Suppose and consider the triples where and . We can construct a array of triples with in the th row and th column. We work modulo .
For example here's a solution for
The number of triples = the number of distinct pairs of integers in and all triples are unique, where the order of the elements is taken into account. (For suppose is in our array, then and for some and thus if this triple is in the th row and th column, and if it is in the th row and th column. So its position is uniquely determined by its elements.
Moreover, if is in the array then none of and can be in the array, for if one of the positions of the elements is fixed the latter three triples cannot obey the construction rule, that its elements increase by since we know that , and increase by
Now consider a pair of integers , which appears in a given triple, . There cannot exist another triple in the array , say, with since is uniquely determined by and
The pair cannot appear in more than three triples, in whatever order (we've already noted that we cannot have both and hence the three and not six), since the remaining element is uniquely determined by and However, it cannot appear in less than three triples since the number of triples equals the number of pairs of distinct integers in so this would mean that another pair must appear in more than three triples, a contradiction.
Thus we have a solution for any odd .
So a solution for is: