Coloring the $n$-smooth numbers with $n$ colors

For which values of $n$ can we color the positive integers with precisely $n$ colors in such a way that for any $a$, the numbers $a,2a,\dots,na$ all get different colors? Pach posed the question around 2008-9. Particular cases appeared in K\"oMaL in April 2010, and the general version appeared in May 2010 on MathOverflow, posted by P\'alv\"olgyi. The question remains open. We discuss the known partial results and investigate a series of related matters attempting to understand the structure of these $n$-satisfactory colorings. Specifically, we show that there is an $n$-satisfactory coloring whenever there is an abelian group operation $\oplus$ on the set $\{1,2,\dots,n\}$ compatible with multiplication in the sense that whenever $i$, $j$ and $ij$ are in $\{1,\dots,n\}$, then $ij=i\oplus j$. This includes in particular the cases where $n+1$ is prime, or $2n+1$ is prime, or $n=p^2-p$ for some prime $p$, or there is a $k$ such that $q=nk+1$ is prime and $1^k,\dots,n^k$ are all distinct modulo $q$ (in which case we call $q$ a strong representative of order $n$). The colorings obtained by this process we call multiplicative. We also show that nonmultiplicative colorings exist for some values of $n$. There is an $n$-satisfactory coloring of $\mathbb Z^+$ if and only if there is such a coloring of the set $K_n$ of $n$-smooth numbers. We identify all $n$-satisfactory colorings for $n\le 5$ and all multiplicative colorings for $n\le 8$, and show that there are as many nonmultiplicative colorings of $K_n$ as there are real numbers for $n=6$ and 8. We show that if $n$ admits a strong representative $q$ then the set of such $q$ has positive natural density in the set of all primes. We show that the question of whether there is an $n$-satisfactory coloring is equivalent to a problem about tilings, and use this to give a geometric characterization of multiplicative colorings.

Prove that for every prime p, there exists a colouring of the positive integers with p−1 colours such that the colours of the numbers {a, 2a, 3a, . . . , (p − 1)a} are pairwise different for every positive integer a. We say that a coloring as required is (p − 1)-satisfactory. To get a feel for the problem, consider for instance the case p = 5. Suppose c is a 4-satisfactory coloring. In particular, 1, 2, 3, 4 have different colors. Note that 6 must have the same color as 1, since c(6) = c(2), c(4) because 2, 4, 6, 8 all have different colors and c(6) = c(3) since 3, 6, 9, 12 all have different colors. It follows that c(8) = c(3). Also, since c(12) = c(3), c(4), c(6), we must have c(12) = c(2). It follows that c(9) = c(4). Similar reasoning allows us to determine the color of many more numbers; the following table shows some of these findings. Here, the coloring is represented by means of 4 rows of integers, with each row representing one of the colors.
1 6 16 36 81 · · · 2 12 27 32 72 · · · 3 8 18 48 108 · · · 4 9 24 54 64 · · · Nothing so far uses that 5 is a prime number, but the relevance of this fact comes into play once we note that the numbers in the i th row are all congruent to i modulo 5, for i = 1, . . . , 4. This suggests how to define a 4-satisfactory coloring compatible with our observations. Indeed, as long as a is not a multiple of 5, we can assign to a the color (a mod 5) and readily observe that if 1 ≤ i < j ≤ 4, then (ai mod 5) = (aj mod 5). We are not quite done yet, as we still need to deal with the multiples of 5. For this, we can begin by noting that 5, 10, 15, 20 have different colors and impose some restrictions as above. . . The reader should promptly realize that this is the same table as before, with each entry multiplied by 5. This suggests that we can define the color of a positive integer n by considering its prime factorization and ignoring powers of 5: letting n = 5 a b where a ≥ 0 and 5 b, we can assign to n the color c(n) = (b mod 5). It is straightforward to verify that this is indeed a 4-satisfactory coloring, and we are done in this case.
The argument suggests an obvious generalization from which the KöMaL problem follows: Theorem 1.1. If p is prime, then there is a (p − 1)-satisfactory coloring.
Proof. Define a coloring c by writing n = p a b where a ≥ 0 and p b, and letting c(n) = (b mod p), so that c uses p − 1 colors and if 1 ≤ i < j < p and n is as indicated, then c(in) = (ib mod p) = (jb mod p) = c(jn).
Although the solution just described makes essential use of the fact that p is prime, it is natural to wonder whether such colorings are possible without this restriction. It is this version of the problem that we discuss in this paper.
1.2. The general question. In May 29, 2010, Dömötör Pálvölgyi posted on Math-Overflow precisely the version just indicated.
Question 1.2. Given any positive integer n, is there is a coloring of the positive integers using n colors and such that for any positive integer a, the numbers a, 2a, . . . , na all have different colors? 2 It was through Pálvölgyi's post that the first-named author became acquainted with the problem. He suggested it to the second-named author, and their partial results became the main content of the latter's master's thesis 3 . Question 1.2 was originally formulated by the third-named author around 2008-2009, motivated by a question of Günter Pilz, see § 1.4. After working on it for a while, he posed several related questions on KöMaL. For instance, besides problem A.506, he also posed problem B.4265 in the April 2010 issue, 4 asking about the case n = 7. Pálvölgyi first saw problem A.506 and became interested in the general version. He contacted the editor of KöMaL in charge of the "A problems" and asked whether they knew the answer for general n. It was not until years later that Pálvölgyi found out that the KöMaL questions and the general version of the problem were originally posed by the third-named author.
Although the general problem remains open, there are enough partial results that we feel it is appropriate to publish this paper now, to further expose the mathematical community at large to question 1.2, and to indicate the current state of affairs and the many additional questions that come out of this exploration. Question 1.2 has connections with number theory and group theory, and an obvious combinatorial core. Some of the ideas we describe benefit from this interaction.
Several results we present are due to others, either from previous research on related topics or through suggestions posted on MathOverflow. We make every attempt to give credit as appropriate.
As suggested above, we call n-satisfactory a coloring as in the statement of question 1.2. The analysis of the case n = 4 in § 1.1 reveals that we can in general restrict our attention to seeking n-satisfactory colorings of the set of n-smooth numbers, that is, the set K n of positive integers whose prime factorization only includes primes less than or equal to n. We call this set the n-core, see definition 2.1, and elaborate on this issue in section 2; briefly, if there is an n-satisfactory coloring of K n , we can color all positive integers by assigning to the number km, where k ∈ K n and gcd(m, n!) = 1 the color of k, and one can quickly check that this is an n-satisfactory coloring of Z + .
We note that, given n, even if question 1.2 has a negative answer for n, strictly fewer than 2n colors suffice to ensure that for any a ∈ K n all numbers ia, 1 ≤ i ≤ n, receive different colors: indeed, letting p be the smallest prime larger than n, we can color K n with p colors as in theorem 1.1, by assigning to m ∈ K n the color (m mod p).
We close this introduction by discussing an application and the original motivation for question 1.2.
Graham's conjecture was finally verified in 1996 by Balasubramanian and Soundararajan via careful analytic estimates of average values of number-theoretic functions associated with the distribution of primes, see [BS96]. Assuming the existence of satisfactory colorings, we obtain a significantly simpler proof. Proof. Argue by contradiction. Accordingly, suppose that there are (m − 1)-satisfactory colorings and that 0 < b 1 < · · · < b m are integers such that 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 of Z + 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.
The relationship highlighted in theorem 1.3 between our question 1.2 and the Balasubramanian-Soundararajan theorem admits a nice graph-theoretic interpretation, that we now proceed to discuss. This connection was first mentioned by Péter Csikvári to the third-named author, and was also noticed independently by Fedor Petrov in MathOverflow 5 and by Bosek, Dębski, Grytczuk, Sokół, Śleszyńska-Nowak and Żelazny, who also arrived independently of us at some of the observations below in their recent paper [BDG + 18] (particularly, see their § 4).
Given a graph G, write χ(G) for its chromatic number, that is, the least cardinal κ such that the set of vertices of G can be colored with κ colors in such a way that adjacent vertices receive different colors. Note that if G admits a clique (complete subgraph) on r vertices, then χ(G) ≥ r, and therefore χ(G) ≥ ω(G), where ω(G) is the clique number of G, that is, the supremum of the cardinalities of the cliques of G.
Given n, consider now the graph G n with the positive integers as vertices and where any two i = j in Z + are connected if and only if max(i, j)/ gcd(i, j) ≤ n (that is, if and only if i, j ∈ {a, 2a, . . . , na} for some a); in [BDG + 18], this graph is denoted B n . Note that G n has many cliques of size n, namely each set {a, 2a, . . . , na} (and possibly others), so that ω(G n ) ≥ n; see figure 1.1. By definition, an n-satisfactory coloring c of Z + is a coloring of G n with n colors, that is, the existence of such a map c is precisely the claim that χ(G n ) is (at most, and therefore equal to) n. By the remark above, this implies that ω(G n ) is (at most, and therefore equal to) n, but this is precisely Graham's conjecture for n + 1, and we have reproved theorem 1.3.
1.4. Pilz's conjecture. Recall that the symmetric difference C D of two sets C, D, is the set of elements that belong to exactly one of C, D, that is, Note that is associative, and so, given sets C 1 , . . . , C m , their symmetric difference m l=1 C i is simply the set of elements that belong to precisely an odd number of sets C l .
If X ⊆ Z + and k ∈ R, we denote by k · X the dilation of X by a factor of k: k · X = {kx : x ∈ X}.
Pilz's conjecture, the original motivation for question 1.2, first appeared in 1992 [Pil92]. For our purposes, it is convenient to phrase it as follows: If n ≥ 1 and A is a finite set of positive integers, then the size of the symmetric difference of the sets A, 2 · A, . . . , n · A is at least n.  [HKP10] and by Pach and C. Szabó [PS11]. The general case remains open.
Theorem 1.4. If there is an n-satisfactory coloring then Pilz's conjecture holds for n under the further assumption that |A| is odd.
Proof. Say that |A| = k, let A = {a j : j ∈ [k]} and, for j ∈ [k], set Note first that the symmetric difference of the sets i · A, i ∈ [n], equals the symmetric difference of the sets B j , j ∈ [k]. The point is that, denoting by χ C (·) the characteristic function of a set C, we have for any element x that For any n-satisfactory coloring, in every B i each color appears exactly once. That is, the sets B 1 , B 2 , . . . , B k contain k numbers from each color class (counted with multiplicity). If k is odd, then this means that their symmetric difference must contain an odd number of elements from each color class (and therefore at least one). But there are n colors.
It was precisely this observation that motivated the third-named author to formulate question 1.2. Sadly, when |A| is even, the trick above does not apply and we do not see a way of establishing the conjecture in full generality.
Pilz formulated in [Pil92] both the general case and the special case of his conjecture where A = [k] for some k (the latter is sometimes called the 1-2-3 conjecture). The paper [PS11] is based on the third-named author master's thesis 6 . For , if there is a k-satisfactory coloring and n is odd, then the same argument gives us that the size of the symmetric difference n i=1 i · A is at least n.
1.5. Organization of this paper. We begin section 2 with some preliminaries, emphasizing the role of what we call the n-core K n . We also include some easy observations on the structure of the set C Kn of n-satisfactory colorings of the ncore. We reformulate question 1.2 as a problem about tilings, and close the section by giving an explicit description of all n-satisfactory colorings for n ≤ 5. In section 3 we explore a suggestion that directly generalizes the approach used to solve the original KöMaL problem (the case where n + 1 is a prime number). This suggestion leads to several interesting number-theoretic questions that we also discuss. In section 4 we discuss a group-theoretic approach to question 1.2 that encompasses the suggestion from section 3. The colorings to which this suggestion applies we call multiplicative; we also characterize these colorings geometrically through the notion of translation invariance, and close the section by listing all multiplicative n-satisfactory colorings for n ≤ 8. We conclude in section 5 by indicating cases where the approach from section 4 fails. This includes a brief review of prior work by Forcade and Pollington [FP90], and also a discussion of nonmultiplicative 6-and 8-satisfactory colorings. The theorem below summarizes the values of n for which a positive answer to question 1.2 is known, see also § 2.5.
Theorem 1.5. Question 1.2 has a positive answer for n, that is, there is an nsatisfactory coloring, in any of the following cases: (1) n + 1 is prime.
