Coloring the n -smooth numbers with n colors Andr es Eduardo - - PowerPoint PPT Presentation

Coloring the n-smooth numbers with n colors

Andr´ es Eduardo Caicedo

Mathematical Reviews, AMS

Mathematics and Computer Science Colloquium, Albion College, March 22, 2018

Joint work with Thomas Chartier and P´ eter Pach.

Caicedo Coloring the n-smooth numbers with n colors

Introduction

Consider the integer grid:

Caicedo Coloring the n-smooth numbers with n colors

Introduction

Consider the integer grid: Can we color it using 3 colors in such a way that all colors appear in any copy of the triomino?

Caicedo Coloring the n-smooth numbers with n colors

Suppose we have such a coloring, and consider an arbitrary triomino on the plane colored as indicated: a b c

Caicedo Coloring the n-smooth numbers with n colors

What can we say about the color of the square indicated below? a

• b

c

Caicedo Coloring the n-smooth numbers with n colors

The square belongs to these two triominos, and it follows that its color must be b: a

• b

c

Caicedo Coloring the n-smooth numbers with n colors

Now notice that the color of the square indicated below must be c:

• a

b b c

Caicedo Coloring the n-smooth numbers with n colors

These two observations and the coloring of a single triomino completely determine the coloring of the whole grid: b c a b c a b c a b c a b c a b b c a b c a c b a

Caicedo Coloring the n-smooth numbers with n colors

Essentially the same argument gives the same result if instead of the whole grid we only consider the first quadrant: b c a b c a b c a b c a b c a b

Caicedo Coloring the n-smooth numbers with n colors

The question I want to discuss throughout this talk can be rephrased as a very similar problem of coloring a grid in such a way that all copies of certain polyomino receive all colors. (Typically, the grid will have many more than 2 dimensions, so we will not be representing it graphically.)

Caicedo Coloring the n-smooth numbers with n colors

The question I want to discuss throughout this talk can be rephrased as a very similar problem of coloring a grid in such a way that all copies of certain polyomino receive all colors. (Typically, the grid will have many more than 2 dimensions, so we will not be representing it graphically.) To rephrase the question in the case of three colors, consider the set K3 of 3-smooth numbers: K3 = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, . . . } This is the set of positive integers whose prime representation only involves 2s and 3s.

Caicedo Coloring the n-smooth numbers with n colors

We can represent K3 graphically on the first quadrant of the integer grid: Moving one square to the right corresponds to multiplying by 2, and moving one square up corresponds to multiplying by 3, starting with 1 in the bottom-left corner: ×2 ×3 27 54 108 216 9 18 36 72 3 6 12 24 1 2 4 8

Caicedo Coloring the n-smooth numbers with n colors

We can now rephrase our coloring question as follows: Can we color the 3-smooth numbers using 3 colors in such a way that for any a the colors of the numbers a, 2a, 3a are all distinct?

Caicedo Coloring the n-smooth numbers with n colors

We can now rephrase our coloring question as follows: Can we color the 3-smooth numbers using 3 colors in such a way that for any a the colors of the numbers a, 2a, 3a are all distinct? The solution we found indicates we must color K3 as follows: ×2 ×3 27 54 108 216 9 18 36 72 3 6 12 24 1 2 4 8

Caicedo Coloring the n-smooth numbers with n colors

The question we have been considering is the case n = 3 of the general problem we will be discussing:

Caicedo Coloring the n-smooth numbers with n colors

The question we have been considering is the case n = 3 of the general problem we will be discussing: Given n, let Kn denote the set of n-smooth numbers, the positive integers whose prime decomposition does not involve any primes larger than n. Question Can we color Kn using n colors in such a way that for any a the numbers a, 2a, . . . , na all have distinct colors?

Caicedo Coloring the n-smooth numbers with n colors

The question we have been considering is the case n = 3 of the general problem we will be discussing: Given n, let Kn denote the set of n-smooth numbers, the positive integers whose prime decomposition does not involve any primes larger than n. Question Can we color Kn using n colors in such a way that for any a the numbers a, 2a, . . . , na all have distinct colors? Let’s say that a coloring is n-satisfactory if it has the property required in the question.

Caicedo Coloring the n-smooth numbers with n colors

Generalities

This question was originally asked by P´ eter Pach, about 10 years

• ago. It remains open.

Caicedo Coloring the n-smooth numbers with n colors

In its original formulation, Pach asked about coloring all positive integers rather than only the n-smooth ones. This makes no difference.

Caicedo Coloring the n-smooth numbers with n colors

In its original formulation, Pach asked about coloring all positive integers rather than only the n-smooth ones. This makes no difference. Theorem There is an n-satisfactory coloring of the positive integers if and

• nly if there is an n-satisfactory coloring of Kn.

Caicedo Coloring the n-smooth numbers with n colors

In its original formulation, Pach asked about coloring all positive integers rather than only the n-smooth ones. This makes no difference. Theorem There is an n-satisfactory coloring of the positive integers if and

• nly if there is an n-satisfactory coloring of Kn.

In fact, if there is at least one n-satisfactory coloring of Kn then there are as many n-satisfactory colorings of the positive integers as there are real numbers.

Caicedo Coloring the n-smooth numbers with n colors

For which n can we prove that there is an n-satisfactory coloring?

Caicedo Coloring the n-smooth numbers with n colors

For which n can we prove that there is an n-satisfactory coloring? Let’s consider an example: n = 4. Can we find a 4-satisfactory coloring of K4 = K3?

