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

TL;DR

This paper investigates the problem of coloring positive integers with n colors such that multiples of any number are differently colored, exploring algebraic structures, existence conditions, and classifying solutions for small n.

Contribution

It establishes conditions for the existence of n-satisfactory colorings using abelian group operations and classifies all such colorings for small n, revealing the structure and density of solutions.

Findings

Existence of n-satisfactory colorings linked to abelian group structures.

Complete classification of n-satisfactory colorings for n ≤ 5.

Identification of the density of primes related to strong representatives.

Abstract

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.

Tables

Table 1. Table 2.1. Linear equations describing an n 𝑛 n -satisfactory coloring c 𝑐 c for n ≤ 31 𝑛 31 n\leq 31 .

$n$	$m\in K_n$	$c\left(m\right)$ modulo $n$
1	1	0
2	$2^{\alpha }$	$\alpha$
3	$2^{\alpha }{3}^{\beta }$	$\alpha +2\beta$
4	$2^{\alpha }{3}^{\beta }$	$\alpha +3\beta$
5	$2^{\alpha }{3}^{\beta }{5}^{\gamma }$	$\alpha +3\beta +4\gamma$
6	$2^{\alpha }{3}^{\beta }{5}^{\gamma }$	$\alpha +3\beta +5\gamma$
7	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }$	$\alpha +3\beta +5\gamma +6\delta$
8	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }$	$\alpha +4\beta +6\gamma +7\delta$
9	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }$	$\alpha +4\beta +6\gamma +7\delta$
10	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }$	$\alpha +4\beta +6\gamma +9\delta$
11	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}$	$\alpha +4\beta +6\gamma +9\delta +10ϵ$
12	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}$	$\alpha +4\beta +9\gamma +7\delta +11ϵ$
13	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }$	$\alpha +4\beta +9\gamma +7\delta +11ϵ+12\zeta$
14	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }$	$\alpha +4\beta +9\gamma +11\delta +7ϵ+13\zeta$
15	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }$	$\alpha +4\beta +9\gamma +11\delta +7ϵ+14\zeta$
16	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }$	$\alpha +5\beta +8\gamma +11\delta +14ϵ+15\zeta$
17	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }$	$\alpha +5\beta +8\gamma +11\delta +14ϵ+15\zeta +16\eta$
18	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }$	$\alpha +5\beta +8\gamma +14\delta +12ϵ+16\zeta +17\eta$
19	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }$	$\alpha +5\beta +8\gamma +14\delta +12ϵ+16\zeta +17\eta +18\theta$
20	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }$	$\alpha +5\beta +12\gamma +8\delta +15ϵ+16\zeta +18\eta +19\theta$
21	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }$	$\alpha +5\beta +12\gamma +15\delta +8ϵ+9\zeta +18\eta +19\theta$
22	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }$	$\alpha +5\beta +12\gamma +15\delta +8ϵ+18\zeta +19\eta +21\theta$
23	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }$	$\alpha +5\beta +12\gamma +15\delta +8ϵ+18\zeta +19\eta +21\theta +22\iota$
24	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }$	$\alpha +5\beta +12\gamma +15\delta +18ϵ+9\zeta +21\eta +22\theta +23\iota$
25	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }$	$\alpha +5\beta +12\gamma +15\delta +18ϵ+9\zeta +21\eta +22\theta +23\iota$
26	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }$	$\alpha +5\beta +12\gamma +15\delta +18ϵ+21\zeta +9\eta +23\theta +25\iota$
27	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }$	$\alpha +5\beta +12\gamma +18\delta +20ϵ+25\zeta +9\eta +16\theta +22\iota$
28	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }$	$\alpha +5\beta +12\gamma +18\delta +21ϵ+25\zeta +9\eta +16\theta +27\iota$
29	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }{29}^{\kappa }$	$\alpha +5\beta +12\gamma +18\delta +21ϵ+25\zeta +9\eta +16\theta +27\iota +28\kappa$
30	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }{29}^{\kappa }$	$\alpha +5\beta +12\gamma +20\delta +26ϵ+28\zeta +9\eta +16\theta +19\iota +23\kappa$
31	$2^{\alpha }{3}^{\beta }{5}^{\gamma }{7}^{\delta }{11}^{ϵ}{13}^{\zeta }{17}^{\eta }{19}^{\theta }{23}^{\iota }{29}^{\kappa }{31}^{\lambda }$	$\alpha +5\beta +12\gamma +20\delta +26ϵ+28\zeta +9\eta +16\theta +19\iota +23\kappa +30\lambda$

Table 2. Table 3.1. Smallest strong representative p = k ​ n + 1 𝑝 𝑘 𝑛 1 p=kn+1 of order n 𝑛 n for n ≤ 33 𝑛 33 n\leq 33 .

$n$	$k$	$p$
$\mathrm{𝟏}$	1	2
$\mathrm{𝟐}$	1	3
$\mathrm{𝟑}$	2	7
$\mathrm{𝟒}$	1	5
$\mathrm{𝟓}$	2	11
$\mathrm{𝟔}$	1	7
$\mathrm{𝟕}$	94	659
$\mathrm{𝟖}$	2	17
$\mathrm{𝟗}$	2	19
$\mathrm{𝟏𝟎}$	1	11
$\mathrm{𝟏𝟏}$	2	23
$\mathrm{𝟏𝟐}$	1	13
$\mathrm{𝟏𝟑}$	198,364	2,578,733
$\mathrm{𝟏𝟒}$	2	29
$\mathrm{𝟏𝟓}$	2	31
$\mathrm{𝟏𝟔}$	1	17
$\mathrm{𝟏𝟕}$	2,859,480	48,611,161
$\mathrm{𝟏𝟖}$	1	19
$\mathrm{𝟏𝟗}$	533,410	10,134,791
$\mathrm{𝟐𝟎}$	2	41
$\mathrm{𝟐𝟏}$	2	43
$\mathrm{𝟐𝟐}$	1	23
$\mathrm{𝟐𝟑}$	2	47
$\mathrm{𝟐𝟒}$	56,610,508	1,358,652,193
$\mathrm{𝟐𝟓}$	1,170,546,910	29,263,672,751
$\mathrm{𝟐𝟔}$	2	53
$\mathrm{𝟐𝟕}$	6,700,156,678	180,904,230,307
$\mathrm{𝟐𝟖}$	1	29
$\mathrm{𝟐𝟗}$	2	59
$\mathrm{𝟑𝟎}$	1	31
$\mathrm{𝟑𝟏}$	27,184,496,610	842,719,394,911
$\mathrm{𝟑𝟐}$	162,802,814,486	5,209,690,063,553
$\mathrm{𝟑𝟑}$	2	67

Table 3. Table 3.2. 4 ​ m 4 𝑚 4m -representatives for m ≤ 25 𝑚 25 m\leq 25 .

$k=4m$	$n$	$p=kn+1$
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

Table 4. Table 3.3. Density of strong representatives of order 5.

$m$	${10}^6$	$2\cdot {10}^6$	$3\cdot {10}^6$	$4\cdot {10}^6$	$5\cdot {10}^6$
$\left|{𝒞}^1\left(m\right)\right|$	626	1203	1757	2314	2838
$\left|{𝒞}^5\left(m\right)\right|$	626	1210	1783	2291	2822
$\left|𝒞\left(m\right)\right|$	1252	2413	3540	4605	5660
$\left|{𝒞}_T\left(m\right)\right|$	19617	37188	54175	70779	87062
$\frac{\left|{𝒞}^1\left(m\right)\right|}{\left|{𝒞}^5\left(m\right)\right|}$	1	0.994215	0.985418	1.010039	1.005670
$\frac{\left|𝒞\left(m\right)\right|}{\left|{𝒞}_T\left(m\right)\right|}$	0.063822	0.064887	0.065344	0.065062	0.065011

Table 5. Table 3.4. Density of strong representatives of order n 𝑛 n , 2 ≤ n ≤ 10 2 𝑛 10 2\leq n\leq 10 .