(3) More generally, there is a strong representative of order n, i.e., a prime p of the form nk + 1 for some k such that 1 k , . . . , n k are pairwise distinct modulo p. (4) Yet more generally, n admits a partial G-isomorphism for some abelian group (G, ⊕) of order n, i.e., there is a bijection h : [n] → G such that whenever a, b ∈ [n], if ab ∈ [n], then h(ab) = h(a) ⊕ h(b). (5) In particular, question 1.2 has a positive answer for all n < 195 and for all n of the form p 2 − p for some prime p. Proof.

The core
An n-coloring of a set X is a coloring of X using exactly n colors. An nsatisfactory coloring is an n-coloring witnessing a positive answer to the nth instance of question 1.2. The nature of these colors is of course irrelevant, but we need some convention since we want to address questions about the number of ncolorings satisfying some property (such as, primarily, being n-satisfactory). There are two natural ways of thinking about n-colorings, and we adopt both in what follows. We will typically consider only colorings of the n-core K n rather than of all of Z + , but what follows applies in either case. In the first approach, we think of an n-coloring as a map c with range [n], and we further adopt the convention that c(i) = i for i ∈ [n]. The second approach is perhaps more natural: rather than thinking of a coloring as a map, we think of it as an equivalence relation, whose classes are precisely the colors. We still adopt functional notation, so we write, for example, c(a) = c(b) to indicate that c assigns the same color to the numbers a and b. Still, on occasion we may stray from these conventions for ease of exposition.
2.1. The core K n and n-appropriate sets.
Definition 2.1. The n-core, or simply the core if n is understood, is the set K n of all positive integers whose prime decomposition only involves primes less than or equal to n. This is the set of numbers usually called n-smooth.
In the literature the notion of n-smooth numbers is typically reserved for the case where n itself is a prime number. We do not impose this requirement so, for instance, K 7 = K 8 = K 9 = K 10 . In the notation of [BDG + 18], K n is denoted N n .
The key reason for considering cores is that there is an n-satisfactory coloring (of Z + ) if and only if there is an n-satisfactory coloring of the n-core. In fact we prove a bit more, indicating that in order to understand the structure of the set of n-satisfactory colorings, attention can be restricted to those of the n-core. Once we establish this fact, we proceed accordingly, which explains the title of this paper.
In particular, restricting attention to colorings of the n-core allows us to address the following question.
Question 2.2. Given n > 1, how many n-satisfactory colorings are there, if any at all?
As we will see, the answer to question 2.2 is c = |R| for colorings of Z + even in cases where there are only finitely many n-satisfactory colorings of K n , see corollary 2.7 and theorem 4.14.
Definition 2.3. Say that X ⊆ Z + is n-appropriate if and only if X is nonempty and contains ix and x/j whenever x ∈ X, i, j ≤ n, and j divides x.
If X is n-appropriate, say that an n-coloring c of X is n-satisfactory if and only if c(ix) = c(jx) whenever x ∈ X and i < j ≤ n. Note that this coincides with the previous notion of n-satisfactory when X = Z + (or X = K n ).
We denote by P n the set of numbers relatively prime to n!, i.e., those positive integers whose prime decomposition only involves prime numbers strictly larger than n.
The notation X =˙ a∈A X a means both that X is the union of the sets X a for a ∈ A, and that the sets X a are pairwise disjoint.
Moreover, if this is the case, then we have A = P n ∩ X.
Proof. Note that K n is n-appropriate and therefore so is a · K n for any a ∈ P n . It follows that any X of the form˙ a∈A a · K n for A ⊆ P n and nonempty is nappropriate as well.
Towards the converse, suppose now that X is n-appropriate. Each m ∈ Z + can be uniquely written in the form m = a m k m where a m ∈ P n and k m ∈ K n . Let First, note that if a = b are in P n , then a · K n and b · K n are pairwise disjoint. Now, if a ∈ A, then there is some x ∈ X such that a = a x . Since X is n-appropriate, h/j ∈ X whenever h ∈ X and j ∈ K n divides h. In particular, a = a x = x/k x ∈ X. Similarly, hi ∈ X whenever h ∈ X and i ∈ K n . Therefore, a · K n ⊆ X. This means that˙ a∈A a · K n ⊆ X.
But, if x ∈ X, then x ∈ a x · K n , and we have thaṫ This proves the equality and establishes the equivalence.
Second, since if a ∈ P n then the only member of P n in a · K n is a itself, it follows that if X =˙ a∈A a · K n for some A ⊆ P n , then in fact A = P n ∩ X, and we are done.
For X n-appropriate, let C X,n be the set of n-satisfactory colorings of X, and denote by C n the set C Z + ,n . We also write C X , C if n is clear from context.
The following proposition shows that C = ∅ if and only if C X = ∅ for some n-appropriate set X if and only if C X = ∅ for all n-appropriate sets X.
In particular, as emphasized earlier, it follows that the question of whether there are any n-satisfactory colorings is really a question about whether there are nsatisfactory colorings of K n . In fact, the proposition shows how the satisfactory colorings of the core completely determine all satisfactory colorings.
(1) Suppose X ⊆ Y are n-appropriate. If C Y = ∅, then C X = ∅. In fact, the restriction c X is in C X for any c ∈ C Y . (2) Given a ∈ P n and c ∈ C a·Kn , if c denotes the n-coloring of K n such that c (k) = c (l) if and only if c(ak) = c(al), then c ∈ C Kn . (3) Given a ∈ P n and c ∈ C Kn , ifĉ a denotes the n-coloring of a · K n such that c a (ak) =ĉ a (al) if and only if c(k) = c(l), thenĉ a ∈ C a·Kn . (4) If X is n-appropriate, then c ∈ C X if and only if for each a ∈ P n ∩ X there is a map c a ∈ C Kn such that c =˙ a∈Pn∩Xĉ a a .
(5) C = ∅ if and only if C X = ∅ for any n-appropriate X if and only if C X = ∅ for some n-appropriate set X. Moreover if X ⊆ Y and both are n-appropriate, then d ∈ C X if and only if d = c X for some c ∈ C Y . Proof.
(2) Given a ∈ P n and c ∈ C a·Kn , if c is defined as in item (2), then c (ib) = c (jb) for any b ∈ K n and any i < j ≤ n because c is n-satisfactory and therefore c(aib) = c(ajb). But this means that c is n-satisfactory as well.
(3) Conversely, if c ∈ C Kn , a ∈ P n , andĉ a is defined as in item (3), then c a (im) =ĉ a (jm) for any m ∈ a · K n and i < j ≤ n since c is satisfactory and therefore c(i(m/a)) = c(j(m/a)). But this means thatĉ a is satisfactory as well.
(4) Suppose that X is n-appropriate. First, X = a∈Pn∩X a · K n , by lemma 2.4. If c ∈ C X , then, by item (1), d = c a · K n ∈ C a·Kn for any a ∈ P n ∩ X, and c a = d ∈ C Kn by item (2). But d =d a , i.e., c = a∈Pn∩Xĉ a a . Conversely, suppose that C Kn = ∅, for each a ∈ P n ∩ X let c a ∈ C Kn and define c = a∈Pn∩Xĉ a a . As shown in lemma 2.4, a · K n ∩ b · K n = ∅ whenever a = b are in P n . From this, and item (3), c is well defined and has domain a∈Pn∩X a · K n , which equals X, again by lemma 2.4. If m ∈ X and i < j ≤ n, then there is a unique a ∈ P n ∩ X such that mi and mj belong to a · K n , and by item (3) it follows that c(mi) = c(mj). This proves that c is satisfactory, and completes the proof of item (4).
(5) Now, if X is n-appropriate, and C X = ∅, then C a·Kn = ∅ for any a in the nonempty set P n ∩ X, by item (1). But then C Kn = ∅, by item (2). It follows from item (4) that C = C Z + = ∅. But then C Y = ∅ for any n-appropriate Y , again by item (1).
Finally, if X ⊆ Y are n-appropriate and c ∈ C Y , then d = c X ∈ C X , by item (1). Conversely, if d ∈ C X , let e ∈ C Kn , which exists as shown above. Let c a = e for a ∈ P n ∩ (Y X). For a ∈ P n ∩ X, let c a = (d a · K n ) . As in item (4), we have that c = a∈Pn∩Yĉ a a ∈ C Y . And, by construction, d = c X. This completes the proof of item (5).
Remark 2.6. The notion of n-appropriate can be extended in a natural way, allowing us to verify that, for instance, there is an n-satisfactory coloring of K n if and only if there is one of Z {0}. More interesting is whether this is also equivalent to the existence of an n-satisfactory coloring of Q {0} or, what is the same, ofK n := {a/b : a, b ∈ K n }. We show below that this is indeed the case, see proposition 2.12. We also suggest a subtler problem in question 2.10. In [BDG + 18], K n is denoted Q n .
When discussing n-satisfactory colorings, proposition 2.5 provides us with the ability to restrict our attention from all of Z + to K n . The relation the proposition details between arbitrary satisfactory colorings and colorings of the core has the following corollary.
Proof. If there is a coloring of the core (thought of as a function with range [n]), there are at least n! ≥ 2 such colorings of any a · K n for a ∈ P n different from 1 (obtained by invoking item (3) of proposition 2.5 and simply permuting the colors). By item (4) of proposition 2.5, there is a bijective correspondence between the elements of C, and the set of functions from P n to C Kn , from which we get that |C| = |C Kn Pn | ≥ n! |Pn| = c. On the other hand, any element of C is a function from Z + to [n] (satisfying certain restrictions), so |C| ≤ |[n] Z + | = n ℵ0 = c, and it follows that |C| = c by the Bernstein-Cantor-Schröder theorem.
Proposition 2.5 and corollary 2.7 give us that if there is an n-satisfactory coloring of K n , then there are as many n-satisfactory colorings of Z + as there are real numbers. However, this abundance of colorings is a distraction since the underlying structure of any satisfactory coloring can be described in terms of what is happening on the core.
2.2. The structure of C Kn . We mention here some easy observations regarding the closure of C Kn under some natural operations.
(1) First, C Kn is a closed subset of the Polish space This topological fact is trivial in some cases, since (as shown in § 2.4) C Kn is sometimes finite, but see § 5.2. The Polish topology of the space [n] Kn is generated by a natural metric: enumerate K n in increasing order as . This metric is complete both in the whole space [n] Kn and in C Kn .
The truth is, we understand very little of the topological structure of C Kn . It is unclear, for instance, whether the following question should have a positive answer.
Question 2.8. Given n ∈ Z + , suppose that C Kn is nonempty. Should it have isolated points?
(2) The following is an immediate but useful observation.
Lemma 2.9. Let ρ be an automorphism of the structure ([n], |), that is, of the Hasse diagram for divisibility on [n]. Extend ρ to a bijection of K n in the natural way: if m = 2 α1 . . . p k α k is the prime factorization of m ∈ K n , then If c is an n-satisfactory coloring of K n , then so isc, wherec(a) =c(b) if and only if c(ρ(a)) = c(ρ(b)).
For an application, see the discussion of the case n = 5 in § 2.4, where it is also shown that the more inclusive condition that ρ be a permutation of the set of primes in [n] is not enough in general.
Proof. Note first that ρ permutes the primes less than or equal to n, and equation (2.1) holds for all m ∈ [n], so that the suggested extension is well defined and maps K n to itself. Since ρ is a permutation on the primes in [n], it follows as well that ρ is surjective on K n . Note also that for any m 1 , m 2 ∈ K n , ρ(m 1 m 2 ) = ρ(m 1 )ρ(m 2 ). Now, if c is satisfactory andc is as indicated, then for i = j in [n] and a ∈ K n , we have thatc(ia) = c(ρ(i)ρ(a)) = c(ρ(j)ρ(a)) =c(ja) since ρ(i), ρ(j) ∈ [n].
(3) Another natural operation on C Kn can be defined by letting c k ∈ C Kn , for c ∈ C Kn and k ∈ K n , be given by c k (l) = c k (m) if and only if c(kl) = c(km).
(Abusing slightly 7 the notation used in proposition 2.5, c k = (c k · K n ) .) We remark that although we concentrate on n-satisfactory colorings throughout the whole paper, on occasion we may consider translations c k of arbitrary n-colorings c, with the understanding that the definition just given applies in general.
