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\title{\bf The Spectrum for\\ $3$-way $k$-homogeneous Latin Trades}%An elementary proof\\ of the reconstruction conjecture}

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\author{Trent G. Marbach\\
\small Department of Mathematics\\[-0.8ex]
\small The University of Queensland\\[-0.8ex] 
\small Brisbane 4072, Australia\\
\small\tt trent.marbach@uqconnect.edu.au\\
\and
Lijun Ji\thanks{Research supported by NSFC grant 11222113, 11431003 (L. Ji)}\\
\small Department of Mathematics\\[-0.8ex]
\small Soochow University\\[-0.8ex]
\small Suzhou 215006, China\\
\small\tt jilijun@suda.edu.cn
}

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\date{ \dateline{Mar 23, 2015}{Sep 17, 2015}{Oct 16, 2015}\\
\small Mathematics Subject Classifications: 05B15}

\begin{document}

\maketitle

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\begin{abstract}
A $\mu$-way $k$-homogeneous Latin trade was defined by Bagheri Gh, Donovan, Mahmoodian (2012), where the existence of $3$-way $k$-homogeneous Latin trades was specifically investigated. 
We investigate the existence of a certain class of $\mu$-way $k$-homogeneous Latin trades with an idempotent like property. We present a number of constructions for $\mu$-way $k$-homogeneous Latin trades with this property, and show that these can be used  to fill in the spectrum of  $3$-way $k$-homogeneous Latin trades for all but $196$ possible exceptions.

\bigskip\noindent \textbf{Keywords:} Latin square; Latin trade; $\mu$-way $k$-homogeneous Latin trade
\end{abstract}


\section{Introduction}




%define LS?

A \emph{partial Latin square of order $m$}, $T=[t(r,c)]$,  is an $m \times m$ array of cells with each cell either filled with an element $t(r,c)$ of $\Omega$ (a set of $m$ symbols) or left empty, such that each symbol of $\Omega$ appears at most once in each row, and at most once in each column. 
In what follows, we typically take $\Omega = [m]=\{1,2,\dots,m\}$ and index the rows and columns of the partial Latin square by $[m]$.
A partial Latin square $T$ has \emph{volume} $s$ if it has precisely $s$ filled cells, where $0 \leq s \leq m^2$.
A partial Latin square with volume $m^2$ is a \emph{Latin square}.
We can represent $T$ as a set of $s$ ordered triples $\{(r,c,t(r,c)) \mid r,c\in [m] \text{ and cell }(r,c) \text{ is not empty}\}$. 
The \emph{back-circulant Latin squares} are defined as $B_m = \{(i,j,i+j\mod{m}) \mid i,j \in [m]\}$.
A \emph{diagonal} of a Latin square $L$ is a set of $m$ cells of $L$ such that each row and each column is represented in the set of cells precisely once. 
A \emph{transversal} of a Latin square is a diagonal that also has each symbol represented precisely once in the diagonal. 
%See \cite{TranLSSurvey2} for a survey of transversals in Latin squares.


The \emph{shape} of $T$ is $\mathcal{S}(T) = \{(r',c') \in [m]\times [m] \mid (r',c',e)\in T \text{, for some } e\in [m]  \}$.
The \emph{$r$-th row set} of $T$ is defined as $\mathcal{R}_{r}(T) = \{ e' \in [m] \mid  (r,c',e')\in T \text{, for some } c'\in [m] \}$.
The \emph{$c$-th column set} of $T$ is defined as $\mathcal{C}_{c}(T) = \{ e' \in [m] \mid (r',c,e')\in T \text{, for some } r' \in [m]  \}$.
The \emph{$e$-th symbol set} of $T$ is defined as $\mathcal{E}_{e}(T) = \{ (r',c') \in [m]\times [m] \mid (r',c',e)\in T \}$.



\begin{definition}
For natural numbers $\mu,m$, $\mu \leq m$, a \emph{$\mu$-way Latin trade of order $m$ on symbol set $\Omega$} is a collection $\mathcal{T} = (T_1, \ldots, T_\mu)$ of $\mu$ partial Latin squares of order $m$ using symbols of $\Omega$ such that:
\label{def1}
\begin{itemize}
\item $\mathcal{S}(T_\alpha) = \mathcal{S}(T_\beta)$,  for each $1 \leq \alpha < \beta \leq \mu$;
\item for each $(r,c)\in \mathcal{S}(T_{\alpha})$ it holds that $t_\alpha(r,c) \neq t_\beta(r,c)$, for every $1 \leq \alpha < \beta \leq \mu$; and 
\item  $\mathcal{R}_r(T_\alpha) =  \mathcal{R}_r(T_\beta)$ and $\mathcal{C}_c(T_\alpha) =  \mathcal{C}_c(T_\beta)$, for each $r,c \in [m]$ and $1 \leq \alpha < \beta \leq \mu$.
%\item  Set-wise, the elements appearing in row (column) $r$ of $T_\alpha$ are the same as those of row (column) $r$ of $T_\beta$, for all $r \in [m]$   and $1 \leq \alpha < \beta \leq \mu$.
\end{itemize}
\end{definition}
%Should we have  a section to explain what these each mean?       "Set-wise, the elements appearing in row (column) $r$"
%change notation here?         $\alpha, \beta \in [\mu], \alpha \neq \beta$

Let  $\mathcal{T} = (T_1, \ldots, T_\mu)$ be a $\mu$-way Latin trade. 
As the shape of each $T_{\alpha}$ is the same, we can define the shape of $\mathcal{T}$ as $\mathcal{S}(\mathcal{T}) =   \mathcal{S}(T_1)$.  
Then the volume of $\mathcal{T}$ is the volume of $T_1$. 
Similarly, the row sets (resp.\ column sets and symbol sets) of each $T_{\alpha}$ are the same, so we can define a row set for row $r$ (resp.\ column set for column $c$ and symbol set for symbol $e$)  of $\mathcal{T}$ as $\mathcal{R}_r(\mathcal{T}) =   \mathcal{R}_r(T_1)$ (resp.\ $\mathcal{C}_c(\mathcal{T}) =   \mathcal{C}_c(T_1)$ and $\mathcal{E}_e(\mathcal{T}) =   \mathcal{E}_e(T_1)$). 

\begin{definition}\label{muWayTrade} %definition of. %we can label these properties with P, and relabel below to be C
For an integer $k\geq 0$, a \emph{$(\mu,k,m)$-Latin trade} on symbol set $\Omega$, $\mathcal{T}=(T_1, \ldots, T_\mu)$, is a $\mu$-way Latin trade of order $m$ on $\Omega$ that has  $k = |\mathcal{R}_r(\mathcal{T})| = |\mathcal{C}_c(\mathcal{T})| = |\mathcal{E}_e(\mathcal{T})|$, for each $r,c,e \in [m]$. Such a $\mu$-way Latin trade is called \emph{$k$-homogeneous}.
\end{definition}

A $(\mu,k,m)$-Latin trade can have $k=0$ in the case that each of the $\mu$ partial Latin squares is empty; otherwise $k$ must satisfy $\mu \leq k \leq m$.

We will require the $(\mu,k,m)$-Latin trades that we investigate to have the property that if $(r,c,e)\in T_{\alpha}$, where $T_{\alpha}$ is one of the partial Latin squares that form a $(\mu,k,m)$-Latin trade, then $r,c,e$ are pairwise distinct. 
With this property, $T_{\alpha} \cup \{(i,i,i) \mid i \in [m]\}$ would form a new partial Latin square that resembles an idempotent Latin square with some unfilled cells. 
%We may not be able to complete $T_{\alpha} \cup \{(i,i,i) \mid i \in [m]\}$ to an idempotent Latin square.

\begin{definition}
A  $\mu$-way Latin trade $\mathcal{T}$ of order $m$ is \emph{idempotent} if $i \notin \mathcal{R}_i (\mathcal{T}) \cup \mathcal{C}_i (\mathcal{T})$ and $(i,i) \notin \mathcal{S} (\mathcal{T})$, for $i \in [m]$. 
\end{definition}


\begin{definition}
A  $(\mu,k,m)$-Latin trade is \emph{circulant} if it can be obtained from the elements of its first row, called the base row (denoted by $\mu-B_m^k$), by simultaneously permuting each of the coordinates cyclically. That is, for each $\alpha$, the cell $(1,c,e) \in T_\alpha$ implies $(1+i,c+i \mod{m},e+i\mod{m}) \in T_\alpha$, for $1 \leq i \leq m-1$. 
\end{definition}

We write the base row as $B=\{(a_1, \ldots, a_\mu)_{c_l} \mid1 \leq l \leq k\}$, where $a_{\alpha}, c_{l} \in [m]$. 
Then the corresponding $\mu$ partial Latin squares can be constructed as $T_\alpha = \{(1+i, c_l + i\mod{m}, a_\alpha +i\mod{m}) \mid 0 \leq i \leq m-1, 1 \leq l \leq k\}$, $\alpha \in [\mu]$.
We will denote an idempotent circulant $(\mu,k,m)$-Latin trade by $\mu-IB_m^k$. 
%Basic definitions and aims. !!we need $\mathcal{U}$ is a $\mu$-way Latin trade on symbol set $(1, \ldots, e)$ in positions $\{(i,j) \mid 1 \leq i \leq r, 1 \leq j \leq c \}$.

%!!shorten $\Omega_1^m=(1, \ldots, m)$


Let $m$ be an integer. 
The spectrum of $\mu$-way homogeneous Latin trades of order $m$, $\mathcal{S}^{\mu}_m$, is the set of values of $k$ such that there exists a $(\mu,k,m)$-Latin trade. 
The spectrum of idempotent $\mu$-way homogeneous Latin trades of order $m$, $\mathcal{IS}^{\mu}_m$, is the set of values of $k$ such that there exists an idempotent $(\mu,k,m)$-Latin trade. 

A previous study of $(\mu,k,m)$-Latin trades \cite{3way} posed the question:
\begin{question}
For given $m$ and $k$, $m \geq k \geq \mu$, does there exist a $(\mu,k,m)$-Latin trade?
\end{question}
The primary goal of this paper is to investigate this question by deducing $\mathcal{S}^{3}_m$. 
However in order to do this, we will use a construction that requires us to first investigate $\mathcal{IS}^{3}_m$. 
Clearly  $\mathcal{IS}^{3}_m \subseteq {\mathcal{S}}^{3}_m$. 
It is known that $\{3, \ldots, m\} \supseteq \mathcal{S}^3_m$, and also that $3 \in \mathcal{S}^3_m$ if and only if $3 | m$ (see \cite{3way}).
In this paper, we show there exists $3$-way $k$-homogeneous Latin trades of order $m$ with $4 \leq k \leq m$ for all but a finite list of possible exceptions.



\section{Literature review}

A $2$-way Latin trade is typically called a Latin bitrade. 
There have been three distinct approaches used to construct $k$-homogeneous Latin bitrades.

The first approach used graph theoretic constructions (see also \cite{cav3homo}, \cite{grannellHomogeneous}, \cite{Partitioning3HomoLatinBitrade}, and  \cite{Perm34Homo}):

\begin{theorem}{\rm\cite{3-homo,4-homo}}
There exists a $(2,p,3m)$-Latin trade when $p=3,4$ and $m \geq 3$.
\end{theorem}



The second approach used block theoretic based constructions:
\begin{theorem}{\rm\cite{Exiskhomo,khomo}}
There exists a $(2,k,m)$-Latin trade when $3 \leq k \leq 37$ and $m \geq k$.
\end{theorem}



The third approach relies on finding pairs of transversals of given intersection in the back-circulant Latin squares:

\begin{theorem} {\rm\cite{NumbTran}}
For each odd $m \neq 5$ and for each $t \in \{0, \ldots, m-3\}\cup\{m\}$, there exists two transversals in $B_m$, $T_1$ and $T_2$, with $|T_1 \cap T_2|=t $. 
When $m=5$ and for each $t \in \{0, 1,5\}$, there exists two transversals in $B_m$, $T_1$ and $T_2$, with $|T_1 \cap T_2|=t $.
\end{theorem}

\begin{lemma} {\rm\cite{NumbTran}}
For $m$ an odd integer, let $T_1$ and $T_2$ be two transversals in $B_m$ such that $|T_1 \cap T_2| = t$. Then there exists a $(2,m-t,m)$-Latin trade.
\end{lemma}





These results lead to the completion of the spectrum problem for homogeneous Latin bitrades:

\begin{theorem}{\rm\cite{Exiskhomo,khomo,3-homo,4-homo,NumbTran}}
There is a $(2,k,m)$-Latin trade for all $3 \leq k \leq m$ and a $(2,2,2m)$-Latin trade, for all $m \geq 1$.
\label{AllBitrades}
\end{theorem}
%\cite{CavSurvey}









The first study of $(\mu,k,m)$-Latin trades for general $\mu$  produced a number of block theoretic constructions \cite{3way} that yielded results for small $k$ when $\mu=3$:

%\bigbreak

\begin{theorem}{\rm\cite{3way}\label{old_3way_results}}
There exist $(3,k,m)$-Latin trades for $m \geq k$ when:\nobreak
\begin{itemize}
\item $k=3$ and $3 \mid m$;
\item $k=4$ and $m \neq 6,7,11$;
\item $5 \leq k \leq 13$;
\item $k = 15$; and 
\item $k=m$.
\end{itemize}
\end{theorem}


\section{Idempotent generalization of basic constructions}

The constructions that have appeared earlier in the literature for $(\mu,k,m)$-Latin trades \cite{3way}  can be used (or modified) for the construction of  idempotent  $(\mu,k,m)$-Latin trades. As many of the constructions differ only trivially from their original appearance, we label the source of the original construction, and give the original proof with an extension if necessary.

%
%\begin{construction} \cite{3way} \label{3mm}
%There exists a $(\mu,\mu,\mu)$-Latin trade, for all $\mu \geq 1$.
%\label{Ais0}
%\end{construction}
%%\begin{proof}
%%Consider the Latin square induced by the group table of the cyclic group of order $\mu$. If we replace the element of each cell with the element increased by a constant mod $\mu$, we obtain $\mu$ Latin squares that form a $(\mu,\mu,\mu)$-Latin trade.
%%\end{proof}


\begin{theorem}{\rm\cite{3way}}%(Addition construction)
\label{addition}
If there exist idempotent $(\mu,k, m_i)$-Latin trades, for $i=1,2$, then there exists an idempotent $(\mu,k, m_1 +m_2)$-Latin trade.
\end{theorem}





\begin{theorem} {\rm\cite{3way}}% (The Direct Product Construction) %Kronecker
\label{multiplication}
If there exist an idempotent $(\mu_1,k_1, m_1)$-Latin trade and a $(\mu_2,k_2, m_2)$-Latin trade, then there exists an idempotent $(\mu_1 \mu_2,k_1 k_2, m_1 m_2)$-Latin trade.
\end{theorem}



\begin{theorem}{\rm\cite{3way}}\label{extendedTheorem}
If $l \neq 2, 6$ and for each $k \in \{k_2,\dots,k_l\}$ there exists a $(\mu, k, p)$-Latin
trade and there exists an idempotent $(\mu, k_1, p)$-Latin
trade, then an idempotent $(\mu, k_1+\cdots+k_l, lp)$-Latin trade exists. (Some $k_i$'s can possibly be zero.)
\end{theorem}
\begin{proof}
For $l \neq 2,6$, there exists two $l \times l$ orthogonal Latin squares. 
Denote these Latin squares by $L_1$ and $L_2$, with elements chosen from the sets $\{e_1, \ldots, e_l\}$ and $\{f_1, \ldots, f_l\}$, respectively. 
We can simultaneously permute the rows and columns of $L_1$ and $L_2$ so the main diagonal of $L_2$ contains only $f_1$, and then re-label the symbols of $L_1$ so that the symbols in cell $(j,j)$ of $L_1$ is $e_j$.
Assume that $L^*$ is the square that is formed by superimposing $L_1$ and $L_2$. 
We replace each $(e_i, f_j) \in L^*$ such that $j \geq 2$ with a $(\mu,k_j,p)$-Latin trade whose elements are from the set $\{(i-1)p +1, \ldots, ip\}$, and when $j=1$ with an idempotent $(\mu,k_1, p)$-Latin trade whose elements are from the set $\{(i-1)p +1, \ldots, ip\}$. 
As a result we obtain a $(\mu, k_1 + \cdots +k_l, lp)$-Latin trade, which we denote as $\mathcal{T}$. 

