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There might already be a post regarding this problem: Guaranteeing an integer lattice point centroid, but I am too dumb to figure out a proof given the discussion on the said post.

Let $S\subset\mathbb{Z}^2$. I am trying to show that if $\# S = 13$, then there always exists four points $z_1,\dots,z_4$ such that

$$\frac{1}{4}\sum_{k=1}^4 z_k\in\mathbb{Z}^2$$

This is the same as saying that for the sum $w = \sum_{k=1}^4z_k$, both coordinates are divisible by $4$. I am fully aware that some form of Pigeonhole principle needs to be applied, but I do not see what the categories that we should consider are.

Were the problem to be that the mean of two points is a lattice point, then this problem is quite simple to conclude for $\# S = 5$: The mean of two lattice points is a lattice point precisely when both the sums of both coordinates are even. Each lattice point can be categorized as $(0,0),(1,0),(0,1),(1,1)$ based on residue modulo $2$. So when we choose any five points, necessarily two of them are in the same class and hence the claim follows.

This is quite a bit trickier with modulo $4$ since it seems that given a class (when we consider the coordinates of points based on their residues modulo $4$) e.g. $(1,1)$, either all other points need to also be in the class $(1,1)$ or if one is, say, $(3, 3)$, then the choices for the other two points are $(2,2),(3,1),(1,3),(0,0)$ such that the sums of the both coordinates (individually) are divisible by $4$. Not so simple...

Wasradin
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    Have you tried constructing a set of 13 points that violates this theorem? It might give you some insight into why it must be true. – CyclotomicField Feb 17 '25 at 18:00
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    It's a two-step pigeonhole argument. Like here or here. – Jyrki Lahtonen Feb 17 '25 at 18:20
  • @CyclotomicField A bit, but I have not spend too much time with it as I do not know how to articulate nor generalize the pattern. It might be easier for me to try to understand the big picture. – Wasradin Feb 17 '25 at 18:36
  • @JyrkiLahtonen I fail to understand what the steps should be. We can use the argument I posted to choose two sets of five points such that in each of them the mean of at least two points is a lattice point. I do not really know how to continue from here. We could "act" on the two means by choosing the remaining three points to obtain, again, five points. But I do not see how this should help. – Wasradin Feb 17 '25 at 18:40
  • Can you characterize when four points have a centroid on a lattice point? – CyclotomicField Feb 17 '25 at 19:13

1 Answers1

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The answer to this is indeed contained in the proposed associated questions, particularly this one, but the answer is only sketched in the comments and it's rather hard to put together. Accordingly, I'll write it out here in greater detail. Given that this is essentially a duplicate, I'll post this as a community solution.

Step $1$: (first application of pigeonhole principle) As noted by the OP, remark that, given any $5$ lattice points we can find a pair such that their midpoint is a lattice point. Indeed, there are only $4$ possible points $\pmod 2$ so with $5$ we must get two which are congruent $\pmod 2$ and that's equivalent to our requirement.

Step $2$: Taking our $13$ points, we first choose a good pair as in step $1$. Then another from the remaining $11$ until we wind up with $5$ pairs of points each with integer midpoint.

Step $3$: (second application of pigeonhole principle) Consider those $5$ pairs. If we choose two of those pairs, the centroid of the $4$ points is, of course, the midpoint of the two midpoints. So, we invoke Step $1$ again to argue that we can choose two pairs from the given $5$ such that the midpoint of their midpoints is still a lattice point.

And we are done.

Might be worth remarking that $13$ is minimal with respect to this property, at least so far as this argument is concerned. The linked quasi-duplicate contains candidates of $12$ points such that no $4$ have integer centroid, but I have not verified those arguments.

lulu
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  • For the counter example, working mod 4, taking 3 points each of $(0,0), (0, 1), (1, 0), (1, 1)$, will yield no 4 whose sum is $(0, 0)$. – Calvin Lin Feb 17 '25 at 20:43
  • @CalvinLin Well, that certainly makes sense. Thanks! – lulu Feb 17 '25 at 20:54
  • @CalvinLin So simple! If you had to arrive at this conclusion from scratch, how would you do it? Is there any systematic way to conclude that it suffices to consider these four representatives? – Wasradin Feb 18 '25 at 07:18
  • @lulu I arrived at this same conclusion when I woke up: Just apply the miniargument repeatedly to get five lattice midpoints of some other 13 lattice points. Then, just apply the said argument again to these five points to conclude the result. – Wasradin Feb 18 '25 at 07:20
  • @CartesianBear This is a pretty standard setup for repeated PHP, esp if you're familiar with the well-known 5 points variant. $\quad$ In terms of coming up with the counterexample, essentially re-apply this approach to determine the structure you want to avoid. We can always find 4 pairs of points that sum up to even coordinates. For any way of getting these 4 pairs, we need A/ them to have different sums mod 4, and B/ The remaining 4 pairs cannot have any 2 that sum up to even coordinates. From there, I found my set. – Calvin Lin Feb 18 '25 at 16:34