$N$	$\pi \left(N\right)$	${\pi }_2\left(N\right)$	${\pi }_2\left(N\right)/\pi \left(N\right)$	${\pi }_3\left(N\right)$	${\pi }_3\left(N\right)/\pi \left(N\right)$	${\pi }_4\left(N\right)$	${\pi }_4\left(N\right)/\pi \left(N\right)$
${10}^2$	25	13	0.52	2	0.08	1	0.04
${10}^3$	168	87	0.51785…	20	0.11904…	10	0.05952…
${10}^4$	1229	625	0.50854…	134	0.10903…	82	0.06672…
${10}^5$	9592	4808	0.50125…	1087	0.11332…	602	0.06276…
${10}^6$	78498	39276	0.50034…	8732	0.11123…	4857	0.06187…

Table 6. Table 3.5. Degree of ℚ ( ζ n , j n : j ∈ [ n ] ∩ ℙ ) : ℚ \mathbb{Q}(\zeta_{n},\sqrt[n]{j}:j\in[n]\cap\mathbb{P}):\mathbb{Q} and expected density of the set of strong representatives of order n 𝑛 n in the set of all primes for n 𝑛 n prime below 10.

$n$	degree	expected density
2	$2^1\cdot 1=2$	$1/2=0.5$
3	$3^2\cdot 2=18$	$1/9=0.\overline{1}$
5	$5^3\cdot 4=500$	$2/125=0.016$
7	$7^4\cdot 6=14406$	$6/2401=0.00249\dots$

Table 7. Table 4.1. A ℤ / 6 ​ ℤ ℤ 6 ℤ \mathbb{Z}/6\mathbb{Z} -satisfactory group.

$\oplus$	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	1	3	5
3	3	6	2	5	1	4
4	4	1	5	2	6	3
5	5	3	1	6	4	2
6	6	5	4	3	2	1

Table 8. Table 4.2. ℤ / 6 ​ ℤ ℤ 6 ℤ \mathbb{Z}/6\mathbb{Z} -coloring strongly represented by 13 = 2 ⋅ 6 + 1 13 ⋅ 2 6 1 13=2\cdot 6+1 .

$\oplus$	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	5	3	1
3	3	6	4	1	2	5
4	4	5	1	3	6	2
5	5	3	2	6	1	4
6	6	1	5	2	4	3

Table 9. Table 4.3. ℤ / 6 ​ ℤ ℤ 6 ℤ \mathbb{Z}/6\mathbb{Z} -coloring strongly represented by 103 = 17 ⋅ 6 + 1 103 ⋅ 17 6 1 103=17\cdot 6+1 .

$\oplus$	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	1	3	5
3	3	6	1	5	4	2
4	4	1	5	2	6	3
5	5	3	4	6	2	1
6	6	5	2	3	1	4

Table 10. Table 4.4. ℤ / 6 ​ ℤ ℤ 6 ℤ \mathbb{Z}/6\mathbb{Z} -coloring strongly represented by 487 = 81 ⋅ 6 + 1 487 ⋅ 81 6 1 487=81\cdot 6+1 .

$\oplus$	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	1	3	5
3	3	6	4	5	2	1
4	4	1	5	2	6	3
5	5	3	2	6	1	4
6	6	5	1	3	4	2

Table 11. Table 4.5. ℤ / 6 ​ ℤ ℤ 6 ℤ \mathbb{Z}/6\mathbb{Z} -coloring strongly represented by 547 = 91 ⋅ 6 + 1 547 ⋅ 91 6 1 547=91\cdot 6+1 ; this is also the ( ℤ / 9 ​ ℤ ) ∗ superscript ℤ 9 ℤ (\mathbb{Z}/9\mathbb{Z})^{*} -coloring given by the proof of theorem 4.26 .

$\oplus$	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	3	1	5
3	3	6	1	5	4	2
4	4	3	5	6	2	1
5	5	1	4	2	6	3
6	6	5	2	1	3	4

Table 12. Table 4.6. ℤ / 7 ​ ℤ ℤ 7 ℤ \mathbb{Z}/7\mathbb{Z} -coloring strongly represented by 659 659 659 .

$\oplus$	1	2	3	4	5	6	7
1	1	2	3	4	5	6	7
2	2	4	6	5	3	7	1
3	3	6	2	7	1	4	5
4	4	5	7	3	6	1	2
5	5	3	1	6	7	2	4
6	6	7	4	1	2	5	3
7	7	1	5	2	4	3	6

Table 13. Table 4.7. ℤ / 7 ​ ℤ ℤ 7 ℤ \mathbb{Z}/7\mathbb{Z} -coloring strongly represented by 1429 1429 1429 .

$\oplus$	1	2	3	4	5	6	7
1	1	2	3	4	5	6	7
2	2	4	6	3	7	5	1
3	3	6	7	5	2	1	4
4	4	3	5	6	1	7	2
5	5	7	2	1	3	4	6
6	6	5	1	7	4	2	3
7	7	1	4	2	6	3	5

Table 14. Table 4.8. ℤ / 7 ​ ℤ ℤ 7 ℤ \mathbb{Z}/7\mathbb{Z} -coloring strongly represented by 2087 2087 2087 .

$\oplus$	1	2	3	4	5	6	7
1	1	2	3	4	5	6	7
2	2	4	6	3	1	7	5
3	3	6	5	7	4	1	2
4	4	3	7	6	2	5	1
5	5	1	4	2	7	3	6
6	6	7	1	5	3	2	4
7	7	5	2	1	6	4	3

Table 15. Table 4.9. ℤ / 7 ​ ℤ ℤ 7 ℤ \mathbb{Z}/7\mathbb{Z} -coloring strongly represented by 3557 3557 3557 .

$\oplus$	1	2	3	4	5	6	7
1	1	2	3	4	5	6	7
2	2	4	6	7	1	5	3
3	3	6	2	5	7	4	1
4	4	7	5	3	2	1	6
5	5	1	7	2	6	3	4
6	6	5	4	1	3	7	2
7	7	3	1	6	4	2	5

Table 16. Table 4.10. ℤ / 7 ​ ℤ ℤ 7 ℤ \mathbb{Z}/7\mathbb{Z} -coloring strongly represented by 17431 17431 17431 .

$\oplus$	1	2	3	4	5	6	7
1	1	2	3	4	5	6	7
2	2	4	6	7	3	1	5
3	3	6	7	1	4	5	2
4	4	7	1	5	6	2	3
5	5	3	4	6	2	7	1
6	6	1	5	2	7	3	4
7	7	5	2	3	1	4	6

Table 17. Table 4.11. ℤ / 7 ​ ℤ ℤ 7 ℤ \mathbb{Z}/7\mathbb{Z} -coloring strongly represented by 21911 21911 21911 .

$\oplus$	1	2	3	4	5	6	7
1	1	2	3	4	5	6	7
2	2	4	6	5	7	1	3
3	3	6	5	1	2	7	4
4	4	5	1	7	3	2	6
5	5	7	2	3	6	4	1
6	6	1	7	2	4	3	5
7	7	3	4	6	1	5	2

Table 18. Table 4.12. ℤ / 8 ​ ℤ ℤ 8 ℤ \mathbb{Z}/8\mathbb{Z} -coloring corresponding to the sequence ( 1 , 5 , 3 , 7 ) 1 5 3 7 (1,5,3,7) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	1	5	3	7
3	3	6	4	5	7	8	2	1
4	4	8	5	7	2	1	6	3
5	5	1	7	2	6	3	8	4
6	6	5	8	1	3	7	4	2
7	7	3	2	6	8	4	1	5
8	8	7	1	3	4	2	5	6

Table 19. Table 4.13. ℤ / 8 ​ ℤ ℤ 8 ℤ \mathbb{Z}/8\mathbb{Z} -coloring corresponding to the sequence ( 1 , 7 , 5 , 3 ) 1 7 5 3 (1,7,5,3) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	1	7	5	3
3	3	6	1	7	8	2	4	5
4	4	8	7	3	2	5	1	6
5	5	1	8	2	7	3	6	4
6	6	7	2	5	3	4	8	1
7	7	5	4	1	6	8	3	2
8	8	3	5	6	4	1	2	7