Caicedo Coloring the n-smooth numbers with n colors

For which n can we prove that there is an n-satisfactory coloring? Let’s consider an example: n = 4. Can we find a 4-satisfactory coloring of K4 = K3? Below, we represent the coloring c by writing the numbers in K4 in 4 rows, with two numbers sharing a row if and only if they have the same color. Remember that the requirement is that for any a, the numbers a, 2a, 3a, 4a all have different colors (that is, they must occupy different rows). We also write c(a) to indicate the color of a.

Caicedo Coloring the n-smooth numbers with n colors

1 2 3 4 How should we color 6?

Caicedo Coloring the n-smooth numbers with n colors

1 2 3 4 Since 6 = 3 × 2, we have c(6) = c(2) and c(6) = c(2 × 2) = c(4). Also, 6 = 2 × 3, so c(6) = c(3). So c(6) = c(1).

Caicedo Coloring the n-smooth numbers with n colors

1 6 2 3 4 Since 8 = 4 × 2, we have c(8) = c(2), c(8) = c(4), c(8) = c(6). So c(8) = c(3).

Caicedo Coloring the n-smooth numbers with n colors

1 6 2 3 8 4 Since 12 = 4 × 3 = 3 × 4 = 2 × 6, we must have c(12) = c(2).

Caicedo Coloring the n-smooth numbers with n colors

1 6 2 12 3 8 4 Since 9 = 3 × 3, c(9) = c(2 × 3), c(4 × 3), c(3), so c(9) = c(4).

Caicedo Coloring the n-smooth numbers with n colors

1 6 2 12 3 8 4 9

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 2 12 3 8 4 9

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 2 12 3 8 18 4 9

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 36 2 12 3 8 18 4 9

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 36 2 12 27 3 8 18 4 9

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 36 2 12 27 3 8 18 4 9 24

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 36 . . . 2 12 27 . . . 3 8 18 . . . 4 9 24 . . . Is there a general pattern?

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 36 . . . 2 12 27 . . . 3 8 18 . . . 4 9 24 . . . At least in this case, yes: Note that the numbers in the ith column are all congruent to i modulo 5.

Caicedo Coloring the n-smooth numbers with n colors

1 6 16 36 . . . 2 12 27 . . . 3 8 18 . . . 4 9 24 . . . And we can easily check that this idea works: Proposition The coloring c(a) = (a mod 5) is a 4-satisfactory coloring of K4.

Caicedo Coloring the n-smooth numbers with n colors

This result suggests a generalization.

Caicedo Coloring the n-smooth numbers with n colors

This result suggests a generalization. Theorem For any prime p, there is a (p − 1)-satisfactory coloring of Kp−1.

Caicedo Coloring the n-smooth numbers with n colors

This result suggests a generalization. Theorem For any prime p, there is a (p − 1)-satisfactory coloring of Kp−1. Proof. Consider the coloring c(a) = (a mod p). ❑

Caicedo Coloring the n-smooth numbers with n colors

It was through this result that I became aware of the question:

Caicedo Coloring the n-smooth numbers with n colors

It was through this result that I became aware of the question: Pach posed this theorem as problem A.506 in the April 2010 issue

• f the Hungarian journal K¨
• MaL (K¨
• z´

episkolai Matematikai ´ es Fizikai Lapok), a Mathematics and Physics journal primarily aimed at High School students.

Caicedo Coloring the n-smooth numbers with n colors

Caicedo Coloring the n-smooth numbers with n colors

A Hungarian combinatorialist, P´ alv¨

• lgyi D¨
• m¨
• t¨
• r, saw the problem

in the magazine, and asked the general case on MathOverflow.

Caicedo Coloring the n-smooth numbers with n colors

I saw the problem on MathOverflow, and suggested it to my student Tommy Chartier as a potential topic for his master’s thesis.

Caicedo Coloring the n-smooth numbers with n colors

Pach formulated the problem in an attempt to solve a question from 1992 on coding theory due to G¨ unter Pilz.

Caicedo Coloring the n-smooth numbers with n colors

Pach formulated the problem in an attempt to solve a question from 1992 on coding theory due to G¨ unter Pilz. Question If A is a finite subset of positive integers, is the size of the symmetric difference of the sets A, 2 · A, . . . , n · A at least n?

Caicedo Coloring the n-smooth numbers with n colors

Question If A is a finite subset of positive integers, is the size of the symmetric difference of the sets A, 2 · A, . . . , n · A at least n?

Caicedo Coloring the n-smooth numbers with n colors

Question If A is a finite subset of positive integers, is the size of the symmetric difference of the sets A, 2 · A, . . . , n · A at least n? To illustrate, consider the case n = 3, A = {1, 2, 3}. We have 2 · A = {2, 4, 6} and 3 · A = {3, 6, 9}.

Caicedo Coloring the n-smooth numbers with n colors

Question If A is a finite subset of positive integers, is the size of the symmetric difference of the sets A, 2 · A, . . . , n · A at least n? To illustrate, consider the case n = 3, A = {1, 2, 3}. We have 2 · A = {2, 4, 6} and 3 · A = {3, 6, 9}. One can check that the symmetric difference of a collection A1, . . . , Am of sets is the set of elements that belong to an odd number of the Ai.

Caicedo Coloring the n-smooth numbers with n colors

Question If A is a finite subset of positive integers, is the size of the symmetric difference of the sets A, 2 · A, . . . , n · A at least n? To illustrate, consider the case n = 3, A = {1, 2, 3}. We have 2 · A = {2, 4, 6} and 3 · A = {3, 6, 9}. One can check that the symmetric difference of a collection A1, . . . , Am of sets is the set of elements that belong to an odd number of the Ai. In the case of A, 2 · A, 3 · A, this is the set {1, 4, 9}, with 3 elements.