Then clearly $(j,j) \notin \mathcal{S}(\mathcal{T})$ as each of the entries on the main diagonal of $\mathcal{T}$ came from an idempotent $(\mu,k_1,p)$-Latin trade. 
Note that in $\mathcal{T}$ a row $r \in \{(i-1)p+1, \ldots, ip\}$ contains cells filled with symbols $e \in \{(i-1)p+1, \ldots, ip\}$ only in columns $c\in \{(i-1)p+1, \ldots, ip\}$, and these filled cells came from an idempotent $(\mu,k,p)$-Latin trade.
So if $(i-1)p+i' \in \mathcal{R}_{(i-1)p+i'}(\mathcal{T})$, $1 \leq i' \leq p$, then there must be a cell in row $(i-1)p+i'$ and column $c$ with $c \in \{(i-1)p+1, \ldots, ip\}$ that contains symbol $(i-1)p+i'$, and this comes from an idempotent $(\mu,k,m)$-Latin trade, say $\mathcal{U}$. 
But then $\mathcal{U}$ would have $i' \in \mathcal{R}_{i'}(\mathcal{U})$, a contradiction as $\mathcal{U}$ is idempotent. 
The analogous result holds for the columns, and $\mathcal{T}$ forms an idempotent $(\mu, k_1+\cdots+k_l, lp)$-Latin trade.
\end{proof}





\begin{theorem} \label{OldTheorem8} \label{advaddition}
Take $k$ and $k'$ to be integers with $k' > {k}$. 
 If for every $k'  \leq l \leq 2k'-1$ there exists an idempotent $(\mu, {k}, l)$-Latin trade, then for any $m \geq k'$ there exists an idempotent $(\mu, {k},m)$-Latin trade.
\end{theorem}
\begin{proof}
For every $m \geq 2k'$, we can write $m = rk' + sl$, for some  $r, s \geq 0$ and $k' + 1 \leq l \leq 2k' - 1$. Since there exist an idempotent
$(\mu, {k}, k')$-Latin trade and an idempotent $(\mu, {k}, l)$-Latin trade, by Theorem  \ref{addition} we conclude that there exists an idempotent $(\mu, {k},m)$-Latin trade.
\end{proof}




%\subsection{Constructing $(\mu,m-1,m)$-Latin trades}



A large set of idempotent Latin squares of order $m$ is a set of $m-2$ idempotent  Latin squares of order $m$, $(L_1, \ldots, L_{m-2})$, such that for $\alpha, \beta$ with $1 \leq \alpha < \beta \leq m-2$ and $i,j \in [m]$,  $L_{\alpha}(i,i) = L_{\beta}(i,i)=i$ and $L_\alpha(i,j) \neq L_\beta(i,j)$ when $i \neq j$.
%The previous results (\cite{LargeSetA}, \cite{LargeSetB}  and \cite{LargeSetC}) are written in terms of large sets of quasigroups, rather than in the equivalent terms of large sets of Latin squares, the later of which we will  find it more convenient to use. 
%This result extends Theorem 3 of \cite{3way}:

\begin{theorem}\label{mMinus1}
For $m\geq 3$, $m \neq 6$, there exists an idempotent $(\mu,m-1,m)$-Latin trade whenever $1 \leq \mu \leq m-2$.
\end{theorem}
\begin{proof}
 It was shown in \cite{LargeSetA} that for $m \neq 6,14,62$ there exists a large set of idempotent Latin squares of order $m$. The cases $m = 14,62$ were solved in \cite{LargeSetB} and \cite{LargeSetC} respectively. 
By taking such a large set and deleting the cells of the main diagonals of each of the idempotent Latin squares, we have an idempotent $(m-2,m-1,m)$-Latin trade for $m\geq 3$, $m\neq 6$. 
Clearly we can remove any number of the resulting partial Latin squares to yield an idempotent $(\mu,m-1, m)$-Latin trade for $1 \leq \mu \leq m-2$. 
\end{proof}





Generalizing from the method of finding pairs of transversals of given intersection in the back-circulant Latin squares, the author of \cite{Marbach3Way} was able to determine the possible intersection sizes of three transversals in the back circulant Latin square:

\begin{theorem}{\rm\cite{Marbach3Way}}\label{trades3}
For odd integer $m\geq 33$ with $m= 18I +9 + 2d$, $0\leq d <9$, and $m= 22I' +11 + 2d'$, $0\leq d' <11$, there exists three transversals of the back circulant Latin square $B_m$, $T_1$, $T_2$, $T_3$, for each $t  \in \{\min(d'+3,d), \ldots,   m\} \setminus \{m-5, \ldots, m-1\}$ such that $S=T_1\cap T_2=T_1\cap T_3=T_2\cap T_3$ and $|S|=t$, except possibly when:
\begin{itemize}
\item $m=51$ and $t=29$,
\item $m=53$ and $t=30$.
\end{itemize}
\end{theorem}

Further, a transformation is provided to construct $(\mu,k,m)$-Latin trades from a collection of $\mu$ transversals of the back circulant Latin square.
\begin{theorem}\rm{\cite{Marbach3Way}}\label{Marbach3Way}
Take $m$ odd and $0 \leq t \leq m$. 
If there exists a set $S\subseteq [m]^3$ with $|S|=t$ and  $\mu$ transversals of $B_m$, $T_1, \ldots, T_{\mu}$ with $T_\alpha \cap T_\beta = S$, for $\alpha,\beta \in [\mu]$ and $\alpha \neq \beta$, then there exists a circulant $(\mu,m-t,m)$-Latin trade.   
\end{theorem}
\begin{proof}
Consider the $\mu$ partial Latin squares defined by $Q_\alpha = \{(i, c+i, r+c+i) \mid i \in [m], (r,c,r+c)\in T_\alpha\setminus S\}$. The set $(Q_1, \ldots, Q_{\mu})$ forms a $(\mu,m-t,m)$-Latin trade that is circulant by definition. 
\end{proof}

This can be generalized for our purposes in the following manner:

\begin{theorem}
Take $m$ odd and $1 \leq t \leq m$. 
If there is a set $S\subseteq [m]^3$ with $|S|=t$ such that there exists $\mu$ transversals of $B_m$, $T_1, \ldots, T_{\mu}$, with $T_\alpha \cap T_\beta = S$, for $\alpha,\beta \in [m]$ and $\alpha \neq \beta$, then there exists an idempotent circulant $(\mu,m-t,m)$-Latin trade.   
\end{theorem}
\begin{proof}
Notice that $T_\alpha^{x,y} = \{(r+x, c+y,r+c+x+y)\mid (r,c,r+c)\in T_\alpha \}$,  for $\alpha \in [\mu]$ and $x,y \in \{0, \ldots, m-1\}$, will define a new collection of $\mu$ transversals of $B_m$ with $T_\alpha^{x,y} \cap T_\beta^{x,y}  = S^{x,y}$ such that $|S^{x,y} | =t$, for $\alpha, \beta \in [\mu]$ and $\alpha \neq \beta$.
As $t \geq 1$, this allows us to assume without loss of generality that $(m,m,m) \in T_\alpha$,  for all $\alpha \in [\mu]$, or equivalently $(m,m,m)\in S$. 
Applying the construction  from Theorem \ref{Marbach3Way} to these transversals of $B_m$, we obtain a circulant $(\mu,m-t,m)$-Latin trade, which we denote as $\mathcal{Q}=(Q_1, \ldots, Q_\mu)$. 
As $(m,m,m) \in S$, then each $(r,c,r+c)\in T_\alpha \setminus S$ has $r+1 \not\equiv 1$, $c+1 \not\equiv 1$, and $r+c+1 \not\equiv 1$. 
Then $\mathcal{Q}$ has its $(1,1)$ cell empty as $c+1 \not\equiv 1$, and the symbol $1$ will not appear in the first row as $r+c+1 \not\equiv 1$. 
The first column of $Q_\alpha$ contains cells $\{(m-c+1,1,r+1) \mid (r,c,r+c)\in T_\alpha \setminus S\}$, for $\alpha \in [\mu]$. 
As $r+1 \not\equiv 1$, the symbol $1$ does not appear in the first column. 
 By the circulant nature, we also have $(i,i) \notin \mathcal{S}(\mathcal{Q})$ and $i \notin \mathcal{R}_i(\mathcal{Q}) \cap\mathcal{C}_i(\mathcal{Q})$, for $i \in [m]$. 
Then $\mathcal{Q}$ is an idempotent circulant $(\mu,m-t,m)$-Latin trade.
\end{proof}

We can then exploit the existence of two \cite{NumbTran} and three \cite{Marbach3Way} transversals of given intersection as:

\begin{theorem}\label{idemptrades2}
For odd integer $m \geq 5$, there exists an idempotent circulant $(2,m-t,m)$-Latin trade for $t \in [m]\setminus \{m-2,m-1\}$ except, perhaps, when $(t,m)=(2,5)$.
\end{theorem}

\begin{theorem}  \label{idemptrades3}
For odd integer $m\geq 33$ with $m= 18I +9 + 2d$, $0\leq d <9$, and $m= 22I' +11 + 2d'$, $0\leq d' <11$, there exists  an idempotent circulant $(3,m-t,m)$-Latin trade, for $t\in \{\max(1,\min(d'+3,d)),\ldots,   m\} \setminus \{m-5,\ldots,  m-1\}$, except, perhaps, when:
\begin{itemize}
\item $m=51$ and $t=29$,
\item $m=53$ and $t=30$.
\end{itemize}
\end{theorem}




\section{New constructions for idempotent $(\mu,k,m)$-Latin trades}



In this section, we will consider block theoretic constructions that are able to determine the spectrum of $(3,k,m)$-Latin trades for all but a small list of values of $k$ and $m$.






%\subsection{Doubling construction}
%
%\begin{construction}
%If there exists a $(3,k,m)$-Latin trade, we construct a $(3,2k,2m)$-Latin trade.
%Take $(U_1,U_2,U_3)$ to be a $(3,k,m)$-Latin trade. We construct $(V_1,V_2,V_3)$, a $(3,2k,2m)$-Latin trade, by $v_{\alpha}(i,j) = u_{\alpha}(i,j)$, $v_{\alpha}(i+m,j+m) = u_{\alpha}(i,j)$, $v_{\alpha}(i+m,j) = u_{\alpha}(i,j)+m$, and $v_{\alpha}(i,j+m) = u_{\alpha}(i,j)+m$, for each $1 \leq i,j \leq m$ and $\alpha \in \{1,2,3\}$.
%It is clear that each row, and each column contain $2k$ filled cells, and each symbol appears $2k$ times.
%\end{construction}





\subsection{Computer search for small orders}

If $B=\{(a_1, \ldots, a_\mu)_{c_l} \mid 1 \leq l \leq k\}$, Algorithm 1 of \cite{3way} can be used to show $B$ is the base row of a $(\mu,k,m)$-Latin trade. If for each $(a_1, \ldots, a_\mu)_{c_l}$ it further holds that:
\begin{itemize}
\item $a_\alpha \neq 1$, for all $\alpha \in [\mu]$;
\item $c_l \neq 1$; and
\item $a_\alpha  \neq c_l$,
\end{itemize} 
then $B$ is the base row of an idempotent $(\mu,k,m)$-Latin trade.
As the result of a computational search, we introduce the following base rows of idempotent $(3,k,m)$-Latin trades:

\begin{itemize}
\item$k=5$
\begin{align*}3-IB^{ 5 }_{7} = &\{(3, 4, 5)_{2}, (5, 7, 4)_{3}, (7, 5, 2)_{4}, (2, 3, 7)_{5}, (4, 2, 3)_{6}\} \\ 
3-IB^{ 5 }_{8} = &\{(3, 4, 6)_{2}, (8, 2, 4)_{3}, (6, 8, 3)_{4}, (4, 6, 2)_{5}, (2, 3, 8)_{6}\} \\ 
3-IB^{ 5 }_{9} = &\{(3, 6, 9)_{2}, (6, 2, 7)_{3}, (2, 7, 3)_{4}, (9, 3, 6)_{5}, (7, 9, 2)_{8}\} \\ 
3-IB^{ 5 }_{11} = &\{(5, 7, 9)_{2}, (7, 2, 8)_{3}, (9, 8, 7)_{4}, (2, 9, 5)_{6}, (8, 5, 2)_{9}\} \\ 
3-IB^{ 5 }_{12} = &\{(4, 5, 8)_{2}, (11, 2, 5)_{3}, (8, 11, 4)_{5}, (5, 8, 2)_{6}, (2, 4, 11)_{8}\} 
 \end{align*}
\item
$k=6$
\begin{align*}3-IB^{ 6 }_{8} = &\{(3, 4, 6)_{2}, (5, 8, 2)_{3}, (8, 2, 5)_{4}, (2, 6, 3)_{5}, (4, 5, 8)_{6}, (6, 3, 4)_{7}\} \\ 
3-IB^{ 6 }_{9} = &\{(3, 6, 7)_{2}, (7, 4, 5)_{3}, (6, 2, 3)_{4}, (4, 7, 6)_{5}, (2, 5, 4)_{6}, (5, 3, 2)_{7}\} \\ 
3-IB^{ 6 }_{10} = &\{(5, 8, 9)_{2}, (9, 2, 4)_{3}, (2, 5, 3)_{4}, (4, 3, 8)_{5}, (3, 9, 2)_{6}, (8, 4, 5)_{7}\} \\ 
3-IB^{ 6 }_{11} = &\{(3, 4, 10)_{2}, (6, 9, 4)_{3}, (10, 2, 6)_{4}, (2, 6, 3)_{5}, (4, 3, 9)_{6}, (9, 10, 2)_{7}\} \\ 
3-IB^{ 6 }_{12} = &\{(3, 9, 11)_{2}, (7, 2, 5)_{3}, (11, 5, 3)_{4}, (2, 7, 9)_{5}, (5, 3, 7)_{6}, (9, 11, 2)_{7}\} \\ 
3-IB^{ 6 }_{13} = &\{(8, 11, 13)_{2}, (13, 8, 9)_{3}, (9, 2, 8)_{4}, (3, 9, 2)_{5}, (2, 3, 11)_{6}, (11, 13, 3)_{7}\} 
 \end{align*}
\item
$k=7$
\begin{align*}
%3-IB^{ 7 }_{9} = &\{(3, 4, 5)_{2}, (5, 6, 7)_{3}, (7, 9, 2)_{4}, (9, 2, 6)_{5}, (2, 7, 3)_{6}, (4, 5, 9)_{7}, \\&(6, 3, 4)_{8}\} \\ 
%3-IB^{ 7 }_{10} = &\{(3, 4, 6)_{2}, (6, 8, 5)_{3}, (8, 7, 2)_{4}, (7, 6, 8)_{5}, (4, 3, 7)_{6}, (2, 5, 4)_{7},\\& (5, 2, 3)_{8}\} \\ 
3-IB^{ 7 }_{m} = &\{(3, 4, 6)_{2}, (7, 6, 5)_{3}, (6, 2, 7)_{4}, (2, 9, 3)_{5}, (9, 7, 2)_{6}, (5, 3, 4)_{7}, (4, 5, 9)_{8}\},\\& \text{ for }m \geq 9  \end{align*}
\item
$k=8$
\begin{align*}
3-IB^{ 8 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 7)_{3}, (7, 8, 2)_{4}, (9, 2, 6)_{5}, (8, 5, 3)_{6}, (4, 3, 9)_{7}, (6, 9, 4)_{8},\\& (5, 7, 8)_{9}\}, \text{ for }m \geq 10 
 \end{align*}
\item 
$k=9$
\begin{align*}3-IB^{ 9 }_{m} = &\{(3, 4, 5)_{2}, (5, 8, 7)_{3}, (7, 2, 9)_{4}, (9, 6, 2)_{5}, (11, 9, 8)_{6}, (2, 11, 3)_{7}, (4, 3, 6)_{8},\\& (6, 5, 4)_{9}, (8, 7, 11)_{10}\}, \text{ for }m \geq 11  
 \end{align*}
%
%$k=9$
%\begin{align*}3-IB^{ 9 }_{11} = &\{(3, 4, 5)_{2}, (5, 6, 4)_{3}, (7, 8, 9)_{4}, (9, 11, 2)_{5}, (11, 2, 8)_{6}, (2, 5, 11)_{7},\\& (4, 9, 3)_{8}, (6, 3, 7)_{9}, (8, 7, 6)_{10}\} \\ 
%3-IB^{ 9 }_{12} = &\{(3, 4, 6)_{2}, (5, 6, 2)_{3}, (8, 10, 9)_{4}, (10, 9, 8)_{5}, (9, 5, 7)_{6}, (2, 8, 4)_{7},\\& (7, 3, 10)_{8}, (6, 2, 3)_{9}, (4, 7, 5)_{10}\} \\ 
%3-IB^{ 9 }_{13} = &\{(3, 4, 6)_{2}, (5, 7, 10)_{3}, (8, 9, 3)_{4}, (10, 6, 8)_{5}, (9, 3, 7)_{6}, (2, 10, 9)_{7},\\& (7, 2, 5)_{8}, (6, 8, 4)_{9}, (4, 5, 2)_{10}\} \\ 
%3-IB^{ 9 }_{m} = &\{(3, 5, 6)_{2}, (5, 8, 2)_{3}, (8, 3, 9)_{4}, (10, 6, 8)_{5}, (9, 10, 7)_{6}, (2, 9, 4)_{7},\\& (7, 2, 10)_{8}, (6, 4, 3)_{9}, (4, 7, 5)_{10}\}, \text{ for }m \geq 14  \end{align*}
\item$k=10$
\begin{align*}3-IB^{ 10 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 5, 9)_{4}, (9, 10, 2)_{5}, (11, 2, 10)_{6}, (10, 11, 3)_{7}, (4, 3, 7)_{8},\\& (7, 8, 11)_{9}, (5, 7, 8)_{10}, (8, 9, 6)_{11}\}, \text{ for }m \geq 12
 \end{align*}

