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 of the Hungarian journal K¨ oMaL ( K¨ oz´ 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¨ olgyi D¨ om¨ ot¨ or, 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 A 1 , . . . , A m of sets is the set of elements that belong to an odd number of the A i . 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 A 1 , . . . , A m of sets is the set of elements that belong to an odd number of the A i . 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 = { a 1 , a 2 , . . . , a k } . The symmetric difference of A, 2 · A, . . . , n · A is also the symmetric difference of the sets B 1 , B 2 , . . . , B k , where B i = { a i , 2 a i , . . . , na i } . ❑ 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 = { a 1 , a 2 , . . . , a k } . The symmetric difference of A, 2 · A, . . . , n · A is also the symmetric difference of the sets B 1 , B 2 , . . . , B k , where B i = { a i , 2 a i , . . . , na i } . For any n -satisfactory coloring, in every B i each color appears exactly once. That is, the sets B 1 , B 2 , . . . , B k 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 = { a 1 , a 2 , . . . , a k } . The symmetric difference of A, 2 · A, . . . , n · A is also the symmetric difference of the sets B 1 , B 2 , . . . , B k , where B i = { a i , 2 a i , . . . , na i } . For any n -satisfactory coloring, in every B i each color appears exactly once. That is, the sets B 1 , B 2 , . . . , B k contains k many numbers from each color class (counted by multiplicity). If k is odd, 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´ o around 2008–2009. That the general version remains open suggests that our 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´ o around 2008–2009. That the general version remains open suggests that our 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 < a 1 < a 2 < · · · < a n are integers, then a i max gcd( a i , a j ) ≥ n. i,j Caicedo Coloring the n -smooth numbers with n colors
In 1970, Ronald Graham conjectured the following: Conjecture If n ≥ 1 , and 0 < a 1 < a 2 < · · · < a n are integers, then a i max gcd( a i , a j ) ≥ n. i,j For example, let n = 4 and a 1 = 2 , a 2 = 3 , a 3 = 4 , a 4 = 6 . We have � a i � gcd( a i , a j ) : 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 < b 1 < · · · < b m are integers such that max i,j b i / gcd( b i , b j ) < 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 < b 1 < · · · < b m are integers such that max i,j b i / gcd( b i , b j ) < m . Suppose i � = j and let M = gcd( b i , b j ) . Let a i = b i /M and a j = b j /M , so a i , a j are both less than m , and a i � = a j . Since b i = a i M and b j = a j M , in any ( m − 1) -satisfactory coloring we must have that b i is colored differently from b j . ❑ 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 < b 1 < · · · < b m are integers such that max i,j b i / gcd( b i , b j ) < m . Suppose i � = j and let M = gcd( b i , b j ) . Let a i = b i /M and a j = b j /M , so a i , a j are both less than m , and a i � = a j . Since b i = a i M and b j = a j M , in any ( m − 1) -satisfactory coloring we must have that b i is colored differently from b j . 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 2 n + 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 2 n + 1 is prime, then there is an n -satisfactory coloring. Proof. Let p = 2 n + 1 and color a ∈ K n with color c ( a ) = ( a 2 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 2 n + 1 is prime, then there is an n -satisfactory coloring. Proof. Let p = 2 n + 1 and color a ∈ K n with color c ( a ) = ( a 2 mod p ) . Note that if a 2 ≡ b 2 (mod p ) then p | a 2 − b 2 = ( 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 2 n + 1 is prime, then there is an n -satisfactory coloring. Proof. Let p = 2 n + 1 and color a ∈ K n with color c ( a ) = ( a 2 mod p ) . Note that if a 2 ≡ b 2 (mod p ) then p | a 2 − b 2 = ( a − b )( a + b ) , which means that either a ≡ b (mod p ) or a ≡ − b (mod p ) . This shows that 1 2 , . . . , n 2 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 2 n + 1 is prime, then there is an n -satisfactory coloring. Proof. Let p = 2 n + 1 and color a ∈ K n with color c ( a ) = ( a 2 mod p ) . Note that if a 2 ≡ b 2 (mod p ) then p | a 2 − b 2 = ( a − b )( a + b ) , which means that either a ≡ b (mod p ) or a ≡ − b (mod p ) . This shows that 1 2 , . . . , n 2 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 ∈ K n 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 1 k , 2 k , . . . , n k are pairwise distinct modulo p , then c ( a ) = ( a k mod p ) is an n -satisfactory coloring of K n . Caicedo Coloring the n -smooth numbers with n colors