Table 20. Table 4.14. ℤ / 8 ​ ℤ ℤ 8 ℤ \mathbb{Z}/8\mathbb{Z} -coloring corresponding to the sequence ( 3 , 1 , 5 , 7 ) 3 1 5 7 (3,1,5,7) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	3	1	5	7
3	3	6	7	1	8	5	4	2
4	4	8	1	7	6	2	3	5
5	5	3	8	6	4	7	2	1
6	6	1	5	2	7	3	8	4
7	7	5	4	3	2	8	1	6
8	8	7	2	5	1	4	6	3

Table 21. Table 4.15. ℤ / 8 ​ ℤ ℤ 8 ℤ \mathbb{Z}/8\mathbb{Z} -coloring corresponding to the sequence ( 3 , 7 , 1 , 5 ) 3 7 1 5 (3,7,1,5) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	3	7	1	5
3	3	6	4	7	2	8	5	1
4	4	8	7	5	6	1	2	3
5	5	3	2	6	1	4	8	7
6	6	7	8	1	4	5	3	2
7	7	1	5	2	8	3	6	4
8	8	5	1	3	7	2	4	6

Table 22. Table 4.16. ℤ / 8 ​ ℤ ℤ 8 ℤ \mathbb{Z}/8\mathbb{Z} -coloring corresponding to the sequence ( 7 , 1 , 3 , 5 ) 7 1 3 5 (7,1,3,5) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	7	1	3	5
3	3	6	5	1	4	7	8	2
4	4	8	1	5	3	2	6	7
5	5	7	4	3	1	8	2	6
6	6	1	7	2	8	3	5	4
7	7	3	8	6	2	5	4	1
8	8	5	2	7	6	4	1	3

Table 23. Table 4.17. ℤ / 8 ​ ℤ ℤ 8 ℤ \mathbb{Z}/8\mathbb{Z} -coloring corresponding to the sequence ( 7 , 5 , 1 , 3 ) 7 5 1 3 (7,5,1,3) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	7	5	1	3
3	3	6	1	5	4	2	8	7
4	4	8	5	3	1	7	2	6
5	5	7	4	1	3	8	6	2
6	6	5	2	7	8	4	3	1
7	7	1	8	2	6	3	5	4
8	8	3	7	6	2	1	4	5

Table 24. Table 4.18. The partial multiplication table determined by the sequence ( 3 , 7 , 5 , 1 ) 3 7 5 1 (3,7,5,1) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	3	7	5	1
3	3	6		7				5
4	4	8	7	1	6	5	3	2
5	5	3		6				7
6	6	7		5				3
7	7	5		3				6
8	8	1	5	2	7	3	6	4

Table 25. Table 4.19. The partial multiplication table determined by the sequence ( 7 , 5 , 3 , 1 ) 7 5 3 1 (7,5,3,1) .

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	7	5	3	1
3	3	6		5				7
4	4	8	5	1	3	7	6	2
5	5	7		3				6
6	6	5		7				3
7	7	3		6				5
8	8	1	7	2	6	3	5	4

Table 26. Table 4.20. Group corresponding to the sequence ( 3 , 7 , 5 , 1 ) 3 7 5 1 (3,7,5,1) with a = 1 𝑎 1 a=1 , 2, 4, or 8.

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	3	7	5	1
3	3	6	$a$	7	$8\oplus a$	$2\oplus a$	$4\oplus a$	5
4	4	8	7	1	6	5	3	2
5	5	3	$8\oplus a$	6	$4\oplus a$	$a$	$2\oplus a$	7
6	6	7	$2\oplus a$	5	$a$	$4\oplus a$	$8\oplus a$	3
7	7	5	$4\oplus a$	3	$2\oplus a$	$8\oplus a$	$a$	6
8	8	1	5	2	7	3	6	4

Table 27. Table 4.21. Group corresponding to the sequence ( 7 , 5 , 3 , 1 ) 7 5 3 1 (7,5,3,1) with a = 1 𝑎 1 a=1 , 2, 4, or 8.

$\oplus$	1	2	3	4	5	6	7	8
1	1	2	3	4	5	6	7	8
2	2	4	6	8	7	5	3	1
3	3	6	$a$	5	$4\oplus a$	$2\oplus a$	$8\oplus a$	7
4	4	8	5	1	3	7	6	2
5	5	7	$4\oplus a$	3	$a$	$8\oplus a$	$2\oplus a$	6
6	6	5	$2\oplus a$	7	$8\oplus a$	$4\oplus a$	$a$	3
7	7	3	$8\oplus a$	6	$2\oplus a$	$a$	$4\oplus a$	5
8	8	1	7	2	6	3	5	4

Table 28. Table 5.1. Groupless n ≤ 500 𝑛 500 n\leq 500 .

195	248	279	311	337	367	394	423	451	480
205	252	283	313	339	368	395	424	452	481
208	253	286	314	340	370	397	425	454	482
211	255	287	317	343	373	399	427	457	484
212	257	289	318	344	374	401	433	458	487
214	258	290	319	347	376	402	434	461	489
217	259	291	322	349	377	403	435	463	492
218	263	294	324	351	379	406	436	465	493
220	264	295	325	353	381	407	437	467	494
227	265	297	327	355	383	409	439	469	496
229	266	298	328	356	385	412	444	471	497
235	267	301	331	357	387	415	445	472	499
242	269	302	332	361	389	416	446	474	500
244	271	304	333	362	390	417	447	475
246	274	305	334	364	391	421	449	477
247	275	307	335	365	392	422	450	479

Equations

$$\underset{i,j}{\max }\frac{a_i}{\gcd (a_i,a_j)}\ge n.$$

$$\underset{i,j}{\max }{b}_i/\gcd (b_i,b_j)<m.$$

$$C△D=(C\cup D)\setminus (C\cap D)=(C\setminus D)\cup (D\setminus C).$$

$$k\cdot X=\{kx:x\in X\}.$$

$$B_j=\{i\cdot a_j:i\in [n]\}.$$

$$\chi_{\bigtriangleup_{i=1}^{n}i\cdot A}(x)=\biggl{(}\sum_{i=1}^{n}\chi_{i\cdot A}(x)\biggr{)}\bmod 2=|\{i\in[n]:\exists j\in[k]\,(x=i\cdot a_{j})\}|\bmod 2$$

$$\chi_{\bigtriangleup_{j=1}^{k}B_{j}}(x)=\biggl{(}\sum_{j=1}^{k}\chi_{B_{j}}(x)\biggr{)}\bmod 2=|\{j\in[k]:\exists i\in[n]\,(x=i\cdot a_{j})\}|\bmod 2,$$

$$|\{(i,j)\in [n]\times [k]:x=i\cdot a_j\}|\mathrm{mod}2.$$

$$C_{k}p+O_k(p^{1-ϵ(k)}),$$

$$C_{k}=\begin{cases}\vspace{1mm}\displaystyle\frac{k-1}{2k^{2}}&\mbox{ if $k$ is odd, and}\\ \displaystyle\frac{k-2}{2k^{2}}&\mbox{ if $k$ is even,}\end{cases}$$