\item
$k=11$
\begin{align*}3-IB^{ 11 }_{m} = &\{(3, 4, 5)_{2}, (5, 6, 8)_{3}, (7, 2, 10)_{4}, (9, 11, 3)_{5}, (11, 10, 2)_{6}, (13, 8, 11)_{7},\\& (2, 13, 9)_{8}, (4, 3, 6)_{9}, (6, 5, 4)_{10}, (8, 7, 13)_{11}, (10, 9, 7)_{12}\}, \text{ for }m \geq 13 
 \end{align*}
%$k=11$
%\begin{align*}3-IB^{ 11 }_{13} = &\{(3, 4, 5)_{2}, (5, 6, 4)_{3}, (7, 2, 6)_{4}, (9, 11, 10)_{5}, (11, 10, 13)_{6},\\& (13, 8, 11)_{7}, (2, 13, 3)_{8}, (4, 3, 2)_{9}, (6, 5, 7)_{10}, (8, 7, 9)_{11}, (10, 9, 8)_{12}\} \\ 
%3-IB^{ 11 }_{14} = &\{(3, 4, 5)_{2}, (2, 6, 7)_{3}, (8, 5, 6)_{4}, (10, 11, 12)_{5}, (12, 10, 11)_{6},\\& (9, 12, 4)_{7}, (11, 2, 9)_{8}, (5, 8, 3)_{9}, (7, 3, 2)_{10}, (4, 7, 10)_{11}, (6, 9, 8)_{12}\} \\ 
%3-IB^{ 11 }_{15} = &\{(3, 4, 5)_{2}, (2, 6, 7)_{3}, (8, 5, 12)_{4}, (10, 11, 6)_{5}, (12, 10, 11)_{6},\\& (9, 12, 4)_{7}, (11, 2, 10)_{8}, (5, 8, 3)_{9}, (7, 3, 9)_{10}, (4, 7, 2)_{11}, (6, 9, 8)_{12}\} \\ 
%3-IB^{ 11 }_{16} = &\{(3, 4, 5)_{2}, (2, 6, 12)_{3}, (8, 5, 6)_{4}, (10, 11, 9)_{5}, (12, 10, 11)_{6},\\& (9, 12, 3)_{7}, (11, 2, 7)_{8}, (5, 8, 10)_{9}, (7, 3, 4)_{10}, (4, 7, 8)_{11}, (6, 9, 2)_{12}\} \\ 
%3-IB^{ 11 }_{m} = &\{(3, 4, 6)_{2}, (2, 6, 8)_{3}, (8, 5, 10)_{4}, (10, 11, 2)_{5}, (12, 10, 9)_{6},\\& (9, 12, 3)_{7}, (11, 2, 7)_{8}, (5, 8, 11)_{9}, (7, 3, 4)_{10}, (4, 7, 12)_{11},\\& (6, 9, 5)_{12}\}, \text{ for }m \geq 17
% \end{align*}
\item$k=12$
\begin{align*}3-IB^{ 12 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 2, 3)_{4}, (8, 9, 11)_{5}, (11, 12, 8)_{6}, (13, 3, 12)_{7},\\& (12, 13, 2)_{8}, (4, 10, 13)_{9}, (7, 5, 6)_{10}, (5, 8, 9)_{11}, (10, 11, 7)_{12}, (9, 7, 10)_{13}\},\\& \text{ for }m \geq 14 
 \end{align*}
\item
$k=13$
\begin{align*}3-IB^{ 13 }_{m} = &\{(3, 4, 5)_{2}, (5, 6, 4)_{3}, (2, 5, 10)_{4}, (9, 10, 12)_{5}, (11, 13, 2)_{6}, (13, 2, 9)_{7},\\& (15, 12, 13)_{8}, (12, 15, 3)_{9}, (4, 3, 8)_{10}, (6, 7, 15)_{11}, (8, 9, 7)_{12}, (10, 11, 6)_{13},\\& (7, 8, 11)_{14}\}, \text{ for }m \geq 15
 \end{align*} 
%$k=13$
%\begin{align*}3-IB^{ 13 }_{15} = &\{(3, 4, 5)_{2}, (5, 6, 4)_{3}, (2, 5, 6)_{4}, (9, 10, 11)_{5}, (11, 12, 13)_{6},\\& (13, 15, 3)_{7},  (15, 2, 12)_{8}, (12, 13, 2)_{9}, (4, 8, 15)_{10}, (6, 3, 8)_{11},\\& (8, 7, 10)_{12}, (10, 9, 7)_{13}, (7, 11, 9)_{14}\} \\ 
%3-IB^{ 13 }_{16} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 2, 8)_{4}, (10, 9, 12)_{5}, (12, 13, 11)_{6},\\& (14, 12, 13)_{7}, (11, 14, 2)_{8}, (13, 8, 7)_{9}, (4, 11, 3)_{10}, (9, 3, 10)_{11},\\& (8, 5, 14)_{12}, (5, 7, 9)_{13}, (7, 10, 6)_{14}\} \\ 
%3-IB^{ 13 }_{17} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 3, 8)_{4}, (10, 9, 11)_{5}, (12, 13, 14)_{6},\\& (14, 12, 3)_{7}, (11, 14, 13)_{8}, (13, 5, 6)_{9}, (4, 11, 12)_{10}, (7, 2, 10)_{11},\\& (9, 7, 2)_{12}, (8, 10, 7)_{13}, (5, 8, 9)_{14}\} \\ 
%3-IB^{ 13 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 3, 8)_{4}, (10, 9, 12)_{5}, (12, 13, 2)_{6},\\& (14, 12, 13)_{7}, (11, 14, 10)_{8}, (13, 5, 14)_{9}, (4, 11, 9)_{10}, (7, 2, 6)_{11},\\& (9, 7, 3)_{12}, (8, 10, 7)_{13}, (5, 8, 11)_{14}\}, \text{ for }m \geq 18 
% \end{align*}
\item
$k=14$
\begin{align*}3-IB^{ 14 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 2, 3)_{4}, (8, 9, 2)_{5}, (10, 3, 13)_{6}, (13, 14, 12)_{7},\\& (15, 13, 14)_{8}, (14, 15, 11)_{9}, (4, 5, 6)_{10}, (7, 12, 15)_{11}, (5, 11, 10)_{12},\\& (11, 7, 8)_{13}, (9, 10, 7)_{14}, (12, 8, 9)_{15}\}, \text{ for }m \geq 16 
 \end{align*}
\item
$k=15$
\begin{align*}3-IB^{ 15 }_{m} = &\{(3, 4, 5)_{2}, (5, 6, 4)_{3}, (2, 5, 6)_{4}, (8, 2, 9)_{5}, (11, 13, 14)_{6}, (13, 15, 12)_{7},\\& (15, 12, 2)_{8}, (17, 14, 15)_{9}, (14, 3, 17)_{10}, (4, 17, 3)_{11}, (6, 7, 10)_{12},\\& (10, 11, 8)_{13}, (9, 10, 7)_{14}, (7, 9, 11)_{15}, (12, 8, 13)_{16}\}, \text{ for }m \geq 17 
 \end{align*}
\item 
$k=16$
\begin{align*}3-IB^{ 16 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 2, 3)_{4}, (8, 9, 2)_{5}, (4, 3, 10)_{6}, (13, 12, 15)_{7}, \\&(15, 16, 13)_{8}, (17, 15, 16)_{9}, (14, 17, 6)_{10}, (16, 5, 17)_{11}, (5, 7, 14)_{12},\\& (7, 14, 8)_{13}, (10, 13, 12)_{14}, (12, 11, 7)_{15}, (11, 8, 9)_{16}, (9, 10, 11)_{17}\},\\& \text{ for }m \geq 18
 \end{align*}
\item
$k=17$
\begin{align*}3 -IB^{ 17 }_{19} = &\{(3, 4, 5 )_2, (5, 6, 4)_3, (2, 5, 6)_4, (8, 2, 3)_5, (10, 11, 12)_6, (13, 14, 15)_7,\\& (15, 17, 19)_8, (17, 13, 14)_9, (19, 16, 17)_{10}, (16, 19, 2)_{11}, (4, 3, 16)_{12},\\& (6, 8, 7)_{13}, (9, 7, 11)_{14}, (11, 9, 10)_{15}, (7, 12, 9)_{16}, (14, 15, 13)_{17},\\& (12, 10, 8)_{18}\}\\
3-IB^{ 17 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 2, 3)_{4}, (8, 9, 7)_{5}, (10, 5, 11)_{6}, (14, 15, 16)_{7},\\& (16, 17, 15)_{8}, (18, 14, 17)_{9}, (15, 16, 14)_{10}, (17, 18, 2)_{11}, (4, 13, 18)_{12},\\& (7, 3, 8)_{13}, (12, 8, 10)_{14}, (5, 11, 13)_{15}, (11, 7, 6)_{16}, (13, 12, 9)_{17},\\& (9, 10, 12)_{18}\}, \text{ for }m \geq 20 
 \end{align*}
\item
$k=18$
\begin{align*}3-IB^{ 18 }_{m} = &\{(3, 4, 5)_{2}, (2, 6, 4)_{3}, (6, 2, 3)_{4}, (8, 9, 2)_{5}, (4, 3, 8)_{6}, (11, 8, 12)_{7},\\& (15, 16, 17)_{8}, (17, 18, 16)_{9}, (19, 15, 18)_{10}, (16, 17, 15)_{11}, (18, 19, 6)_{12},\\& (5, 7, 19)_{13}, (7, 5, 9)_{14}, (9, 14, 13)_{15}, (12, 11, 7)_{16}, (14, 13, 10)_{17},\\& (13, 10, 14)_{18}, (10, 12, 11)_{19}\}, \text{ for }m \geq 20  
 \end{align*}
\end{itemize}






%%
%%Let $N_k^m$ be the set of all functions $f:[k] \rightarrow [m]$ and $S_k$ to be the set of all permutations of order $k$.
%%Consider the quintuples $(\tau_1, \tau_2, \sigma_1, \sigma_2, \sigma_3) \in N_k^m \times N_k^m \times S_k \times S_k \times S_k$ such that:
%%\begin{enumerate}
%%\item $\tau_1( \sigma_i(x)) \neq \tau_1( \sigma_i(y))$ for all $x,y \in [k]$, $i \in \{1,2,3\}$
%%\item For each $x \in [k]$ and $i,j\in \{1,2,3\}$, $i \neq j$, there exists some $y \in [k]$ such that $\tau_1( \sigma_i(x)) -\tau_2(x) = \tau_1( \sigma_j(y))-\tau_2(y)$. 
%%\end{enumerate} 
%%Such quintuples are  equivalent to a row of a circulant $(3,k,m)$-Latin trade. Further, if for each $x \in [k]$:
%%\begin{enumerate}
%%\item $\tau_1( x) \neq m$;
%%\item $\tau_2( x) \neq m$; and
%%\item $\tau_1( \sigma_i(x)) -\tau_2(x) \neq m$,
%%\end{enumerate} 
%%then the equivalent row is the $m$th row of an idempotent $(3,k,m)$-Latin trade. 
%%We have created a depth first search for quintuplets with both these sets of properties \cite{prog:mine}, the results of which appear as base rows in Table \ref{tab:baseIdem}. 


\begin{theorem}\label{new_3way_results}
There exist idempotent $(3,k,m)$-Latin trades for $m \geq k+1$ when:
\begin{itemize}
%\item $k=3$ and $3 \mid m$.
\item $k=4$ and $5|m$;
\item $k=5$, except when $m=6$ and  perhaps when $m= 10, 13$; and
\item $6 \leq k \leq 18$.
\end{itemize}
\end{theorem}
\begin{proof}
The previously stated base rows, along with the idempotent $(3, m-1, m)$-Latin trades of Theorem \ref{mMinus1} complete the cases for $7 \leq k \leq 18$. 
For $k=6$, we can use the previously stated base rows along with an idempotent $(3, 6, 7)$-Latin trade from Theorem \ref{mMinus1} with Theorem \ref{advaddition}. 
For $k=5$, there exists idempotent $(3,5,m)$-Latin trades for $m \in \{7,8,9,11,12\}$.
Using these $(3,5,m)$-Latin trades in Theorem \ref{addition} will, after two iterations, yield the required $(3,5,m)$-Latin trades for Theorem \ref{advaddition} with $k' = 14$. 
For $k=4$, there exists an idempotent $(3,4,5)$-Latin trade by Theorem \ref{mMinus1}, on which we can repeatedly apply Theorem \ref{addition} to get the result. 
If an idempotent  $(3,5,6)$-Latin trade existed, then adding the cells $\{(i,i,i) \mid 1 \leq i \leq n\}$ to each of the partial Latin squares of the idempotent $(3,5,6)$-Latin trade would yield an ordered triple of idempotent Latin squares of order $6$ that are pairwise disjoint in each cell not on the main diagonal.   
Using a computer we searched for such ordered triples of idempotent Latin squares by testing all possible triples of idempotent Latin squares of order $6$. 
None existed, and so there does not exists an idempotent $(3,5,6)$-Latin trade. 
\end{proof}

We conjecture that the two unresolved cases with $k=5$ and  $m= 10, 13$ both exist.

