We can say that an $n$-by-$n$ square is regular provided that:
Each of the integers from $0$ to $n^2 − 1$ appears in exactly one cell, and each cell contains only one integer (so that the square is filled), and
If we express the entries in base-$n$ form, each base-$n$ digit occurs exactly once in the units’ position, and exactly once in the $n$’s position.
Example with 3-by-3 regular square:
The square is regular because each of the ternary digits $0$, $1$, and $2$ appears exactly once in the units’ and $3$’s position in each row and each column.
Can someone help me construct an example of a 4-by-4 regular square, showing my answer in both decimal and base-$4$ notations.