$$X=\dot{\bigcup _{a\in A}}a\cdot K_n.$$

$$\dot{\bigcup _{a\in A}}a\cdot K_n\subseteq X.$$

$$\dot{\bigcup _{a\in A}}a\cdot K_n\supseteq X.$$

$$\mathrm{div}(c,a)(k)=\mathrm{div}(c,a)(l)\text{mbox}ifandonlyifc(ak)=c(al)$$

$$\mathrm{div}(c,a)(k)=\pi \circ c(ak)$$

$$\mathrm{mult}(c,a)(ak)=\mathrm{mult}(c,a)(al)\text{mbox}ifandonlyifc(k)=c(l)$$

$$c=\dot{\bigcup _{a\in P_n\cap X}}{\pi }_a\circ \mathrm{mult}(c_a,a).$$

$$\mathrm{div}(c,a)(ib)\ne \mathrm{div}(c,a)(jb)$$

$$\mathrm{mult}(c,a)(im)\ne \mathrm{mult}(c,a)(jm)$$

$$c=\bigcup _{a\in P_n\cap X}{\pi }_a\circ \mathrm{mult}(c_a,a).$$

$$c\in \{f\in [n{]}^{K_n}:\forall i\in [n](f(i)=i)\}\cap \bigcap _{a\in K_n}\bigcap _{1\le i<j\le n}\{g\in [n{]}^{K_n}:g(ia)\ne g(ja)\},$$

$$\{g\in [n{]}^{K_n}:g(b)\ne g(c)\}=\bigcup _{\begin{array}{c}(\alpha ,\beta )\in [n]\times [n]\\ \alpha \ne \beta \end{array}}\{g\in [n{]}^{K_n}:g(b)=\alpha \text{mbox}andg(c)=\beta \}$$

$$\rho (m)=\rho (2{)}^{{\alpha }_1}\dots \rho (p_k{)}^{{\alpha }_k}.$$

$$t(j)+b_1=t(i)+b_2.$$

$$D_m=\{(a_1,\dots ,a_{\pi (n)}):|a_i|\le m\text{mbox}foralli\}.$$

$$d^{\pi (n)\cdot m+a-1}={d^{\pi (n)\cdot m+a}}_{p_a}$$

$$D_l=\{(a_1,\dots ,a_{\pi (n)})\in {\mathbb{O}}_n:0\le a_i\le l\text{mbox}foralli\in [\pi (n)]\}$$

$$c(2^{\alpha })=c(2^{\alpha \mathrm{mod}2})$$

$$c(2^{\alpha }{3}^{\beta })=c(2^{\alpha +2\beta \mathrm{mod}3})$$

Full text

Specifically, we show that there is an $n$-satisfactory coloring whenever there is an abelian group operation $\oplus$ on the set $\{1,2,\dots,n\}$ that is 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 $n\leq 5$ and all multiplicative colorings for $n\leq 8$, 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 it admits infinitely many and in fact the set of such $q$ has positive natural density in the set of all primes.

We also 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.

1. 1 Introduction

2. 1.1 A problem from KöMaL

3. 1.2 The general question

4. 1.3 The Balasubramanian–Soundararajan theorem

5. 1.4 Pilz’s conjecture

6. 1.5 Organization of this paper

7. 2 The core

8. 2.1 The core $K_{n}$ and $n$-appropriate sets

9. 2.2 The structure of $C_{K_{n}}$

10. 2.3 Tilings

11. 2.4 Satisfactory colorings with $n\leq 5$

12. 2.5 A table of linear equations

13. 3 Generalizing the approach for $p$ prime

14. 3.1 Strong representatives

15. 3.2 $k$-representatives

16. 3.3 Examples

17. 3.4 Density of strong representatives

18. 3.5 Asymptotics of coincidences

19. 4 Multiplicative colorings

20. 4.1 Multiplicativity

21. 4.2 Partial $G$-isomorphisms

22. 4.3 Translation invariance

23. 4.4 A multiplicative coloring of $p^{2}-p$

24. 4.5 Multiplicative colorings for $n\leq 8$

25. 5 Groupless numbers and nonmultiplicative colorings

26. 5.1 Groupless numbers

27. 5.2 Nonmultiplicative 6-satisfactory colorings

28. 5.3 Nonmultiplicative 8-satisfactory colorings

29. 6 Open questions

1. Introduction

1.1. A problem from KöMaL

The following was posed by the third-named author as problem A.506 in the April 2010 issue of the Hungarian journal KöMaL (Középiskolai Matematikai és Fizikai Lapok), a mathematics and physics journal primarily aimed at high school students111See https://www.komal.hu/feladat?a=honap&h=201004&t=mat&l=en.:

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 $\{a,2a,3a,\dots,(p-1)a\}$ 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 $c(6)\neq c(2),c(4)$ because $2,4,6,8$ all have different colors and $c(6)\neq c(3)$ since $3,6,9,12$ all have different colors. It follows that $c(8)=c(3)$. Also, since $c(12)\neq c(3),c(4),c(6)$, 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.

[TABLE]

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 $i^{\mathrm{th}}$ row are all congruent to $i$ modulo 5, for $i=1,\dots,4$. 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 $(a\bmod 5)$ and readily observe that if $1\leq i<j\leq 4$, then $(ai\bmod 5)\neq(aj\bmod 5)$. 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. For instance, $c(30)=c(5)$, $c(40)=c(15)$, $c(60)=c(10)$, $c(45)=c(20)$, etc., as illustrated in the table below.

[TABLE]

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 $a\geq 0$ and $5\nmid b$, we can assign to $n$ the color $c(n)=(b\bmod 5)$. 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 $a\geq 0$ and $p\nmid b$, and letting $c(n)=(b\bmod p)$, so that $c$ uses $p-1$ colors and if $1\leq i<j<p$ and $n$ is as indicated, then $c(in)=(ib\bmod p)\neq(jb\bmod p)=c(jn)$. ∎

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 MathOverflow precisely the version just indicated.

Question 1.2**.**

Given any positive integer $n$, is there a coloring of the positive integers using $n$ colors such that for any positive integer $a$, the numbers $a,2a,\dots,na$ all have different colors?222See https://mathoverflow.net/q/26358/

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 thesis333See http://scholarworks.boisestate.edu/td/231/.

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 in KöMaL. For instance, besides problem A.506, he also posed problem B.4265 in the April 2010 issue,444See https://www.komal.hu/feladat?a=honap&h=201004&t=mat&l=en. 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 as well as a clearly 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 $k\in K_{n}$ and $\gcd(m,n!)=1$ the color of $k$, and one can quickly check that this is an $n$-satisfactory coloring of $\mathbb{Z}^{+}$.

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 $a\in K_{n}$ all numbers $ia$, $1\leq i\leq n$, 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 $m\in K_{n}$ the color $(m\bmod p)$. If question 1.2 turns out to have a negative answer, it seems worth studying the following natural variant:

Question 1.3**.**

Assuming that question 1.2 has a negative answer for $n$, can we find a better bound than the smallest prime larger than $n$ on the number of colors required to ensure a positive answer?

We close this introduction by discussing an application and the original motivation for question 1.2.

1.3. The Balasubramanian–Soundararajan theorem

In 1970, Ronald Graham [Gra70] conjectured the following:

If $n\geq 1$, and $0<a_{1}<a_{2}<\cdots<a_{n}$ are integers, then

[TABLE]

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.

Theorem 1.4**.**

If there is an $(m-1)$-satisfactory coloring, then Graham’s conjecture holds for $n=m$.

Proof.

Argue by contradiction. Accordingly, suppose that there are $(m-1)$-satisfactory colorings and that $0<b_{1}<\cdots<b_{m}$ are integers such that

[TABLE]

Suppose $i\neq j$ 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 $a_{i}\neq a_{j}$. Since $b_{i}=a_{i}M$ and $b_{j}=a_{j}M$, in any $(m-1)$-satisfactory coloring of $\mathbb{Z}^{+}$ 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 1.4 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 MathOverflow555See https://mathoverflow.net/q/26358/ 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 $\chi(G)$ for its chromatic number, that is, the least cardinal $\kappa$ such that the set of vertices of $G$ can be colored with $\kappa$ colors in such a way that adjacent vertices receive different colors. Note that if $G$ admits a clique (complete subgraph) on $r$ vertices, then $\chi(G)\geq r$, and therefore $\chi(G)\geq\omega(G)$, where $\omega(G)$ is the clique number of $G$, that is, the supremum of the cardinalities of the cliques of $G$.