Most of the colorings we consider in this paper are multiplicative (see § 4.1). For them, this operation is uninteresting: c = c k for any k ∈ K n whenever c is multiplicative. However, the operation may generate new colorings otherwise, see § 5.2. It also suggests the following natural problem.
Question 2.10. Given an n-satisfactory coloring c and k ∈ K n , is there is an n-satisfactory coloring d such that d k = c? In that case, how many such colorings d are there?
(4) Because C Kn is closed in [n] Kn , it is also closed under a construction coming from applications of König's infinity lemma. We discuss this construction in the next subsection, once the appropriate notation and terminology have been introduced, see remark 2.15.
2.3. Tilings. The switch from Z + to the set K n allows us to restate question 1.2 as a problem about tilings (this restatement is also mentioned by Pálvölgyi on his post in MathOverflow, and is the subject of [BDG + 18, § 4]).
To simplify the description, consider for now the case n = 3. In this case, the problem lives in the integer grid. Identify m = 2 a 3 b ∈ K 3 with the point (a, b) in the first quadrant of the integer lattice or, equivalently, the unit square with sides parallel to the axes and bottom left corner at (a, b). Now, a coloring is 3-satisfactory if and only if for any such pair (a, b), the pairs (a, b), (a + 1, b) (corresponding to 2m) and (a, b + 1) (corresponding to 3m) all receive different colors. The question of whether there is a 3-satisfactory coloring of K 3 becomes the question of whether we can assign to each unit square in the first quadrant one of three colors in such a way that all translates of the triomino consisting of the three unit squares in the bottom left corner of the quadrant contain tiles of all colors.
The case n = 3 is simple enough that one can easily see that there is a unique way of accomplishing this, illustrated in figures 2.1 and 2.2. (The names of the colors in figure 2.1 are chosen so that the color of each i = 1, 2, 3 is i itself.) Before proceeding, the reader may enjoy verifying that, similarly, there is a unique 4-satisfactory tiling of the first quadrant, as illustrated in figures 2.3 and 2.4.   Further cases are harder to illustrate, as they correspond in general to tilings of the first orthant of Z π(n) where, as usual, π(·) denotes the prime counting function (and in general lack the periodicity displayed in these two examples, but see remark 4.5). These tilings use unit "cubes" of n possible colors as tiles. More interestingly, we can instead restate question 1.2 as a problem about tilings with translates of the π(n)-dimensional polyomino corresponding to the set {1, 2, . . . , n} as tiles. We proceed now to explain this connection.
Given n, work in Z π(n) . As suggested above, we identify each member of K n with the tuple of its prime exponents: Any m ∈ K n can be written in a unique way as m = π(n) i=1 p αi i , where 2 = p 1 < · · · < p π(n) are the primes less than or equal to n, listed in increasing order, and the α i are nonnegative integers. We identify m with the tuple t(m) = (α 1 , . . . , α π(n) ) in the first orthant O n of Z π(n) , noting that t : K n → O n is a bijection, and let T n = {t(i) : i ∈ [n]}. Note that t turns multiplication into vector addition in the sense that t(kk ) = t(k) + t(k ) for any k, k ∈ K n . We will find several maps with similar properties in what follows, see for instance definition 4.1, where we call them multiplicative. Given A, C ⊆ Z π(n) , say that A tiles C (or, equivalently, that C can be tiled by A) if and only if there is a set B ⊆ Z π(n) such that C is the direct sum of A and B, that is, (1) C = A + B := {a + b : a ∈ A, b ∈ B}, and in fact (2) any c ∈ C admits a unique decomposition as a sum of a member of A and a member of B, that is, there is a unique pair (a, b) ∈ A × B with c = a + b. Also, say that A essentially tiles C if and only if C can be covered by a set that can be tiled by A (in which case, we call such a tiling of a superset of C an essential tiling of C by A). We remark that, as we did above, we may identify without further comment points (α i : i ∈ [π(n)]) in Z π(n) with the corresponding π(n)-dimensional For a fixed value of n, consider now the following two statements: (i) There is an n-satisfactory coloring of K n . (ii) T n essentially tiles the orthant O n . We have the following result.
Proposition 2.11. With notation as above, (i) and (ii) are equivalent.
Proof. To see that (ii) implies (i), consider a tiling by T n of a superset C of O n , say C = T n + B, the sum being direct. Note that via this direct sum, each element of C, and therefore each x ∈ O n , belongs to exactly one tile, that is, a unique copy of T n . There is a unique m ∈ K n such that x = t(m), where t is the map described above. Color m with the position of x within this tile. In other words, let the color classes be the preimages under t of the sets a + B for a ∈ T n . We must argue that this coloring is n-satisfactory. Indeed, given k ∈ K n and i, j ∈ [n], suppose that ki and kj receive the same color, that is, there are α ∈ T n and b 1 , b 2 ∈ B such that t(ki) = α + b 1 and t(kj) = α + b 2 . By the multiplicative property of t, it follows that t(j) + b 1 = t(i) + b 2 . Since the sum T n + B is direct, this means that (b 1 = b 2 and) t(i) = t(j), thus i = j, and the coloring is indeed n-satisfactory; see figure 2.5. To see that, conversely, (i) implies (ii), consider an n-satisfactory coloring c. Letting B be the image under the map t of one of the color classes, note that the sum T n + B is direct. Indeed, suppose that i, j ∈ [n], and k, k ∈ K n are such that t(k), t(k ) ∈ B and t(i) + t(k) = t(j) + t(k ), that is, ik = jk . By removing common factors if necessary, we may further assume that i, j are relatively prime. This means that there is a positive integer k such that k = jk and k = ik , and observe that k ∈ K n . The assumption that B is the image of a color class then means that c(jk ) = c(ik ), and since c is n-satisfactory, then i = j and so also k = k . This proves that the sum T n + B is indeed direct.
Let now x ∈ O n be sufficiently far from the boundary of O n , in the sense that all of x − t(1), . . . , x − t(n) are themselves in O n (equivalently, if x = t(m), then all of m, m/2, . . . , m/n are positive integers), and fix a color class B . We claim that for some i ∈ [n], we have that x − t(i) ∈ B . Otherwise, by the pigeonhole principle, for some i = j, both in [n], it must be that x − t(i) and x − t(j) are in the same color class B . But then the decompositions x = t(i) + (x − t(i)) = t(j) + (x − t(j)) contradict that the sum T n + B is direct, as shown in the previous paragraph.
We have shown that for any color class B , the sum T n +B is direct and contains a translate of O n , for instance x 0 +O n , where x 0 = t(lcm([n])). Setting B = B −x 0 , then T n + B is a direct sum and covers O n , as desired; see figure 2.6.  (iii) Z π(n) can be tiled by T n . Obviously, (iii) implies (ii) (and therefore (i)), and it is natural to ask whether the converse holds. We argue below that this is indeed the case. Note that the proof of proposition 2.11 shows that (iii) is equivalent to the following statement (cf. remark 2.6): (iv) There is an n-satisfactory coloring ofK n = {a/b : a, b ∈ K n }.
Proof. LetĜ n be the graph with set of verticesK n where two points x, y are connected if and only if there is an m ∈K n such that x, y ∈ {im : i ∈ [n]} (cf. § 1.3). It is enough to argue that χ(Ĝ n ) = n, since this is equivalent to (iv). But this is a consequence of compactness (in the form of the de Bruijn-Erdős theorem [dBE51]): Given any finite subgraph G ofĜ n , by multiplying all vertices by an appropriate k ∈ K n we see that G is isomorphic to a finite subgraph of G n and is therefore n-colorable, since (i) is equivalent to the assertion that χ(G n ) = n.
Essentially the same argument was also noted in [BDG + 18], whereĜ n is denoted W n . Incorporating into the argument the proof of the de Bruijn-Erdős theorem in the countable case reveals a subtlety worth pointing out, as it leads to the interesting question 2.13 below. To help see the connection, we rephrase the proof just given using directly the integer grid rather than the accompanying graph.
For each positive integer m, let D m be the hypercube Each D m admits a coloring d m that is "partially n-satisfactory" in the sense that any copy of the polyomino T n completely contained in D m receives n colors, namely, consider an n-satisfactory coloring of K n , seen as a coloring of the orthant O n , and a cube D m of the same size as D m but completely contained in O n . Now define d m simply by translating D m onto D m and copying the given coloring. Naturally, the colorings d m are not compatible in general. To obtain an actual n-satisfactory coloring ofK n , that is, a coloring of the whole integer grid where any copy of T n receives n colors, we need an additional step, which amounts to a standard application of König's infinity lemma. Explicitly: enumerate the points in the grid as x 1 , x 2 , . . . , and note that from some point on, x 1 is in the domain of d m . Also, since only n colors are possible, there is a subsequence of the d m that always assigns to x 1 the same color. Passing to a subsubsequence, we can also fix the color assigned to x 2 . Going to yet a further subsequence, we can fix the color of x 3 . Recursively carrying this procedure out produces a "limit" n-coloring d of the whole grid that is in addition satisfactory, since for any two points x, y with x = t(ik), y = t(jk) for i = j in [n] and some k ∈K n , for m large enough (say, m ≥ m 0 ) the tile T n + t(k) is completely contained in D m and so all associated colorings d m assign to x, y different colors. Say x = a r and y = a s with r < s. Since the color that d assigns to x is d m (x) for infinitely many m, and the color that it assigns to y is d m (y) for a subsequence of these m, in particular there is such an m with m > m 0 and therefore d(x) = d(y), as needed. The subtlety we referred to above is that, naturally, the resulting coloring d needs not be compatible with the initial coloring c of the orthant; in fact, for any k ∈ K n , d may not be compatible with any of the translates c k (that is, with the original coloring of any of the translates O n + t(k)) and the question remains whether we can further impose this compatibility requirement. Say that a tiling T n + B of Z π(n) essentially extends a tiling T n + B of a superset of O n if and only if any tile of T n + B completely contained in O n is a tile of T n + B (so that, if at all, only partial tiles covering a part of the boundary of the orthant could in principle change).
(1) Does any n-satisfactory coloring of K n extend to one ofK n ?
(2) If T n essentially tiles O n via a tiling T n + B, is there a tiling by T n of all of Z π(n) that essentially extends it?
Question 2.13 has a positive answer for n = 3, 4 but seems delicate in general. It obviously has a positive answer for n if there is a multiplicative n-satisfactory coloring (see remark 4.5). Question 2.10 (whether for any n-satisfactory c and any k ∈ K n we can find an n-satisfactory d such that d k = c) is a close relative; we briefly explore the latter in a particular case in § 5.2. In terms of tilings, question 2.10 is asking whether we can extend any tiling by T n of (a superset of) O n to one of O n − t(k). If this is always possible for a given n, it provides us with a positive answer to question 2.13.
Lemma 2.14. For any n, a positive answer to question 2.10 implies a positive answer to question 2.13.
Proof. Let p 1 < · · · < p π(n) be the primes in [n]. Starting with an n-satisfactory coloring c = d 0 of K n , iteratively extend the corresponding coloring of the orthant "one layer" in each dimension, i.e., find n-satisfactory colorings d 1 , d 2 , . . . such that d 0 = d 1 p1 and, in general, d π(n)·m+a−1 = d π(n)·m+a pa for any nonnegative m and any a ∈ [π(n)]. If d j = d j+1 p , we can think of d j+1 as extending the domain of d j to the set dom(d j+1 ) = {a/p : a ∈ dom(d j )}. The colorings d j agree on their common domains as j increases, and any m ∈K n is eventually included in these domains. This means that there is a unique "limit" n-satisfactory coloring d of all ofK n obtained by this process.
Remark 2.15. The second proof we gave of proposition 2.12 illustrates an application of König's lemma that can be interpreted as a construction that the space C Kn is closed under. We promised in (4) of § 2.2 to explain this construction here.
