orthogonal Latin squares
Given two Latin squares^{} ${L}_{1}=(A,B,{C}_{1},{f}_{1})$ and ${L}_{2}=(A,B,{C}_{2},{f}_{2})$ of the same order $n$, we can combine them coordinatewise to form a single square, whose cells are ordered pairs of elements from ${C}_{1}$ and ${C}_{2}$ respectively. Formally, we can form a function $f:A\times B\to {C}_{1}\times {C}_{2}$ given by
$$f(i,j)=({f}_{1}(i,j),{f}_{2}(i,j)).$$ 
This function $f$ says that we have created a new square $A\times B$, whose cell $(i,j)$ contains the ordered pair of values, the first coordinate of which corresponds to the value in cell $(i,j)$ of ${L}_{1}$, and the second to the value in cell $(i,j)$ of ${L}_{2}$. We may write the combined square ${L}_{1}*{L}_{2}$.
For example,
$$\left(\begin{array}{cccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill & \hfill d\hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill d\hfill & \hfill c\hfill & \hfill b\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill a\hfill & \hfill d\hfill & \hfill c\hfill \end{array}\right)*\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right)=\left(\begin{array}{cccc}\hfill (a,1)\hfill & \hfill (b,2)\hfill & \hfill (c,3)\hfill & \hfill (d,4)\hfill \\ \hfill (c,4)\hfill & \hfill (d,3)\hfill & \hfill (a,2)\hfill & \hfill (b,1)\hfill \\ \hfill (d,2)\hfill & \hfill (c,1)\hfill & \hfill (b,4)\hfill & \hfill (a,3)\hfill \\ \hfill (b,3)\hfill & \hfill (a,4)\hfill & \hfill (d,1)\hfill & \hfill (c,2)\hfill \end{array}\right)$$ 
In general, the combined square is not a Latin square unless the original two squares are equivalent^{}: ${f}_{1}(i,j)={f}_{1}(k,\mathrm{\ell})$ iff ${f}_{2}(i,j)={f}_{2}(k,\mathrm{\ell})$. Nevertheless, the more interesting aspect of pairing up two Latin squares (of the same order) lies in the function $f$:
Definition. We say that two Latin squares are orthogonal^{} if $f$ is a bijection^{}.
Since there are ${n}^{2}$ cells in the combined square, and ${C}_{1}\times {C}_{2}={n}^{2}$, the function $f$ is a bijection if it is either onetoone or onto. It is therefore easy to see that the two Latin squares in the example above are orthogonal.
Remarks.

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The combined square is usually known as a GraecoLatin square, originated from statisticians Fischer and Yates.
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(Bose) If $n\ge 3$, then ${L}_{1},\mathrm{\dots},{L}_{m}$ form a complete set of pairwise orthogonal Latin squares of order $n$ iff there exists a finite projective plane of order $n$.
References
 1 H. J. Ryser, Combinatorial Mathematics, The Carus Mathematical Monographs, 1963
Title  orthogonal Latin squares 

Canonical name  OrthogonalLatinSquares 
Date of creation  20130322 16:04:47 
Last modified on  20130322 16:04:47 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  12 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 62K10 
Classification  msc 05B15 
Synonym  mutually orthogonal Latin squares 
Synonym  MOLS 
Synonym  pairwise orthogonal Latin squares 
Defines  complete set of Latin squares 