>>14185709Ok i have some progress - a construction when N=6k+1 is prime:
Pick a primitive element g of Z_N. Divide all the residues into three sets:
S1 = [1, g^3, g^6, ... g^6k-3]
S2 = [g, g^4, g^7, ... g^6k-2]
S3 = [g^2, g^5, g^8, ... g^6k-1]
Now some parity counting shows you that you can pick a, b, c from S1, S2, S3 respectively, such that a+b+c=0.
And then you notice that ag^3 + bg^3 + cg^3=0 is another triple, and you can repeat this.
To give an example: 2 is a primitive root mod 13, we have the lists
[1, 8, 12, 5]
[2, 3, 11, 10]
[4, 6, 9, 7]
we find one triple (1, 3, 9), then we can generate the rest: (8, 11, 7), (12, 10, 4), (5, 2, 6)
You may look at pic related (the definition is not exactly what you are looking for. However when you look at the triplets listed at the bottom, these fit your problem, after adding the same triples negated mod n)
link for the screenshot:
https://books.google.pl/books?id=CViPe4WOJ3UC&pg=PA203Also you might look at this:
https://en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problemhttps://en.wikipedia.org/wiki/Steiner_system steiner triples, resolvable steiner systems look related, a more intensive google search might give you something
although i'm afraid i might be circling back to your original combinatorics problem
