4D Moser sets with d at least 2 have at most 42 points
Proof: Assume that such a set exists. We have already proved that it contains no point of type e. So we can assume that. This set does not contain such a point.
This set since it has 43 points must have at least 18 points of type c. See http://michaelnielsen.org/polymath1/index.php?title=Maple_calculations. Since there is a point of type d we can slice the set so that one side cube has its center slot filled and so must have 13 points or less. The other cubes must sum to 30. The other side cube must not have 16 points since then by the list of Pareto optimizers its distribution would be (4, 12, 0, 0) and it would contribute no points of type c to the Moser set. Then the other side cube could contribute at most 3 points of type c. So for there to be 18 the central cube must contribute 15 which is not possible.
This means the center cube must have at least 15 points for there to be a total of 43. Now we will show that the configuration has at most one point of type d.
Assume that there are more than one. Then no two can have the same c-statistic or we can slice and get two side cubes with center points filled which give each a total of at most 13 points which would force the center slice to have 17 points which is impossible.
So we can get a slice that has one point of type d in a side cube and the other in the center cube. That means the side cube must have its center slot filled and it must have 13 points, the center cube contains one point of type c(type d in the overall Moser set) and thus must have 15 points or less as the only Pareto optimal set with 16 points contains none of type c. The remaining side cube must have 15 points or more.
The only distribution which works is 13 points in one side cube and 15 in the other two. Now since there is no Moser three set with one point of type c and a total of fifteen points. There must be at least two points of type c in the center cube in giving at least 3 for the overall set. From this and the Pareto optimizers we see that the distribution of the side slice with 13 points must be (3, 6,3,1) or (4,6,2,1) and of the center and side cubes (3,9,3,0) or (4,9,2,1) This means we have at least 17 points of type b. Every one of these points appears in the center cube in exactly one slice. There are 4 slices so there is one slice with 5 of these points in the center cube.
Now since we have at least three points of type d and we can’t have two of them in the side cubes we must have at least one in the center cube. This means that that cube must have at most 13 points. No if either of the side cubes has a point of type d its total will be at most 13 and the third cube must be 17 which gives a contradiction so all three center points must be in the center cube.
This lowers the total of the center cube to at most 12. This in fact forces the center cube to have distribution (5,4,3,0) or that distribution with one point removed.. This also forces the sides cubes to have total 31. This means that one will have 16 points and thus have statistics (4,12,0,0). The other will have 15 points and hence have 9 point s of type b giving a total of 25 of type b if the center cube has 12 points.
If it has 11 points then it will have the distribution (5, 3, 3 as that is the only subset of the Pareto optimizers satisfying the conditions. Then both side cubes will have 16 points which means that neither will contain points of type c which means that there must be 18 points of type c in the center cube which gives a contradiction.
So the center cube has 12 points and the total number of points of type b is 25 so since each point of type b occurs in one slice and there are four possible slices there must be one slice with 6 points of type b since we get a contradiction as above if there are less than 11 points. So there must be 12 points the only distribution that works is (6,6,0,0).
But then the middle slice contains none of the points of type d and one must be on a slide slice lowering the total of that slice to 13 or less forcing the other side slice to be above 18. So we have reached a contradiction in assuming that we have more than one point of type d the remaining case is that there is exactly one point of type d.
If there is exactly one slice so it is in one of the side cubes. That cube will have at most 13 points. The other side cube can have at most 15 points because if it has 16 it will have no points of type c and the other side cube will have at most 3 of type c and since the total size is 18 we must have 15 of type c in the center cube which gives a contradiction.
So the center cube must have at least 15 points. Furthermore since it has no points of type c since we have our single point of type d in one of the side cubes it must be a subset of (4,12,0,0) and must have at least 11 points of type b. To avoid forming a monochromatic line the points of type b of the side cubes must sum to 13 or less.
If the side cube without the center cube has 15 or more points it must have at least 9 points of type b forcing at most 4 of type be in the other side cube which forces its size to 11 or less which forces both of the other cubes to have more than 16 which contradicts the fact that no side cube can have more than 15 points proved above. So the side cube without the center point must have at most 14 points. The other side cube can have at most 13 points so the center cube must have 16 points and since it cannot have 17 we have the distribution (13,16,14).
This forces the center cube to have all its points of type b there are 12 of these. From the Pareto optimizers the side cube with 13 points with the center spot filled must have at exactly 6 points of type b. We will get a line unless the remaining side cube has more than six points of type b. The only distributions with at least 14 points that satisfies this is (3,6,5,0) and (2,6,6,0). From the Pareto optimal statistics we can limit the possibilities for the other cubes of slice and get the following:
So one center cube has statistics (4,12,0,0) the side cube with thirteen points and the center filled has distribution (3,6,3,1) or (4,6,2,1) and final cube has statistics (3,6,5,0) and (2,6,6,0). So the statistics of the Moser set must be (6,16, 21,1,0), (7,16, 20,1,0), or (8,16, 19,1,0). So the exact number of points of type b must be 16.
Now slice the set so that the point of type d is in the center slice. Then the center cube contains a point of type c and has at most 10 points of type b. It must have at most 14 points that means that the other two cubes must total at least 29. If one is 16 then it contains no points of type c and since the center cube contributes at most 10 the other side cube must have 8 such points. This gives a contradiction which means one side cube must be of size 15 and the other of size 14. The one of size 15 contains at most three of type c. The center contains at most 10 so the remaining cube must have 5 or more.
Again from the Pareto statistics we can limit the statistics of each of the cubes the side slice of size 15 must be (3,9,3,0) (4,9,2,0) otherwise it will have no points of type c and we will get a contradiction as above. The side slice must have at least 5 points of type c and at least 14 points so it must have statistics (2,6,6,0) or (3,6,5,0). The center slice must have 1 point of type c in its statistics and at least 9 points of type b in its statistics which are statistics of type in the whole Moser set. Thus it statistics are either (4,9,1,0) which is a Pareto optimal set with one point removed or (3,10,1,0). Then the distributions are either (5, 19,18,1), (5,18,19,1), (6,19,17,1,0),(6,18,18,1,0), (7,19,16,1,0) or (7,18,17,1,0). Since we must have 18 points of type c the only possible distributions are (5, 19,18,1), (5,18,19,1) or (5,18,19,1). But we have shown the exact number of points of type b must be 16 which is a contradiction and we are done. So every Moser set with d=4 cannot contain a point with three two’s.