Caicedo Coloring the n-smooth numbers with n colors

Theorem If A is an odd-sized subset of the positive integers, and there is an n-satisfactory coloring, then the symmetric difference of A, 2 · A, . . . , n · A has size at least n.

Caicedo Coloring the n-smooth numbers with n colors

Theorem If A is an odd-sized subset of the positive integers, and there is an n-satisfactory coloring, then the symmetric difference of A, 2 · A, . . . , n · A has size at least n. Proof. (Sketch) Let A = {a1, a2, . . . , ak}. The symmetric difference of A, 2 · A, . . . , n · A is also the symmetric difference of the sets B1, B2, . . . , Bk, where Bi = {ai, 2ai, . . . , nai}. ❑

Caicedo Coloring the n-smooth numbers with n colors

Theorem If A is an odd-sized subset of the positive integers, and there is an n-satisfactory coloring, then the symmetric difference of A, 2 · A, . . . , n · A has size at least n. Proof. (Sketch) Let A = {a1, a2, . . . , ak}. The symmetric difference of A, 2 · A, . . . , n · A is also the symmetric difference of the sets B1, B2, . . . , Bk, where Bi = {ai, 2ai, . . . , nai}. For any n-satisfactory coloring, in every Bi each color appears exactly once. That is, the sets B1, B2, . . . , Bk contains k many numbers from each color class (counted by multiplicity). ❑

Caicedo Coloring the n-smooth numbers with n colors

Theorem If A is an odd-sized subset of the positive integers, and there is an n-satisfactory coloring, then the symmetric difference of A, 2 · A, . . . , n · A has size at least n. Proof. (Sketch) Let A = {a1, a2, . . . , ak}. The symmetric difference of A, 2 · A, . . . , n · A is also the symmetric difference of the sets B1, B2, . . . , Bk, where Bi = {ai, 2ai, . . . , nai}. For any n-satisfactory coloring, in every Bi each color appears exactly once. That is, the sets B1, B2, . . . , Bk contains k many numbers from each color class (counted by multiplicity). If k is

• dd, then this means that the symmetric difference must contain

an odd number of elements from each color class, and therefore at least 1. But there are n color classes. ❑

Caicedo Coloring the n-smooth numbers with n colors

Pilz’s question remains open. The case where A = {1, 2, . . . , k} was independently solved by P.-Y. Huang, W.-F. Ke and Pilz and by Pach and C. Szab´

• around

2008–2009. That the general version remains open suggests that

• ur coloring problem is nontrivial.

Caicedo Coloring the n-smooth numbers with n colors

Pilz’s question remains open. The case where A = {1, 2, . . . , k} was independently solved by P.-Y. Huang, W.-F. Ke and Pilz and by Pach and C. Szab´

• around

2008–2009. That the general version remains open suggests that

• ur coloring problem is nontrivial.

Here is another application:

Caicedo Coloring the n-smooth numbers with n colors

In 1970, Ronald Graham conjectured the following:

Caicedo Coloring the n-smooth numbers with n colors

In 1970, Ronald Graham conjectured the following: Conjecture If n ≥ 1, and 0 < a1 < a2 < · · · < an are integers, then max

i,j

ai gcd(ai, aj) ≥ n.

Caicedo Coloring the n-smooth numbers with n colors

In 1970, Ronald Graham conjectured the following: Conjecture If n ≥ 1, and 0 < a1 < a2 < · · · < an are integers, then max

i,j

ai gcd(ai, aj) ≥ n. For example, let n = 4 and a1 = 2, a2 = 3, a3 = 4, a4 = 6. We have

• ai

gcd(ai, aj) : 1 ≤ i ≤ j ≤ 4

• = {1, 2, 3, 4}

and Graham’s conjecture holds (in fact, we have equality) in this case.

Caicedo Coloring the n-smooth numbers with n colors

Graham’s conjecture was finally verified in 1996 by R. Balasubramanian and K. Soundararajan. A very simple proof can be obtained from the existence of satisfactory colorings:

Caicedo Coloring the n-smooth numbers with n colors

Graham’s conjecture was finally verified in 1996 by R. Balasubramanian and K. Soundararajan. A very simple proof can be obtained from the existence of satisfactory colorings: Theorem If there is an (m − 1)-satisfactory coloring then Graham’s conjecture holds for n = m.

Caicedo Coloring the n-smooth numbers with n colors

Graham’s conjecture was finally verified in 1996 by R. Balasubramanian and K. Soundararajan. A very simple proof can be obtained from the existence of satisfactory colorings: Theorem If there is an (m − 1)-satisfactory coloring then Graham’s conjecture holds for n = m. Proof. Argue by contradiction. Accordingly, suppose that there are (m − 1)-satisfactory colorings and that 0 < b1 < · · · < bm are integers such that maxi,j bi/ gcd(bi, bj) < m. ❑

Caicedo Coloring the n-smooth numbers with n colors

Graham’s conjecture was finally verified in 1996 by R. Balasubramanian and K. Soundararajan. A very simple proof can be obtained from the existence of satisfactory colorings: Theorem If there is an (m − 1)-satisfactory coloring then Graham’s conjecture holds for n = m. Proof. Argue by contradiction. Accordingly, suppose that there are (m − 1)-satisfactory colorings and that 0 < b1 < · · · < bm are integers such that maxi,j bi/ gcd(bi, bj) < m. Suppose i = j and let M = gcd(bi, bj). Let ai = bi/M and aj = bj/M, so ai, aj are both less than m, and ai = aj. Since bi = aiM and bj = ajM, in any (m − 1)-satisfactory coloring we must have that bi is colored differently from bj. ❑