Consider a finite coloring c of K n , seen as a coloring of O n . For each l ∈ N, let be the π(n)-dimensional cube with sides of length l. There are only finitely many colorings d of D l that are realized by c in the sense that for some x ∈ O n , the coloring c (x + D l ), seen as a coloring of D l in the natural way, coincides with d, that is, d(y) = c(x + y) for any y ∈ D l . Define an infinite finitely branching tree T as follows: start with the empty coloring of the empty set (seen as the only node of T at level −1) and, for each l ∈ N, use as nodes of the lth level of T the colorings d realized by c on infinitely many distinct copies of D l (that is, those d for which there are infinitely many x as above). Use as immediate successors of such a coloring d the colorings d at the (l + 1)st level that extend d in the sense that d D l = d.
By König's lemma, the tree admits an infinite branch, that is, a sequence of colorings (d l : l ∈ N) such that dom(d l ) = D l for each l, and the colorings are compatible in the sense that d l = d l+1 D l for each l. The union of these colorings is a coloring d of O n and, just as in the second proof of proposition 2.12, if c is n-satisfactory, then so is d.
Note that we could define another finitely branching tree T by being more generous and considering all colorings that are realized rather than only those that are realized infinitely often, but this version realizes as branches many colorings we already had access to by other procedures (for instance, starting with c, all colorings c k , k ∈ K n , appear as branches of T ), while restricting attention to T may potentially result in different colorings. Moreover, even if, say, c itself appears as a branch through T , this now reveals something about the structure of c.
The combinatorial fact behind both the argument just given and the second proof of proposition 2.12 is that in order to show that there are n-satisfactory colorings of K n it is enough to argue that there are partially n-satisfactory colorings of the cubes D l for all l (in the sense mentioned earlier, that any copy of the polyomino T n completely contained in D l receives all colors). However, we do not see at the moment a scenario allowing us to verify the latter without directly exhibiting the former.
A related matter is whether the operations described in the proof of proposition 2.11 are inverses of each other. We formulate this as a question about the proof of the equivalence between (iii) and (iv) above.
Question 2.16. Given an n-satisfactory coloring c ofK n , let B be the image under t of a color class of c. The proof of proposition 2.11 shows that the sum T n + B is a tiling of Z π(n) . From this tiling we can define an n-satisfactory coloring c with color classes the preimages under t of the translates t(i) + B, i ∈ [n]. Is c = c?
The answer is positive for multiplicative colorings, see remark 4.24. Also, note the order in which we consider the operations: if instead we start with a tiling, define a coloring from it, and use the coloring to derive a tiling, we simply return to the original tiling.
Instead of (iii) and (iv) we could consider (i) and (ii). The situation here (where we only consider the orthant O n rather than the whole Z π(n) ) is somewhat more delicate: now from an essential tiling of O n we get an n-satisfactory coloring of K n just as before, but from such a coloring c we only get a tiling of a subset of the orthant, and it was only by translation that we got a tiling of a superset in the proof of proposition 2.11. However, the process of translation may effectively change even well-behaved colorings (see for instance theorem 5.3). On the other hand, c gives us not just one, but n partial tilings of O n , and any point in the orthant belongs to at least one of the resulting direct sums T n + B. Any of these sums defines a partial n-satisfactory coloring of K n .
Question 2.17. In the setting just described, are the resulting partial colorings compatible? If they are, their union gives us a coloring c of K n . Is c = c?
We can also ask whether the partial tilings can be extended to essential tilings in compatible ways. We address question 2.17 in remark 5.4. 2.4. Satisfactory colorings with n ≤ 5. For n ≤ 5 it is easy to give an explicit description of all n-satisfactory colorings. For each n < 5 there is exactly one such coloring, and there are precisely two for n = 5. As we will see in § 5.2, such an explicit listing is no longer possible even for n = 6. We use N for the set of natural numbers, including 0.
• n = 2. Similarly, there is only one 2-satisfactory coloring c of K 2 = {2 a : a ∈ N}: presented as an equivalence relation with 2 classes c(1) = c(2), we have This introduces a recurring theme: we could describe the coloring simply as c(2 α ) = (α mod 2), that is, the coloring described this way has precisely the same classes as the one in equation (2.2).
Note that 2 + 1 = 3 is prime, so c(m) = (m mod 3) is a 2-satisfactory coloring of K 2 , as shown in § 1.1. One can easily verify that (as expected) this coloring also coincides with the one in equation (2.2).
• n = 3. Suppose now that c is a 3-satisfactory coloring of for all α, β ∈ N. Conversely, it is easy to check that equation (2.3) together with the requirement that c(1), c(2), c(3) are distinct describes a 3-satisfactory coloring of K 3 . Naturally, this is the coloring indicated in figures 2.1 and 2.2, which in turn can be described (up to the name of the colors used) by saying that for nonnegative integers α, β, the unit square with bottom left corner at (α, β) has color (α + 2β mod 3), so that as before, the coloring of a number m = 2 α 3 β ∈ K 3 is given as a linear equation in the exponents of the prime factorization of m.
• n = 4.  Again by uniqueness and the result of § 1.1 (since 4+1 = 5 is prime), the coloring can also be described by c(m) = (m mod 5).

2.5.
A table of linear equations. We close the section by providing in table 2.1 a nonexhaustive list of linear equations verifying a positive solution to question 1.2 for n ≤ 31: given such an n, let k = π(n) and let p 1 < · · · < p k be the prime numbers less than or equal to n. We exhibit coefficients a 1 , . . . , a k such that the n-coloring c of K n given by is n-satisfactory. In particular, note that the entry for n = 7 provides a positive solution to KöMaL problem B.4265.
We identify these coefficients by a naive greedy algorithm, where for each i we choose a i as small as possible so that no repeated colors occur among the numbers in [n] of the form j≤i p βi i . The point is, of course, that such a linear coloring c is n-satisfactory if and only if it is injective on [n]. For instance, for n = 7, the coloring indicated in table 2.1 is given by so that if c(m) = (k mod 7), say, then and c is indeed 7-satisfactory. We revisit this approach and provide additional context through the notion of partial isomorphism in § 4.2.
Theorem 3.1. If n, k are positive integers such that p = kn + 1 is prime and 1 k , 2 k , . . . , n k are distinct modulo p, then c(m) = (m k mod p), m ∈ K n , is an n-satisfactory coloring of K n .
Proof. We begin noting that there are exactly n pairwise incongruent nonzero kth power residues modulo kn + 1. For i = j ∈ [n] and a ∈ K n , we note that c(ia) = c(ja) since the hypothesis implies that a k i k ≡ a k j k (mod p).
In particular, we recover the proof of problem A.506 from KöMaL given in the introduction since the assumption that 1, 2, . . . , n are distinct modulo n + 1 is trivially valid. We also have the following consequence (and note that there are infinitely many n such that 2n + 1 is prime).
Proof. It is enough to verify that 1 2 , . . . , n 2 are pairwise incongruent modulo p. This is trivial since This leads us to the following definition.
Definition 3.3 (Strong representatives). A satisfactory n-coloring c admits a strong representation if and only if there exists a prime p of the form kn + 1 for some positive integer k such that 1 k , . . . , n k are pairwise distinct modulo p, and c(m) = (m k mod p) for all m ∈ K n . In this case, we call p a strong representative of order n (for c). If some satisfactory n-coloring admits a strong representation, we also say that n admits a strong representative.
Whenever it applies, theorem 3.1 allows us to exhibit satisfactory colorings with a simple structure. However, given n, even if there are primes p = kn + 1 as required by the theorem, identifying them is not necessarily feasible. For instance, the smallest strong representative of order 32 is p = 5, 209, 690, 063, 553. Table 3.1 lists for n ≤ 33 the smallest strong representative of order n.
Remark 3.4. In terms of notions introduced below, all colorings obtained through strong representatives are multiplicative, and in fact are Z/nZ-colorings. However, there are satisfactory colorings that are nonmultiplicative (see theorem 5.5), multiplicative colorings that are not Z/nZ-colorings (see table 4.20 for a = 1), and Z/nZ-colorings that do not admit a strong representative (see table 4.12).

k-representatives.
Definition 3.5. Let k ∈ Z + . A prime p of the form kn + 1 is a k-representative if and only if p is a strong representative of order n, that is, the numbers 1 k , . . . , n k are distinct modulo p.  Note that, in general, the roles of k and n cannot be interchanged. If p = k + 1 is prime, it is trivially a k-representative. Our goal in this subsection is to show that for every k > 2 there are only finitely many n such that p = kn + 1 is a k-representative. In fact, we will show that for some values of k there are no such n.
We begin by discussing the case k = 3; this case was also the subject of KöMaL problem B.4401 in November 2011, 9 proposed by the third-named author.
We claim that at least one of these two situations must happen for some x ∈ [n]. Note first that −3 is a quadratic residue modulo p: where q p denotes the Legendre symbol.
It follows that the equation y 2 = −3 has two solutions, one in the first half of the interval [1, p − 1]. If y is actually in the first third, we are done, we get x = y − 1 ∈ [n]. Suppose otherwise. Note that either y or p − y is odd. Call it z, and note that z ≤ 2p/3, and therefore x = (z − 1)/2 is at most The case when k is a multiple of 4 can also be treated by elementary means.
The key is Fermat's result that an odd prime p is a sum of two squares if and only if p ≡ 1 (mod 4).
Theorem 3.7. If k is a multiple of 4 and p = kn + 1 is a k-representative, then p < k 2 , so in particular, there are only finitely many k-representatives.
Proof. Suppose p = kn + 1 is a k-representative. By Fermat's result, there are integers x and y with 1 ≤ x < y such that p = x 2 + y 2 . Note that if p ≥ k 2 then p 2 /k 2 = p · p k 2 ≥ p > y 2 , so x < y ≤ p/k = n + 1/k and therefore in fact x < y ∈ [n], but x 2 ≡ −y 2 (mod p), so x k ≡ y k (mod p).
The bound on p found in the theorem allows us to identify for any given value of k that is a multiple of 4 all the possible values of p that are k-representatives by a quick exhaustive search. Table 3.2 lists these values p = 4mn + 1 for all k = 4m ≤ 100.
We now proceed to the general case. The key observation is that if p is prime and G ≤ (Z/pZ) * is nontrivial, then g∈G g = 0. Indeed, let S = g∈G g and let h ∈ G be different from the identity. The map g → hg is a permutation of G, and we have S = g∈G hg = hS.
Suppose now that (n, k) = (1, 1) and p = kn + 1 is prime. The observation, applied to the case where G is the group of kth powers of nonzero elements of (Z/pZ) * , gives us that if p is a k-representative, then n i=1 i k ≡ 0 (mod p). But this sum is a polynomial P k (x) (of degree k + 1) with rational coefficients evaluated at x = n. If kx+1 is not a factor of P k (x), then, by applying the division algorithm and clearing out denominators, there are integers a, b, c with c = 0, such that aP k (x) has integer coefficients and aP k (x) + b(kx + 1) = c.
For x = n we have that p = kn + 1 divides P k (n) (by the observation) and therefore p | c, and there are only finitely many possibilities for p, all of which lie among the k = 4m n p = kn + 1  4  1  5  8  none  none  12  1 and 3 13 and 37  16  1  17  20  none  none  24  none  none  28  1  29  32  none  none  36  1  37  40  1  41  44  none  none  48  none  none  52  1  53  56  none  none  60  1 and 3 61 and 181  64  none  none  68  none  none  72  1  73  76  none  none  80  3  241  84  5  421  88  1  89  92  none  none  96  1  97  100 1 101 prime factors of c. The only remaining issue is how to prove that indeed kx + 1 is not a factor of P k (x). We circumvent this obstacle by arguing instead that also n i=1 i 2k ≡ 0 (mod p), and the polynomial P 2k (n) is now B 2k+1 (n+1) for B 2k+1 (x) the (2k+1)st Bernoulli polynomial, for which all its rational roots are known, and we can proceed as above.
Theorem 3.8. If k > 2, then only finitely many primes are k-representatives.
The argument we have been outlining was suggested by Darij Grinberg and Gergely Harcos 10 .
Note also that for k = 1, 2 we have that n i=1 i = n(n + 1) 2 and n i=1 i 2 = n(n + 1)(2n + 1) 6 , so the argument above fails (as it should) since kn + 1 is in both cases a factor of the corresponding polynomial.