Given $n$, consider now the graph $\mathcal{G}_{n}$ with the positive integers as vertices and where any two $i\neq j$ in $\mathbb{Z}^{+}$ are connected if and only if $\max(i,j)/\gcd(i,j)\leq n$ (that is, if and only if $i,j\in\{a,2a,\dots,na\}$ for some $a$); in [BDG*+*18], this graph is denoted $B_{n}$. Note that $\mathcal{G}_{n}$ has many cliques of size $n$, namely each set $\{a,2a,\dots,na\}$ (and possibly others), so that $\omega(\mathcal{G}_{n})\geq n$; see figure 1.1.

By definition, an $n$-satisfactory coloring $c$ of $\mathbb{Z}^{+}$ is a coloring of $\mathcal{G}_{n}$ with $n$ colors, that is, the existence of such a map $c$ is precisely the claim that $\chi(\mathcal{G}_{n})$ is (at most, and therefore equal to) $n$. By the remark above, this implies that $\omega(\mathcal{G}_{n})$ is (at most, and therefore equal to) $n$, but this is precisely Graham’s conjecture for $n+1$, and we have reproved theorem 1.4.

1.4. Pilz’s conjecture

Recall that the symmetric difference $C\triangle D$ of two sets $C,D$, is the set of elements that belong to exactly one of $C,D$, that is,

[TABLE]

Note that $\triangle$ is associative, and so, given sets $C_{1},\dots,C_{m}$, their symmetric difference $\bigtriangleup_{l=1}^{m}C_{l}$ is simply the set of elements that belong to precisely an odd number of sets $C_{l}$.

If $X\subseteq\mathbb{Z}^{+}$ and $k\in\mathbb{R}$, we denote by $k\cdot X$ the dilation of $X$ by a factor of $k$:

[TABLE]

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 $n\geq 1$ and $A$ is a finite set of positive integers, then the size of the symmetric difference of the sets $A,2\cdot A,\dots,n\cdot A$ is at least $n$.

For $m$ a positive integer, it will be convenient in what follows to write $[m]$ for the set $\{1,2,\dots,m\}$. The particular case of Pilz’s conjecture where $A=[k]$ for some $k\in\mathbb{Z}^{+}$ was eventually established independently during the academic year 2008–2009 by P.-Y. Huang, W.-F. Ke and G. F. Pilz [HKP10] and by Pach and C. Szabó [PS11]. The general case remains open.

Theorem 1.5**.**

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 $A=\{a_{j}:j\in[k]\}$ and, for $j\in[k]$, set

[TABLE]

Note first that the symmetric difference of the sets $i\cdot A$, $i\in[n]$, equals the symmetric difference of the sets $B_{j}$, $j\in[k]$. The point is that, denoting by $\chi_{C}(\cdot)$ the characteristic function of a set $C$, we have for any element $x$ that

[TABLE]

and

[TABLE]

and both expressions coincide since both equal

[TABLE]

For any $n$-satisfactory coloring, in every $B_{j}$ each color appears exactly once. That is, the sets $B_{1},B_{2},\dots,B_{k}$ 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’s master’s thesis666See http://web.cs.elte.hu/blobs/diplomamunkak/mat/2009/pach$\underline{\ }$peter$\underline{\ }$pal.pdf.

For $A=[k]$, if there is a $k$-satisfactory coloring and $n$ is odd, then the same argument gives us that the size of the symmetric difference $\bigtriangleup_{i=1}^{n}i\cdot A$ 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 $C_{K_{n}}$ of $n$-satisfactory colorings of the $n$-core. 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 $n\leq 5$. In section 3 we explore an idea 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 $n\leq 8$. 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 final section 6 lists several of the remaining open problems. We proceed to list some additional details.

The theorem below summarizes the values of $n$ for which a positive answer to question 1.2 is known, see also § 2.5.

Theorem 1.6**.**

Question 1.2 has a positive answer for $n$, that is, there is an $n$-satisfactory coloring, in any of the following cases:

1. (1)

$n+1$* is prime.* 2. (2)

$2n+1$* is prime.* 3. (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 $1^{k},\dots,n^{k}$ are pairwise distinct modulo $p$. 4. (4)

Yet more generally, $n$ admits a partial $G$-isomorphism for some abelian group $(G,\oplus)$ of order $n$, i.e., there is a bijection $h\!:[n]\to G$ such that whenever $a,b\in[n]$, if $ab\in[n]$, then $h(ab)=h(a)\oplus h(b)$. In particular: 5. (5)

For all $n<195$, and 6. (6)

For all $n$ of the form $p^{2}-p$ for some prime $p$.

Proof.

(1) This is theorem 1.1.

(2) See corollary 3.2.

(3) See theorem 3.1.

(4) See theorem 4.10.

(5) See theorem 5.1 or [FP90].

(6) See theorem 4.26.

That (3) generalizes (1) and (2) is explained in § 3.1. That (4) generalizes (3) is explained in § 4.1. That (5) follows from (4) is explained in § 5.1. That (6) follows from (4) is shown in the proof of theorem 4.26. ∎

We feel that although question 1.2 was the guiding influence for much of the research reported in this paper, the topic will not be concluded even when the question is settled completely. Indeed, much of the paper is devoted to exploring how many $n$-satisfactory colorings there are for a given $n$ and, more generally, to studying the structure of these colorings.

Theorem 1.7**.**

If there is an $n$-satisfactory coloring, then there are as many such colorings as there are real numbers. Any $n$-satisfactory coloring of $\mathbb{Z}^{+}$ is determined by a sequence of $n$-satisfactory colorings of the core $K_{n}$ and a sequence of permutations of $[n]$.

On the other hand, there are values of $n$ for which there are only finitely many $n$-satisfactory colorings of the core, and others for which there are again as many such colorings as there are real numbers.

Proof.

See item (4) of proposition 2.5, where the precise way in which $n$-satisfactory colorings correspond to a sequence of colorings of the core and a sequence of permutations is described. From this, the number of $n$-satisfactory colorings is easily obtained, see corollary 2.7.

In § 2.4 we show that for $n\leq 5$ there are only finitely many $n$-satisfactory colorings of the core. In § 5.2 and § 5.3 we show that there are as many $n$-satisfactory colorings of the core as there are real numbers for $n=6,8$. ∎

This result shows that the study of the structure of $n$-satisfactory colorings should really focus on the core, and we orient our efforts accordingly. Particular attention is paid to colorings with special structure.

Definition 4.1.

An $n$-satisfactory coloring $c$ of $\mathbb{Z}^{+}$ or $K_{n}$ is multiplicative if and only if there is an abelian group structure $([n],\oplus)$ such that $c(ab)=c(a)\oplus c(b)$ for all $a,b$.

Note that for any given $n$ there are only finitely many such group structures. Nevertheless, a version of theorem 1.7 holds for this case as well.

Theorem 1.8**.**

There is a multiplicative coloring of $K_{n}$ if and only if there are as many such colorings of $\mathbb{Z}^{+}$ as there are real numbers.

For any $n$, there are only finitely many multiplicative colorings of $K_{n}$.

Proof.

See theorem 4.14 and corollary 4.4. ∎

We explore $n$-satisfactory multiplicative colorings throughout the paper, and in particular in section 4. We list all of them for $n\leq 5$ in § 2.4 and for $6\leq n\leq 8$ in § 4.5. In theorem 4.26 we show that there are such colorings if $n=p^{2}-p$ for some prime $p$. The class of multiplicative colorings to which we devote most attention is the following, already encountered in theorem 1.6.

Definition 3.3.

A strong representative of order $n$ is a prime $p$ of the form $kn+1$ for some $k$ such that $1^{k},\dots,n^{k}$ are pairwise distinct modulo $p$. If there is such a prime $p$, we say that $n$ admits a strong representative.

Theorem 1.9**.**

If $p=kn+1$ is a strong representative of order $n$, then, up to renaming of the colors, the map $c(a)=(a^{k}\bmod p)$ is a multiplicative $n$-satisfactory coloring.

Proof.

See the beginning of § 4.1. ∎

As indicated above, this gives us infinitely many examples of values of $n$ admitting $n$-satisfactory colorings. since it applies in particular when $n+1$ is prime (so $k=1$) and when $2n+1$ is prime (so $k=2$). On the other hand, if $k>2$, examples are harder to come by.

Theorem 1.10**.**

If $k>2$, then there are only finitely many $n$ such that $p=kn+1$ is a strong representative of order $n$.

Proof.

There are no such primes $p$ when $k=3$, by theorem 3.6.

If $k$ is a multiple of 4, then any such prime $p$ must satisfy $p<k^{2}$, so there are only finitely many such $n$, by theorem 3.7.

For the general case, see theorem 3.8. The argument uses the theory of Bernoulli polynomials and is due to Grinberg and Harcos. ∎

The proof of theorem 1.10 provides us with an algorithm to find, for each $k>2$, all primes $p=kn+1$ that are strong representatives of order $n$. This is illustrated in § 3.3 with some examples.

For fixed $k>2$, in the brief § 3.5 we include two results by Elkies on the asymptotic number of coincidences $a^{k}\equiv b^{k}\pmod{p}$ with $1\leq a<b\leq n$ as the prime $p=nk+1$ increases. That such coincidences occur is a consequence of theorem 1.10, and we feel that the inclusion of these observations rounds up the picture, as it provides a quantitative measure of how badly large values of $p$ of the form $kn+1$ fail to be strong representatives of order $n$. In particular, we have the following.

Theorem 1.11** (Elkies).**

For $k>2$, the number of coincidences $a^{k}\equiv b^{k}\pmod{p}$ for $p$ of the form $kn+1$ and sufficiently large, and distinct $a,b\in[n]$ is

[TABLE]

where

[TABLE]

and $\epsilon(k)=1/\phi(k)$, where $\phi$ is Euler’s totient function.

Proof.

See theorem 3.19. ∎

The study of strong representatives is interesting in its own right. We devote section 3 to it. In particular, using Chebotarëv’s theorem and tools of algebraic number theory, we prove the following.

Theorem 1.12**.**

If $n$ admits a strong representative $p$, then it admits infinitely many, and in fact the set of such primes is of positive natural density among all primes.

Proof.

See theorem 3.17. ∎

Besides this result, we also collect some related numerical data in § 3.4. Part of the interest in this result is that early numerical explorations (while the second-named author was working on his master’s thesis) suggested that the collection of strong representatives of order $n$ is very sparse, see for instance table 3.1, and this result indicates that the opposite is indeed true.

So far, our description of the results listed above emphasizes the number- and group-theoretic aspects of our work. We also bring to bear some geometric and combinatorial ideas, by showing that the existence of $n$-satisfactory colorings is equivalent to the existence of certain tilings. To state the equivalence, recall that $\pi(n)$ is the number of prime numbers less than or equal to $n$.

Say that a set $A\subseteq\mathbb{Z}^{\pi(n)}$ tiles another such set $C$ if and only if there is a $B$ such that $C$ is the direct sum of $A$ and $B$. Let $2=p_{1}<\dots<p_{\pi(n)}$ list the primes in $[n]$ in increasing order. Define $T_{n}$ be the image of $[n]$ under the map that sends $p_{1}^{\alpha_{1}}\cdots p_{\pi(n)}^{\alpha_{\pi(n)}}$ in $K_{n}$ to $(\alpha_{1},\dots,\alpha_{\pi(n)})$ in the nonnegative orthant $\mathbb{O}_{n}$ of $\mathbb{Z}^{\pi(n)}$.

Theorem 1.13**.**

There is an $n$-satisfactory coloring of $K_{n}$ if and only if $T_{n}$ tiles a superset of $\mathbb{O}_{n}$.

Proof.

See proposition 2.11. ∎

A compactness argument shows that we can replace $\mathbb{O}_{n}$ with $\mathbb{Z}^{\pi(n)}$ itself, and $K_{n}$ with the quotient field $\hat{K}_{n}=\{a/b:a,b\in K_{n}\}$, see proposition 2.12 and the remarks immediately preceding it.

Theorem 1.13 transforms the problem of finding satisfactory colorings into a geometric question. The approach is fruitful, as it was essential to the results in § 5.2 and § 5.3.

Using tilings we also obtain an elegant characterization of multiplicative colorings. If $c$ is a coloring of $K_{n}$ and $k\in K_{n}$, let $c_{k}$ be the coloring where two numbers $m,m^{\prime}\in K_{n}$ receive the same color if and only if $c(km)=c(km^{\prime})$.

Definition 4.15.

A coloring $c$ of $K_{n}$ is translation invariant if and only if $c_{k}=c$ for all $k\in K_{n}$.

This admits a natural geometric description, see § 4.3.

Theorem 1.14**.**

An $n$-satisfactory coloring is translation invariant if and only if it is multiplicative.

Proof.

See theorem 4.25. ∎

We admit we understand very little those colorings that are not multiplicative. We show examples in § 5.2 and § 5.3, but more is needed. In particular, whether question 1.2 admits a positive answer depends essentially on whether there are nonmultiplicative $n$-satisfactory colorings for various $n$, such as $n=195$. The point is that there are various $n$ which do not admit multiplicative colorings. This is briefly discussed in § 5.1, which reviews the work of Forcade and Pollington [FP90]. These numbers $n$ we call groupless. Several questions we ask suggest ways of trying to understand some of the structure of nonmultiplicative colorings, see in particular question 2.8, which refers to the topology of the collection of $n$-satisfactory colorings, described in item (1) of § 2.2.

2. The core

An $n$-coloring of a set $X$ is a coloring of $X$ using exactly $n$ colors. An $n$-satisfactory coloring is an $n$-coloring witnessing a positive answer to the $n^{\mathrm{th}}$ 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 $n$-colorings 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 $\mathbb{Z}^{+}$, 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 $i\in[n]$. The point of this convention is to avoid overcounting when looking at the number of $n$-satisfactory colorings for fixed $n$. For instance, as we will see in § 2.4, there is precisely one 3-satisfactory coloring of $K_{3}$, but without the convention it would seem as if there are six.

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 $\mathbb{N}_{n}$.

The key reason for considering cores is that there is an $n$-satisfactory coloring (of $\mathbb{Z}^{+}$) 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 $\mathfrak{c}=|\mathbb{R}|$ for colorings of $\mathbb{Z}^{+}$ 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 $X\subseteq\mathbb{Z}^{+}$ is $n$-appropriate if and only if $X$ is nonempty and contains $ix$ and $x/j$ whenever $x\in X$, $i,j\leq n$, and $j$ divides $x$.

If $X$ is $n$-appropriate, say that an $n$-coloring $c$ of $X$ is $n$-satisfactory if and only if $c(ix)\neq c(jx)$ whenever $x\in X$ and $i<j\leq n$. Note that this coincides with the previous notion of $n$-satisfactory when $X=\mathbb{Z}^{+}$ (or $X=K_{n}$). Considering colorings as maps, if $1\in X$ we add the restriction mentioned earlier that $n$-satisfactory colorings must be the identity on $[n]$.

Note that we are insisting that if $X$ is $n$-appropriate, $1\in X$, and $c$ is $n$-satisfactory on $X$, then $c$ is the identity on $[n]$, while we impose no such restrictions on the satisfactory colorings of other appropriate sets; for instance, one could wonder why we do not ask that if $a$ is the minimum of $X$, then $c(ai)=i$ for all $i\in[n]$. The reason for this convention is that we want that if $X,Y$ are disjoint and $n$-appropriate, then the union of $n$-satisfactory colorings of $X$ and $Y$ is an $n$-satisfactory coloring of $X\cup Y$, and any $n$-satisfactory coloring of $X\cup Y$ is obtained this way.

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 $n$. In the literature, these numbers are referred to as $n$-rough. Note that 1 is $n$-rough for any $n$.

The notation $X=\dot{\bigcup}_{a\in A}X_{a}$ means both that $X$ is the union of the sets $X_{a}$ for $a\in A$, and that the sets $X_{a}$ are pairwise disjoint.

