[Math] Overlaying Latin squares of order 4

combinatoricslatin-square

Here are two latin squares overlayed upon each other to make one latin square, if you will. One "sub-latin" square is labeled with $1,2,3,4$ while the other is represented with $a,b,c,d$. There must be an $a$ corresponding to each of $1,2,3,4$ as you will see(and so on) and there must be $1$ corresponding to each $a,b,c,d$ as well and so on. 

                     (1,a)  (2,d)  (3,b)  (4,c)
                     (2,c)  (1,b)  (4,d)  (3,a)
                     (4,b)  (3,c)  (2,a)  (1,d)
                     (3,d)  (4,a)  (1,c)  (2,b)

Can we find three latin squares of order four such that any two of them will overlie on each other perfectly? How many different such Latin squares can we have?

EDIT:
Is there a method to creating mutually orthogonal squares to any extent?

Best Answer

You’re asking about mutually orthogonal Latin squares. The answer to the specific question is that yes, it is possible to find three mutually orthogonal Latin squares of order $4$, and that is the maximum. You’ll find a little more information and a few references at OEIS A001438.

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