Caicedo Coloring the n-smooth numbers with n colors

Graham’s conjecture was finally verified in 1996 by R. Balasubramanian and K. Soundararajan. A very simple proof can be obtained from the existence of satisfactory colorings: Theorem If there is an (m − 1)-satisfactory coloring then Graham’s conjecture holds for n = m. Proof. Argue by contradiction. Accordingly, suppose that there are (m − 1)-satisfactory colorings and that 0 < b1 < · · · < bm are integers such that maxi,j bi/ gcd(bi, bj) < m. Suppose i = j and let M = gcd(bi, bj). Let ai = bi/M and aj = bj/M, so ai, aj are both less than m, and ai = aj. Since bi = aiM and bj = ajM, in any (m − 1)-satisfactory coloring we must have that bi is colored differently from bj. This is impossible, since it would mean the coloring uses at least m colors. ❑

Caicedo Coloring the n-smooth numbers with n colors

Prime numbers

We saw earlier that if n + 1 is prime, then there is an n-satisfactory coloring. We can extend the idea as follows:

Caicedo Coloring the n-smooth numbers with n colors

Prime numbers

We saw earlier that if n + 1 is prime, then there is an n-satisfactory coloring. We can extend the idea as follows: Theorem If 2n + 1 is prime, then there is an n-satisfactory coloring.

Caicedo Coloring the n-smooth numbers with n colors

Prime numbers

We saw earlier that if n + 1 is prime, then there is an n-satisfactory coloring. We can extend the idea as follows: Theorem If 2n + 1 is prime, then there is an n-satisfactory coloring. Proof. Let p = 2n + 1 and color a ∈ Kn with color c(a) = (a2 mod p). ❑

Caicedo Coloring the n-smooth numbers with n colors

Prime numbers

We saw earlier that if n + 1 is prime, then there is an n-satisfactory coloring. We can extend the idea as follows: Theorem If 2n + 1 is prime, then there is an n-satisfactory coloring. Proof. Let p = 2n + 1 and color a ∈ Kn with color c(a) = (a2 mod p). Note that if a2 ≡ b2 (mod p) then p | a2 − b2 = (a − b)(a + b), which means that either a ≡ b (mod p) or a ≡ −b (mod p). ❑

Caicedo Coloring the n-smooth numbers with n colors

Prime numbers

We saw earlier that if n + 1 is prime, then there is an n-satisfactory coloring. We can extend the idea as follows: Theorem If 2n + 1 is prime, then there is an n-satisfactory coloring. Proof. Let p = 2n + 1 and color a ∈ Kn with color c(a) = (a2 mod p). Note that if a2 ≡ b2 (mod p) then p | a2 − b2 = (a − b)(a + b), which means that either a ≡ b (mod p) or a ≡ −b (mod p). This shows that 12, . . . , n2 are pairwise inequivalent modulo p and that any nonzero square modulo p is one of them. ❑

Caicedo Coloring the n-smooth numbers with n colors

Prime numbers

We saw earlier that if n + 1 is prime, then there is an n-satisfactory coloring. We can extend the idea as follows: Theorem If 2n + 1 is prime, then there is an n-satisfactory coloring. Proof. Let p = 2n + 1 and color a ∈ Kn with color c(a) = (a2 mod p). Note that if a2 ≡ b2 (mod p) then p | a2 − b2 = (a − b)(a + b), which means that either a ≡ b (mod p) or a ≡ −b (mod p). This shows that 12, . . . , n2 are pairwise inequivalent modulo p and that any nonzero square modulo p is one of them. In terms of our coloring, this says that we are using precisely n colors, and that if a ∈ Kn and 1 ≤ i < j ≤ n, then (ia)2 ≡ (ja)2 (mod p), so that c(ia) = c(ja). ❑

Caicedo Coloring the n-smooth numbers with n colors

The argument generalizes as follows:

Caicedo Coloring the n-smooth numbers with n colors

The argument generalizes as follows: Theorem If p = kn + 1 is prime, and 1k, 2k, . . . , nk are pairwise distinct modulo p, then c(a) = (ak mod p) is an n-satisfactory coloring of Kn.

Caicedo Coloring the n-smooth numbers with n colors

The argument generalizes as follows: Theorem If p = kn + 1 is prime, and 1k, 2k, . . . , nk are pairwise distinct modulo p, then c(a) = (ak mod p) is an n-satisfactory coloring of Kn. Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero kth powers modulo p.

Caicedo Coloring the n-smooth numbers with n colors

The argument generalizes as follows: Theorem If p = kn + 1 is prime, and 1k, 2k, . . . , nk are pairwise distinct modulo p, then c(a) = (ak mod p) is an n-satisfactory coloring of Kn. Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero kth powers modulo p. Our assumption says that these powers are precisely 1k, . . . , nk.

Caicedo Coloring the n-smooth numbers with n colors

The argument generalizes as follows: Theorem If p = kn + 1 is prime, and 1k, 2k, . . . , nk are pairwise distinct modulo p, then c(a) = (ak mod p) is an n-satisfactory coloring of Kn. Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero kth powers modulo p. Our assumption says that these powers are precisely 1k, . . . , nk. If 1 ≤ i < j ≤ n and a ∈ Kn then ik ≡ jk (mod p), so c(ia) = c(ja). ❑

Caicedo Coloring the n-smooth numbers with n colors

The argument generalizes as follows: Theorem If p = kn + 1 is prime, and 1k, 2k, . . . , nk are pairwise distinct modulo p, then c(a) = (ak mod p) is an n-satisfactory coloring of Kn. Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero kth powers modulo p. Our assumption says that these powers are precisely 1k, . . . , nk. If 1 ≤ i < j ≤ n and a ∈ Kn then ik ≡ jk (mod p), so c(ia) = c(ja). ❑ Note that the assumption that 1k, . . . , nk are distinct modulo p always holds if k = 1, 2.

Caicedo Coloring the n-smooth numbers with n colors

There is, however, a serious problem verifying the assumption if k > 2. For example, consider the case k = 3, and look for primes

• f the form 3n + 1.

Caicedo Coloring the n-smooth numbers with n colors

There is, however, a serious problem verifying the assumption if k > 2. For example, consider the case k = 3, and look for primes

• f the form 3n + 1.

n = 2. We have 23 = 8 ≡ 1 = 13 (mod 7).

Caicedo Coloring the n-smooth numbers with n colors

There is, however, a serious problem verifying the assumption if k > 2. For example, consider the case k = 3, and look for primes

• f the form 3n + 1.

n = 2. We have 23 = 8 ≡ 1 = 13 (mod 7). n = 4. We have 33 = 27 ≡ 1 = 13 (mod 13).

Caicedo Coloring the n-smooth numbers with n colors

There is, however, a serious problem verifying the assumption if k > 2. For example, consider the case k = 3, and look for primes

• f the form 3n + 1.

n = 2. We have 23 = 8 ≡ 1 = 13 (mod 7). n = 4. We have 33 = 27 ≡ 1 = 13 (mod 13). n = 6. We have 33 = 27 ≡ 8 = 23 (mod 19).

Caicedo Coloring the n-smooth numbers with n colors

There is, however, a serious problem verifying the assumption if k > 2. For example, consider the case k = 3, and look for primes

• f the form 3n + 1.

n = 2. We have 23 = 8 ≡ 1 = 13 (mod 7). n = 4. We have 33 = 27 ≡ 1 = 13 (mod 13). n = 6. We have 33 = 27 ≡ 8 = 23 (mod 19). n = 10. We have 53 = 125 ≡ 1 = 13 (mod 31).

Caicedo Coloring the n-smooth numbers with n colors

There is, however, a serious problem verifying the assumption if k > 2. For example, consider the case k = 3, and look for primes

• f the form 3n + 1.

n = 2. We have 23 = 8 ≡ 1 = 13 (mod 7). n = 4. We have 33 = 27 ≡ 1 = 13 (mod 13). n = 6. We have 33 = 27 ≡ 8 = 23 (mod 19). n = 10. We have 53 = 125 ≡ 1 = 13 (mod 31). n = 12. We have 103 = 1000 ≡ 1 = 13 (mod 37).

Caicedo Coloring the n-smooth numbers with n colors

In general, we have

Caicedo Coloring the n-smooth numbers with n colors

In general, we have Theorem Suppose p = 3n + 1 is prime. Then there is an i, 2 < i ≤ n, such that i3 ≡ 8 or i3 ≡ 1 (mod p).

Caicedo Coloring the n-smooth numbers with n colors

In general, we have Theorem Suppose p = 3n + 1 is prime. Then there is an i, 2 < i ≤ n, such that i3 ≡ 8 or i3 ≡ 1 (mod p). That is, we can never verify the assumption that p = 3n + 1 is prime and 13, . . . , n3 are distinct modulo p.

Caicedo Coloring the n-smooth numbers with n colors

This is not to say that it is never the case that k > 2, p = kn + 1 is prime and 1k, . . . , nk are distinct modulo p. In this holds, we say that p is a strong representative of order n.

Caicedo Coloring the n-smooth numbers with n colors

This is not to say that it is never the case that k > 2, p = kn + 1 is prime and 1k, . . . , nk are distinct modulo p. In this holds, we say that p is a strong representative of order n. For example:

Caicedo Coloring the n-smooth numbers with n colors

This is not to say that it is never the case that k > 2, p = kn + 1 is prime and 1k, . . . , nk are distinct modulo p. In this holds, we say that p is a strong representative of order n. For example: p = 659 = 94 × 7 + 1 is the smallest strong representative of

• rder 7. (In particular, there is a 7-satisfactory coloring.)

Caicedo Coloring the n-smooth numbers with n colors

This is not to say that it is never the case that k > 2, p = kn + 1 is prime and 1k, . . . , nk are distinct modulo p. In this holds, we say that p is a strong representative of order n. For example: p = 659 = 94 × 7 + 1 is the smallest strong representative of

• rder 7. (In particular, there is a 7-satisfactory coloring.)

p = 2, 578, 733 = 198, 364 × 13 + 1 is the smallest strong representative of order 13.

Caicedo Coloring the n-smooth numbers with n colors

This is not to say that it is never the case that k > 2, p = kn + 1 is prime and 1k, . . . , nk are distinct modulo p. In this holds, we say that p is a strong representative of order n. For example: p = 659 = 94 × 7 + 1 is the smallest strong representative of

• rder 7. (In particular, there is a 7-satisfactory coloring.)

p = 2, 578, 733 = 198, 364 × 13 + 1 is the smallest strong representative of order 13. p = 1, 358, 652, 193 = 56, 610, 508 × 24 + 1 is the smallest strong representative of order 24.

Caicedo Coloring the n-smooth numbers with n colors

This is not to say that it is never the case that k > 2, p = kn + 1 is prime and 1k, . . . , nk are distinct modulo p. In this holds, we say that p is a strong representative of order n. For example: p = 659 = 94 × 7 + 1 is the smallest strong representative of

• rder 7. (In particular, there is a 7-satisfactory coloring.)

p = 2, 578, 733 = 198, 364 × 13 + 1 is the smallest strong representative of order 13. p = 1, 358, 652, 193 = 56, 610, 508 × 24 + 1 is the smallest strong representative of order 24. After an extensive computer search, we could not find any strong representative of order 34, although we know there are infinitely many.