Proof. Let B(t, x) = te tx e t −1 . The Bernoulli polynomials B m (x) are defined as follows using the power series expansion in terms of t of B(t, x): It is well known that each B m (x) is a polynomial in x of degree m with rational coefficients, and for all positive integers n, see for instance [Was97]. Writing  Suppose p = kn + 1 is a k-representative. We claim that 1 2k + 2 2k + · · · + n 2k ≡ 0 (mod p).
To see this, notice that there are precisely p−1 d = n/ gcd(2, n) incongruent (2k)th power residues modulo p, where d = gcd(2k, p − 1) = k gcd(2, n). If n is odd, this is precisely n, which means that the numbers 1 2k , . . . , n 2k are all distinct and are precisely all the nonzero (2k)th powers. If n is even, this means that each nonzero (2k)th power appears exactly twice among these numbers. In either case, it follows that the sum is zero by the same argument as above. Since it must be the case that (kn + 1) | B 2k+1 (n + 1). By Inkeri's theorem 3.9, since k > 2, the polynomial kx + 1 is relatively prime to the polynomial B 2k+1 (x + 1). But then there must be polynomials u, v ∈ Q[x] such that (In fact, v is a constant.) Multiplying this identity by an appropriate integer constant L = L 1 L 2 , it follows that there are polynomialsǔ = Lu,B 2k+1 = L 1 B 2k+1 , andv = L 2 v, all in Z[x] such that (kx + 1) ·ǔ(x) +B 2k+1 (x + 1) ·v(x) = L.
Since B 2k+1 (n + 1) ≡ 0 (mod kn + 1), evaluating the last displayed equation at x = n gives us that p = kn + 1 | L. But there are only finitely many such p.
Note that this argument does not supersede theorems 3.6 or 3.7. For theorem 3.7 in particular, note that the bound obtained there is in general much smaller than the bound L found in the proof of theorem 3.8, which depends on the size of the denominator of B 2k+1 (x + 1).
Let us illustrate this result with some examples, for which it suffices to consider n i=1 i k rather than the sum of (2k)th powers. Example 3.10.
• k = 3. Recall that Clearly, if 3n + 1 is prime, it does not divide n 2 (n + 1) 2 , and it follows that no prime is a 3-representative. This provides another solution to KöMaL problem B.4401.
Although it ends up not making a significant difference here, using theorem 5.21, we only had to consider primes not exceeding 64.
3.3. Density of strong representatives. All satisfactory n-colorings with n ≤ 5 admit strong representations: first, for each n ≤ 4 there is exactly one satisfactory n-coloring, as shown in § 2.4; moreover, n + 1 is prime for n = 1, 2, 4, and 2n + 1 is prime for n = 3. For n = 5, there are precisely two satisfactory n-colorings, which we labeled c 1 and c 5 so that c i (6) = c i (i), see § 2.4. Note that 2n + 1 = 11 is prime; the corresponding coloring is c 5 since 6 2 = 36 ≡ 25 = 5 2 (mod 11). Similarly, 421 = 84 · 5 + 1 is prime, and is a strong representative of order 5 for c 1 since Strong representatives are hardly unique. For instance, any prime is a strong representative of order 1. The case n = 2 is more interesting.
Fact 3.11. A prime p is a strong representative of order 2 if and only if p ≡ ±3 (mod 8). In particular, there are infinitely many such primes.
Proof. The first sentence is a restatement of the supplementary law for quadratic reciprocity: a prime p = 2k + 1 is a strong representative of order 2 if and only if 2 k = 2 p−1 2 ≡ 1 (mod p), which is equivalent to asserting that 2 is not a square modulo p. The supplementary law tells us that (1) If n admits a strong representative p, does it admit infinitely many?
(2) If the answer to item (1) is positive, is the set of such primes of positive natural density among all primes? (3) Further, suppose n admits a strong representative. For a satisfactory ncoloring c, is the set of strong representatives of order n for c of positive natural density, and is this density independent of c?
We present some numerical data for n ≤ 10. Consider, for instance, n = 5. First, none of the 17 primes of the form 5n + 1 in the interval [12, 420] is a strong representative of order 5, and neither are any of the 11 such primes in the interval [422, 700]. However, additional strong representatives eventually appear. Similarly, one can check that p = 2311 = 462 · 5 + 1 is a strong representative of order 5 for c 5 .
Given a real x, denote by C 1 (x) and C 5 (x) the sets of primes p ≤ x that are strong representatives of order 5 for c 1 and c 5 , respectively, and let C(x) = C 1 (x) ∪ C 5 (x). Also, write C T (x) for the set of all primes p ≤ x of the form 5n+1. Table 3.3 provides some numerical evidence suggesting a positive answer to item (3) of question 3.12 for n = 5. m 10 6 2 · 10 6 3 · 10 6 4 · 10 6 5 ·  Table 3.3. Density of strong representatives of order 5. For x ≥ 0, denote by π n (x) the number of strong representatives of order n less than or equal to x. In table 3.4 we provide data suggesting the density of strong representatives of order n in the set of primes, for n ≤ 10.
We now mention some remarks explaining that items (1) and (2) of question 3.12 admit a positive answer. First, we recall a well-known observation.
Lemma 3.14. Suppose that n and p are prime. If not all numbers are nth powers modulo p then p ≡ 1 (mod n).
Proof. Indeed, all numbers are nth powers modulo n. For p of the form nk + a with 1 < a < n, let α ∈ [n] be the multiplicative inverse of 1 − a modulo n, and note that x α(p−1)+1 ≡ x (mod p) for any x and that α(p − 1) + 1 ≡ 0 (mod n).
Many of the intricacies of the general case seem to be present already for n = 3, so we consider this case first in some detail.
Theorem 3.15. The set of primes that are strong representatives of order 3 has natural (asymptotic) density 1/9 in the set of all primes.
Proof. A prime p = 3k + 1 is a strong representative of order 3 if and only if 2 k ≡ 1 (mod 3), 3 k ≡ 1 (mod 3), and 2 k ≡ 3 k (mod 3), and this is equivalent to asserting that 2, 3, and 12 = 2 3 · 3/2 are not cubes modulo p (note that the fact that p ≡ 1 (mod 3) follows from the assertion that 2 is not a cube modulo p, by lemma 3.14). This indicates that the key technical result needed to determine whether question 3.12 holds is Chebotarëv's theorem, see [Lan94, Theorem VIII.10] 11 . We recall the 11 Lang states the result in terms of Dirichlet density, but the same conclusion holds for natural density, see [Lan94, § XV.5].
theorem and some basic facts from algebraic number theory that should allow us to apply it in the case at hand. Denote by P the set of integral primes. Recall that a set A ⊆ P has natural density δ in P if and only if lim n→∞ |A ∩ [n]|/π(n) exists and equals δ.
Theorem 3.16 (Chebotarëv). Let L/k be a Galois extension with Galois group G = Gal(L/k), and let C be a conjugacy class of G. The set of primes p of k that are unramified in L and for which the Frobenius symbol σ p of p in G is in C has natural density |C|/|G|.
For instance, in the example under consideration, that 2, 3, 12 are not cubes modulo p means that has no roots modulo p. This suggests to consider L = Q( 3 √ 2, 3 √ 3, ζ 3 ), the splitting field of f over k = Q, where ζ 3 denotes a primitive cubic root of unity. Note that [L : Q] = 18. In fact, we can quickly check that G is the generalized dihedral group for the elementary abelian group of order 9, that is, any automorphism in G is determined by its action on 3 √ 2, 3 √ 3, and ζ 3 ; the former two correspond to independent copies of Z/3Z, while the latter corresponds to the abelian group of order 2, which acts on (Z/3Z) 2 via the inverse map. We can list G as where π a,b,c is the field automorphism of L that maps 3 √ 2 to ζ a √ 3, and ζ 3 to ζ c 3 . We then look at those primes p that do not ramify over L. For this, note that L is the compositum of Q( 3 √ 2), Q( 3 √ 3), Q(ζ 3 ), and therefore an integral prime ramifies over L if and only if it ramifies in one of these fields, so the only such primes are 2, 3. In particular, all strong representatives p of order 3 are unramified in L.
Although we do not need it explicitly, we briefly explain what remains undescribed from the statement of theorem 3.16, namely the Frobenius, of Artin symbol. Suppose an integral prime p is unramified in L and q is a prime of L lying over p. The Frobenius is the unique σ ∈ G such that for all α ∈ L, where N L/Q (·) is the norm. The choice of σ = σ p depends on q, but any two such choices are conjugate, which explains why we consider conjugacy classes rather than individual members of the Galois group.
What we really need of Chebotarëv's result is the following application, actually due to Frobenius, see [SL96]: suppose g ∈ Q[x] is monic and K is its splitting field. Given an integral prime p, denote by F p the field of p elements. In F p [x], g factors into irreducible polynomials, say g = g 1 · · · g m . Letting n i denote the degree of g i for i ∈ [m], we can associate to g the partition Π p = Π p (g) = (n 1 , . . . , n m ) of deg(g). Letting G = Gal(K/Q), we can identify G with a group of permutations of the roots of g in K. Fixing an ordering of the roots, we can write each σ ∈ G as a product of disjoint cycles, say σ = τ 1 · · · τ l . Letting t i denote the length of τ i , we can associate to σ its cycle pattern Λ σ = (t 1 , . . . , t l ).
Frobenius's theorem states that the set of p unramified in K with associated partition Π p has natural density in the set of primes equal to the fraction of σ ∈ G whose associated cycle pattern Λ σ coincides with Π p .
In the case under consideration, the condition on 2, 3, 12 means that we are looking at those integral primes p with Π p (f ) = (3, 3, 3). Since each π a,b,c fixes (setwise) the sets {ζ s √ 12, and we need that jc + a = j, kc + b = k and lc + 2a + b = l for any j, k, l, where the inequalities are all modulo 3. There are two cases, depending on c. First, if c = −1, the first condition says that a − j = j, or a = 2j, but this is impossible to satisfy simultaneously for all j. Second, if c = 1, what we need is that a + j = j, k + b = k and l + 2a + b = l, that is, a, b and 2a + b should all be different from 0, and the last requirement is equivalent to asking that a = b. There are precisely two members of G that satisfy all these conditions, namely π 1,−1,1 and π −1,1,1 .
This means that the set of strong representatives of order 3 has natural density 2/18 = 1/9. Remark 3.17. Note that the value 1/9 was to be expected: with notation as in the proof above, since |G| = 18, if the set of strong representatives of order 3 was to have a natural density r at all, r would have to be a rational number of the form a/18 for some a ∈ [18], and table 3.4 strongly suggests that, indeed, r = 1/9 = 0.1.
Note also that our argument in particular established the existence of strong representatives of order 3 (although, of course, there are much simpler proofs of this assertion); the point is that this is a benefit that does not automatically generalize, as there are primes n for which there are no strong representatives of order n, such as n = 211, see table 5.1 (that is, we cannot remove in item (1) of question 3.12 the hypothesis that strong representatives of order n exist).
The same approach works in general: given n, that a prime p = nk +1 is a strong representative of order n means that i k ≡ j k (mod p) whenever i < j are in [n], that is, (j/i) k ≡ 1 (mod p), which means that j ·i n−1 is not an nth power modulo p, or, what is the same, that the polynomial x n − i · j n−1 has no roots modulo p. This translates, just as in the example above, into a condition on a conjugacy class in the Galois group of certain Galois extension of Q, namely L = Q(ζ n , n √ j : j ∈ [n] ∩ P), the splitting field over Q of the polynomial where ζ n denotes a primitive nth root of unity. The condition is in general messier than in the case n = 3, since many different factorization patterns may occur for f in F p [x] that are compatible with f not having roots in F p .
As before, L is the compositum of the fields Q( n √ j), Q(ζ n ), for j ∈ [n] ∩ P, so any integral prime that ramifies in L divides n!, and in particular any strong representative of order n is unramified in L. If n is prime, it follows from lemma 3.14 that any p for which f has no roots modulo p is automatically congruent to 1 modulo n. We also expect that, if n itself is prime, the Galois group of the extension should be given by where (Z/nZ) * denotes the group of units modulo n, a group of order φ(n) = n − 1. We have verified this by direct computation for small values of n. If n is not prime, however, there may be unexpected relations between the various n √ j and ζ n . For instance, for n = 8 and ζ 8 = e 2π √ −1/8 , we have ζ 8 + ζ −1 8 = √ 2 or, if n = 2p where p is prime and p ≡ 1 (mod 4), then √ p ∈ Q(ζ n ). Thus in general the Galois group may be a proper subgroup of the semidirect product indicated above (determining whether this is indeed the case involves Kummer theory). Still, we can ensure that the primes we consider are congruent to 1 modulo n: note that there is a natural projection from the Galois group of the extension to Gal(Q(ζ n )/Q), and the congruence condition means that this projection maps the automorphisms we are interested in counting to the identity of this smaller Galois group.
