## Prologue

Consider the following problems:

**P3:** What is the maximum density of a set A in without a 3-term AP? (AP=arithmetic progression.)

This is the celebrated Cap Set problem and we reported here in 2016 the breakthrough results by Croot-Lev-Pach-Ellenberg-Gijswijt (CLPEG) asserting that .

**P4:** What is the maximum density of a set A in without a 4-term AP?

**P5:** What is the maximum density of a set A in without a 5-term AP?

Moving from 3 term AP to 4 term AP represents a major leap for such problems. In some cases moving from 4 to 5 also represents a major leap. Usually the techniques that work for continue to work for , that is, I am not aware of cases where moving from 5 to 6 also represent a major leap. In other words, we do not have reasons to think, so far, that progress on P6 (below) will be considerably harder than progress on P5.

**P6:** What is the maximum density of a set A in without a 6-term AP?

And, of course, we can move on to **P7, P8, P9,….** ,

It is known that the density of a set without -term AP for every fixed m goes to zero with n. This was proved by Furstenberg and Katznelson using ergodic theoretical tools. Following the CLPEG breakthrough we can certainly believe that the density that guarantees m-term AP in for every fixed m is exponentially small in n. I vaguely remember that the best known bounds for P4 are polylogarithmically small with n and were achieved by Green and Tao using tools from Gowers’ work on 4AP-free sets over the integers.

### A little more before we move on

To have some general notation let denote the maximal size of a set with no *k* distinct elements in arithmetic progression. It is reasonable to believe that for some , that is that sets in avoiding terms AP are exponentially small. To be more precise let

We can believe that the limits exist, that , and we may also guess (wildly but safely) that they satisfy .

These problems, as well as the analogous problems over the integers, (Roth’s and Szemeredi’s theorems and later developments), and the related density Hales Jewett problem, form a holy grail of additive combinatorics and they attracted a lot of attention also on this blog.

Here is a partial list of posts: 2008 Pushing the Behrend bound; 2009A, 2009B 2009C Attempts to relate the problem to the Frankl-Wilson Theorem (and the polynomial method) and to Frankl-Rodl theorem and also to find counterexamples; 2009D polymath1; 2009P public opinion poll about the true behavior; 2010 The logarithmic bound for Roth is reached; 2011 cap sets and matrix multiplications; 2016a, 2016b The CLPEG breakthrough; 2016W A polynomial method workshop; 2020 The logarithmic bound for Roth is broken.

Let’s move on to the new result by Péter Pál Pach and Richárd Palincza

## The absolutely stunning result by Péter Pál Pach and Richárd Palincza

As we already mentioned, so far we do not have reasons to expect that proving density results for 6-terms AP in (or ) will be much harder than the question over . **However**, what Pach and Palincza proved is as follows:

**Theorem**: sets avoiding 6-term arithmetic progressions in have size at most .

The big surprise is that the problem P6 over integers modulo six is **much easier** than P4 and P5. The proof is based on a simple combinatorial reasoning showing that

The paper arXived on September 2020 is

Péter Pál Pach and Richárd Palincza: Sets avoiding six-term arithmetic progressions in are exponentially small

Click and enjoy this beautiful, stunning and **very simple** proof.

Congratulations to Péter and Richárd!

Very nice. Is your statement of “much easier” simply referring to the fact

that $5.709/6\approx 0.95$ for P6 while the corresponding ratio is more like $0.9$ for P4 and P5?

Dear Kodlu, the statement “much easier” refers to the difficulty of proving an upper bound (or even exponentially-small upper bound). We have reasons to think that this is considerably harder for 4 compared to 3. And that it is as hard or harder for 5 compared to 4. And until the new result we could have thought that this is at least as hard for 6 compared to 4 and 5. I don’t think the difficulty of finding a proof is very related to the precis value of the upper bound.

(I also edited the wording of the post to make the point clearer.)Thank you for the clarification

Tau –> Tao ……

GK: thanksKind of reminds of the situation in topology, the following is quoted from the wikipedia entry on the h-cobordism theorem:

“Before Smale proved this theorem, mathematicians became stuck while trying to understand manifolds of dimension 3 or 4, and assumed that the higher-dimensional cases were even harder. The h-cobordism theorem showed that (simply connected) manifolds of dimension at least 5 are much easier than those of dimension 3 or 4. The proof of the theorem depends on the “Whitney trick” of Hassler Whitney, which geometrically untangles homologically-tangled spheres of complementary dimension in a manifold of dimension >4. An informal reason why manifolds of dimension 3 or 4 are unusually hard is that the trick fails to work in lower dimensions, which have no room for untanglement.”

Thanks for your comment, Eitan! This is a nice case to remember. Here the situation is somewhat different since AP of length 6 (over (Z/mZ)^n when 6|m) are ( in view of the new result) surprisingly easy (as easy as 3) but this does not say that AP of length larger than 6 are easy. (So we do not have statements about AP of length 10 over Z/10Z or of length 15 over Z/15Z reducing these problems to questions about smaller length AP). Regarding AP of length 4,5 and 6 the results give exponential improvement over (Z/mZ)^n when m is 0 mod 6. Now we know exponential improvement in (Z/6Z)^n for the cases of 4- 5- and 6- terms AP. Are 5-terms AP over (Z/5Z)^n and 4-term AP over (Z/4Z)^n so much harder? (And by “harder” I refer to how hard it is to prove things). BTW I plan a series of posts based on our open problem session so I’ll probably ask you more details about your probךem.

https://arxiv.org/abs/2012.08426

Thanks!, did you check the argument?

There’s a mistake unfortunately. At the bottom of page 5 a set is claimed to be an AP of length but it isn’t.