 




\subsection{Extended multiplication construction}



\begin{lemma} \label{specLem3}                                            %Define $\Gamma_m = \{0,4,5, \ldots, m-1,m\}$. 
Take $n\geq 3$ and $m \geq 4$. If $n=6$, let $y$ be a positive integer with $\mu < y \leq \frac{m}{4}$. If $n \neq 6$,  let $y$ be a positive integer with $\mu < y \leq \frac{m}{2}$.  If there exists an idempotent $(\mu,k,m)$-Latin trade for each $k \in \{0,y,y+1, \ldots, m-1\}$, then  there exists an idempotent  $(\mu,k,mn)$-Latin trade for each $k \in \{0,y,y+1, \ldots, mn-1\}$.
\end{lemma}
\begin{proof}
In the case that $n \neq 6$, Theorem \ref{extendedTheorem} yields idempotent $(\mu,k,mn)$-Latin trades for $k \in \{\sum_{i=1}^n y_i \mid y_1 \in  \{0, y, y+1, \ldots, m-1\}\text{ and } y_i \in \{0, y, y+1, \ldots, m\}, 2 \leq i \leq n \} = \{0,y, y+1, \ldots, mn-1\}$ ($y_i$, with $2 \leq i \leq n$, may equal $m$ as there exists a $(3,m,m)$-Latin trade by Theorem \ref{old_3way_results}). 

Now we consider the case when $n=6$. 
Applying Theorem \ref{multiplication} using an idempotent $(\mu,k,m)$-Latin trade and a $(1,2,2)$-Latin trade (which exists by Theorem \ref{AllBitrades}) yields an idempotent $(\mu,2k,2m)$-Latin trade for each $k$ with $y \leq k \leq m-1$.
Applying Theorem \ref{addition} to the idempotent $(\mu,k,m)$-Latin trades yields idempotent $(\mu,k,2m)$-Latin trades for each $k$ with $y \leq k \leq m-1$.
Considering these, along with the existence of an idempotent $(\mu,2 m-1,2m)$-Latin trade from Theorem \ref{mMinus1}, yields 
idempotent $(\mu,k',2m)$-Latin trades for $k' \in \Gamma= \Gamma_1 \cup \Gamma_2$, where $\Gamma_1 =  \{0 \}\cup \{y, y+1, \ldots, m-1\}$ and $\Gamma_2 = \{2y,2y+2, \ldots, 2m-2\}\cup\{2m-1\}$. 
Applying Theorem \ref{extendedTheorem} with $l=3$, using the above idempotent $(\mu,k',2m)$-Latin trades with $k' \in \Gamma$ along with $(\mu,2m,2m)$-Latin trades and $(\mu,m,2m)$-Latin trades that exist by Theorem \ref{old_3way_results} and Theorem \ref{addition},  yields idempotent $(\mu,k'',6m)$-Latin trades for each $k'' \in \{\sum_{i=1}^3 y_i \mid y_1 \in \Gamma \text{ and } y_2, y_3 \in \Gamma \cup \{m,2m\} \}$. 

It holds that  $\{\sum_{i=1}^3 y_i \mid y_1 \in  \Gamma_2 \text{ and } y_2,y_3 \in  \Gamma_2\cup \{2m\}  \} = \{6y, 6y+2, \ldots, 4y+2m-4, 4y+2m-2\} \cup \{4y+2m-1, 4y+2m, \ldots, 6m-1\}$, 
and also that  $ \{\sum_{i=1}^3 y_i \mid y_1 \in  \Gamma_1 \text{ and } y_2,y_3 \in  \Gamma_1\cup\{m\} \} =  \{0 \}\cup \{y, y+1, \ldots, m\}\cup \{2y, 2y+1, \ldots, 2m\}\cup \{3y, 3y+1, \ldots, 3m-1\}= \{0 \}\cup \{y, y+1, \ldots, 3m-1\}$ as $y \leq m/4$ implies both $m\geq2y$ and $2m \geq 3y$. 
Then it holds that  $\{\sum_{i=1}^3 y_i \mid y_1 \in \Gamma \text{ and } y_2, y_3 \in \Gamma\cup\{m,2m\} \}\supseteq %\{\sum_{i=1}^3 y_i \mid y_i \in \Gamma_1 \} \cup \{\sum_{i=1}^3 y_i \mid y_i \in \Gamma_2 \}=
\{0 \}\cup \{y, y+1, \ldots, 3m-1\} \cup \{4y+2m-1, \ldots, 6m-1\}=   \{0 \}\cup \{y,\ldots, 6m-1\}$, as $m \geq 4y$, and so the proof is complete.
\end{proof}


There does not exist a pair of orthogonal Latin squares of order $2$, so we do not have a similar result to this lemma when $n=2$. 
This leaves us with two cases that are of particular interest, as  they are not covered by Lemma  \ref{specLem3}: $m=p$ and $m=2 p$, for $p$ a prime.
The next two subsection contain constructions that will be used to fill the spectrum $\mathcal{IS}^3_m$ for certain $m=p$ and $m=2 p$.  






















\subsection{Packing construction}

The following theorem uses $\mu$-way Latin trades of order $\lambda$ and volume $s$ in the construction of  $(\mu,s,m)$-Latin trades, for certain integers $m\geq \lambda^2 + 2\lambda +1$. %, for $m$ of the form $m = \lambda \cdot (\lambda+a) +b$ such that $gcd(m,\lambda) = gcd(\lambda,b)=1$, where $a$ and $b$ are integers with $0 \leq b < \lambda$ and $a \geq b+1$.
Afterwards we will modify the resulting structures to yield  idempotent $(\mu,s,m)$-Latin trades. 
The $3$-way intersection problem for Latin squares has been studied previously, and this will yield the $3$-way Latin trades we need in order to apply this construction, which we detail later.


\begin{theorem} \label{sHomozResultFirst}
Suppose there exists a $\mu$-way Latin trade of volume $s$ and of order $\lambda$. For every $m = \lambda  (\lambda+a) +b$, where $0 < b < \lambda$, $a \geq b+1$, and $\gcd(m,\lambda) = \gcd(\lambda,b) = 1$, there exists a  $(\mu,s,m)$-Latin trade.%\label{SHomogenize}
\end{theorem}

In order to prove this theorem, we construct the $(\mu,s,m)$-Latin trade as follows.






\begin{construction} 
\label{SHomogenizeConst}
Suppose there exists a  $\mu$-way Latin trade $\mathcal{U}=(U_1, \ldots, U_\mu)$ of volume $s$, order $\lambda$, and using  symbol set $\Omega=[\lambda]$. 
Let $m = \lambda (\lambda+a) +b$, where $a$ and $b$ are integers with $0 < b < \lambda$, $a \geq b+1$, and $\gcd(m,\lambda)=\gcd(\lambda,b)=1$.
%TODO - from here
Let $U_\alpha[f]$ be the array obtained from $U_\alpha$ by replacing each occurrence of symbol $i$ with symbol $b_i+f\pmod{m}$, where  $b_i =i  (\lambda+a-1)$ for each $1 \leq i \leq \lambda$, $\alpha \in [\mu]$, and $0 \leq f < m$.

We construct $R_\alpha$, an $m \times m$ array of cells for each $\alpha \in [\mu]$.
For each $f \in \{0, \ldots, m-1\}$, consider the $\lambda \times \lambda$ block of cells within $R_\alpha$ given by  $B_f = \{(i,j) \mid \lambda  f < i \leq \lambda  (f+1), f < j \leq f+\lambda\}$, with $i,j$ taken $\text{mod } m$, but $0$ is identified with $m$ such that $1 \leq i,j \leq m$. 
This placement of the blocks of cells in $R_\alpha$ is displayed visually in Figure \ref{fig:SHomoz}. 
Notice $B_f \cap B_g = \varnothing$ for each $0 \leq f,g <m$, $f \neq g$.
%The set of foremost cells of each $B_e$ combine to give the transversal $T= \{(\lambda  e, e, e)\mid 0 \leq e <m\}$ of order $m$. 
Place $U_\alpha[f]$ into the cells $B_f$ of $R_{\alpha}$, for each $0 \leq f <m$, and leave every other cell empty. 
Then cell $(\lambda f +i',f+j') \in B_f$ is filled in $R_{\alpha}$ with the symbol $U_{\alpha}[f](i',j')$, for $1 \leq i',j' \leq \lambda$.
\end{construction}

The following proof verifies that the $R_\alpha$  form a $(\mu,s,m)$-Latin trade.


\begin{figure} 
\begin{center}
\includegraphics{HomogPicture.pdf} %[width=1.0\textwidth]
\caption{An illustrative example of the placement of blocks in Construction \ref{SHomogenizeConst}}  \label{fig:SHomoz}
\end{center}
\end{figure}



\begin{proof}



We will show that the collection of $\mu$ $m \times m$ arrays $\mathcal{R}=(R_1, \ldots, R_\mu)$ defined by Construction \ref{SHomogenizeConst} form a $(\mu,s,m)$-Latin trade. 
We begin by showing that $\mathcal{R}$ forms a $\mu$-way Latin trade of order $m$.
This amounts to showing that each $R_{\alpha}$ forms a partial Latin square, as then  by construction it is clear  that the $\mu$ partial Latin squares form a $\mu$-way Latin trade of order $m$. % satisfying  the three conditions of Definition \ref{def1}. 
To this end, we must verify that any symbol appears in a  column of $R_{\alpha}$ at most once, and that any symbol appears in a row of $R_{\alpha}$  at most once.

To show the symbols that appear in a column of $R_{\alpha}$ are distinct, we will consider a specific symbol  $b_{\lambda} = \lambda(\lambda+a-1)$.
This symbol appears only in the block of cells $B_{j  (\lambda + a -1)}$ of $R_{\alpha}$, for $0 \leq j \leq \lambda-1$. 
The columns of the block of cells $B_{j (\lambda + a -1)}$ of $R_{\alpha}$ are exactly those columns $c$ with  $j(\lambda + a-1)  < c \leq j (\lambda + a-1) +\lambda$.
%When $i=0$, these rows are  $1 \leq r \leq \lambda$. 
For $0 \leq j \leq \lambda-1$, the sets of integers $\{j(\lambda + a-1)+1, \ldots,  j (\lambda + a-1) +\lambda\}$ are each disjoint, and in the range $1$ to $m$. 
That is to say that two distinct block of cells $B_{f}$ and $B_{f'}$ containing $b_{\lambda}$ do not intersect column-wise, and as a column within each block of cells $B_f$ of $R_{\alpha}$ can contain $b_{\lambda}$ at most once, we can conclude that $b_{\lambda}$ appears in each column of $R_{\alpha}$ at most once.
By construction, $(r,c,b_{\lambda})\in R_{\alpha}$ if and only if $(r+\lambda, c+1, b_{\lambda} +1) \in R_{\alpha}$, and so symbol $b_{\lambda}+1$ appears in each column of $R_\alpha$ at most once. 
Repeating this argument, we see that every symbol will appear in each column of $R_{\alpha}$  at most once. 



To show the symbols that appear in a row of $R_{\alpha}$ are distinct, %rows of what?
we consider a specific symbol $a-b-1 = b_{\lambda} + (\lambda + a -1) -m$, which we denote as $b_{\lambda+1}$.
This symbol appears in the block of cells $B_{j  (\lambda + a -1)}$ of $R_{\alpha}$, for $1 \leq j \leq \lambda$. 
The rows of the subsquare  $B_{j (\lambda + a -1)}$ of $R_{\alpha}$ are exactly those rows $r$ with  $j\lambda(\lambda + a-1)  < r \leq j \lambda(\lambda + a-1)  +\lambda$, or equivalently $m-j(\lambda+b) < r \leq m-j(\lambda+b)+\lambda$ once we consider $r$ to be taken modulo $m$. 
For $1 \leq j \leq \lambda$, the sets of integers $\{m-j(\lambda+b) +1, \ldots,  m-j(\lambda+b) +\lambda\}$ are each disjoint.
 As $\lambda(\lambda+b) < \lambda(\lambda+a)  < m$,  these sets of integers only contain values in the range $1$ to $m$. 
That is to say that two distinct blocks of cells $B_{f}$ and $B_{f'}$ that both contain  $b_{\lambda+1}$ do not intersect row-wise, so we can conclude that $b_{\lambda+1}$ appears in each row of $R_{\alpha}$ at most once. 
By construction, $(r,c,b_{\lambda+1})\in R_{\alpha}$ if and only if $(r+\lambda, c+1, b_{\lambda+1} +1) \in R_{\alpha}$, and so $b_{\lambda+1} +1$ appears in each row of $R_\alpha$ at most once. 
Repeating this argument, we see that every symbol will appear in each row of $R_{\alpha}$  at most once. 
We have now shown that the $R_{\alpha}$ form a $\mu$-way Latin trade of order $m$. 

Now  we show that $R_{\alpha}$ is $s$-homogeneous, for each $\alpha \in [\mu]$.
For $\alpha \in [\mu]$, the construction filled each of the $m$ blocks $B_f$ of $R_{\alpha}$ with $s$ cells, for $0 \leq f < m$. 
As no overlap occurs between the blocks $B_f$, $R_{\alpha}$ was filled by precisely $sm$ filled cells. 
By construction, if cell $(r,c)$ is filled in $R_{\alpha}$, then it holds that $(r,c,e)\in R_{\alpha}$  if and only if $(r+\lambda,c+1,e+1)\in R_{\alpha}$. 
Then row $r$ (column $c$, symbol $e$)  must contain the same number of filled cells as row $r+\lambda$ (column $c+1$, symbol $e+1$). 
We can repeat this argument $m-1$ times to show each row and column will have the same number of filled cells, and that each each symbol will have the same number of occurrences in $R_{\alpha}$  (for the row case, we have used the assumption that $gcd(m,\lambda)=1$).
Then this implies the $sm$ filled cells are spread evenly amongst the $m$ rows, columns, and symbols. 
This gives $s$ filled cells per row, $s$ filled cells per column, and $s$ occurrences per symbol.

This shows that each $R_{\alpha}$ is $s$-homogeneous, and so the proof is complete.
\end{proof}





\begin{example}
We demonstrate this technique using a $2$-way Latin trade of volume $s=7$ and of order $\lambda=3$ given by the pair of partial Latin squares:
\begin{center}
 \begin{tabular}{| c | c | c|c|c| c | c | c | c|c|c| c |  }
    \hline
    1&2&3 \\
        \hline
    3&&1 \\
        \hline
    2&3& \\
        \hline
  \end{tabular} \ \ \ \ \ \ 
 \begin{tabular}{| c | c | c|c|c| c | c | c | c|c|c| c |  }
    \hline
    2&3&1 \\
        \hline
    1&&3 \\
        \hline
    3&2& \\
        \hline
  \end{tabular}
\end{center}

We will take $a=2$ and $b=1$, giving the order of the resulting trade as $\lambda(\lambda+a)+b = 16$. Then $b_1=4$, $b_2=8$, and $b_3=12$.
Using the first of the above partial Latin squares of order $3$ and volume $7$ in the construction gives a partial Latin square of order $16$ that is $7$-homogeneous: 

\begin{center}
\small
 \begin{tabular}{| c | c | c|c|c| c | c | c | c|c|c| c |c|c|c| c | c |  c | c | }
    \hline
    4&8&12 	 &&&1&&9 			&&&2&6&		&&&\\
        \hline
    12&&4 	&&&13&1&			&&&&15&3&7	&&\\
        \hline
    8&12& 	&&&& 10&14&2		&&&7&&15		&&\\
        \hline

    &5&9&13 	&&& 2&&10			&&&3&7&		&&\\
        \hline
    &13&&5 	&&& 14&2&			&&&&16&4&8	&\\
        \hline
    &9&13& 		&&&&11&15&3		&&&8&&16	&\\
        \hline

    &&6&10&14 	&&&3&&11			&&&4&8&	&\\
        \hline
    &&14&&6 		&&&15&3&			&&&&1&5&9\\
        \hline
    &&10&14& 	&&&&12&16&4			&&&9&&1\\
        \hline

    &&&7&11&15 	&&&4&&12				&&&5&9	&\\
        \hline
    10&&&15&&7 	&&&16&4&				&&&&2&6\\
        \hline
    2&&&11&15& 	&&&&13&1&5			&&&10&\\
        \hline

    &&&&8&12&16 		&&&5&&13			&&&6&10\\
        \hline
    7&11&&&16&&8 		&&&1&5&			&&&&3\\
        \hline
    &3&&&12&16& 		&&&&14&2&6		&&&11\\
        \hline

    11&&&&&9&13&1 		&&&6&&14			&&&7\\
        \hline
  \end{tabular}
\normalsize
\end{center}

We can use the second partial Latin square of order $3$ and volume $7$ to construct a similar partial Latin square of order $16$ that is $7$-homogeneous. 
Together, these partial Latin squares form a $2$-way $7$-homogeneous Latin trade of order $16$.

