OK, my proof is based in logic, not a formula.
4 rounds equals 12 playing partners for each player (3 each round).
You want each to play the other eleven, which means you can only duplicate one playing partner over all 4 rounds.
All told, 12 players must have exactly 1 duplicate each, spread over 4 rounds.
However, you can't even have two rounds of totally unique foursomes (in fact, at least half of the players must repeat a playing partner from the previous day).
Proof:
there are 3 groups each day - A, B, C
players 1,2,3,4 play in group A first day
where will they play day 2?
put 1 in A, 2 in B, 3 in C ... player 4 must go into a group with someone he played the previous round with.
Go ahead and try to design 2 totally unique rounds.
btw, you could make this work with 16 players, I think.
What this means is there are at least 6 duplicates each round after the first, for a minmum of 18 duplicates. Since your premise only allows for 12 duplicates, you can't win.
I'll take payment in beer ...
Show your work.