theta$ are $ax+b_y$, $y\in GF(3)$, and the first component mappings for $\phi$ are $Ax+B_y$, $y\in GF(3)$, then, as in Theorem~\ref{thm:Orthog}, $C_{\theta,\phi,y}=(aB_y-Ab_y)/(a(a-A))$, and $C_{\phi,\theta,y}=(Ab_y-aB_y)/(A(A-a))$. First let us note an easily established upper bound. \begin{theorem}\label{thm:UpperwL3q} $\omega(\mathcal{L}_{3}(q))\le q-2$. \end{theorem} \begin{proof} Let $\theta_1,\dots,\theta_n$ be a pairwise orthogonal set of orthomorphisms in $\mathcal{L}_3(q)$, and let the first component mappings of $\theta_i$ be $a_ix+b_{i,y}$ for $i=1,\dots,n$. Then $0,1,a_1,\dots,a_n$ are distinct elements of $GF(q)$. The result follows. \end{proof} In~\cite{Evans:2002} it was shown that $\omega(\mathbb{Z}_{3n})\ge 2$ for all $n\ge 5$, $n$ not divisible by $2$ or $3$. We will show that this result also holds for $\omega(\mathcal{L}_3(q))$ when the characteristic of $GF(q)$ is not $2$ or $3$. In the process we will show how the orthogonal pair of orthomorphisms, constructed in~\cite{Evans:2002}, can be obtained by solving the difference equations~(\ref{eqn:IsOrtho}) and adjacency difference equations~(\ref{eqn:Orthog}): this should shed light on the construction used in~\cite{Evans:2002}. \begin{theorem}\label{thm:3qboundpodd} If $q>2$ is a power of a prime greater than $3$ then $\omega(\mathcal{L}_{3}(q))\ge 2$. \end{theorem} \begin{proof} Let $p$ be the characteristic of $GF(q)$. Let $\theta,\phi\colon GF(q)\times GF(3)$ be bijections; let $\theta$ have first component mappings $\theta_{1,0}(x)=2x$, $\theta_{1,1}(x)=2x-2$, and $\theta_{1,2}(x)=2x+2$; and let $\phi$ have first component mappings $\phi_{1,0}(x)=-x$, $\phi_{1,1}(x)=-x-2$, and $\phi_{1,2}(x)=-x+2$. These are the first component mappings used in ~\cite{Evans:2002} for $\mathbb{Z}_{3n}$, as opposed to $GF(q)^+\times GF(3)^+$. Simple calculations show that $C_{\theta,1}=1=C_{\phi,1}$ and $C_{\theta,2}=-1=C_{\phi,2}$. Thus the difference equations~(\ref{eqn:IsOrtho}) for $\theta$ are identical to the difference equations~(\ref{eqn:IsOrtho}) for $\phi$. Hence, we may impose the conditions $\mathrm{sum}_{\theta}=\mathrm{sum}_{\phi}=\mathrm{sum}$, $\mathrm{def}_{\theta}=\mathrm{def}_{\phi}=\mathrm{def}$, and $\mathrm{discr}_{\theta}=\mathrm{discr}_{\phi}=\mathrm{discr}$. We will also impose the condition $\mathrm{discr}=-\mathrm{def}$: we leave it to the reader to show that this condition can be derived from the difference equations~(\ref{eqn:IsOrtho}). Then the difference equations~(\ref{eqn:IsOrtho}) for both $\theta$ and $\phi$ reduce to \begin{equation}\label{eqn:3qbounda} \mathrm{sum}(x+1)-\mathrm{sum}(x)=-\mathrm{def}(x)-\mathrm{def}(x+1)+1. \end{equation} Equation~(\ref{eqn:3qbounda}) has a solution if and only if, for each