\end{example}







\begin{theorem} \label{sHomozIdemResult}
Suppose there exists a $\mu$-way Latin trade of volume $s$ and of order $\lambda$, with $\lambda \geq 1$. For every $m = \lambda  (\lambda+a) +b$, where $0 < b < \lambda$, $a \geq b+1$, and $gcd(m,\lambda) = gcd(\lambda,b) = 1$, there exists a circulant idempotent $(\mu,s,m)$-Latin trade.%\label{SHomogenize}
\end{theorem}
\begin{proof}
Consider the $\mu$ arrays $R_1, \ldots, R_\mu$ from Construction \ref{SHomogenizeConst}. 
Define the array $\bar{R}_{\alpha}$ by the set of ordered triples $\bar{R}_{\alpha}=\{(\sigma_1(r),c,\sigma_2(e))  \mid (r,c,e) \in R_{\alpha}\}$ with $\sigma_1(r) = \lambda^{-1}\cdot (r-1)-1 \pmod{m}$ and $\sigma_2(e)=e-2(\lambda+a-1) \pmod{m}$, for each $\alpha \in [\mu]$, where $\lambda^{-1}$ is the unique inverse of $\lambda \pmod{m}$ which exists by the assumption that $gcd(m,\lambda) = 1$. 
Both $\sigma_1$ and $\sigma_2$ are permutations of $[m]$.
As the $R_{\alpha}$ form a $(\mu,s,m)$-Latin trade, the $\bar{R}_{\alpha}$ also form a $(\mu,s,m)$-Latin trade as the three properties of Definition \ref{def1} are invariant under permutation swaps of the rows, columns, and symbols. 

Since $(r,c,e)\in R_{\alpha}$ implies $(r + \lambda, c+1, e+1)\in R_{\alpha}$ by construction, it follows that $(r,c,e)\in \bar{R}_{\alpha}$ implies $(r + 1, c+1, e+1)\in \bar{R}_{\alpha}$, and so $\bar{R}_{\alpha}$ is a circulant $(\mu,s,m)$-Latin trade.

We show that $1 \notin \mathcal{R}_1(\bar{R}_{\alpha})$, $1 \notin \mathcal{C}_1 (\bar{R}_{\alpha})$, and $(1,1) \notin \mathcal{S} (\bar{R}_{\alpha})$.
Noting that $\sigma_1^{-1}(1) = 2\lambda +1$ and $\sigma_2^{-1}(1) = 2 \lambda + 2a -1$, this is equivalent to showing $2\lambda+2a-1 \notin \mathcal{R}_{2\lambda+1}(R_{\alpha})$, $2\lambda+2a-1 \notin \mathcal{C}_1 (R_{\alpha})$, and $(2\lambda+1,1) \notin \mathcal{S} (R_{\alpha})$. 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%//What if less than 3 such blocks??//

We first show that $2\lambda+2a-1 \notin \mathcal{R}_{2\lambda+1}(R_{\alpha})$. 
The symbol $2 \lambda + 2 a -1$ appears only in the blocks $B_{j(\lambda + a -1)  + b + 3 \lambda + 2a -1}$ of $R_\alpha$ for $0 \leq j \leq \lambda-1$, hence it only appears within the rows $T=\cup_{j=0}^{\lambda-1}\{\lambda (j (\lambda+a-1) +b +3\lambda +2a -1) +1, \ldots, \lambda (j (\lambda+a-1) +b +3\lambda +2a -1) +\lambda\}$. If we perform a change in variables, sending $j$ to $\lambda -2-j$, then $T = \cup_{j=0}^{\lambda-2}\{\lambda+1+j(\lambda+b), \ldots, 2\lambda + j(\lambda+b)\} \cup \{m-b+1, \ldots, m\} \cup \{1, \ldots, \lambda-b\}$, which does not contain $2\lambda+1$. %double checked
Then the symbol $2\lambda + 2 a -1$ does not appear in the row $2\lambda+1$. 

%//The symbol $2\lambda+2a-1$ appears in the blocks $B_{1-(\lambda+a-1)}$ and  $B_1$ of $R_{\alpha}$, and hence it may appear in the rows $\{\lambda(2-\lambda-a)+1, \ldots, \lambda(2-\lambda-a)+\lambda \}=\{2\lambda+b+1, \ldots, 2\lambda+b+\lambda\}$ and $\{\lambda+1, \ldots, 2\lambda\}$ respectively, and in the columns $\{m-\lambda-a, \ldots, m-a\}$ and $\{2, \ldots, \lambda+1\}$ respectively.  
%If we assume for the sake of contradiction that $2\lambda+2a-1 \in \mathcal{R}_{2\lambda+1}(R_{\alpha})$, then there must be a third block $B_e$ of $R_{\alpha}$ that intersects row $2\lambda+1$, contains symbol $2\lambda+2a-1$, and is disjoint to the blocks  $B_{1-(\lambda+a-1)}$ and $B_1$. But then the $\lambda$ rows of $B_e$ must have index greater than or equal to $2\lambda+1$ and less than or equal to $2\lambda+b$,  meaning $B_e$ has to have less than or equal to $b$ rows, which is impossible as $b<\lambda$.   


Secondly we show that $2\lambda+2a-1 \notin \mathcal{C}_1 (R_{\alpha})$.  
The symbol $2 \lambda + 2 a -1$ appears exactly in the blocks $B_{j(\lambda + a -1) +b+3\lambda+2a-1}$ of $R_{\alpha}$, for $0 \leq j \leq \lambda-1$. 
These blocks only use the columns $\cup_{j=0}^{\lambda-1} \{j(\lambda + a -1) +b+3\lambda+2a, \ldots, j(\lambda + a -1) +b+3\lambda+2a+\lambda-1\} = \cup_{j=0}^{\lambda-3} \{j(\lambda + a -1) +b+3\lambda+2a, \ldots, j(\lambda + a -1) +b+3\lambda+2a+\lambda-1\} \cup \{2, \ldots, \lambda+1\} \cup \{\lambda+a+1, \ldots, \lambda+a+\lambda\}$. As such, the column with index $1$ does not contain $2\lambda+2a-1$.

Thirdly we show that  $(2\lambda+1,1) \notin \mathcal{S} (R_{\alpha})$. 
Suppose for the sake of contradiction that $(2\lambda+1,1) \in \mathcal{S} (R_{\alpha})$.
Then there must be some block $B_{f'}$ that contains cell $(2\lambda+1,1)$, $0 \leq f' \leq m-1$. 
As $B_1 = \{(i,j) \mid \lambda+1 \leq i \leq 2 \lambda, 2 \leq  j \leq 1+\lambda\}$, $B_{f'}$ cannot contain the cell $(2\lambda,2)$. Then $B_{f'}$ must either have exactly the rows $\{2\lambda+1, \ldots, 3\lambda\}$, or have exactly the columns $\{m-\lambda+2, \ldots, m\}\cup \{1\}$. 

The former implies $B_{f'}$ contains exactly the same rows as $B_2$. 
As the first rows are the same, $f' \lambda  +1 \equiv 2\lambda+1 \mod{m}$, and as $\gcd(\lambda,m)=1$, we have $f'=2$. 
But then $B_2$ must contain cell $(2\lambda+1,1)$, which when we look at the columns implies $\lambda+2 \geq m+1$. As $m \geq \lambda(\lambda+2)+1$ and $\lambda \geq 1$, this is impossible.

%//The former is impossible, as then $B_{e'}$ contains exactly the same rows as $B_2$ but is also distinct from $B_2$ as the columns of $B_{e'}$ and $B_2$ are different, which implies $gcd(m,\lambda) \neq 1$. 
The later implies $B_{f'} = B_{1-\lambda} = B_{m+1-\lambda}$. 
Then each of the $m$ rows are represented at least once in the set of $\lambda+1$ blocks $\{B_{m+1-\lambda}, B_{m+2-\lambda}, \ldots, B_{m-1}  \}\cup \{B_0, B_1\}$, but these $\lambda+1$ blocks only use $\lambda$ rows each, and so $\lambda (\lambda+1)$ rows in total. 
This implies $m \leq \lambda(\lambda+1)$, which contradicts the fact that $m=\lambda(\lambda+a)+b$ and that $a > b \geq 1$.

Then none of the cases are possible, forming a contradiction, and so $(2\lambda+1,1) \notin \mathcal{S} (R_{\alpha})$.
This completes the proof.  
\end{proof}








The three-way intersection problem for Latin squares has been studied in \cite{3waySpectrum}, where the authors consider three Latin squares $L_1, L_2, L_3$ of order $n$ with common intersection $P = L_1 \cap L_2= L_1 \cap L_3= L_2 \cap L_3$. 
The collection of partial Latin squares $(L_1\setminus P, L_2\setminus P, L_3\setminus P)$ forms a $3$-way Latin trade of volume $n^2 - |P|$ and of any order $n'$ greater than or equal to $n$ (as we allow empty rows and columns in $3$-way Latin trades). 
We can thus interpret the results of  \cite{3waySpectrum} in terms of $3$-way Latin trades and combine them with Theorem \ref{sHomozIdemResult} to yield: %(note that a $\mu$-way trade of volume $s$ and volume $\lambda$:
\begin{theorem} \label{spanned_results}
For $\lambda \geq 3$, there exists circulant idempotent $(3,k,m)$-Latin trades for $m = \lambda  (\lambda+a) +b$, where $0 < b < \lambda$, $\gcd(\lambda,b)=1$, and $a \geq b+1$, and:
\begin{itemize}
\item $k\in \{0,9\}$, for $\lambda=3$;
\item $k \in \{0,9,12,15,16\}$, for $\lambda=4$;
\item $ k \in \{0,9,12,15,16\}$ or $18 \leq k \leq 25$, for $\lambda=5$;
\item $k \in \{0,9,12\}$ or $15 \leq k \leq \lambda^2$, for $\lambda\geq 6$.
 
\end{itemize}
%
%\begin{itemize}
%\item $k\in \{0,9\}$, for $\lambda=3$;
%\item $k \in \{0,9,12,15,16\}$, for $\lambda=4$;
%\item $18 \leq k \leq 25$, for $\lambda=5$;
%\item $k=17$ and $26 \leq k \leq 36$, for $\lambda=6$;
%\item $(\lambda-1)^2-1 \leq k \leq \lambda^2$, for $\lambda\geq 7$.
% 
%\end{itemize}
\end{theorem}
















\subsection{Construction via $RPBD$s} 


\begin{definition}
A \emph{$(v,M, \lambda)$ pairwise balanced design}, denoted $PBD(v,M,\lambda)$, is a pair $(V,B)$, with $V$ a set of $v$ symbols and $B$ a set of subsets of $V$ (each subset is called a block) with sizes from $M$, such that each pair of elements of $V$ can be found in exactly $\lambda$ blocks of $B$.
\end{definition}

\begin{definition}
A \emph{resolvable $(v,M, \lambda, n)$ pairwise balanced design}, which we denote by $RPBD(v,M,\lambda,n)$, is a pair $(V,B)$ along with $n$ resolution classes $R_1, \ldots, R_n$, such that  $(V,B)$ is a $PBD(v,M,\lambda)$, the sets $R_1, \ldots, R_n$ partition $B$,  and each symbol appears in precisely one block of each resolution class.
\end{definition}



\begin{theorem} \label{RPBD2Trade}
Suppose there exists a $RPBD(v,M,1,n+1)$, $(V,B)$, with resolution classes $R_1, \ldots, R_n$ and $R_{\infty}$. 
Suppose there exists integers $d_i \geq 1$, $1 \leq i \leq n$, such that for each $b \in R_i$ there exists an idempotent $(\mu,|b|-d_i,|b|)$-Latin trade and integer $d_{\infty}\geq 0$ such that for each $b \in R_{\infty}$ there exists a $(\mu,|b|-d_{\infty},|b|)$-Latin trade. 
Then there exists a $(\mu, v+n-\sum_{i=1}^n d_i - d_{\infty}, v)$-Latin trade.
\end{theorem}

The following construction was suggested by Prof.\ L.\ Zhu, which is a modification of a construction for sets of idempotent Latin squares (see \cite{WallisCombDesigns}, page 188). After the construction, we will give a proof to show that the construction yields Theorem \ref{RPBD2Trade}.

\begin{construction} \label{constZhu}
Take $(V,B)$, a $RPBD(v,M,1,n+1)$ with resolution classes $R_1, \ldots, R_n$ and $R_{\infty}$. 
Suppose there are integers $d_i \geq 1$, $1 \leq i \leq n$, such that for each $b \in R_i$ there exists an idempotent $(\mu,|b|-d_i,|b|)$-Latin trade, and integer $d_{\infty}\geq 0$ such that for each $b \in R_{\infty}$ there exists a $(\mu,|b|-d_{\infty},|b|)$-Latin trade. 
We impose an arbitrary total ordering $<$ on $V$.
For a block $b\subseteq V$, define $b^h$ to be the $h$th smallest symbol in $b$ when $b$ is considered under the ordering imposed on $V$, for $1 \leq h \leq |b|$.
That is $\{b^1, \ldots, b^{|b|}\} = b$ and $b^i < b^{i+1}$ for $1 \leq i \leq |b|-1$.

We construct  $\mu$ $v \times v$ arrays, $T_1, \ldots, T_{\mu}$, each with rows and columns indexed by $V$. 
For each $1 \leq i \leq n$  and each block $b \in R_i$, let $\mathcal{S}=(S_1, \ldots, S_{\mu})$ be an idempotent $(\mu,|b|-d_i,|b|)$-Latin trade on the set of symbols $\Omega = [|b|]$.
For each block $b \in R_{\infty}$, let $\mathcal{S}=(S_1, \ldots, S_{\mu})$ be a $(\mu,|b|-d_{\infty},|b|)$-Latin trade on the set of symbols $\Omega = [|b|]$.
Whenever $(r,c,e)\in S_\alpha$ put $(b^r, b^c, b^e)$ into $T_{\alpha}$, for $1 \leq r,c \leq |b|$. 
Note that if $(r,c)$ is empty in $S_\alpha$ for any $1 \leq r,c \leq |b|$, then $(b^r, b^c)$ is left empty in $T_\alpha$. 
%See Figure \ref{fig:Zhu1} and Figure \ref{fig:Zhu2} for an example of this filling.
\end{construction}

%
%\begin{figure}
%\begin{center}
%\includegraphics[width=0.8\textwidth]{ZhuPicture.pdf} 
%\caption{An illustration for Construction \ref{constZhu}. We construct a $(\mu,k,m)$-Latin trade $\mathcal{T}$ and include in the example the $n$ blocks of a resolution class $b_{1i}\in R_1$, where in particular we have $b_{11}=\{x_1, \ldots, x_r\}$, $b_{12}=\{y_1, \ldots, y_r\}$, and $b_{1n}=\{z_1, \ldots, z_r\}$,  and one block of a different resolution class $b=\{x_r, y_s,a_1,a_2\}\in R_2$.} \label{fig:Zhu1}
%\end{center}
%\end{figure}
%
%
%\begin{figure}
%\begin{center}
%\includegraphics[width=0.8\textwidth]{ZhuPicturePt2.pdf} 
%\caption{An example of the transformation we apply to a partial Latin square  in Construction \ref{constZhu} for a block $\{x_r, y_s, a_1, a_2\}$. Note that the resulting square would fill the black-centered cells of Figure \ref{fig:Zhu1}. }\label{fig:Zhu2}
%\end{center}
%\end{figure}
%



\begin{proof}
Consider the $\mu$ $v \times v$ arrays $\mathcal{T}=(T_1, \ldots, T_{\mu})$ from Construction \ref{constZhu}. We will show that $\mathcal{T}$ is a $(\mu, v+n-\sum_{i=1}^n d_i - d_{\infty}, v)$-Latin trade.