Caicedo Coloring the n-smooth numbers with n colors

There is a good reason why finding values of k > 2 that lead to strong representatives of a given order is hard:

Caicedo Coloring the n-smooth numbers with n colors

There is a good reason why finding values of k > 2 that lead to strong representatives of a given order is hard: Theorem If k > 2 then there are only finitely many primes p such that p = kn + 1 for some n and p is a strong representative of order n.

Caicedo Coloring the n-smooth numbers with n colors

There is a good reason why finding values of k > 2 that lead to strong representatives of a given order is hard: Theorem If k > 2 then there are only finitely many primes p such that p = kn + 1 for some n and p is a strong representative of order n. We illustrate the proof with an example (k = 6). The key idea was suggested on MathOverflow by Darij Grinberg. First, we need some notation.

Caicedo Coloring the n-smooth numbers with n colors

There is a good reason why finding values of k > 2 that lead to strong representatives of a given order is hard: Theorem If k > 2 then there are only finitely many primes p such that p = kn + 1 for some n and p is a strong representative of order n. We illustrate the proof with an example (k = 6). The key idea was suggested on MathOverflow by Darij Grinberg. First, we need some notation. Let p be prime. Denote by Z∗

p the collection of nonzero numbers

modulo p. A subgroup of Z∗

p is a nonempty set G ⊆ Z∗ p such that

1 ∈ G. xy ∈ G whenever x, y ∈ G.

Caicedo Coloring the n-smooth numbers with n colors

There is a good reason why finding values of k > 2 that lead to strong representatives of a given order is hard: Theorem If k > 2 then there are only finitely many primes p such that p = kn + 1 for some n and p is a strong representative of order n. We illustrate the proof with an example (k = 6). The key idea was suggested on MathOverflow by Darij Grinberg. First, we need some notation. Let p be prime. Denote by Z∗

p the collection of nonzero numbers

modulo p. A subgroup of Z∗

p is a nonempty set G ⊆ Z∗ p such that

1 ∈ G. xy ∈ G whenever x, y ∈ G. G is nontrivial if G = {1}.

Caicedo Coloring the n-smooth numbers with n colors

For example, if p = kn + 1 is prime and n, k > 1, then the collection of nonzero kth powers modulo p is a nontrivial subgroup

• f Z∗

p:

Caicedo Coloring the n-smooth numbers with n colors

For example, if p = kn + 1 is prime and n, k > 1, then the collection of nonzero kth powers modulo p is a nontrivial subgroup

• f Z∗

p:

Clearly the product of two nonzero kth powers is again a kth power, and it is nonzero modulo p since p is prime.

Caicedo Coloring the n-smooth numbers with n colors

For example, if p = kn + 1 is prime and n, k > 1, then the collection of nonzero kth powers modulo p is a nontrivial subgroup

• f Z∗

p:

Clearly the product of two nonzero kth powers is again a kth power, and it is nonzero modulo p since p is prime. For instance, for p = 19 = 3 × 6 + 1, the group of nonzero cubes is {1, 23, 33, . . . (mod 19)} = {1, 7, 8, 11, 12, 18 (mod 19)}. Note for example that 7 × 8 ≡ 18 (mod 19), 11 × 11 ≡ 7 (mod 19) and so on.

Caicedo Coloring the n-smooth numbers with n colors

The key observation is that if p is prime and G is a nontrivial subgroup of Z∗

p, then

• g∈G

g = 0.

Caicedo Coloring the n-smooth numbers with n colors

The key observation is that if p is prime and G is a nontrivial subgroup of Z∗

p, then

• g∈G

g = 0. Indeed, let S =

g∈G g and let 1 = h ∈ G. The map g → hg is a

permutation of G (because it is injective and G is finite), and we have S =

g∈G hg = hS.

Caicedo Coloring the n-smooth numbers with n colors

The key observation is that if p is prime and G is a nontrivial subgroup of Z∗

p, then

• g∈G

g = 0. Indeed, let S =

g∈G g and let 1 = h ∈ G. The map g → hg is a

permutation of G (because it is injective and G is finite), and we have S =

g∈G hg = hS.

If moreover p = kn + 1 is a strong representative of order n, then the group of kth powers is just {1k, . . . , nk (mod p)}, and we conclude that n

i=1 ik ≡ 0 (mod p).

Caicedo Coloring the n-smooth numbers with n colors

Now, for any k, m

i=1 ik is a polynomial in m of degree k + 1. For

instance,

Caicedo Coloring the n-smooth numbers with n colors

Now, for any k, m

i=1 ik is a polynomial in m of degree k + 1. For

instance,

m

• i=1

i = m(m + 1) 2 ,

Caicedo Coloring the n-smooth numbers with n colors

Now, for any k, m

i=1 ik is a polynomial in m of degree k + 1. For

instance,

m

• i=1

i = m(m + 1) 2 ,

m

• i=1

i2 = m(m + 1)(2m + 1) 6 , etc.

Caicedo Coloring the n-smooth numbers with n colors

Now, for any k, m

i=1 ik is a polynomial in m of degree k + 1. For

instance,

m

• i=1

i = m(m + 1) 2 ,

m

• i=1

i2 = m(m + 1)(2m + 1) 6 , etc. In particular, for k = 6, we have

m

• i=1

i6 = m(m + 1)(2m + 1)(3m4 + 6m3 − 3m + 1) 42 .

Caicedo Coloring the n-smooth numbers with n colors

n

• i=1

i6 = n(n + 1)(2n + 1)(3n4 + 6n3 − 3n + 1) 42 .