orbit $O$ of $x\mapsto x+1$, \[ \sum_{x\in O}\mathrm{def}(x)=-p. \] Now $C_{\theta,\phi,1}=-1$, $C_{\theta,\phi,2}=1$, $C_{\phi,\theta,1}=2$ and $C_{\phi,\theta,2}=-2$, and the adjacency difference equations for $\theta$ and $\phi$ reduce to \begin{align}\label{eqn:3qboundb} \mathrm{sum}(x-1)-\mathrm{sum}(x+2) \\= \mathrm{discr}_{\theta,\phi}(x)-\mathrm{def}(x-1)+\mathrm{def}(x+2) \notag\\=\mathrm{discr}_{\theta,\phi}(x+1)-\mathrm{def}(x+2)+\mathrm{def}(x-1).\notag \end{align} Equation~(\ref{eqn:3qbounda}) can be used repeatedly to compute $\mathrm{sum}(x+m)-\mathrm{sum}(x+n)$ for all integers $m$ and $n$. Doing so yields \begin{align} \mathrm{sum}(x+2)-\mathrm{sum}(x-1) = -\mathrm{discr}_{\theta,\phi}(x)+\mathrm{def}(x-1)-\mathrm{def}(x+2)\label{equa3}\\ = -\mathrm{discr}_{\theta,\phi}(x+1)+\mathrm{def}(x+2)-\mathrm{def}(x-1)\notag\\ = -\mathrm{def}(x+2)+\mathrm{def}(x+1)+\mathrm{def}(x)-\mathrm{def}(x-1).\notag \end{align} Thus \[ \mathrm{discr}_{\theta,\phi}(x)=-\mathrm{def}(x+1)-\mathrm{def}(x)-\mathrm{def}(x-1). \] Hence $\theta$ can be orthogonal to $\phi$ if and only if $\sum_{x\in O} \mathrm{def}(x)=-p$, for each orbit $O$ of $x\mapsto x+1$, i.e., $\theta,\phi\in\mathcal{L}_3(q)$; and \[ \mathrm{def}(x)+\mathrm{def}(x+1)+\mathrm{def}(x+2)\ne 0 \] for all $x\in GF(q)$. Such a $\mathrm{def}$ exists. For example set $\mathrm{def}(w),\mathrm{def}(w+1),\ldots,\mathrm{def}(w+p-1)=1,-1,1,-1,\ldots,1,-1,\epsilon$, where $\epsilon\equiv -p\pmod{3}$ and $w$ is a given element of an orbit $O$ of $x\mapsto x+1$: this is the solution that was used in~\cite{Evans:2002}. \end{proof} In~\cite{Evans:2007} the result that $\omega(\mathbb{Z}_{3n})\ge 2$ for all $n\ge 5$, $n$ not divisible by $2$ or $3$, was improved to $\omega(\mathbb{Z}_{3n})\ge 3$ for all $n\ge 5$, $n$ not divisible by $2$ or $3$, $n\ne 7,17$. This yields a better lower bound for $\omega(\mathcal{L}_3(q))$ when the characteristic of $GF(q)$ is not $2$, $3$, $7$, or $17$. \begin{theorem}\label{thm:wL3qge3} If the characteristic of $GF(q)$ is not $2$, $3$, $7$, or $17$, then \[ \omega(\mathcal{L}_3(q))\ge 3. \] \end{theorem} \begin{proof} Let $p$ be the characteristic of $GF(q)$. Let $\theta,\phi,\gamma\colon GF(q)\times GF(3)$ be bijections; let $\theta$ have first component mappings $\theta_{1,0}(x)=2x$, $\theta_{1,1}(x)=2x+2$, and $\theta_{1,2}(x)=2x-2$; let $\phi$ have first component mappings $\phi_{1,0}(x)=-x$, $\phi_{1,1}(x)=-x+2$, and $\phi_{1,2}(x)=-x-2$; let $\gamma$ have first component mappings $\gamma_{1,0}(x)=((p+1)/2)x$, $\gamma_{1,1}(x)=((p+1)/2)x+2$, and $\gamma_{1,2}(x)=((p+1)/2)x-2$. These are the first component mappings used in ~\cite{Evans:2007} for $\mathbb{Z}_{3n}$, as opposed to $GF(q)^+\times GF(3)^+$. Simple