During the construction, any single cell $(r,c,e)\in T_\alpha$ with $|\{r,c,e\}|\leq 2$ must have been constructed using some $x \in R_{\infty}$, as the other blocks were replaced during the construction by idempotent $(\mu,k,m)$-Latin trades, which would imply $|\{r,c,e\}|=3$ by the definition of idempotent $\mu$-way Latin trades. 

Suppose the construction filled two cells $(r_1, c_1, e_1)$ and $(r_2, c_2, e_2)$ of $T_\alpha$ such that the two cells have two of three indices the same. Let $a$ and $b$ be the values of the two identical indices (for example if the two cells we are observing are $(r,c,e_1),(r,c,e_2)\in T_{\alpha}$ with $e_1 \neq e_2$, then $a=r$ and $b=c$). 
If distinct blocks $x$ and $y$ were used respectively to construct $(r_1,c_1,e_1)$ and $(r_2, c_2, e_2)$, then $\{a,b\} \subseteq x \cap y$. 
By the definition of a $PBD(v,M,1)$, $|x\cap y|\leq 1$, and so $a=b$. Then $|\{r_1,c_1,e_1\}|\leq 2$ and $|\{r_2,c_2,e_2\}|\leq 2$, and so $x,y \in R_\infty$, which implies $|x\cap y|=0$ as $x$ and $y$ are distinct blocks in the same resolution class.
This forms a contradiction, as $a \in x\cap y$. 
So any two filled cells that have two of three indices the same were both filled during construction using the same block.

As we filled $\mu$-way Latin trades into $T_\alpha$ from these  blocks, it follows that no cell was filled twice, each row contains each symbol at most once, and  each column contains each symbol at most once.
Then each $T_\alpha$ is a partial Latin square.


%//We show that no cell was filled twice during the construction of $\mathcal{T}$. 
%Suppose, for the sake of contradiction, a cell $(r,c)$ of $T_{\alpha}$ was filled twice.
%Then there exists at least two distinct blocks $x,y\in B$ with $(r,c) = (x^i,x^j)= (y^{{i'}},y^{{j'}})$, for some $i,j,{i'}, {j'}$ with $1 \leq i,j \leq |x|$ and $1 \leq{i'}, {j'} \leq |y|$. 
%Then it holds that $r,c \in x \cap y$.
%By the definition of a $PBD(v,M,1)$, $|x \cap y| \leq 1$, which implies $r=c$.
%
%During the construction, any two cells of $T_\alpha$ that shared two of three indices (say for example $(r,c,e_1)$ and $(r,c,e_2)$) must have been constructed using a block of the resolution class $R_\infty$, as the blocks of other resolution classes were replaced by idempotent $(\mu,k,m)$-Latin trades, in which any pair of cells have at most one of the three indices the same.
%If $x\in R_{\infty}$ and $y\in R_{\infty}$, then their intersection is $x \cap y = \varnothing$ by definition, contradicting our assumption.
%Then each filled cell $(r,c)$ of $T_{\alpha}$ was filled once for each $ \alpha \in [\mu]$.
%
%Now we show that each array $T_{\alpha}$ is actually a partial Latin square. 
%This requires us to show each symbol appears at most once per row and at most once per column. 
%To show the former, suppose for the sake of contradiction that $(r,c_1,e), (r,c_2,e) \in T_\alpha$. 
%If $\{r,c_1,c_2\}$ appeared in one block of $B$, then  thepartial Latin square that was used to replace this one block had a symbol that appeared twice in the same row. 
%This is impossible, and so there must exists two distinct blocks $x,y \in B$ with $(r,e) = (x^i,x^j) = (y^{i'},y^{j'})$ for some $i,j,{i'}, {j'}$ with $1 \leq i,j \leq |x|$ and $1 \leq{i'}, {j'} \leq |y|$.
%We saw previously that this lead to a contradiction, as it also does here. 
%Then each symbol appears at most once per row. The equivalent result for columns holds similarly. 
%Then each $T_{\alpha}$ forms a partial Latin square.

To see $\mathcal{T}$ forms a $\mu$-way Latin trade, it is enough to note that $\mathcal{S}(T_{\alpha})$ must be the same for each $ \alpha \in [\mu]$; that each filled cell $(r,c)$ was filled differently in each $T_{\alpha}$, $ \alpha \in [\mu]$; and that each row (resp.\ column) contains setwise the same symbols, each of which are clear from the construction.
Then the $\mu$ arrays $T_{\alpha}$ form a $\mu$-way Latin trade of order $v$. 

We are left to show $\mathcal{T}$ is $(v+n-\sum_{i=1}^n d_i  -d_{\infty})$-homogeneous. 
To show that there are  $v+n-\sum_{i=1}^n d_i  -d_{\infty}$ filled cells in each row, 
observe that for each symbol $r$ and each resolution class $R_i$, there is precisely one block $b_i$ such that $r \in b_i \in R_i$,  for each $i$ with $1 \leq i \leq n$ or $i=\infty$. 
%Then row $r$ only intersects with the blocks of cells $(b_i \times b_i)$, for $1 \leq i \leq n$ and $i=\infty$. 
Then any filled cells in row $r$ are in the cells $b_i \times b_i$  for some $i$ with $1 \leq i \leq n$ or $i=\infty$.
%//TODO
There are $|b_i|-d_i$ filled cells in the intersection of row $r$ and the block of cells $b_i \times b_i$, for $1 \leq i \leq n$ and $i=\infty$, showing row $r$ has a total of $\sum_{i=1}^n ( |b_i| -d_i )+(|b_{\infty}| -d_{\infty} ) = v+n-\sum_{i=1}^n d_i  -d_{\infty}$ filled cells.
The proof is analogous for the number of filled cells per column and for the number of occurrences of each symbol.
This shows $\mathcal{T}$ is a $\mu$-way Latin trade of order $v$ that is $(v+n-\sum_{i=1}^n d_i -d_{\infty})$-homogeneous, so we are done.
\end{proof}




\begin{theorem} \label{RPBD2TradeIdempotent}
Suppose there exists a $RPBD(v,M,1,n+1)$, $(V,B)$ with resolution classes $R_1, \ldots, R_n$ and $R_{\infty}$. 
Suppose there exists integers $d_i \geq 1$, $1 \leq i \leq n$, such that for each $b \in R_i$ there exists an idempotent $(\mu,|b|-d_i,|b|)$-Latin trade and integer $d_{\infty}\geq 1$ such that for each $b \in R_{\infty}$ there exists an idempotent $(\mu,|b|-d_{\infty},|b|)$-Latin trade. 
Then there exists an idempotent $(\mu, v+n-\sum_{i=1}^n d_i - d_{\infty}, v)$-Latin trade.
\end{theorem}
\begin{proof}

Consider the $\mu$ $v \times v$ arrays $\mathcal{T}=(T_1, \ldots, T_{\mu})$ from Construction \ref{constZhu} using idempotent $(\mu,|b|-d_i,|b|)$-Latin trades to fill in the squares $b \times b$, for blocks $b \in R_{\infty}$. 

The proof of Theorem \ref{RPBD2Trade} shows  $\mathcal{T}$ is a $(\mu, v+n-\sum_{i=1}^n d_i - d_{\infty}, v)$-Latin trade. 
We show that $\mathcal{T}$ is idempotent. 
Assume for the sake of contradiction that $\mathcal{T}$ is not idempotent. 
Then there exists a $(r,c,e)\in T_\alpha$ with at least two of the three indices $r,c,e$ the same. This only occurs when the block $x$ with $\{r,c,e\} \subseteq x$ and $x \in R_i$ was used along with a non-idempotent $(\mu,k,|x|)$-Latin trade to construct the cell $(r,c,e) \in T_{\alpha}$.
But there is no such $x$ as each of the $(\mu,k,|x|)$-Latin trades are idempotent. 
Then $\mathcal{T}$ is idempotent. 
\end{proof}



We wish to choose an $RPBD(v,M,1,n+1)$ with resolution classes $R_1, \ldots, R_n$ and $R_{\infty}$ such that there will exist idempotent $(\mu,|b|-d_i,|b|)$-Latin trades for each $b\in R_i$, for some $d_i \geq 1$ and $i \in \{1, \ldots, n, \infty\}$. 
By making $M$ contain as few values as possible, we can limit the number of idempotent $(\mu,k,m)$-Latin trades that are required to exist, as $|b| \in M$. 
A resolvable transversal design is a  $RPBD(\alpha n,\{\alpha,n\},1,n+1)$, and so suits our purposes as $|M|\leq2$.
We are able to modify the resolvable transversal design by removing elements in order to yield $RPBD(v,M,1,n+1)$ such that $v$ can be any positive integer, while $M$ contains as few values as possible.


\begin{definition}
A \emph{transversal design $TD(\alpha,n)$} of order $n$ and block size $\alpha$, is a triple $(V,G,B)$ such that:
\begin{enumerate}
\item $V$ is a set of $\alpha n$ elements;
\item $G$ is a partition of $V$ into $\alpha$ subsets (called the groups), each of size $n$;
\item $B$ is a collection of subsets of $V$ (called the blocks), each of size $\alpha$ ; and
\item every unordered pair of elements of $V$ appears in precisely one block of $B$, or one group of $G$, but not both.
\end{enumerate}
\end{definition}



\begin{definition}
A \emph{resolvable transversal design $RTD(\alpha,n)$} of order $n$ and block size $\alpha$, is a triple $(V,G,B)$  such that $B$ can be partitioned into $n$ resolution classes  $R_1, \ldots, R_n$, such that each  $R_i$ is a partition of $V$ into $n$ classes.
\end{definition}

The following two lemmata are well known (See III.3.2 and III.3.3 in \cite{combinatorialHandbook}).

\begin{lemma}
A $RTD(\alpha,n)$ is equivalent to a $TD(\alpha+1,n)$.
\end{lemma}

\begin{lemma} \label{RTDexists}
For $n$ a prime power and $\alpha \leq n$, there exists a $TD(\alpha+1,n)$ and hence there exists a $RTD(\alpha,n)$.
\end{lemma}



\begin{construction}\label{PBD5} 
Consider a $RTD(\alpha,n)$ $(V,G,B)$ with  resolution classes $R_1, \ldots, R_n$, and let $G=\{G_1, \ldots, G_{\alpha}\}$.
We take $ 0 \leq x \leq n$, $0 \leq \gamma \leq \alpha$ and $0 \leq u \leq n-x$. 
We will form a $RPBD(v,M,1,n+1)$, $(\hat{V}, \hat{B})$, by deleting a set of $(n-x)\gamma+u$ points, which we label as $\bar{V}$. 
The points $\bar{V}$ that we delete will be $n-x$ points from $G_{i}$ for each $i$ with $\alpha- \gamma+1\leq i \leq \alpha$, and $u$ points of $G_{\alpha-\gamma}$.
Each point that was removed from a group is also removed from any block that contains it. 
This gives point set $\hat{V} = V \setminus \bar{V}$, block set $\hat{B} = \{b \setminus \bar{V} \mid b\in G \cup B\}$, and $n+1$  resolution classes $\hat{R}_i = \{b \setminus \bar{V} \mid b \in R_i \}$ for $1\leq i \leq n$ and $\hat{R}_\infty = \{b\setminus \bar{V} \mid b \in G\}$.


This results in a $RPBD(n\alpha - n\gamma+x\gamma-u,M,1,n+1)$ with $M = \{\alpha-(\gamma+1), \ldots, \alpha \} \cup \{x,n-u,n\}$.

\end{construction}




%//NOTE: why do we have $x=5$? Explain that this is because it covers more odd cases during computation. MAYBE we dont need to explain this?
It will be useful to summarize  the results of this section, which yield the following lemma:


\begin{lemma} \label{ZhuSummary}
Take $n$ a prime power and positive integers $\alpha$, $x$, $\gamma$, and $u$ such that $\alpha \leq n$, $ 0 \leq x \leq n$, $0\leq \gamma \leq  \alpha$, and $0 \leq u \leq n-x$.
Suppose there exists integers $d_i \geq 1$, $1 \leq i \leq n$ and $d_{\infty}\geq 1$, such that for each $b$ with $\alpha - (\gamma +1) \leq b \leq \alpha$  there exists an idempotent $(3,b-d_i,b)$-Latin trade when $1\leq i \leq n$, and for each $b \in \{x,n-u,n\}$ there exists an idempotent $(3,b-d_{\infty},b)$-Latin trade. Then there exists an idempotent $(3, v+n-\sum_{i=1}^n d_i - d_{\infty}, v)$-Latin trade, where $v=n\alpha - n\gamma+x\gamma-u$.
\end{lemma}
\begin{proof}
For these values of $n$, $\alpha$, $x$, $\gamma$, and $u$, Lemma \ref{RTDexists} gives us a $RTD(\alpha,n)$, which we can use in Construction \ref{PBD5} to yield a resolvable pairwise balanced design that can be used in Theorem \ref{RPBD2TradeIdempotent} along with the given idempotent $(3,b-d_i,b)$-Latin trades that have been assumed to exist, for $i \in \{1,\ldots, n, {\infty}\}$, to yield the result.
\end{proof}

%TODO
%\begin{theorem}
%Suppose there exists a $TD(\alpha+1,n+1)$
%Suppose there exists a $RPBD(v,M,1)$ with resolution classes $R_1, \ldots, R_n$. Suppose there exists integers $k_i$, $1 \leq i \leq n$, such that for each $b \in R_i$ there exists a $(\mu,|b|-k_i,|b|)$-Latin trade. Then there exists a $(\mu, \sum_{i=1}^n k_i, v)$-Latin trade.
%\end{theorem}
%\begin{proof}
%\end{proof}

%Assume $(V, \mathcal{G},\mathcal{B})$ is a $TD(\alpha+1,n)$ with groups $\mathcal{G}=\{G_1, \ldots, G_{\alpha+1}\}$ and block set $\mathcal{B}$. We remove group $G_{\alpha+1} \in \mathcal{G}$ and form a $RTD(\alpha,n)$ with resolution classes $R_i$ such that $\{b_1, \ldots, b_\alpha\} \in R_i$ if and only if $ i \in G_{\alpha+1}$ and $ \{b_1, \ldots, b_\alpha, i \} \in \mathcal{B} $. We also take $R_0$ to be the resolution class defined as the set $\mathcal{G} \setminus G_{\alpha+1}$.  

%We further remove $u_i$ points from the $i$th group (removing the points from any blocks in which they appear), for $1 \leq i \leq n- (\mu+2)$, insisting $0 \leq u_i \leq n-\mu$. For practicality, we assume $u_i<n-\mu$ implies $u_{i+1}=0$. Define $\gamma$ to be the integer with $u_{\gamma+2}=0$, $0<u_{\gamma+1} \leq n-\mu$ and $u_i=n-\mu$ for $1 \leq i \leq \gamma$.  Now we relabel each remaining point with the values $1$ to $m=\alpha \cdot n - \gamma \cdot (n-\mu) -u_{\gamma+1}$. Define $\mathcal{U}$ to be this resulting structure.

%Fix integers $d_0 \geq 0$ and $d_i \geq 1$, for $1 \leq i \leq n$, such that for each block $b \in R_i$ there exists some $(\mu,\|b\|-d_i,\|b\|)$-Latin trade. We also insist that if $d_i \geq 1$, that the diagonal cells $(j,j)$ of all the $(\mu,\|b\|-d_i,\|b\|)$-Latin trades are empty, $1 \leq i \leq \|b\|$. 

%For $0 \leq i \leq n$ and each block $b=\{b_1, \ldots, b_s\} \in R_i$ we place a $(\mu,\|b\|-d_i,\|b\|)$-Latin trade (with diagonal cells empty if $d_i \geq 1$) on element set $\Omega=b$ in positions $b \times b$ of $\mathcal{T}$. 