Caicedo Coloring the n-smooth numbers with n colors

n

• i=1

i6 = n(n + 1)(2n + 1)(3n4 + 6n3 − 3n + 1) 42 . Suppose then that p = 6n + 1 is a strong representative of order n. It follows that p must divide 3n4 + 6n3 − 3n + 1.

Caicedo Coloring the n-smooth numbers with n colors

n

• i=1

i6 = n(n + 1)(2n + 1)(3n4 + 6n3 − 3n + 1) 42 . Suppose then that p = 6n + 1 is a strong representative of order n. It follows that p must divide 3n4 + 6n3 − 3n + 1. But 432(3n4+6n3−3n+1) = (216n3+396n2−66n−205)(6n+1)+637,

Caicedo Coloring the n-smooth numbers with n colors

n

• i=1

i6 = n(n + 1)(2n + 1)(3n4 + 6n3 − 3n + 1) 42 . Suppose then that p = 6n + 1 is a strong representative of order n. It follows that p must divide 3n4 + 6n3 − 3n + 1. But 432(3n4+6n3−3n+1) = (216n3+396n2−66n−205)(6n+1)+637, so p must divide 637 = 72 · 13, and n = 1 or n = 2.

Caicedo Coloring the n-smooth numbers with n colors

n

• i=1

i6 = n(n + 1)(2n + 1)(3n4 + 6n3 − 3n + 1) 42 . Suppose then that p = 6n + 1 is a strong representative of order n. It follows that p must divide 3n4 + 6n3 − 3n + 1. But 432(3n4+6n3−3n+1) = (216n3+396n2−66n−205)(6n+1)+637, so p must divide 637 = 72 · 13, and n = 1 or n = 2. Since 26 = 64 ≡ −1 ≡ 1 (mod 13), it follows that p = 7 and p = 13 are the only such primes.

Caicedo Coloring the n-smooth numbers with n colors

Partial homomorphisms

Let’s consider a 4-satisfactory coloring c.

Caicedo Coloring the n-smooth numbers with n colors

Partial homomorphisms

Let’s consider a 4-satisfactory coloring c. By arguments like the ones we saw before we have that for any a ∈ K4, c(6a) = c(a) and c(8a) = c(3a).

Caicedo Coloring the n-smooth numbers with n colors

Partial homomorphisms

Let’s consider a 4-satisfactory coloring c. By arguments like the ones we saw before we have that for any a ∈ K4, c(6a) = c(a) and c(8a) = c(3a). It follows that c(16a) = c(8 · 2a) = c(3 · 2a) = c(a), and thus

Caicedo Coloring the n-smooth numbers with n colors

Partial homomorphisms

Let’s consider a 4-satisfactory coloring c. By arguments like the ones we saw before we have that for any a ∈ K4, c(6a) = c(a) and c(8a) = c(3a). It follows that c(16a) = c(8 · 2a) = c(3 · 2a) = c(a), and thus c(2α3β) = c(2α+3β mod 4).

Caicedo Coloring the n-smooth numbers with n colors

Partial homomorphisms

Let’s consider a 4-satisfactory coloring c. By arguments like the ones we saw before we have that for any a ∈ K4, c(6a) = c(a) and c(8a) = c(3a). It follows that c(16a) = c(8 · 2a) = c(3 · 2a) = c(a), and thus c(2α3β) = c(2α+3β mod 4). Conversely, it is easy to check that coloring 2α3β according to (α + 3β mod 4) is 4-satisfactory.

Caicedo Coloring the n-smooth numbers with n colors

This suggests the following:

Caicedo Coloring the n-smooth numbers with n colors

This suggests the following: Theorem Let p1, . . . , pk be the primes less than or equal to n. Given constants a1, . . . , ak, define the map c on Kn by c

• i

pαi

i

• = (a1α1 + · · · + akαk mod n).

If c is injective on {1, . . . , n}, then it is an n-satisfactory coloring.

Caicedo Coloring the n-smooth numbers with n colors

For example:

Caicedo Coloring the n-smooth numbers with n colors

For example: For n = 5, we can define c on Kn by c(2α3β5γ) = (α + 3β + 4γ mod 5).

Caicedo Coloring the n-smooth numbers with n colors

For example: For n = 5, we can define c on Kn by c(2α3β5γ) = (α + 3β + 4γ mod 5). For n = 13, we can define c on Kn by c(2α3β5γ7δ11ε13ζ) = (α + 4β + 9γ + 7δ + 11ε + 12ζ mod 13).

Caicedo Coloring the n-smooth numbers with n colors

For example: For n = 5, we can define c on Kn by c(2α3β5γ) = (α + 3β + 4γ mod 5). For n = 13, we can define c on Kn by c(2α3β5γ7δ11ε13ζ) = (α + 4β + 9γ + 7δ + 11ε + 12ζ mod 13). For n = 24, we can define c by c(2α3β5γ7δ11ε13ζ17η19θ23ι) = (α + 5β + 12γ + 15δ + 18ε + 9ζ + 21η + 22θ + 23ι mod 24).

Caicedo Coloring the n-smooth numbers with n colors

For example: For n = 5, we can define c on Kn by c(2α3β5γ) = (α + 3β + 4γ mod 5). For n = 13, we can define c on Kn by c(2α3β5γ7δ11ε13ζ) = (α + 4β + 9γ + 7δ + 11ε + 12ζ mod 13). For n = 24, we can define c by c(2α3β5γ7δ11ε13ζ17η19θ23ι) = (α + 5β + 12γ + 15δ + 18ε + 9ζ + 21η + 22θ + 23ι mod 24). At least for small values of n, it is easy to build by hand examples

• f such linear combinations by means of a greedy algorithm with

backtracking.