Lemma 2.4**.**

A set $X\subseteq\mathbb{Z}^{+}$ is $n$-appropriate if and only if there is a nonempty set $A\subseteq P_{n}$ such that

[TABLE]

Moreover, if this is the case, then we have $A=P_{n}\cap X$.

Proof.

Note that $K_{n}$ is $n$-appropriate and therefore so is $a\cdot K_{n}$ for any $a\in P_{n}$. It follows that any $X$ of the form $\dot{\bigcup}_{a\in A}a\cdot K_{n}$ for $A\subseteq P_{n}$ and nonempty is $n$-appropriate as well.

Towards the converse, suppose now that $X$ is $n$-appropriate. Each $m\in\mathbb{Z}^{+}$ can be uniquely written in the form $m=a_{m}k_{m}$ where $a_{m}\in P_{n}$ and $k_{m}\in K_{n}$. Let $A=\{a_{x}:x\in X\}$. We claim that $X=\dot{\bigcup}_{a\in A}a\cdot K_{n}$.

First, note that if $a\neq b$ are in $P_{n}$, then $a\cdot K_{n}$ and $b\cdot K_{n}$ are pairwise disjoint. Now, if $a\in A$, then there is some $x\in X$ such that $a=a_{x}$. Since $X$ is $n$-appropriate, $h/j\in X$ whenever $h\in X$ and $j\in K_{n}$ divides $h$. In particular, $a=a_{x}=x/k_{x}\in X$. Similarly, $hi\in X$ whenever $h\in X$ and $i\in K_{n}$. Therefore, $a\cdot K_{n}\subseteq X$. This means that

[TABLE]

But, if $x\in X$, then $x\in a_{x}\cdot K_{n}$, and we have that

[TABLE]

This proves the equality and establishes the equivalence.

Second, if $a\in P_{n}$, then the only member of $P_{n}$ in $a\cdot K_{n}$ is $a$ itself. It follows that if $X=\dot{\bigcup}_{a\in A}a\cdot K_{n}$ for some $A\subseteq P_{n}$, then in fact $A=P_{n}\cap X$, 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 $C_{\mathbb{Z}^{+},n}$. We also write $C_{X},C$ if $n$ is clear from context.

The following proposition shows that $C\neq\emptyset$ if and only if $C_{X}\neq\emptyset$ for some $n$-appropriate set $X$ if and only if $C_{X}\neq\emptyset$ 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 $n$-satisfactory colorings of $K_{n}$. In fact, the proposition shows how the satisfactory colorings of the core completely determine all satisfactory colorings.

Proposition 2.5**.**

Let $n\in\mathbb{Z}^{+}$.

1. (1)

Suppose $X\subseteq Y$ are $n$-appropriate. If $C_{Y}\neq\emptyset$, then $C_{X}\neq\emptyset$. In fact, thinking of colorings as maps, the restriction $c\upharpoonright X$ is in $C_{X}$ for any $c\in C_{Y}$. 2. (2)

Given $a\in P_{n}$ and $c\in C_{a\cdot K_{n}}$, if $\operatorname{div}(c,a)$ denotes the $n$-coloring of $K_{n}$ such that

[TABLE]

for all $k,l\in K_{n}$, then $\operatorname{div}(c,a)\in C_{K_{n}}$. Considering colorings as maps with range $[n]$,

[TABLE]

for all $k\in K_{n}$, where $\pi$ is the permutation of $n$ such that $\pi(c(ai))=i$ for all $i\in[n]$. 3. (3)

Given $a\in P_{n}$ and $c\in C_{K_{n}}$, if $\operatorname{mult}(c,a)$ denotes the $n$-coloring of $a\cdot K_{n}$ such that

[TABLE]

for all $k,l\in K_{n}$, then $\operatorname{mult}(c,a)\in C_{a\cdot K_{n}}$. Considering colorings as maps, $\operatorname{mult}(c,a)(a\cdot k)=c(k)$ for all $k\in K_{n}$. However, note that if $a\neq 1$, then for any permutation $\pi$ of $[n]$, $\pi\circ\operatorname{mult}(c,a)$ is also in $C_{a\cdot K_{n}}$. 4. (4)

If $X$ is $n$-appropriate, then a map $c$ is in $C_{X}$ if and only if for each number $a\in P_{n}\cap X$ there is a map $c_{a}\in C_{K_{n}}$ and a permutation $\pi_{a}$ of $[n]$ such that $\pi_{1}$ is the identity and

[TABLE] 5. (5)

$C\neq\emptyset$* if and only if $C_{X}\neq\emptyset$ for any $n$-appropriate $X$ if and only if $C_{X}\neq\emptyset$ for some $n$-appropriate set $X$. Moreover if $X\subseteq Y$ and both are $n$-appropriate, then $d\in C_{X}$ if and only if $d=c\upharpoonright X$ for some $c\in C_{Y}$.*

Proof.

(1) This is clear.

(2) Given $a\in P_{n}$ and $c\in C_{a\cdot K_{n}}$, if $\operatorname{div}(c,a)$ is defined as in item (2), then

[TABLE]

for any $b\in K_{n}$ and any $i<j\leq n$ because $c$ is $n$-satisfactory and therefore $c(aib)\neq c(ajb)$. But this means that $\operatorname{div}(c,a)$ is $n$-satisfactory as well.

Considering colorings as functions, $c$ is a map with range $[n]$ and the only obstacle for $k\mapsto c(ak)$ to be an $n$-satisfactory coloring is the additional restriction we have imposed that such a map must be the identity on $[n]$, which explains why we may need to precompose it with a permutation to achieve this.

(3) Conversely, if $c\in C_{K_{n}}$, $a\in P_{n}$, and $\operatorname{mult}(c,a)$ is defined as in item (3), then

[TABLE]

for any $m\in a\cdot K_{n}$ and any $i<j\leq n$ since $c$ is satisfactory and therefore $c(i(m/a))\neq c(j(m/a))$. But this means that $\operatorname{mult}(c,a)$ is satisfactory as well. Considering colorings as maps, the inequalities just indicated are maintained under any permutation $\pi$ of $[n]$, so if $a\neq 1$, then $\pi\circ\operatorname{mult}(c,a)$ is also satisfactory.

(4) Suppose that $X$ is $n$-appropriate. First, $X=\bigcup_{a\in P_{n}\cap X}a\cdot K_{n}$, by lemma 2.4. If $c\in C_{X}$, then, by item (1), $d=c\upharpoonright a\cdot K_{n}\in C_{a\cdot K_{n}}$ for any $a\in P_{n}\cap X$, and $c_{a}\coloneqq\operatorname{div}(d,a)\in C_{K_{n}}$ by item (2). Writing $\pi_{a}^{-1}$ for the permutation as in item (2), we have that $d=\pi_{a}\circ\operatorname{mult}(c_{a},a)$, and therefore

[TABLE]

Conversely, suppose that $C_{K_{n}}\neq\emptyset$. For each $a\in P_{n}\cap X$ let $c_{a}\in C_{K_{n}}$ and $\pi_{a}$ be a permutation of $[n]$, with $\pi_{1}$ being the identity if $1\in X$. Define

[TABLE]

As mentioned in lemma 2.4, $a\cdot K_{n}\cap b\cdot K_{n}=\emptyset$ whenever $a\neq b$ are in $P_{n}$. From this, and item (3), $c$ is well defined and has domain $\bigcup_{a\in P_{n}\cap X}a\cdot K_{n}$, which equals $X$, again by lemma 2.4. If $m\in X$ and $i<j\leq n$, then there is a unique $a\in P_{n}\cap X$ such that $mi$ and $mj$ belong to $a\cdot K_{n}$, and by item (3) it follows that $c(mi)\neq c(mj)$. This proves that $c$ is satisfactory, and completes the proof of item (4).