%TODO: does M need more explaining?



%Prof.\ Lie Zhu suggested to modify a construction from (Combinatorial designs, W.D.Wallis, pg189), which deals with the construction of pairs of idempotent MOLS. The basic modification of W.D. Wallis's Theorem 10.7 we will use is as follows.




















\section{Result when $\mu=3$}




We will develop an inductive proof for the existence of idempotent $(3,k,m)$-Latin trades for $m >194$, however we will require the knowledge of the existence of a great deal of base cases. To this end, we will use a computer program to combine the results so far stated in this paper to deduce the spectrum $\mathcal{IS}^3_m$ for $m \leq 2^{18}$.
We will create two computer programs, Program A and Program B, which are implemented in C++ \cite{prog1}. 

We begin by finding the spectrum $\mathcal{IS}^3_m$ for $m \leq 5618$ using Program A.
The value $5618=2\cdot 53^2$ was chosen as there was some difficulty  filling in the spectrum $\mathcal{IS}^3_{5618}$ later on, stemming from the fact that not enough is known about $\mathcal{IS}^3_{2\cdot53}$, and so Lemma \ref{specLem3} cannot be used to fill in the spectrum  $\mathcal{IS}^3_{5618}$.   
We split the computation into four parts. 
Parts $1$, $2$, and $4$ are straightforward to program, but there is some complications with Part $3$. 
We begin with a $(5618+1)\times (5618+1)$ array of booleans $A=[a_{k,m}]$, where we set $a_{k,m}=false$ for each $0 \leq k,m \leq 5618$. 
When we find that there does exist an idempotent $(3,k,m)$-Latin trade, we set $a_{k,m}=true$. 

PART 1:
As there trivially exists an idempotent $(3,0,m)$-Latin trade, we set $a_{0,m}=true$ for each $0 \leq m \leq 5618$. 
As the existence of idempotent $(3,k,m)$-Latin trades in Theorem \ref{mMinus1}, Theorem \ref{idemptrades3}, Theorem \ref{new_3way_results}, and Theorem \ref{spanned_results} do not depend on the existence of smaller idempotent $(3,k',m')$-Latin trades, we set $a_{k,m}=true$ for these values.


We can use the idempotent $(2,k_1,2m'+1)$-Latin trades of Theorem \ref{idemptrades2} when $3 \leq k_1 < 2m'+1$, $2m'+1 \geq 5$, and $(k,2m'+1) \neq (3,5)$, along with the $(2,k_2,m_2)$-Latin trades of Theorem \ref{AllBitrades} for $2 \leq k_2 \leq m_2$, with $k_i =2$ only if $m_i$ is even,  with Theorem \ref{multiplication} to yield an idempotent $(3,k_1 k_2,(2m'+1) m_2)$-Latin trade. 
This does not depend on the existence of smaller idempotent $(3,k,m)$-Latin trades, so we set $a_{k_1 k_2,(2m'+1) m_2}=true$ under these conditions. 

PART 2:
Theorem \ref{addition} and Theorem \ref{extendedTheorem} each require the knowledge of  the existence of smaller idempotent $(3,k,m)$-Latin trades, however we can gather this information from what we have stored in $A$. 
Theorem \ref{extendedTheorem} also uses the existence of non-idempotent $(3,m,m)$-Latin trades, and a non-idempotent $(3,5,6)$-Latin trade (an example of a $(3,5,6)$-Latin trade is shown in the next section). 


PART 3:
Programming Lemma \ref{ZhuSummary} is not completely straightforward, as the time required can be quite large if not done with due care. 
We implement Lemma \ref{ZhuSummary} twice. 
The first implementation uses $d_\infty=1$, and the second implementation uses $\gamma=0$. 
Both of these restrictions speed up the computation immensely, and the values not covered by one are covered by the other.  
By first looping over $\alpha$ and $\gamma$, we can store the values of $d_i$ such that there exists an idempotent $(3,b-d_i,b)$-Latin trade for each $b$ with $\alpha - \gamma-1\leq b \leq \alpha$. Then we are able to find the possible values of  $\sum_{i=1}^n d_i$ without much extra computation as we increase $n$, by storing the previously computed values of $\sum_{i=1}^{n-1} d_i $.

PART 4:
We once again apply the procedure for Theorem \ref{addition}, which fills in a couple of gaps in the spectrum introduced incidentally in Part 3.


Performing this computation gives the following lemma:

\begin{lemma} \label{mCompSmall}
For $14 \leq m \leq 5618$, there exists idempotent $(3,k,m)$-Latin trades for $5 \leq k \leq m$ except, perhaps, for those values in Table \ref{tab:unknownExistence}.
\end{lemma}






We need to extend the base results further, which we achieve by way of another computer program, Program B. 
We begin with an array of $2^{18}+1$ booleans $B = [b_{m}]$, where we set $b_m = false$ for each $0 \leq m \leq 2^{18}$. 
When we find that there exists idempotent $(3,k,m)$-Latin trades for every $5 \leq k \leq m$, we set $b_m = true$. 
We begin by setting the values of $b_m$ to be true when the values $a_{k,m}$  each are true for $5 \leq k \leq m$. 
Then Lemma \ref{specLem3} with $y=5$ tells us that we can set $b_{nm}$ to be true whenever $m \geq 10$, $b_m$ is true, $n \geq 3$, and if $n =6$ then $m \geq 20$. 


In the case that $m$ is a prime or twice a prime, we can apply Theorem \ref{new_3way_results} and Theorem \ref{spanned_results} to yield idempotent $(3,k,m)$-Latin trades for $5 \leq k \leq l_1$, and we can apply Lemma \ref{ZhuSummary}  with $x=7$ to yield idempotent $(3,k,m)$-Latin trades for $l_2 \leq k \leq m$, for some integers $l_1,l_2$.
To save computation, we only consider Lemma \ref{ZhuSummary} with $\gamma \in \{0,1\}$ and $d_{\infty}=1$.
Note that this means $(3,b-d_{\infty},b)$-Latin trades always exist for $b\geq7$, as stated in Theorem \ref{mMinus1}. 
Our program checks if $b_{\alpha-2}=b_{\alpha-1}=b_{\alpha}=true$ for each $\alpha \geq 16$ and $n \geq \alpha$.
In this case, we assume $d_i \leq \alpha-7$. 
Then the conditions of a $(3,\alpha-(\gamma +1)-d_i,\alpha-(\gamma +1))$-Latin trade existing hold independently of whether $\gamma=0$ or $\gamma=1$. 
Then it will be more convenient to write $u' = \gamma(n-7)+u$. 
If so, for each $m=n \alpha-u'$ with $n \geq \alpha$, $0 \leq u' \leq 2(n-7)$, and $m$ a prime or twice a prime, we know that there exists an idempotent $(3,k,m)$-Latin trade for each $k$ with $m + n - n(\alpha-7)-1 = 8n-u'-1 \leq k \leq m-1$. 
To find the existence of idempotent $(3,k,m)$-Latin trades with $ 5 \leq k <8n-u'-1$, we find the greatest $\lambda$ with $m = \lambda(\lambda + a) +b$, $a>b$, $\gcd(m,\lambda)=1$, and $\lambda \geq 5$. 
Then  Theorem \ref{spanned_results} yields the existence of idempotent $(3,k,m)$-Latin trades with $ 18 \leq k \leq \lambda^2$. 
There exists idempotent $(3,k,m)$-Latin trades for $5 \leq k \leq 17$ by Theorem \ref{new_3way_results}.
If $8n-u'-1 \leq \lambda^2$, then there exists an idempotent $(3,k,m)$-Latin trade for $5 \leq k \leq m-1$, and so we set $b_m$ to be true.

Performing this computation gives the following lemma:

\begin{lemma} \label{mCompLarge}
For $14 \leq m \leq 2^{18}$, there exists idempotent $(3,k,m)$-Latin trades for $5 \leq k \leq m$ except, perhaps, for those values in Table \ref{tab:unknownExistence}.
\end{lemma}


%//TODO - make program available online

%
%!!!!!!!!!!!!!!Should we take $r\geq 9$ as this is what we have?
%
%Take $\alpha,\gamma \in \mathbb{Z}^+$, $\alpha > \gamma+6$, and define  $D_{\alpha \gamma}=\{1\} \cup \{2 \leq d \leq \alpha - \gamma - 5 \mid d \ even\}$. 
%The following will be useful in proving a proceeding lemma:
%\begin{lemma}  \label{dIdentity}
%Take $\alpha,\gamma \in \mathbb{Z}^+$, $\alpha > \gamma+6$. It holds that:
%\[
%\{d_{\infty} + \sum_{i=1}^{n} d_i \mid d_{\infty} \in \{0,1\} \  and \ d_i \in D_{\alpha \gamma}\} \supseteq \{n, n+1, \ldots, n(\alpha-\gamma-6)+1\}
%\]
%\end{lemma}
%\begin{proof}
%Let $E=\{d_{\infty} + \sum_{i=1}^{n} d_i \mid d_{\infty} \in \{0,1\} \  and \ d_i \in D_{\alpha \gamma}\}$. 
%Take $E_1=\{d_{\infty} + \sum_{i=1}^{n} d_i \mid d_{\infty} \in \{0,1\} \  and \ d_i \in D_{\alpha \gamma}\}$. 
%In the case that $\alpha - \gamma - 5$ is even, $E_1 = \{2n, 2n+1, \ldots, n (\alpha - \gamma - 5)+1\}$, and in the case that it is odd,  $E_1 = \{2n, 2n+1, \ldots, n (\alpha - \gamma - 6)+1\}$.
%Then $E_1 \supseteq \{2n, 2n+1, \ldots, n (\alpha - \gamma - 6)+1\}$. 
%Take $E_2=\{d_{\infty} + \sum_{i=1}^{n} d_i \mid d_{\infty} \in \{0,1\} \  and \ d_i \in D_{\alpha \gamma}\}$. Then $E_2 = \{n, n+1, \ldots, 2n+1\}$. 
%It is clear that $E \supseteq E_1 \cup E_2 \supseteq   \{2n, 2n+1, \ldots, n (\alpha - \gamma - 6)+1\} \cup \{n, n+1, \ldots, 2n+1\}$. The result follows.
%\end{proof}
%
%
%//TODO - note that $D_{\alpha}$ is unnecessary here.



We have been able to apply Lemma \ref{ZhuSummary} in this computation as we have been able to run a procedure to check which integers $n$ are prime powers. 
In order to create a theoretic construction, we restrict the prime powers that we use, so that $n$ is of the form $2^p$, for an integer $p$. 
We are then able to show, despite this restriction, that Lemma \ref{ZhuSummary} can yield a large portion of the spectrum of $(3,k,m)$-Latin trades for all $m \geq 2^{18}$. 

\begin{lemma} \label{m_zhu}
Take $p\geq 10$. 
Suppose there exists idempotent $(3,k',m')$-Latin trades for $5 \leq k' \leq m'-1$ and $2^{p-2}-6 \leq m' \leq 2^{p}$. % such that $m'-k' \in D_{\alpha \gamma}$. 
 Then there exists idempotent $(3,k,m)$-Latin trades for $14 \cdot 2^p \leq k \leq m$ and  $2^{2p-2}< m \leq 2^{2p}$.
\end{lemma}
\begin{proof}
Take $\alpha = n=2^p$, $p \geq 10$, $0 \leq \gamma \leq  \gamma_{max}$, $\gamma_{max}=2^p-2^{p-2}+5$,  $x=7$, and $0 \leq u \leq n-7$.
We assume the existence of idempotent $(3,k',m')$-Latin trades when $5 \leq k' \leq m'-1$ and $\alpha - (\gamma_{max}+1) \leq m' \leq \alpha$, noting that $\alpha - (\gamma_{max}+1) = 2^{p-2}-6$.
There exists a $(3,m'-1,m')$-Latin trade for $m' \in \{7,n-u,n\}$ by Theorem \ref{mMinus1}, as $m' \geq 7$.

Then Lemma \ref{ZhuSummary} with these idempotent $(3,k',m')$-Latin trades yields an idempotent $ (3,m + n -\sum_{i=1}^n d_i - d_{\infty},m)$-Latin trade with $m=n \alpha - \gamma(n-7)-u$, where $1 \leq d_i \leq \alpha - (\gamma+1)-5$ for $1 \leq i\leq n$, and $d_\infty =1$.
Taking $d_i \leq \alpha - (\gamma+1) -5$ assures us that $m' - d_i \geq 5$ for each $m'$ with $\alpha - (\gamma+1) \leq m' \leq \alpha$, and so an idempotent $(3,m'-d_i,m')$-Latin trade exists by our assumptions.

Then this procedure yields an idempotent $(3,k,m)$-Latin trade for each $k \in \{m+n-n(\alpha-(\gamma+1)-5)-1, \ldots, m-1\} = \{7 \gamma + 7n - u -1, \ldots, m-1\} \supseteq \{14n, \ldots, m-1\}$, which holds for each $m= n\alpha - \gamma(n-7) -u$ within $n \alpha -(\gamma_{max}+1)(n-7) \leq m \leq n \alpha$.
In particular, it holds that $n \alpha -(\gamma_{max}+1)(n-7) = 2^{2p}-(2^p-2^{p-2}+6)(2^p-7) = 2^{p-2}-3\cdot 2^{p-2}+42 \leq 2^{2p-2}$ and so there exists an idempotent $(3,k,m)$-Latin trade for $2^{2p-2} \leq m \leq 2^{2p}$ and $14\cdot2^p \leq k \leq m-1$, showing the result.
\end{proof}




%/*
%\begin{lemma} \label{NewTheorem8}
%Consider natural numbers $a, k, k'$ with $k \leq k'$. If for every $k' + 1 \leq l \leq k'+a$ there exists a $(\mu, k, l)$-Latin trade, then for any $m \geq \lceil \frac{k-1}{a} \rceil k$ there exists a $(\mu, k,m)$-Latin trade.
%\end{lemma}
%\begin{proof}
%$\alpha$ st $\alpha\geq \lceil \frac{k-1}{a} \rceil$, then $\alpha (k + a) \geq (\alpha+1)k-1$.
%For each $m' \in \{\alpha k, \ldots, \alpha (k + a) \}$ there exists integers $\alpha'$ and $\beta'$ such that $m' = \sum_{i=1} ^{\alpha'}(k + \beta'-1) + \sum_{i=\alpha'} ^{\alpha}(k + \beta')$, where $\beta' \leq a$.
%As there exists a $(\mu,k,K+\beta'-1)$-Latin trade and a $(\mu,k,K+\beta')$-Latin trade
%\end{proof}
%*/


\begin{theorem} \label{thm:final}
For $m \geq 5$, there exists an idempotent $(3,k,m)$-Latin trade for $5 \leq k \leq m-1$ except possibly for those values in Table \ref{tab:unknownExistence} and when $(k,m) \in \{(5,6),(5,10),(5,13)\}$.
\end{theorem}
\begin{proof}
Define $P(r)$ to be the statement ``There exists idempotent $(3,k,m)$-Latin trades for each $5 \leq k \leq m-1$ and $2^{r-1}-6 \leq m \leq 2^r$''.  