Via Chebotarëv's theorem (or, rather, Frobenius's theorem), if the requirements imposed on the automorphisms in G are at all satisfiable, then a positive proportion of all primes are strong representatives modulo n. But that the requirements are satisfiable is precisely the claim that there is at least one such prime. We have proved: Theorem 3.18. If there is a strong representative of order n, then the set of such primes has positive natural density in the set of all primes.
Given an n-satisfactory coloring c, this analysis can be further extended to capture in addition that p is a strong representative for c. This is slightly more delicate, but the point is that if c admits a strong representative, then it is multiplicative, in the sense of section 4, and such a c is completely determined by the tuple of values (c(ij) : i ≤ j ∈ [n]), see for instance corollary 4.4. This translates into yet a further condition on a conjugacy class, and the problem of computing the natural densities becomes a purely group-theoretic question.
Note that as long as no hidden relations are present (in particular, we expect this to be the case for n prime), the extension has degree [L : Q] = n π(n) · φ(n). For instance, for n = 2, the extension is of degree 2, and thus table 3.4 suggests the density is 1/2, as we indeed verified in fact 3.11. Table 3.5 shows the degree of the corresponding extension for each prime n with 2 ≤ n ≤ 10 and the density that table 3.4 suggests accordingly. For n = 4 no additional relations occur either. In that case, the extension has degree 4 2 · 2 = 32, and the expected density is 1/16 = 0.0625. 12 Remark 3.19. Chebotarëv's theorem admits an effective version. It follows that the expected densities can be verified not just by a combinatorial analysis of the relevant Galois groups, but simply by determining the sizes of these groups, and extending the entries in table 3.4 to a sufficiently large number, allowing us to compare the results with rationals of the form a/m where m is the size of the corresponding group.
3.4. Asymptotics of coincidences. Fix k > 2. For primes p = kn+1 sufficiently large, theorem 3.8 shows the existence of coincidences a k ≡ b k (mod p) with 1 ≤ a < b ≤ n. We close this section by showing that, in fact, the number of such coincidences is asymptotically proportional to p.
The result is due to Noam D. Elkies 13 , and what follows is closely based on his argument.
Theorem 3.20 (Elkies). For k > 2, the number of coincidences a k ≡ b k (mod p) for p of the form kn + 1 and sufficiently large and distinct a, if k is odd, and k − 2 2k 2 if k is even, Proof. First, for a, b nonzero and distinct modulo p, that a k ≡ b k (mod p) is equivalent to saying that b ≡ ma (mod p) where m = 1 is a kth root of unity: m k ≡ 1 (mod p). Since we are only interested in the case where a, b ∈ [n], for k even we further exclude m = −1. Fix m, and consider the nonzero vectors (a, b) in Z 2 defined by the relation b ≡ ma (mod p), a, b ∈ [n]. Note that for any such vector, p | a k − b k , and the latter factors into homogeneous polynomials in a, b of degree at most φ(k), none of which is zero, and therefore the length of the vector is Ω(p (k) ). This means that the solutions to the equation b ≡ ma (mod p) with a, b ∈ [n] are the lattice points in the square with sides parallel to the axis of side length n ≈ p/k and bottom left corner at the origin. This number can be readily estimated as p −1 (p/k) 2 = p/k 2 , with an error bound proportional to the fraction side length length of smallest such vector = O(p 1− (k) ).
The total of such coincidences is now obtained by summing these estimates over all k −1 or k −2 possible values of m, and then dividing by 2 (since each coincidence has been counted twice in the above, as both (a, b) and (b, a)).
As Elkies remarks (at the post linked to in footnote 13), the same approach as for the previous theorem allows one to estimate the number of coincidental triples, or quadruples, etc. Care must be taken "with subsets of the kth roots of unity that have integer dependencies, but at least when k is prime there are no dependencies except that all k of them sum to zero". Elkies further indicates that for j < k the number of j-element subsets of n with the same kth power is asymptotic to k j p/k j+1 , while there are no such subsets with j = k because the sum of all k solutions of a k ≡ c (mod p) vanishes (for any c). By an inclusion-exclusion argument one then obtains the estimate indicated in theorem 3.21.

Multiplicative colorings
4.1. Multiplicativity. As shown in the previous section, any n-satisfactory coloring for n ≤ 5 admits strong representatives. Colorings with strong representatives are very special: fix some n, and suppose that c is a satisfactory coloring of K n admitting a strong representative p = kn + 1. Let The group G is isomorphic to Z/nZ. The map h : for any a, b ∈ K n , where ab is the usual product of a and b and h(a) · h(b) is the product in G. for all a, b ∈ K n . In this case, we say that c is a G-coloring.
The following observation should be immediate.
Fact 4.2. If a satisfactory coloring of K n is both a G 1 -coloring and a G 2 -coloring, then G 1 ∼ = G 2 .
Note that if G is as in definition 4.1, then G is abelian, and consequently we adopt additive notation in what follows, so h is a kind of discrete logarithm but, rather than referring to it this way, we also say that h is multiplicative.  In effect, if p 1 < · · · < p π(n) are the primes less than or equal to n and s = π(n), then the multiplicity requirement gives us Since there are only finitely many group structures on [n], we are done. In fact, all these group structures can be efficiently identified, from the classification theorem for finite abelian groups.
Remark 4.5. Note that whenever an n-satisfactory coloring is multiplicative as witnessed by a group (G, ·), the corresponding tiling of O n by unit blocks of n colors is periodic, which explains the patterns observed in figures 2.2 and 2.4. Indeed, this periodicity is simply a consequence of the fact that x n is the identity of G for any x ∈ G. In turn, this also gives periodicity of the tiling by T n , see § 4.3.
In what follows, given an abelian group (G, ⊕), we will denote α-fold sums of the form g ⊕ · · · ⊕ g α times by g ⊕α rather than αg as above. We extend the notation to include negative values of α (including α = −1).
Remark 4.6. Note that not every abelian group structure on [n] gives rise to a multiplicative coloring. For example, if n = 4, then ⊕ is given by a ⊕ b = ab (mod 5) and, in particular, ([4], ⊕) is isomorphic to Z/4Z and not to Z/2Z × Z/2Z.
The case when G ∼ = Z/nZ, as in the case of a strong representation, deserves special attention. This is a good point to reiterate what we mentioned in remark 3.4. Perhaps surprisingly, the answer to question 4.7 is negative, as we show below (nevertheless, we can answer the question affirmatively at the cost of replacing strong representations with a weak variant, see remark 4.13). Although so far our examples and results have only exhibited multiplicative colorings, it should be pointed out that not every satisfactory coloring is multiplicative, see § 5.2 for a dramatic example. Similarly, not every multiplicative coloring is a Z/nZ-coloring. Examples are presented below. Nonmultiplicative colorings seem more difficult to analyze, and we do not understand them well. In what follows, we restrict our attention to the multiplicative case.

4.2.
Partial G-isomorphisms. The following notion has appeared before in the literature, in particular in connection with Graham's conjecture, and goes back at least to Galovich and Stein [GS81], who talk of KM logarithms, for Kummer and Mills. In MathOverflow, Ewan Delanoy 14 considered the case G = Z/nZ. Though not identical, it is closely related to the concept of Freiman homomorphism in additive combinatorics, see [TV06].  h(b). If G = Z/nZ, we simply call h a partial isomorphism.
Remark 4.9. We require G to be abelian as our goal is to relate partial Gisomorphisms to satisfactory colorings, see theorem 4.10 below, and the natural argument requires that G is abelian. But the question of whether there are partial G-isomorphisms where G is not abelian is interesting in its own right. This seems to be open in general, but for n odd the answer is negative, as shown by K. A. Chandler [Cha88].
Since h is a bijection, it induces a group operation ⊕ on [n] such that ([n], ⊕) ∼ = (G, +) and ⊕ extends the partial graph of multiplication on [n]. Our use of the phrase isomorphism here is perhaps further justified by noting that if h is a partial G-isomorphism, then h(1) = h(1 · 1) = h(1 ⊕ 1) = h(1) + h(1), and it follows that h(1) = 0 G . Theorem 4.10. If h : [n] → G is a partial G-isomorphism, then h can be uniquely extended to a multiplicative mapĥ : Proof. Let h : [n] → G be a partial G-isomorphism. Letting p 1 , . . . , p s be the primes less than or equal to n, a mapĥ : K n → G extends h and is multiplicative if and only if for any a 1 , . . . , a s ∈ N, we havê This proves the existence and uniqueness of the extensionĥ. Moreover, if 1 ≤ i < j ≤ n and a ∈ K n , then h(ia) =ĥ(i) +ĥ(a) =ĥ(j) +ĥ(a) =ĥ(ja) becauseĥ [n] = h is a bijection. But then, letting c = h −1 •ĥ, we have that c : K n → [n] is a G-coloring.
Remark 4.11. Note the similarity between this argument and the proof of corollary 4.4.
Obviously, ifĥ : K n → G is multiplicative and h =ĥ [n] is a bijection, then h is a partial G-isomorphism. Therefore, if c : K n → [n] is a G-coloring as witnessed by the bijection ϕ : [n] → G, then ϕ is a partial G-isomorphism as, by definition, h = ϕ • c is multiplicative, and ϕ = h [n].
This shows that the problem of building G-colorings of K n is equivalent to the problem of building partial G-isomorphisms or, equivalently, G-satisfactory groups: We say that the G-coloring resulting from extending ⊕ as in theorem 4.10 is associated to ([n], ⊕).
There is a two-fold advantage on building G-satisfactory groups rather than partial G-isomorphisms: first, the extension to a G-coloring is immediate. Second, and more significantly, different partial G-isomorphisms may give rise to the same G-coloring, as the notion is only uniquely determined up to automorphisms of G. and (1, 2, 3, 4, 5, 6) h2 −→ (0, 4, 5, 2, 1, 3), then both give rise to the Z/6Z-coloring strongly represented by 7 = 1 · 6 + 1, and this coloring is associated to the G-satisfactory group shown in table 4.1.
Remark 4.13. We are now in a position to explain how Z/nZ-colorings or, equivalently, partial isomorphisms are closely related to strong representations. In fact, we can prove that any Z/nZ-coloring admits a "weak" representation. Let h : [n] → Z/nZ be a partial isomorphism. As before, let p 1 , . . . , p s be the primes less than or equal to n. Extend h to a map from K n to Z/nZ as in the proof of theorem 4.10. Denote the extension again by h.
By Dirichlet's theorem, there are primes P of the form kn + 1. For any such P , let g be a primitive root modulo P , i.e., a generator of (Z/P Z) * . In other words, the powers g ki are precisely the kth power residues modulo P . Invoking again Dirichlet's theorem, for each p i we can find a prime q i such that q i ≡ g h(pi) (mod P ). Now for x ∈ K n define d : where of course the products are computed modulo P . The point is that if i, j ∈ [n], then d(i) = d(j) because h(i) = h(j), h being a bijection. If 0 ≤ h(i) < h(j) < n, then 0 ≤ kh(i) < kh(j) < kn, and g kh(i) = g kh(j) , since g is a primitive root. It follows that d(ix) = d(i)d(x) = d(j)d(x) = d(jx), and d defines a Z/nZ-coloring.
Note how close the coloring given by equation (4.2) is to the colorings described in definition 3.3. Strong representations are the particular case where we can choose P for which we can take q i = p i for all i.
Partial isomorphisms are easy to construct "by hand" for small values of n. Examples of partial isomorphisms for all n ≤ 31 are given in table 2.1. In appendix B of the second-named author's master's thesis 15 , this is extended to all n ≤ 54. The authors of [FP90] have verified their existence for all n < 195.