Caicedo Coloring the n-smooth numbers with n colors

These maps are examples of what we call partial homomorphisms

• f order n. The name comes from the fact that they can be used

to define certain group structures on {1, 2, . . . , n}.

Caicedo Coloring the n-smooth numbers with n colors

These maps are examples of what we call partial homomorphisms

• f order n. The name comes from the fact that they can be used

to define certain group structures on {1, 2, . . . , n}. All partial homomorphisms of order n are n-satisfactory colorings. This approach generalizes the previous approach via strong representatives.

Caicedo Coloring the n-smooth numbers with n colors

These maps are examples of what we call partial homomorphisms

• f order n. The name comes from the fact that they can be used

to define certain group structures on {1, 2, . . . , n}. All partial homomorphisms of order n are n-satisfactory colorings. This approach generalizes the previous approach via strong representatives. Fact If p is a strong representative of order n, the n-satisfactory coloring corresponding to p comes from a partial homomorphism.

Caicedo Coloring the n-smooth numbers with n colors

These maps are examples of what we call partial homomorphisms

• f order n. The name comes from the fact that they can be used

to define certain group structures on {1, 2, . . . , n}. All partial homomorphisms of order n are n-satisfactory colorings. This approach generalizes the previous approach via strong representatives. Fact If p is a strong representative of order n, the n-satisfactory coloring corresponding to p comes from a partial homomorphism. There are, however, partial homomorphisms not induced by strong representatives, so this method is more general.

Caicedo Coloring the n-smooth numbers with n colors

Theorem In p is prime and n = p2 − p then there is a partial homomorphism

• f order n.

Caicedo Coloring the n-smooth numbers with n colors

Theorem In p is prime and n = p2 − p then there is a partial homomorphism

• f order n.

Unfortunately, we also have the following negative result.

Caicedo Coloring the n-smooth numbers with n colors

Theorem In p is prime and n = p2 − p then there is a partial homomorphism

• f order n.

Unfortunately, we also have the following negative result. Theorem (Forcade-Pollington) Not all n admit partial homomorphisms. The first n for which none exists is n = 195.

Caicedo Coloring the n-smooth numbers with n colors

Theorem In p is prime and n = p2 − p then there is a partial homomorphism

• f order n.

Unfortunately, we also have the following negative result. Theorem (Forcade-Pollington) Not all n admit partial homomorphisms. The first n for which none exists is n = 195. Accordingly, n = 195 is the first number for which we do not know if n-satisfactory colorings exist. Either building a 195-satisfactory coloring or showing that there are none will require new ideas.

Caicedo Coloring the n-smooth numbers with n colors

Counting n-satisfactory colorings

I close with some figures:

Caicedo Coloring the n-smooth numbers with n colors

Counting n-satisfactory colorings

I close with some figures: If there is an n-satisfactory coloring of Kn, there are as many n-satisfactory colorings of the positive integers as there are real numbers. In what follows, we only count colorings of Kn.

Caicedo Coloring the n-smooth numbers with n colors

Counting n-satisfactory colorings

I close with some figures: If there is an n-satisfactory coloring of Kn, there are as many n-satisfactory colorings of the positive integers as there are real numbers. In what follows, we only count colorings of Kn. For n ≤ 4 there is exactly one n-satisfactory coloring.

Caicedo Coloring the n-smooth numbers with n colors

Counting n-satisfactory colorings

I close with some figures: If there is an n-satisfactory coloring of Kn, there are as many n-satisfactory colorings of the positive integers as there are real numbers. In what follows, we only count colorings of Kn. For n ≤ 4 there is exactly one n-satisfactory coloring. For n = 5 there are precisely two.

Caicedo Coloring the n-smooth numbers with n colors

Counting n-satisfactory colorings

I close with some figures: If there is an n-satisfactory coloring of Kn, there are as many n-satisfactory colorings of the positive integers as there are real numbers. In what follows, we only count colorings of Kn. For n ≤ 4 there is exactly one n-satisfactory coloring. For n = 5 there are precisely two. For n ≤ 5 all n-satisfactory colorings are partial homomorphisms.

Caicedo Coloring the n-smooth numbers with n colors

For any n there are only finitely many partial homomorphisms

• f order n.

Caicedo Coloring the n-smooth numbers with n colors

For any n there are only finitely many partial homomorphisms

• f order n.

There are precisely five partial homomorphisms of order 6.

Caicedo Coloring the n-smooth numbers with n colors

For any n there are only finitely many partial homomorphisms

• f order n.

There are precisely five partial homomorphisms of order 6. There are precisely six partial homomorphisms of order 7.

Caicedo Coloring the n-smooth numbers with n colors

For any n there are only finitely many partial homomorphisms

• f order n.

There are precisely five partial homomorphisms of order 6. There are precisely six partial homomorphisms of order 7. For n ≤ 7 all partial homomorphisms of order n admit strong representatives.

Caicedo Coloring the n-smooth numbers with n colors

There are precisely fourteen partial homomorphisms of order

• 8. Of these, only 4 admit strong representatives.

Caicedo Coloring the n-smooth numbers with n colors

There are precisely fourteen partial homomorphisms of order

• 8. Of these, only 4 admit strong representatives.

The least n for which there is no partial homomorphisms of

• rder n is n = 195. (See OEIS A204811 for the list of such n.)

Caicedo Coloring the n-smooth numbers with n colors

Theorem There are as many 6-satisfactory colorings not coming from a partial homomorphism as there are real numbers.

Caicedo Coloring the n-smooth numbers with n colors

×2 ×3 81 162 324 6481296 27 54 108 216 432 9 18 36 72 144 3 6 12 24 48 1 2 4 8 16 Many thanks!

Caicedo Coloring the n-smooth numbers with n colors