Lemma \ref{mCompLarge}  shows $P(r)$ is true for $9 \leq r \leq 18$.
Assume for the sake of strong induction that $P(r)$ is true for $9 \leq r <R$, with $R \geq 19$. 
Then $P(\lceil \frac{R}{2} \rceil-1)$ and $P(\lceil \frac{R}{2} \rceil)$ are true, as $9 \leq \lceil \frac{R}{2} \rceil-1 < R$.
This makes the premise of  Lemma \ref{m_zhu} with $p=\lceil \frac{R}{2} \rceil$ true, and so there exists an idempotent $(3,k,m)$-Latin trade for $14 \cdot 2^{\lceil \frac{R}{2} \rceil}\leq k \leq m-1$ and $2^{R-1} \leq m \leq 2^{R}$, as $2^{2\lceil \frac{R}{2}\rceil-2} \leq 2^{R-1}$ and $2^R \leq 2^{2\lceil \frac{R}{2}  \rceil}$.
 As $P(R-1)$ is true, we can apply Theorem \ref{OldTheorem8} with $k' = 2^{R-2}$ and for each $k \in \{5, \ldots, 2^{R-2} -1\}$, which yields idempotent $(3,k,m)$-Latin trade for $5 \leq k \leq 2^{R-2}-1$ and  $2^{R-1} \leq m \leq 2^{R}$. 
As $14 \cdot 2^{\lceil \frac{R}{2} \rceil} \leq 2^{R-2}-1$, this shows there exists an idempotent $(3,k,m)$-Latin trade for $5 \leq k \leq m-1$ and  $2^{R-1} \leq m \leq 2^{R}$, and so $P(R)$ is true. 
By strong induction, $P(r)$ is true for $r \geq 9$.
Theorem \ref{new_3way_results} and Lemma \ref{mCompLarge} complete the result when $5 \leq m < 2^{8}-6$.
\end{proof}


\section{Results}

As $\mathcal{IS}^{\mu}_m \subseteq \mathcal{S}^{\mu}_m$,  the results of Theorem \ref{thm:final} also yield the existence of $(3,k,m)$-Latin trades for identical values of $k$ and $m$. 
There are a few more non-idempotent $(3,k,m)$-Latin trades that we can find.

There does not exist a large set of idempotent Latin squares of order $n = 6$, however 
there does exist a $(4,5,6)$-Latin trade given by:

\ 

 \begin{tabular}{| c | c | c|c|c| c |  }
    \hline
    $(2,3,4,5)$ & $\bullet$ & $(1,4,5,3)$ & $(5,2,1,4)$ & $(3,5,2,1)$ & $(4,1,3,2)$  \\
    \hline
    $\bullet$ & $(3,2,5,4)$ & $(6,3,4,5)$ &  $(4,5,2,6)$ & $(5,6,3,2)$ & $(2,4,6,3)$ \\
        \hline
    $(1,4,5,3)$ & $(4,5,3,6)$ & $(5,1,6,4)$  & $\bullet$ &  $(6,3,1,5)$ & $(3,6,4,1)$  \\
    \hline
    $(4,5,2,1)$ & $(5,6,4,2)$ & $\bullet$  & $(1,4,6,5)$ &  $(2,1,5,6)$ & $(6,2,1,4)$  \\
        \hline
    $(5,1,3,2)$ & $(6,3,2,5)$ & $(3,5,1,6)$  & $(2,6,5,1)$ &  $(1,2,6,3)$ & $\bullet$  \\
        \hline
    $(3,2,1,4)$ & $(2,4,6,3)$ & $(4,6,3,1)$  & $(6,1,4,2)$ &  $\bullet$ & $(1,3,2,6)$\\
        \hline
  \end{tabular}

\ 

Here, the partial Latin squares have been concatenated, so that cell $(r,c)$ has been filled with the ordered $4$-tuple given by $(t_1(r,c),t_2(r,c),t_3(r,c),t_4(r,c))$, where the four partial Latin squares $T_i = [t_i(r,c)]$, $1 \leq i \leq 4$, form the $(4,5,6)$-Latin trade.
Then this yields a non-idempotent $(3,5,6)$-Latin trade.

Applying Theorem 7 of \cite{3way} to the combination of a $(3,5,6)$-Latin trade and a $(3,5,7)$-Latin trade yields a $(3,5,13)$-Latin trade. 
There exists a $(3,m,m)$-Latin trade for $m \geq 3$ by Theorem \ref{old_3way_results}, and a $(3,m,2m)$-Latin trade by applying Theorem 7 of \cite{3way} to the combination of two $(3,m,m)$-Latin trades.
Then the primary result of this paper, combined with previous results \cite{3way}, can be written as the following theorem:

%///
%There exists a $(3,5,13)$-Latin trade by applying Theorem 7 of \cite{3way} containing a $(3,5,6)$-Latin trade and a $(3,5,7)$-Latin trade.
%There exists a $(3,m,m)$-Latin trade for $m \geq 3$ by Theorem \ref{old_3way_results}, and a $(3,m,2m)$-Latin trade by applying Theorem 7 of \cite{3way} containing two $(3,m,m)$-Latin trades.
%Then the primary result of this paper, combined with previous results \cite{3way}, can be written as the following theorem:

\begin{theorem}
For $m \geq 4$ there exists a $(3,k,m)$-Latin trade for $4 \leq k \leq m$ except, perhaps, for those unstared values in Table \ref{tab:unknownExistence} and for $(k,m) =(4,11) $, and except for those values with $(k,m) \in \{(4,6),(4,7)\}$.  For $m \geq 3$, there exists a $(3,3,m)$-Latin trade only when $3 | m$. 
\end{theorem}

This leaves us with $194$ exceptions for which we do not know if a $(3,k,m)$-Latin trade exists, and $2$ exceptions when we know that there does not exist a    $(3,k,m)$-Latin trade.

%%%%%%FOR THESIS%%%%%%%%%%%%%%%

%We note that this $(4,5,6)$-Latin trade can be reduced to a $(3,5,6)$-Latin trade by removing one of the partial Latin squares, the later being an unknown case in \cite{3way}.
%Finding circulant $(\mu,k,m)$-Latin trades is the typical first method of approach of finding $(\mu,k,m)$-Latin trades, as the search space is much smaller. 
%A particular difficulty in finding this $(4,5,6)$-Latin trade was that there does not exist a circulant $(4,5,6)$-Latin trade, nor a circulant $(3,5,6)$-Latin trade, which was seen during a preliminary brute force search for small $(3,k,m)$-Latin trades. 
%
%
%The above $(4,5,6)$-Latin trade was found using the following search technique:
% Any $(4,5,6)$-Latin trade on symbol set $[6]$ can have its rows and columns permuted to have $r$th row set $R_r(\mathcal{T}) = [6]\setminus {r}$ and $c$th column set  $C_c(\mathcal{T}) = [6]\setminus {c}$, $r,c \in [6]$. 
%We construction four arrays of size $6 \times 6$.  
%We select a set of $6$ cells such that each columns and each row is represented precisely once, and we leave these $6$ cells unfilled in each of the $4$ arrays.  
%For each of the four arrays, the cell $(r,c)$ can only contain symbols from the set $R_r(\mathcal{T})  \cap C_c(\mathcal{T})$. 
%Running through each of the cells $(r,c)$, we assign a symbol from $R_r(\mathcal{T})  \cap C_c(\mathcal{T})$ to each of the four arrays, making sure that the arrays are disjoint,  and that any symbol appears at most once per row and at most one per column in any single array.
%If no such assignment is possible, we backtrack to a previously assigned cell, and make the next consecutive assignment in that cell. 


%TODO




\begin{table}
\begin{center}
\begin{tabular}{| c | c | r |  } 
    \hline
    $m$ & $k$\\
    \hline
    $22$ & $ 19 $\\
    \hline
    $23$ & $ 19,20, 21$ \\
    \hline
    $26$ & $ 19,21,23 $ \\
    \hline
    $29$ & $19, 20, 21,22,23,24,25,26,27$ \\
    \hline
    $31$ & $ 19, 20,21,22, 23,24, 25,26,27,28,29$ \\
    \hline
    $34$ & $19,21,23,25,27,29,31$ \\
    \hline
    $37$ & $33,34,35$ \\
    \hline
    $38$ & $19^*,21,23,25,27,29,31,33,35$\\
    \hline
    $41$ & $35,36,37,38,39$ \\
    \hline
    $43$ & $36,37,38,39,40,41$ \\
    \hline
    $46$ & $27,29,31,33,35,37,39,41,43$\\
    \hline
    $53$ & $50,51$ \\
    \hline
    $58$ &$29^*,31,33,35,37,39,41,43,45,47,49,51,53,55$\\
    \hline
    $59$ & $55,56,57$ \\ %51\leq k\leq 67 \text{ such that } k \text{ is odd}
    \hline
    $62$ & $31^*,33,35,37,39,41,43,45,47,49,51,53,55,57,59$ \\
    \hline
    $74$ & $ 35,37^*,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71 $ \\
    \hline
    $82$ & $41^*,43,45,47,49,51,53,55,57,59,61,63,65 $ \\
    \hline
    $86$ & $51, 53, 55, 57, 59, 61, 63, 65, 67, 69$ \\
    \hline
    $94$ & $51, 53, 55, 57, 59, 61, 63, 65$ \\
    \hline
    $106$ & $53^*, 55, 57, 59, 61, 63, 65, 67, 69$ \\
    \hline
    $122$ & $59, 61, 63, 65, 67, 69, 71, 73, 75, 77$ \\
    \hline
    $134$ & $67^*, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93$ \\
    \hline
    $146$ & $83, 85, 87, 89, 91, 93, 95, 97, 99, 101$\\
    \hline
    $158$ & $83, 85, 87, 89, 91$ \\
    \hline
    $194$ & $97^*, 99, 101$ \\
    %\hline
   % $142$ & $\{73,75 $ \\      
    \hline
  \end{tabular} \caption{Values where no idempotent $(3,k,m)$-Latin trade is known to exist with $6 \leq k \leq m-1$. The starred values indicate when a non-idempotent $(3,k,m)$-Latin trade is known to exist.}  \label{tab:unknownExistence}
\end{center}
\end{table} 





\section{Future work}

Given the relative success of finding base rows from Theorem \ref{new_3way_results}, where the program terminated rather early within the search space, it  seems reasonable that  $(3,k,m)$-Latin trades with values in Table \ref{tab:unknownExistence} could exist, and we can use this as evidence towards a conjecture:  
\begin{conjecture}
There exists a $(3,k,m)$-Latin trade exactly when $k=3$ and $3 |m$, and when $4 \leq k \leq m$, except in the cases that $(k,m)\in \{(4,6),(4,7),(4,11)\}$.
\end{conjecture}

It also seems that similar techniques used in this paper could be used to fill in the spectrum of $(4,k,m)$-Latin trades. 
In addition, it may be of interest to investigate the spectrum of circulant $(\mu,k,m)$-Latin trades.

A $\mu$-way Latin trade $(Q_1, \ldots, Q_{\mu})$ can be said to be \emph{primary} if there is no $\mu$-way Latin trade $(R_1, \ldots, R_{\mu})$ such that $R_{\alpha}\subsetneq Q_{\alpha}$. A $\mu$-way Latin trade is said to be \emph{minimal} if there is no partial Latin square $R \subsetneq Q_1$ such that there exists a $2$-way trade $(R,R')$.
Primary $(2,k,m)$-Latin trades were conjectured to exist for $3 \leq k \leq m$ in \cite{NumbTran}. 
It would be of interest to investigate primary and minimal $(\mu,k,m)$-Latin trades in the future.

%Generalization of Construction \ref{PBD5}.

\subsection*{Acknowledgements}
The first author would like to thank  the hospitality of Soochow University where partial research was done during his visit. The authors would also like to thank Prof.\ L.~Zhu for many suggestions, and D.~Donovan for extensive proofreading.





%
%\begin{table}[c]
%\begin{center}
%\begin{tabular}{| c | c | r |  } 
%
%    \hline
%    $m$ & $k$ such that existence is unknown or known not to exist \\
%    \hline
%    $6$ & $ 4 $ \\
%    \hline
%    $7$ & $ 4 $ \\
%    \hline
%    $11$ & $ 4 $ \\
%    \hline
%    $17$ & $ 14 $ \\
%    \hline
%    $19$ & $ 14,16,17 $ \\
%    \hline
%    $22$ & $ 17,19 $\\
%    \hline
%    $23$ & $ 14,16,17,18,19,20,21 $ \\
%    \hline
%    $26$ & $ 17,19,21,23 $ \\
%    \hline 				
%    $29$ & $17 \leq k \leq 27$ \\
%    \hline 
%    $31$ & $16\leq k \leq 29$\\
%    \hline
%    $34$ & $ 19,21,23,25,27,29,31 $ \\
%    \hline
%    $37$ & $33\leq k \leq 35$ \\
%    \hline
%    $38$ & $ 21,23,25,27,29,31,33,35 $ \\
%    \hline
%    $41$ & $35 \leq k \leq 39$ \\
%    \hline
%    $43$ & $36 \leq k \leq 41$ \\
%    \hline
%    $46$ & $27 \leq k \leq 41  \text{ such that } k \text{ is odd}$\\
%    \hline
%    $58$ & $31 \leq k \leq 55 \text{ such that } k \text{ is odd}$ \\
%    \hline
%    $59$ & $55 \leq k \leq 57$ \\
%    \hline
%    $62$ & $33 \leq k \leq 49  \text{ such that } k \text{ is odd}$\\
%    \hline
%    $74$ & $39\leq k\leq 63 \text{ such that } k \text{ is odd}$ \\
%    \hline
%    $82$ &$43\leq k\leq 65 \text{ such that } k \text{ is odd}$\\
%    \hline
%    $86$ & $51\leq k\leq 67 \text{ such that } k \text{ is odd}$ \\
%    \hline
%    $94$ & $ 51,53 $ \\
%    %\hline
%   % $142$ & $ 73,75 $ \\      
%    \hline
%
%  \end{tabular} \caption{Values such that no $(3,k,m)$-Latin trade is known to exists}  \label{tab:unknownExistence}
%\end{center}
%\end{table}
%
%
%
%\begin{center}
%\begin{tabular}{| c | c | r |  } 
%    \hline
%    $m$ & $k$ such that existence is unknown or known not to exist \\
%    \hline
%    $6$ & $ 4 $ \\
%    \hline
%    $7$ & $ 4 $ \\
%    \hline
%    $11$ & $ 4 $ \\
%    \hline
%    $22$ & $ 17,19 $\\
%    \hline
%    $26$ & $ 17,19,21,23 $ \\
%    \hline
%    $34$ & $ 19,21,23,25,27,29,31 $ \\
%    \hline
%    $37$ & $33,35$ \\
%    \hline
%    $38$ & $ 21,23,25,27,29,31,33,35 $ \\
%    \hline
%    $41$ & $35 \leq k \leq 39$ \\
%    \hline
%    $43$ & $36 \leq k \leq 41$ \\
%    \hline
%    $46$ & $27 \leq k \leq 41  \text{ such that } k \text{ is odd}$\\
%    \hline
%    $58$ & $31 \leq k \leq 55 \text{ such that } k \text{ is odd}$ \\
%    \hline
%    $59$ & $55 \leq k \leq 57$ \\
%    \hline
%    $62$ & $33 \leq k \leq 49  \text{ such that } k \text{ is odd}$\\
%    \hline
%    $74$ & $39\leq k\leq 63 \text{ such that } k \text{ is odd}$ \\
%    \hline
%    $82$ &$43\leq k\leq 65 \text{ such that } k \text{ is odd}$\\
%    \hline
%    $86$ & $51\leq k\leq 67 \text{ such that } k \text{ is odd}$ \\
%    \hline
%    $94$ & $ 51,53 $ \\
%    %\hline
%   % $142$ & $\{73,75 $ \\      
%    \hline
%
%  \end{tabular}
%\end{center}





%
%10:   5 
%13:   5 
%22:   19 
%23:   19 20 21 
%26:   19 21 23 
%29:   19 20 21 22 23 24 25 26 27 
%31:   19 20 21 22 23 24 25 26 27 28 29 
%34:   19 21 23 25 27 29 31 
%37:   33 34 35 
%38:   19 21 23 25 27 29 31 33 35 







%Printing those m that DO NOT have a complete spectrum, from m=0 to $m=2^12$















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% \bibliographystyle{plain} 
% \bibliography{myBibFile} 
% If you use BibTeX to create a bibliography
% then copy and past the contents of your.bbl file into your.tex file


%HERE
%\section*{References}
%\bibliographystyle{plain}
%\bibliography{ref}


\newpage

\SquashBibFurther

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\end{document}