The argument generalizes as follows: Theorem If p = kn + 1 is prime, and 1 k , 2 k , . . . , n k are pairwise distinct modulo p , then c ( a ) = ( a k mod p ) is an n -satisfactory coloring of K n . Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero k th 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 1 k , 2 k , . . . , n k are pairwise distinct modulo p , then c ( a ) = ( a k mod p ) is an n -satisfactory coloring of K n . Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero k th powers modulo p . Our assumption says that these powers are precisely 1 k , . . . , n k . Caicedo Coloring the n -smooth numbers with n colors
The argument generalizes as follows: Theorem If p = kn + 1 is prime, and 1 k , 2 k , . . . , n k are pairwise distinct modulo p , then c ( a ) = ( a k mod p ) is an n -satisfactory coloring of K n . Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero k th powers modulo p . Our assumption says that these powers are precisely 1 k , . . . , n k . If 1 ≤ i < j ≤ n and a ∈ K n then i k �≡ j k (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 1 k , 2 k , . . . , n k are pairwise distinct modulo p , then c ( a ) = ( a k mod p ) is an n -satisfactory coloring of K n . Proof. (Sketch) On general grounds, if p = kn + 1 is prime, there are precisely n distinct nonzero k th powers modulo p . Our assumption says that these powers are precisely 1 k , . . . , n k . If 1 ≤ i < j ≤ n and a ∈ K n then i k �≡ j k (mod p ) , so c ( ia ) � = c ( ja ) . ❑ Note that the assumption that 1 k , . . . , n k 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 of the form 3 n + 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 of the form 3 n + 1 . n = 2 . We have 2 3 = 8 ≡ 1 = 1 3 (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 of the form 3 n + 1 . n = 2 . We have 2 3 = 8 ≡ 1 = 1 3 (mod 7) . n = 4 . We have 3 3 = 27 ≡ 1 = 1 3 (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 of the form 3 n + 1 . n = 2 . We have 2 3 = 8 ≡ 1 = 1 3 (mod 7) . n = 4 . We have 3 3 = 27 ≡ 1 = 1 3 (mod 13) . n = 6 . We have 3 3 = 27 ≡ 8 = 2 3 (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 of the form 3 n + 1 . n = 2 . We have 2 3 = 8 ≡ 1 = 1 3 (mod 7) . n = 4 . We have 3 3 = 27 ≡ 1 = 1 3 (mod 13) . n = 6 . We have 3 3 = 27 ≡ 8 = 2 3 (mod 19) . n = 10 . We have 5 3 = 125 ≡ 1 = 1 3 (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 of the form 3 n + 1 . n = 2 . We have 2 3 = 8 ≡ 1 = 1 3 (mod 7) . n = 4 . We have 3 3 = 27 ≡ 1 = 1 3 (mod 13) . n = 6 . We have 3 3 = 27 ≡ 8 = 2 3 (mod 19) . n = 10 . We have 5 3 = 125 ≡ 1 = 1 3 (mod 31) . n = 12 . We have 10 3 = 1000 ≡ 1 = 1 3 (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 = 3 n + 1 is prime. Then there is an i , 2 < i ≤ n , such that i 3 ≡ 8 or i 3 ≡ 1 (mod p ) . Caicedo Coloring the n -smooth numbers with n colors
In general, we have Theorem Suppose p = 3 n + 1 is prime. Then there is an i , 2 < i ≤ n , such that i 3 ≡ 8 or i 3 ≡ 1 (mod p ) . That is, we can never verify the assumption that p = 3 n + 1 is prime and 1 3 , . . . , n 3 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 1 k , . . . , n k 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 1 k , . . . , n k 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 1 k , . . . , n k 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 order 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 1 k , . . . , n k 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 order 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 1 k , . . . , n k 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 order 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 1 k , . . . , n k 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 order 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 k th powers modulo p is a nontrivial subgroup of 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 k th powers modulo p is a nontrivial subgroup of Z ∗ p : Clearly the product of two nonzero k th powers is again a k th power, and it is nonzero modulo p since p is prime. Caicedo Coloring the n -smooth numbers with n colors
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