calculations show that $C_{\theta,1}=-1=C_{\phi,1}$, $C_{\theta,2}=1=C_{\phi,2}$, $C_{\gamma,1}=-8$, $C_{\gamma,2}=-8$, $C_{\theta,\phi,1}=1$, $C_{\theta,\phi,2}=-1$, $C_{\phi,\theta,1}=-2$, $C_{\phi,\theta,2}=2$, $C_{\theta,\gamma,1}=1$, $C_{\theta,\gamma,2}=-1$, $C_{\gamma,\theta,1}=4$ and $C_{\gamma,\theta,2}=-4$, $C_{\gamma,\phi,1}=4$, $C_{\gamma,\phi,2}=-4$, $C_{\phi,\gamma,1}=-2$, and $C_{\phi,\gamma,2}=2$. We will impose the conditions $\mathrm{sum}_{\theta}=\mathrm{sum}_{\phi}=\mathrm{sum}$, $\mathrm{def}_{\theta}=\mathrm{def}_{\phi}=\mathrm{def}$, $\mathrm{discr}_{\theta}=\mathrm{discr}_{\phi}=-\mathrm{def}$, and $\mathrm{discr}_{\gamma}=-\mathrm{def}_{\gamma}$. Let $\mathrm{sum}$, $\mathrm{def}$, $\mathrm{sum}_{\gamma}$, and $\mathrm{def}_{\gamma}$, restricted to $\{0,1,\dots,p-1\}$, be as defined in~\cite{Evans:2007}. Note that in~\cite{Evans:2007} $\mathrm{sum}$ was denoted $f$, $\mathrm{def}$ was denoted $g$, $\mathrm{sum}_{\gamma}$ was denoted $f'$, and $\mathrm{def}_{\gamma}$ was denoted $g'$. Let $0,w_1,\dots,w_{(q/p)-1}$ be representatives of the orbits of $x\mapsto x+1$ in $GF(q)$, and let us extend the definitions of $\mathrm{sum}$, $\mathrm{def}$, $\mathrm{sum}_{\gamma}$, and $\mathrm{def}_{\gamma}$, by setting $\mathrm{sum}(w_i+j)=\mathrm{sum}(j)$, $\mathrm{def}(w_i+j)=\mathrm{def}(j)$, $\mathrm{sum}_{\gamma}(w_i+j)=\mathrm{sum}_{\gamma}(j)$, and $\mathrm{def}_{\gamma}(w_i+j)=\mathrm{def}_{\gamma}(j)$ for $i=1,\dots,(q/p)-1$ and $j=0,1,\dots,p-1$. These functions solve the difference equations~(\ref{eqn:IsOrtho}) for $\theta$, $\phi$, and $\gamma$. As in the proof of Theorem~\ref{thm:3qboundpodd}, $\mathrm{discr}_{\theta,\phi}$, $\mathrm{discr}_{\theta,\gamma}$, and $\mathrm{discr}_{\gamma,\phi}$ can be determined from the adjacency difference equations~(\ref{eqn:Orthog}). These functions satisfy the adjacency difference equations~(\ref{eqn:Orthog}) for $\theta$ and $\phi$, for $\theta$ and $\gamma$, and for $\gamma$ and $\phi$. Hence, $\theta$, $\phi$, and $\gamma$ are pairwise orthogonal orthomorphisms in $\mathcal{L}_3(q)$. \end{proof} Similarly, when the characteristic of $GF(q)$ is $2$, we can find solutions to the difference equations~(\ref{eqn:IsOrtho}) and the adjacency difference equations~(\ref{eqn:Orthog}) that yield orthogonal pairs of orthomorphisms in $\mathcal{L}_3(q)$. \begin{theorem}\label{thm:3qboundp2} If $q>2$ is even, then $\omega(\mathcal{L}_{3}(q))\ge 2$. \end{theorem} \begin{proof} There are two cases to consider, $q$ an even power of $2$, and $q$ an odd power of $2$. \vspace{12 pt} \noindent\textbf{Case 1.