Show that any $n \times n$ array based on $\{1,2,..,n\}$ is a Latin Square if and only if it is simultaneously orthogonal to R and C.

Question

I've been reading Design Theory by Zhe-Xian Wan, but I have been stuck on where to begin with this question - Page 95, Exercises 4.7, Question 4.1.

Could someone please give me a hint?

Let

\begin{equation} R = \begin{pmatrix} 1 & 1 & ... & 1\\ 2 & 2 & ... & 2 \\ \vdots & \vdots & & \vdots \\ n & n & ... & n \end{pmatrix} \end{equation}

\begin{equation} C = \begin{pmatrix} 1 & 2 & ... & n\\ 1 & 2 & ... & n \\ \vdots & \vdots & & \vdots \\ 1 & 2 & ... & n \end{pmatrix} \end{equation}

be $n \times n$ arrays.

Show that any $n \times n$ array based $\{1,2,...,n\}$ is a Latin Square if and only if it is simultaneously orthogonal to $R$ and $C$.

Answer 1

Orthogonal latin squares are defined here and here. The proof should follow from definition.

A latin square $M$ is such that $m_{ij} \neq m_{i'j}$ and $m_{ij} \neq m_{ij'}$ for $i\neq i'$ and $j\neq j'$. So if $(m_{ij},r_{ij}) = (m_{kl},r_{kl})$, then $m_{ij}=m_{kl}$, in which case $k\neq i$ and $l\neq j$ unless $i,j=k,l$. And the same business applies with ordered pairs $(m_{ij},c_{ij})$. So if $M$ is a latin square, it is orthogonal to $R$ and $C$.

Conversely, suppose $M\perp R$ and $M\perp C$. If $m_{ij}=m_{i'j}$ for $i\neq i'$, then $(m_{ij},c_{ij}) = (m_{i'j},c_{i'j})$. Similarly, if $m_{ij}=m_{ij'}$ for $j\neq j'$, then $(m_{ij},r_{ij}) = (m_{ij'},r_{ij'})$. So $M$ is a latin square.