There a n teams of n people, where n is a natural number greater than 1. In one round, each person from each team is grouped with one person from each of the other teams. This forms n sets of n people, where each set contains one person from each team. Is is possible do go for n rounds such that no two people are in the same set twice?
This problem was suggested to me by a teacher who used this scheme for some class exercise. It may be a generalization of the question “How do I rotate 3 groups of 4 people into teams....”
If n is a prime number, the answer is yes, and there is a simple formula for generating the sets. This uses the fact that the integers modulo n is a field when n is prime. Similarly, the answer is yes if n is a power of a prime, since we can match the numbers 1 to n with the elements of the finite field of n elements. But even for n = 6, the number of combinations gets quite large.
Another formulation of the problem: For any natural number n, can we find an n by $n^2$ matrix $a_{i,j}$ where each entry is a number from 1 to n and for any two rows $i_1$ and $i_2$ all of the ordered pairs $(a_{i_1,j},a_{i_2,j})$ are different? Can we do this is such a way, when you look at the n n-by-n submatrices (which correspond to the “rounds”), the entries in each row of each submatrix are distinct (i.e. the row is a permutation of 1..n)?
This is equivalent to finding a system of $n-1$ mutually orthogonal Latin squares (MOLS). Proof:
It is currently unknown whether there are any $n$ which are not prime powers for which $n-1$ MOLS of order $n$ exists. This is equivalent to finding a finite projective plane whose order is not a power of a prime.