} Let $q$ an even power of $2$, let $a$ be a solution to $x^{2}+x+1=0$ in $GF(q)$, and let $A=a+1$. Let $\theta,\phi\colon GF(q)\times GF(3)$ be bijections; let $\theta$ have first component mappings $\theta_{1,0}(x)=ax$, $\theta_{1,1}(x)=ax+1$, and $\theta_{1,2}(x)=ax$;and let $\phi$ have first component mappings $\phi_{1,0}(x)=Ax$, $\phi_{1,1}(x)=Ax$, and $\phi_{1,2}(x)=Ax+1$. Simple calculations show that $C_{\theta,1}=1=C_{\phi,2}$, $C_{\theta,2}=0=C_{\phi,1}$, $C_{\theta,\phi,1}=a^2+1$, $C_{\theta,\phi,2}=1$, $C_{\phi,\theta,1}=1$, and $C_{\phi,\theta,2}=a^2$. By Theorem~\ref{thm:dim1Orthos}, $\mathrm{def}_{\phi}(x)=-1$ for all $x\in GF(q)$, $\mathrm{discr}_{\phi}(x)=1$ for all $x\in GF(q)$, and $\mathrm{sum}_{\phi}$ is constant on orbits of $x\mapsto x+1$. A similarly proof to that of Theorem~\ref{thm:dim1Orthos} shows that $\mathrm{def}_{\theta}(x)=-1$ for all $x\in GF(q)$, $\mathrm{discr}_{\theta}(x)=1$ for all $x\in GF(q)$, and $\mathrm{sum}_{\theta}$ is constant on orbits of $x\mapsto x+1$. Let us impose the condition $\mathrm{sum}_{\theta}=\mathrm{sum}_{\phi}=\mathrm{sum}$. It is clear that $\theta,\phi\in\mathcal{L}_3(q)$ if and only if $\mathrm{sum}(0)=0$, $\mathrm{sum}$ is constant on orbits of $x\mapsto x+1$, and $\mathrm{def}_{\theta}(x)=\mathrm{def}_{\phi}(x)=-1$ for all $x\in GF(q)$. With these conditions on $\theta$ and $\phi$, the adjacency difference equations~(\ref{eqn:Orthog}) are \[ \mathrm{sum}(x+a^2+1)-\mathrm{sum}(x+1)=\mathrm{discr}_{\theta,\phi}(x), \] and \[ \mathrm{sum}(x+1)-\mathrm{sum}(x+a^2)=-\mathrm{discr}_{\theta,\phi}(x). \] These two equations are equivalent as $\mathrm{sum}$ is constant on orbits of $x\mapsto x+1$. It follows from these equations that $\theta$ and $\phi$ are orthogonal if $\mathrm{sum}(x+a^2)\ne \mathrm{sum}(x)$ for all $x\in GF(q)$. Set $H=\{0,1,a^2,1+a^2\}$. $H$ is a subgroup of $GF(q)^+$. let $0=w_0,w_1,\dots,w_{(q/4)-1}$ be a system of coset representatives for $H$ in $GF(q)^+$, and set $\mathrm{sum}(w_i)=\mathrm{sum}(w_i+1)=0$ and $\mathrm{sum}(w_i+a^2)=\mathrm{sum}(w_i+a^2+1)=1$ for $i=0,\dots,(q/4)-1$. With this choice of $\mathrm{sum}$, $\theta$ and $\phi$ are orthogonal orthomorphisms in $\mathcal{L}_3(q)$. \vspace{12 pt} \noindent\textbf{Case 2.} Let $q$ be an odd power of $2$, let $a^{2}+a+1\ne 0$ and let $A=a+1$. Let $\theta,\phi,\gamma\colon GF(q)\times GF(3)$ be bijections; let $\theta$ have first component mappings $\theta_{1,0}(x)=ax$, $\theta_{1,1}(x)=ax+a^2+a$, $\theta_{1,2}(x)=ax$; and let $\phi$ have first component mappings $\phi_{1,0}(x)=Ax$, $\phi_{1,1}(x)=Ax$, $\phi_{1,2}(x)=Ax+a^{2}+a$. Simple calculations show that $C_{\theta,1}=1=C_{\phi,2}$, $C_{\theta,2}=0=C_{\phi,1}$, $C_{\theta,\phi,1}=a^2+1$, $C_{\theta,\phi,2}=a^2+a$, $C_{\phi,\theta,1}=a^2+a$, and $C_{\phi,\theta,2}=a^2$. As in the case $q$ an even power of $2$, we can show that $\mathrm{def}_{\theta}(x)=-1$ for all $x\in GF(q)$, $\mathrm{discr}_{\theta}(x)=1$ for all $x\in GF(q)$, $\mathrm{def}_{\phi}(x)=-1$ for all $x\in GF(q)$, $\mathrm{discr}_{\phi}(x)=1$ for all $x\in GF(q)$, $\mathrm{sum}_{\theta}$ is constant on orbits of $x\mapsto x+1$, and $\mathrm{sum}_{\phi}$ is constant on orbits of $x\mapsto x+1$. As in the case $q$ an even power of $2$, we will impose the condition $\mathrm{sum}_{\theta}=\mathrm{sum}_{\phi}=\mathrm{sum}$. It is clear that $\theta,\phi\in\mathcal{L}_3(q)$ if and only if $\mathrm{sum}(0)=0$, $\mathrm{sum}$ is constant on orbits of $x\mapsto x+1$, and $\mathrm{def}_{\theta}(x)=\mathrm{def}_{\phi}(x)=-1$ for all $x\in GF(q)$. With these conditions on $\theta$ and $\phi$, the adjacency difference equations~(\ref{eqn:Orthog}) are \[ \mathrm{sum}(x+a^2+1)-\mathrm{sum}(x+a^2+a)=\mathrm{discr}_{\theta,\phi}(x), \] and \[ \mathrm{sum}(x+a^2+a)-\mathrm{sum}(x+a^2)=-\mathrm{discr}_{\theta,\phi}(x). \] These two equations are equivalent as $\mathrm{sum}$ is constant on orbits of $x\mapsto x+1$. It follows from these equations that $\theta$ and $\phi$ are orthogonal if $\mathrm{sum}(x+a)\ne \mathrm{sum}(x)$ for all $x\in GF(q)$. Set $H=\{0,1,a,1+a\}$. $H$ is a subgroup of $GF(q)^+$. let $0=w_0,w_1,\dots,w_{(q/4)-1}$ be a system of coset representatives for $H$ in $GF(q)^+$, and set $\mathrm{sum}(w_i)=\mathrm{sum}(w_i+1)=0$ and $\mathrm{sum}(w_i+a)=\mathrm{sum}(w_i+a+1)=1$ for $i=0,\dots,(q/4)-1$. With this choice of $\mathrm{sum}$, $\theta$ and $\phi$ are orthogonal orthomorphisms in $\mathcal{L}_3(q)$. \end{proof} The smallest example, where $q=4$, is given in Figure~\ref{fig:2inL34}, where an element of $\mathbb{Z}_{2}\times \mathbb{Z}_{2}\times \mathbb{Z}_{3}$ is denoted $ijk$, $i,j=0,1$, and $k=0,1,2$. A pair of orthogonal orthomorphisms of $\mathbb{Z}_{2}\times \mathbb{Z}_{2}\times \mathbb{Z}_{3}$ was first found via a computer search (see~\cite{DulmageEtal:1959}): subsequent computer searches (see~\cite{BoseEtal:1960} or~\cite{JohnsonEtal:1961}) improved this to four pairwise orthogonal orthomorphisms. \begin{figure}[htbp] \begin{center} \[ \begin{array}{|c|cccc|cccc|cccc|} \hline x& 000& 010&100&110& 001&011&101&111& 002&012&102&112\\ \hline \theta(x)& 000&111&010&101& 100&012&110&002& 001&112&011&102\\ \hline \phi(x)& 000&101&110&011& 002&100&112&010& 102&001&012&111\\ \hline \end{array} \] \caption{Orthogonal orthomorphisms in $\mathcal{L}_3(4)$.} \label{fig:2inL34} \end{center} \end{figure} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{99} \bibitem{Abel/Colbourn/Dinitz:2007} R. J. R. Abel, C. J. Colbourn, and J. H.Dinitz. \newblock Mutually orthogonal latin squares (MOLS). In: Colbourn, C.J., Dinitz, J.H. 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