Given n, define M and M Kn as the sets of multiplicative colorings of Z + and of K n , respectively. In corollary 4.4 we showed that M Kn is finite. We now show that restricting attention to colorings in M does not affect the computation of the number of satisfactory colorings (corollary 2.7).
Proof. As in corollary 2.7, it is enough to show that n ℵ0 ≤ |M |. Let c : K n → [n] be a multiplicative coloring associated to the G-satisfactory group ([n], ⊕). To each prime p assign a number a p ∈ [n] with the only restriction that a p = p if p ∈ n. Now define c : Z + → [n] as follows: if m ∈ Z + , let i p bi i be its prime factorization, and set It is immediate that any c defined this way is multiplicative and extends c, and that different sequences (a p : p prime) give rise to different c , and therefore we have associated n ℵ0 colorings in M to each c ∈ M Kn .

Translation invariance.
In this subsection we show that multiplicativity of a coloring, an algebraic condition, is equivalent to translation invariance, a geometric condition. This helps elucidate the relation between multiplicativity of colorings and periodicity of the corresponding tilings. We have organized the presentation to highlight how far the assumption of translation invariance alone takes us, with the equivalence itself established at the end.
Recall that if c is a coloring of K n and k ∈ K n , then c k is the coloring where two numbers m, m ∈ K n receive the same color precisely when c(km) = c(km ). For any k ∈ K n , we can naturally identify O n and O n + t(k). This induces a coloring of O n from one of O n + t(k). If we start with c, the resulting coloring is precisely c k . That c is translation invariant means that, for any k ∈ K n , this coloring is again c. Clearly, multiplicative colorings are translation invariant. Notice that translation invariance is a strong requirement on a coloring: the color classes must all look the same, no matter from where we start to look at them. We illustrate this with   Arguably, a coloring c of K n deserves to be called periodic if there is a k ∈ K n larger than 1 and such that c(km) = c(m) for all m ∈ K n . Our actual definition is somewhat more stringent.
Definition 4.17. Let c be a coloring of K n , k ∈ K n be larger than 1, and l ∈ Z + . Say that c is periodic in the direction of k with period l if and only if c(k l m) = c(m) for all m ∈ K n .
Say that c is periodic if and only if it is periodic in every direction.
The following result is an immediate consequence of the existence of the orders o(k), k ∈ K n , for translation invariant colorings.
Corollary 4.18. If c is a translation invariant n-coloring of K n , then c is periodic, with period n in every direction.
Translation invariant n-satisfactory colorings are particularly well-behaved.
Lemma 4.19. For any N , any translation invariant N -coloring of K n admits a unique extension to such a coloring ofK n ; moreover, the extension c is translation invariant in the strong sense that c k = c for all k ∈K n . If the original coloring is in addition n-satisfactory, then so is the extension.
Proof. Let c be translation invariant. We define an extension, that we also denote by c, in the natural way: given m, m Similarly, if a, m 1 , m 2 ∈ K n , then We argue that the extension is translation invariant in the strong sense indicated above: let k ∈K n . We must show that c k = c, that is, that if a, b ∈K n , then To see that the extension we defined is the only possible translation invariant extension using the same colors, suppose c is such an extension of c, and that a, b, i ∈ K n are such that c (a/b) = c (i) = c(i). By translation invariance of c , this is equivalent to asserting that c (bi) = c (a), that is c(bi) = c(a) = c(ab o(b) ) which, again by translation invariance, is in turn equivalent to c (i) = c(i) = c(ab o(b)−1 ). This shows that the extension defined above is indeed the only possible one.
For c a translation invariant coloring of K n , call its extension toK n constructed in the proof of lemma 4.19 the canonical extension of c. The resemblance between classes in translation invariant colorings, mentioned above and illustrated in figure 4.1, is even stronger once we pass from the coloring to its canonical extension. Doing so eliminates the "boundary" of O n given by the coordinate axes. In the absence of such a frame of reference, the color classes are entirely indistinguishable from one another. Remark 4.21. With notation as in the proof of fact 4.20, let k ∈K n , and write its prime factorization as where the p i are the primes in [n], listed in increasing order, P is the set of primes p i that appear in k with positive exponent α i , while N is the set of such primes present in k with negative exponent. Since c is multiplicative, and using the convention that c(i) = i for i ∈ [n], we have that that is, equation (4.1) still holds, regardless of the sign of the exponents.
For l finite, a tiling of Z l by T , say T + B where the sum is direct, is periodic if and only if there is a finite index subgroup Λ of Z l such that B + Λ = B.
In § 2.3 we defined the natural bijective map t : K n → O n associating to a point k in K n the point in O n whose coordinates are the exponents of the prime factorization of k. The same definition gives us an extension of this map toK n that we again denote by t and is now a bijection with Z π(n) . The proof of proposition 2.11 gives us that if B is the image under t of any of the color classes of the canonical extension of c, then the sum T n + B is direct and tiles Z π(n) .
Lemma 4.22. Let c be a translation invariant n-satisfactory coloring and let B be the image under t of a color class of the canonical extension of c. The tiling T n + B of Z π(n) by T n is periodic. In particular, this holds for multiplicative colorings.
Proof. Let p 1 < · · · < p π(n) be the primes less than or equal to n. For i ∈ [π(n)] In fact, we can prove a bit more for multiplicative colorings.
Lemma 4.23. Let c be a multiplicative n-satisfactory coloring as witnessed by the group (G, ⊕), let B be the image under t of a color class of the canonical extension of c, and let Λ be the image under t of the color class of 1. We have that Λ is a finite index subgroup of Z π(n) and that B + Λ = Λ. In fact, Z π(n) /Λ ∼ = G and the images of the color classes are the cosets of Λ in Z π(n) .
Note that, in particular, Z π(n) = T n + Λ, and the sum is direct.
Proof. That Λ is a group is clear from the fact that c is multiplicative: first, 0 = t(1) ∈ Λ; if c(k) = c(1) then c(k −1 ) = c(k) ⊕(−1) = c(1), so Λ is closed under inverses; and since c(k 1 k 2 ) = c(k 1 ) ⊕ c(k 2 ), then Λ is closed under products. That it has finite index in Z π(n) follows from the fact that it contains the group we denoted by Λ in the proof of lemma 4.22. That B + Λ = B is exactly as in that proof.
In fact, let B be the image under t of the color class of i ∈ [n], and note that T n ∩ B = {t(i)}. We claim that B = t(i) + Λ, from which it follows that Z π(n) /Λ ∼ = G and that the images of the color classes are the cosets of Λ. Clearly B ⊇ t(i) + Λ, and if t(k) ∈ B, then c(k/i) = c(k) ⊕ c(i) ⊕(−1) = c(1), so that t(k) = t(i) + (t(k) − t(i)) ∈ t(i) + Λ.    To see this, note that c(ij) = c(c(ij)) and therefore, by translation invariance, c(ijk) = c(c(ij)k). Similarly, c(ijk) = c(ic(jk)), and we are done.
Finally, we must check that for any a, b ∈ K n , c(ab) = c(a) ⊕ c(b). For this, note that c(a) = c(c(a)) and where the last equality is by definition.
We close by reminding the reader of the periodic tiling conjecture of Jeffrey Lagarias and Yang Wang, see [LW96], the relevant version of which in our setting asks whether, given any finite set T ⊆ Z l , if T tiles Z l , then it also does so periodically. With l = π(n) and T = T n , this asks whether the existence of an n-satisfactory coloring implies the existence of a periodic one. We remark that many of the examples of 6-satisfactory colorings constructed in theorem 5.5 below are periodic, which shows that a periodic coloring needs not be translation invariant. However, note that in general, no period in the direction of 3 of the periodic examples exhibited in that result is a factor of 6. 4.4. A multiplicative coloring of p 2 − p. In this short subsection we argue that question 1.2 has a positive answer for n = p 2 − p with p prime, by exhibiting a multiplicative coloring of n.
Theorem 4.26. For any prime p, there is a G-coloring of K p 2 −p , where G = (Z/p 2 Z) * .
The group G in the theorem is isomorphic to Z/(p 2 − p)Z, but visualizing it as indicated was essential to identifying this coloring.
Proof. For i ∈ [p 2 − p] such that p i, let g i ≡ i (mod p 2 ), and for j ∈ [p − 1], let g pj ≡ p 2 − j (mod p 2 ), so that g 1 , g 2 , . . . , g p 2 −p lists the elements of a reduced residue system modulo p 2 .
We claim that g ij = g i g j , whenever ij ≤ n. If p i, j, then the statement is clear. If p i, but j = pj , then g i g j ≡ i(p 2 − j ) ≡ −ij ≡ g ij (mod p 2 ). Finally, if p | i, j, then ij is too large. For G = (Z/p 2 Z) * , this shows that the map g : [p 2 − p] → G given by i → g i is a partial G-isomorphism, and therefore there is a G-coloring of K p 2 −p , by theorem 4.10. 4.5. Multiplicative colorings for n ≤ 8. In this subsection we list all G-satisfactory groups for 6 ≤ |G| ≤ 8, thus determining all multiplicative colorings with at most eight colors. The construction is relatively simple in each case. For instance, we use the fact that if ([n], ⊕) is G-satisfactory for some G, then {2 ⊕ a : a ∈ [n], 2a > n} coincides with the set of odd integers in [n], and repeatedly make use of the fact that ⊕ must be associative and commutative, and that the order in ([n], ⊕) of any element a must divide n. In a sense, identifying these colorings is akin to solving a Sudoku puzzle.
Of the remaining eight sequences, six of them determine ⊕ uniquely as shown below. In all cases, the resulting group is Z/8Z-satisfactory and 2 is a generator. As we will see below, none of the associated colorings is strongly representable.
The remaining two sequences do not contain sufficient information to determine ⊕. What they determine of the multiplication table is shown in table 4.18 for (3, 7, 5, 1) and in table 4.19 for (7, 5, 3, 1).
Note that in both cases we have 2 ⊕4 = 1. We conclude by observing that the value of 3 ⊕ 3 = a completely determines the tables, and any of the four options for a (namely, 1, 2, 4, or 8) is possible, see tables 4.20 and 4.21.

Groupless numbers and nonmultiplicative colorings
In this section we recall results proving that not all numbers admit multiplicative colorings, and argue that not all satisfactory colorings are multiplicative.
In particular, this refutes the natural conjectures that a careful probabilistic argument or even a more careful appeal to Chebatorëv's theorem than the one in § 3.3 would prove the existence of multiplicative colorings for all n. 16 The motivation for this result was Graham's conjecture, discussed in § 1.3. The proof follows from the work initiated by R. W. Forcade, J. W. Lamoreaux, and A. D. Pollington when they posed the following question in 1986 [FLLP86].
Question 5.2. Is it possible, changing only those products that exceed n, to make the set [n] into a multiplicative group?
In our terminology, this is asking whether G-satisfactory groups exist for all values of n. In their article, they conjecture that the answer to question 5.2 is affirmative. In their discussion, they also ask (in different terms) whether strong representatives exist for all values of n.
In 1990, perhaps surprisingly, Forcade and Pollington answered question 5.2 negatively [FP90]. To do so they employed an exhaustive search algorithm that identified 195 as the least value of n for which there are no G-satisfactory groups.
Say that n is groupless if it admits no G-satisfactory group. Table 5.1 lists all groupless n ≤ 500. The data for the table was supplied by Rodney Forcade. This is sequence OEIS A204811 in the Online Encyclopedia of Integer Sequences 17 .
In [BM12], S. R. Blackburn and J. F. McKee study partial Z/nZ-isomorphisms, in the context of constructing what they call k-radius sequences over a finite alphabet. In their paper, our partial isomorphisms are dubbed bijective logarithms of length n or, simply, logarithms of length n. Several references where they are studied are provided in their section 5.1. Their theorem 5.1, for which they further refer to [Mil63, theorem 3], which makes essential use of Chebotarëv's theorem, seems particularly relevant to our question 3.12.
In [BM12,   5.2. Nonmultiplicative 6-satisfactory colorings. We close the paper by proving that nonmultiplicative colorings exist, and in fact there may be many of them.
Here we treat the case of n = 6 colors. In § 5.3 we consider n = 8.
Recall that question 2.10 asks, for a given n-satisfactory coloring c and k ∈ K n , to find all n-satisfactory colorings d with d k = c (meaning that two numbers m, m receive the same color under c if and only if km and km receive the same color under d), that is, all extensions of a given coloring (or, in the sense introduced just before proposition 2.11, a given essential tiling) of O n to one of O n − t(k).
