<s>[INST] For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles? There exists a Latin square of order $8$ which can be partitioned into $2 \times 4$ subrectangles:
$$
\begin{bmatrix}
\color{red} 1 & \color{red} 2 & \color{red} 3 & \color{red} 4 & \color{purple} 5 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 \\
\color{red} 2 & \color{red} 3 & \color{red} 4 & \color{red} 1 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 & \color{purple} 5 \\
7 & 8 & \color{blue} 1 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & 5 & 6 \\
8 & 5 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & \color{blue} 1 & 6 & 7 \\
\color{pink} 5 & \color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{green} 1 & \color{green} 2 & \color{green} 3 & \color{green} 4 \\
\color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{pink} 5 & \color{green} 2 & \color{green} 3 & \color{green} 4 & \color{green} 1 \\
\color{orange} 3 & \color{orange} 4 & \color{brown} 5 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{orange} 1 & \color{orange} 2 \\
\color{orange} 4 & \color{orange} 1 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{brown} 5 & \color{orange} 2 & \color{orange} 3 \\
\end{bmatrix}
$$
If we take the row-symbol parastrophe of this Latin square (i.e., replace each entry $(i,j,l_{ij})$ with $(l_{ij},j,i)$), then the entry colors define a decomposition of $K_{8,8}$ into $2$-regular spanning subgraphs of $K_{4,4}$.  I would like to generalize this.
Question: For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles?
(Note: We assume $a \leq b$.  Subrectangles must have $b$ symbols.  We don't assume that the boundaries of the subrectangles align.)
Observations:


*

*It's trivially possible when


*

*$a$ divides $b$ (construct a Latin square with blocks that are $b \times b$ subsquares)

*$a=1$ or $b=n$.


*The first non-trivial case is when $a=2$, $b=3$, and $n=6$.  If my code is correct, then it is impossible (by exhaustive search).  (I'm tempted to think this is just because the parameters are too small.)  The best possible is $4$ subrectangles, which is straightforward to construct.

*My code found a random Latin square which gave an $a=2$, $b=5$, and $n=10$ example:
$$
\begin{bmatrix}
\color{red} 7 & \color{red} 1 & 9 & 5 & \color{red} {10} & 4 & 8 & \color{red} 2 & 6 & \color{red} 3 \\
\color{red} 3 & \color{red} {10} & 5 & 8 & \color{red} 2 & 6 & 4 & \color{red} 1 & 9 & \color{red} 7 \\
\hline
\color{blue} 9 & \color{blue} 6 & 4 & 7 & 1 & \color{blue} 2 & \color{blue} {10} & \color{blue} 3 & 5 & 8 \\
\color{blue} 6 & \color{blue} 3 & 7 & 1 & 5 & \color{blue} 9 & \color{blue} 2 & \color{blue} {10} & 8 & 4 \\
\hline
\color{pink} 4 & 5 & 1 & \color{pink} 2 & \color{pink} 8 & 3 & 9 & \color{pink} 7 & \color{pink} {10} & 6 \\
\color{pink} 2 & 9 & 6 & \color{pink} {10} & \color{pink} 4 & 1 & 3 & \color{pink} 8 & \color{pink} 7 & 5 \\
\hline
\color{purple} 1 & 2 & \color{purple} 8 & 3 & 7 & \color{purple} {10} & 6 & \color{purple} 5 & 4 & \color{purple} 9 \\
\color{purple} 8 & 4 & \color{purple} {10} & 6 & 3 & \color{purple} 5 & 7 & \color{purple} 9 & 2 & \color{purple} 1 \\
\hline
\color{brown} 5 & 8 & \color{brown} 2 & 9 & 6 & 7 & \color{brown} 1 & 4 & \color{brown} 3 & \color{brown} {10} \\
\color{brown} {10} & 7 & \color{brown} 3 & 4 & 9 & 8 & \color{brown} 5 & 6 & \color{brown} 1 & \color{brown} 2 \\
\end{bmatrix}
$$
 [/INST] I eventually co-authored a paper which includes this topic.  The non-trivial results are:


*

*There exists a Latin square of order $n$ which decomposes into $2 \times (n/2)$ subrectangles for all even $n \not\in \{2,6\}$.

*There exists a Latin square of order n which decomposes into $3 \times (n/3)$ subrectangles if and only if $3$ divides $n$ and $n > 9$.

*For sufficiently large n, there exists a Latin square of order $n$ which partitions into $d \times (n/d)$ subrectangles if and only if $d$ divides $n$.

Akbari, Marbach, Stones, Wu, Balanced equi-n-squares Electron. J. Combin, 2020. (pdf)

The smallest cases we didn't resolve are

*

*$20 \times 20$ Latin squares into $4 \times 5$ subrectangles;

*$28 \times 28$ Latin squares into $4 \times 7$ subrectangles; and

*$30 \times 30$ Latin squares into $5 \times 6$ subrectangles.

</s>