A simple way to state this problem is as a pair consisting of the order of the matrix and an upper bound on the entries (disallowing negative entries). For Guy’s variant (see [2]), where 0 entries are also disallowed and the determinants of the two matrices are constrained to be equal, the number of unique solutions are as follows (as shown in [1]):
Order, UBound | No. Solutions | Order, UBound | No. Solutions |
3, 3 | 0 | 3, 4 | 0 |
3, 5 | 0 | 3, 6 | 0 |
3, 7 | 0 | 3, 8 | 0 |
3, 9 | 3 | 3, 10 | 4 |
3, 11 | 5 | 4, 3 | 0 |
4, 4 | 0 | 4, 5 | 13 |
Of course, this table can be extended indefinitely. The difficulty of finding all solutions increases rapidly with the size of the domains.
The references page gives pointers to several solutions, both individuals and parametric families. Some examples include:
2 | 3 | 2 | 2 | 3 | 5 | 2 | 3 | 6 | 5 | 7 | 6 | 8 | 7 | 8 | 10 | 7 | 12 | ||||||
4 | 2 | 3 | , | 3 | 2 | 3 | , | 3 | 2 | 3 | , | 6 | 4 | 7 | , | 12 | 11 | 7 | , | 4 | 2 | 7 | |
9 | 6 | 7 | 9 | 5 | 7 | 17 | 11 | 16 | 17 | 16 | 20 | 17 | 15 | 16 | 17 | 12 | 20 |