Theorem 5.3. Let c be the 6-satisfactory coloring with strong representative 7, that is, c(m) = (m mod 7) for m ∈ K 6 . There are exactly six 6-satisfactory colorings d such that d 5 = c. In particular, there are nonmultiplicative 6-satisfactory colorings.
The coloring c is particularly nicely behaved, which simplifies the analysis that follows. As a Z/6Z-coloring, it is determined by table 4.1. The reader may consider just as well the 6-satisfactory coloring c with strong representative 487, that is, c (m) = (m 81 mod 487), for which the result also holds with a similar, but slightly more involved, geometric analysis than the one we suggest below for c. As a Z/6Zcoloring, c is determined by table 4.4.
. . . 5 3 6 5 3 6 4 1 2 4 1 2 6 5 3 6 5 3 2 4 1 2 4 1 3 6 5 3 6 5 1 2 4 1 2 4 4 1 2 4 1 2 6 5 3 6 5 3 2 4 1 2 4 1 3 6 5 3 6 5 1 2 4 1 2 4 5 3 6 5 3 6 6 5 3 6 5 3 2 4 1 2 4 1 3 6 5 3 6 5 1 2 4 1 2 4 5 3 6 5 3 6 4 1 2 4 1 2 We think of the problem of finding d as that of extending the coloring in figure 5.1 one extra layer down in the axis corresponding to the prime 5, so that dom(d) = 1 5 · K 6 and also the points in the 2-dimensional grid corresponding to γ = −1 are colored. The condition we should maintain is that if a copy of the 3-dimensional polyomino T 6 (depicted in figure 5.2) is completely contained in the extended orthant, then it must contain all colors. This leads to some restrictions, that we illustrate in figure 5.3, where for each i ∈ [6] we show in the grid for γ = −1 the places where color i is forced (as a result of i not being present in any of the other 5 places within a tile, including the tile's top place in the grid for γ = 0), as well as those where it is forbidden (because i is the color of the top place in a tile; we indicate this by graying out the region occupied by the bottom layer of the tile). Since c is periodic, it is easy to verify that the pattern suggested in figure 5.3 indeed continues.
The point is that these conditions do not determine d entirely. On the grid corresponding to γ = −1, all colors are determined except for those in the bottom row, corresponding to β = 0, see figure 5.4.
As for what colors d must assign to points in that row, what we see is that if d(5 −1 ) = u, d(2 · 5 −1 ) = v, d(4 · 5 −1 ) = w, then {u, v, w} = {3, 5, 6}, and d still satisfies that d(8 · x) = d(x) for any x in its domain. But there are precisely six 6-colorings d satisfying these requirements, and any of them is 6-satisfactory, as claimed.
Note that the effort in the proof of theorem 5.3 came in showing that the six colorings we identified are all the satisfactory colorings d with d 5 = c. A direct verification would have sufficed if all we wanted was to show that there are at least six such colorings; again, thinking of them as having domain 1 5 · K 6 , all of them are given by equation (5.1) except for β = 0, γ = −1, where they are given as described in the last paragraph of the proof.
Remark 5.4. We can extend the argument of theorem 5.3 in a few ways. For instance, their periodicity in the direction of 2 allows us to consider these colorings For σ ∈ [6] <N denote its length by |σ|. Starting with d ∅ = c, we can associate to each finite string σ ∈ [6] <N a coloring d σ with the following properties: (1) dom(d σ ) = a 2 n 5 m : a ∈ K 6 , n ∈ N, 0 ≤ m ≤ |σ| and d σ is 6-satisfactory on its domain. (3) Given d σ , d σ i is completely determined by its values on numbers of the form 2 α 5 −|σ|−1 , α ∈ Z, and in turn these values are given by the permutation associated to i corresponding to {3, 5, 6} if σ is even and to {1, 2, 4} if σ is odd, as follows: if the permutation is (u 0 , u 1 , u 2 ), then d σ i (2 α 5 −|σ|−1 ) = u j , where α ≡ j (mod 3).
(The proof of this is a straightforward extension of that of theorem 5.3, we omit the details.) In turn, this implies that ifK 6 = {2 α 3 β 5 γ : α, γ ∈ Z, β ∈ N}, then |CK 6 | = c, since to each infinite sequence x ∈ [6] Z + we can associate the coloring all these colorings are different, and all are 6-satisfactory. Unfortunately, the argument does not seem to allow for a straightforward extension that would permit us to further extend the domains of these colorings to all of K 6 .
The colorings so obtained have a further application, namely, they imply that question 2.17 has a negative answer. Indeed, for generic x, use d x to obtain a partial tiling by T 6 of the image under t ofK 6 , note that this can be extended to an essential tiling, and let (d x ) be the coloring coming from this tiling. We see that (d x ) = d x . Moreover, this is not an issue of the behavior of partial tiles at the boundary, as this tiling is actually quite tame. In fact, it suffices to take x so that for γ = −1 we use the permutation (3, 5, 6) and for any other γ we use either (3, 6, 5) or (1, 2, 4). The point is that tiles contained in the orthant (so γ ≥ 0) are colored in a certain pattern (copying the coloring of T 6 itself) while tiles involving points using the (3, 5, 6) permutation are colored differently.
That the analysis in the proof of theorem 5.3 ended up working so neatly is because, in addition to the equation c(8m) = c(m), the coloring c also satisfies that for any fixed β, γ in N, the set {c(2 α 3 β 5 γ ) : α ∈ N} is either {1, 2, 4} or {3, 5, 6}, and this only depends on the parity of β+γ. This is most readily apparent geometrically: consider an essential tiling of O 6 by T 6 induced by c. For each fixed γ, the trace on this tiling on the plane grid corresponding to γ shows up in horizontal strips of height two, as can be seen in figure 5.5. This suggests a natural approach towards strengthening the conclusion that there are nonmultiplicative colorings, that we now proceed to present.
Proof. Write O 6 = {(α, β, γ) : α, β, γ ∈ N}, identifying each point (α, β, γ) with the number 2 α 3 β 5 γ ∈ K 6 . We will define a family of 6-satisfactory colorings d by specifying certain essential tilings of O 6 . For each of them, as in the example just discussed, see also figure 5.5, the trace of the tiling on any plane γ = γ 0 is naturally organized along horizontal strips of height two (which, in particular ensures that any such coloring d satisfies d(8m) = d(m) for all m), and that there are many such colorings comes from the fact that different strips are independent of each other.
We follow our usual convention that d(i) = i for i ∈ [6], meaning that T 6 itself is one of the tiles we use (equivalently, if the tiling is T 6 + B ⊃ O 6 , the sum being direct, then 0 ∈ B). For the examples we consider, the trace of the tiling on a strip has one of six possible types, but it is enough for our purposes to only describe three of them. Each description refers to figure 5.5 and the (essential) tiling induced by c depicted there; for instance, to be of type 1 means to be exactly as the tiling of H 2 (2) shown in figure 5.5. Accordingly, in the descriptions below we omit the sentence "shown in figure 5.5" each time.
A tiling of a strip is of type • 1 if and only if it is (precisely) the tiling of H 2 (2), • 2 if and only if it is the tiling of H 0 (2), and • 3 if and only if it is the tiling of H 4 (2). (All tilings here are actually essential tilings, and we omit the word "essential" in what follows.) Note that if in a 6-satisfactory coloring a strip H k (γ) with k + γ even is of type j ∈ [3], then H k−1 (γ + 1) is of type j + 1 (using cyclic notation, so that 4 is identified with 1). This includes the case k = 0 in cases where γ is even.
We are ready to describe the colorings, they are of the form d x for x ∈ [3] Z + , where d x is defined as follows: the tiling induced by the coloring d x has the following trace on the plane γ = 0: H 0 (0) has type 1 (as required by our convention). For each k > 0, H 2k (0) has type x(k). For γ > 0, the traces are recursively defined according to the last sentence of the previous paragraph.
It is immediate from the construction that if x = x then d x = d x , so we have defined 3 ℵ0 = c colorings of K 6 , and that all of them are 6-satisfactory, since what we have actually done is to give d x via an explicit essential tiling of O 6 . Note that, for x 1 the constant function taking the value 1, the coloring d x 1 just described is the Z 6 -coloring with strong representative 103 and determined by table 4.3. Also, note that the set {d x : x ∈ [3] Z + } not only has the same size as the reals but is in fact a perfect subset of C K6 .
There are infinitely many ways of choosing x so that the resulting d x is periodic in the direction of 3 (and therefore periodic), and we can moreover ensure that this period is not a factor of 6, in fact, o(3) can be taken to be arbitrarily large. As remarked in § 4.3, this indicates that the existence of periodic n-satisfactory colorings is not enough to ensure the existence of translation invariant (i.e., multiplicative) ones.
Remark 5.6. We previously obtained a different proof of theorem 5.5 that used the fact that C K6 is closed in its natural topology. The argument proceeded in two stages. First, theorem 5.3 was extended using a variant of the construction in remark 5.4 to obtain an infinite countable family of 6-satisfactory colorings such that for each c in the family there were six colorings d in the family with d 5 = c. The members of the family were arranged as nodes on a complete senary tree, in such a way that the colorings along each branch converged, and the limit colorings so produced were pairwise different. Further, all these colorings d satisfy that d(8m) = d(m) and d(3m) = d(10m) for any m ∈ K 6 . Lon Mitchell also found an elegant proof; his key insight was that one could get rid of the topological argument and instead define directly in a combinatorial way the colorings that the previous limit process had produced. 5.3. Nonmultiplicative 8-satisfactory colorings. We adapt the proof of theorem 5.5 to show that the result applies to C K8 as well.
We begin by describing the four possible types a strip H k (γ 0 , δ 0 ) may be depending on the trace of the tiling on the strip, for which we refer to figure 5.6 (as in the n = 6 case, more types are possible, but these are the only ones we consider). All our tilings are periodic in the sense that the associated coloring d satisfies d(16m) = d(m) for any m ∈ K 8 .  • 4 if and only if it is the depicted tiling with a = −3. We will define the colorings we are interested in by describing the types of the strips H 2k (0, 0), and extending this to a coloring of all of K 8 recursively by the rule that if k + γ 0 + δ 0 is even and H k (γ 0 , δ 0 ) is of type i, then H k−1 (γ 0 + 1, δ 0 ) is of type i + 2 and H k−1 (γ 0 , δ 0 + 1) is of type i + 3, using cyclic notation modulo 4.
We must verify that this procedure is well-defined, specifically, that the recursion just described assigns a unique color to any point in O 8 once the colors of points in the (intersection of the orthant with the) plane γ = δ = 0 are specified. For this, first note that the recursion describes how to assign colors to points in H k (γ, δ) for k of the same parity as γ + δ so, in particular, fixing the values of γ and δ, for any point x in the resulting plane there is a unique k ≥ −1 with x ∈ H k (γ, δ) and k of the relevant parity. Now, by induction on γ + δ, let i be the type of H k+γ+δ (0, 0), and check that the rules specify that the type of H k (γ, δ) is precisely i + 2γ + 3δ (using the cyclic convention), so the color assigned to x is indeed unambiguous.
Finally, we define the colorings d x for x ∈ [4] Z + by setting H 0 (0, 0) to be of type 1 and, for k > 0, H 2k (0, 0) to be of type x(k). The colorings so described are pairwise different and are 8-satisfactory since, by construction, d x (16m) = d x (m) for any m while d x (m), d x (2m), d x (4m), d x (8m) are pairwise distinct, and they are all different from d x (3m), d x (6m), d x (12m), d x (24m) and, moreover, d x (5m) = d x (12m) and d x (7m) = d x (24m). Note that, as in the n = 6 case, the set {d x : x ∈ [4] Z + } is a perfect subset of C K8 .
Note that for x 1 the constant function taking the value 1, the coloring d x 1 is the Z/2Z × Z/4Z-coloring determined by table 4.21 for a = 1.
Remark 5.8. We could have presented the proof purely algebraically by building the colorings d x recursively, while requiring they satisfy the rules indicated in the last paragraph, but we chose the geometric presentation as it seems more intuitive.