Constructing a trio of MOLS of order $n=2^m$?

Question

Is it possible to have a trio of $3$ MOLS, each of which is of order $n=2^m$?

If so, then how to construct them?

So far I have only found ways to construct a pair of two MOLS $X,Y$, but not a trio of three $X,Y,Z$.

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I read this but it seems to talk about finding a pair, not a trio.

Answer 1

We have a set of 3 MOLS, if any two of them form a MOLS pair, by definition.

If this is what you are asking, yes they exist when $n=p^m,$ with $p$ a prime. For your case $p=2.$ They are constructed from Finite fields. As soon as $m\geq 2$ you have 3 MOLS.

See here in Wikipedia for a description of the construction. You will need to construct $GF(2^m)$ via a primitive element $\gamma \in GF(2^m)$ and follow the construction given there. This actually guarantees a full set of MOLS, i.e., $n-1$ of them.