(5) Now, if $X$ is $n$-appropriate, and $C_{X}\neq\emptyset$, then $C_{a\cdot K_{n}}\neq\emptyset$ for any $a$ in the nonempty set $P_{n}\cap X$, by item (1). But this implies that $C_{K_{n}}\neq\emptyset$, by item (2). It follows from item (4) that $C=C_{\mathbb{Z}^{+}}\neq\emptyset$. Thus, $C_{Y}\neq\emptyset$ for any $n$-appropriate $Y$, again by item (1).

Finally, if $X\subseteq Y$ are $n$-appropriate and $c\in C_{Y}$, then $d=c\upharpoonright X\in C_{X}$, by item (1). Conversely, if $d\in C_{X}$, let $e\in C_{K_{n}}$, which exists as shown above. Let $c_{a}=e$ and $\pi=\operatorname{id}$ for $a\in P_{n}\cap(Y\smallsetminus X)$. For $a\in P_{n}\cap X$, let $c_{a}=\operatorname{div}(d\upharpoonright a\cdot K_{n},a)$ and $\pi_{a}$ be the permutation such that $d\upharpoonright a\cdot K_{n}=\pi_{a}\circ\operatorname{mult}(c_{a},a)$ . As in item (4), we have that $c=\bigcup_{a\in P_{n}\cap Y}\pi_{a}\circ\operatorname{mult}(c_{a},a)\in C_{Y}$. And, by construction, $d=c\upharpoonright X$. 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 $\mathbb{Z}\smallsetminus\{0\}$. More interesting is whether this is also equivalent to the existence of an $n$-satisfactory coloring of $\mathbb{Q}\smallsetminus\{0\}$ or, what is the same, of $\hat{K}_{n}\coloneqq\{a/b:a,b\in K_{n}\}$. 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], $\hat{K}_{n}$ is denoted $\mathbb{Q}_{n}$.

When discussing $n$-satisfactory colorings, proposition 2.5 provides us with the ability to restrict our attention from all of $\mathbb{Z}^{+}$ to $K_{n}$. The relation the proposition details between arbitrary satisfactory colorings and colorings of the core has the following corollary.

Corollary 2.7**.**

For $n>1$, if $C_{K_{n}}\neq\emptyset$, then $|C|=\mathfrak{c}$.

Proof.

If there is a coloring of the core (thought of as a function with range $[n]$), then there are at least $n!\geq 2$ such colorings of any $a\cdot K_{n}$ for $a\in P_{n}$ different from $1$, obtained by invoking item (3) of proposition 2.5 and varying the permutation $\pi$. By item (4) of proposition 2.5, there is a bijective correspondence between the elements of $C$, and the set of functions with domain $P_{n}$ that pick for each $a\in P_{n}$ a member of $C_{K_{n}}$ and a permutation of $[n]$ (with the permutation being the identity if $a=1$), from which we get that $|C|\geq n!^{|P_{n}\smallsetminus\{1\}|}=\mathfrak{c}$.

On the other hand, any element of $C$ is a function from $\mathbb{Z}^{+}$ to $[n]$ (satisfying certain restrictions), so $|C|\leq|[n]^{\mathbb{Z}^{+}}|=n^{\aleph_{0}}=\mathfrak{c}$, and it follows that $|C|=\mathfrak{c}$ by the Cantor–Schröder–Bernstein 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 $\mathbb{Z}^{+}$ 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 $C_{K_{n}}$

We mention here some easy observations regarding the closure of $C_{K_{n}}$ under some natural operations.

(1) First, $C_{K_{n}}$ is a closed subset of the Polish space $[n]^{K_{n}}$ of functions from $K_{n}$ to $[n]$ under the product topology (with $[n]$ discrete): $c\in C_{K_{n}}$ if and only if

[TABLE]

and note that for any distinct $b,c\in K_{n}$,

[TABLE]

is a finite union of closed sets.

This topological fact is trivial in some cases, since (as shown in § 2.4) $C_{K_{n}}$ is sometimes finite, but see § 5.2. The Polish topology of the space $[n]^{K_{n}}$ is generated by a natural metric: enumerate $K_{n}$ in increasing order as $\{k_{i}:i\in\mathbb{Z}^{+}\}$. The distance between two distinct colorings $c,c^{\prime}$ is $1/N$, where $N$ is the least index of a disagreement, that is, $c(k_{i})=c^{\prime}(k_{i})$ for all $i<N$, but $c(k_{N})\neq c^{\prime}(k_{N})$. This metric is complete both in the whole space $[n]^{K_{n}}$ and in $C_{K_{n}}$.

The truth is, we understand very little of the topological structure of $C_{K_{n}}$. It is unclear, for instance, whether the following question should have a positive answer.

Question 2.8**.**

Given $n\in\mathbb{Z}^{+}$, suppose that $C_{K_{n}}$ is nonempty. Should it have isolated points?

(2) The following is an immediate but useful observation.

Lemma 2.9**.**

Let $\rho$ be an automorphism of the structure $([n],|)$, that is, of the Hasse diagram for divisibility on $[n]$. Extend $\rho$ to a bijection of $K_{n}$ in the natural way: if $m=2^{\alpha_{1}}\dots p_{k}^{\alpha_{k}}$ is the prime factorization of $m\in K_{n}$, then

[TABLE]

If $c$ is an $n$-satisfactory coloring of $K_{n}$, then so is $\tilde{c}$, where $\tilde{c}(a)=\tilde{c}(b)$ if and only if $c(\rho(a))=c(\rho(b))$.

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 $\rho$ be a permutation of the set of primes in $[n]$ is not enough in general.

Proof.

Note first that $\rho$ permutes the primes less than or equal to $n$, and equation (2.1) holds for all $m\in[n]$, so that the suggested extension is well defined and maps $K_{n}$ to itself. Since $\rho$ is a permutation on the primes in $[n]$, it follows as well that $\rho$ is surjective on $K_{n}$. Note also that for any $m_{1},m_{2}\in K_{n}$, $\rho(m_{1}m_{2})=\rho(m_{1})\rho(m_{2})$.

Now, if $c$ is satisfactory and $\tilde{c}$ is as indicated, then for $i\neq j$ in $[n]$ and $a\in K_{n}$, we have that $\tilde{c}(ia)=c(\rho(i)\rho(a))\neq c(\rho(j)\rho(a))=\tilde{c}(ja)$ since $\rho(i),\rho(j)\in[n]$. ∎

(3) Another natural operation on $C_{K_{n}}$ can be defined by letting $c_{k}\in C_{K_{n}}$, for $c\in C_{K_{n}}$ and $k\in K_{n}$, be given by $c_{k}(l)=c_{k}(m)$ if and only if $c(kl)=c(km)$. (Abusing slightly777In proposition 2.5 we require $k\in P_{n}$, but $P_{n}\cap K_{n}=\{1\}$. the notation used in proposition 2.5, $c_{k}=\operatorname{div}(c\upharpoonright k\cdot K_{n},k)$.) 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 $k\in K_{n}$ 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 $k\in K_{n}$, 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 $C_{K_{n}}$ is closed in $[n]^{K_{n}}$, it is also closed under a construction coming from applications of Kőnig’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 $\mathbb{Z}^{+}$ 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 $m=2^{a}3^{b}\in K_{3}$ 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 $\mathbb{Z}^{\pi(n)}$ where, as usual, $\pi(\cdot)$ 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 $\pi(n)$-dimensional polyomino corresponding to the set $\{1,2,\dots,n\}$ as tiles. We proceed now to explain this connection.

Given $n$, work in $\mathbb{Z}^{\pi(n)}$. As suggested above, we identify each member of $K_{n}$ with the tuple of its prime exponents: any $m\in K_{n}$ can be written in a unique way as $m=\prod_{i=1}^{\pi(n)}p_{i}^{\alpha_{i}}$, where $2=p_{1}<\dots<p_{\pi(n)}$ are the primes less than or equal to $n$, listed in increasing order, and the $\alpha_{i}$ are nonnegative integers. We identify $m$ with the tuple $t(m)=(\alpha_{1},\dots,\alpha_{\pi(n)})$ in the first orthant $\mathbb{O}_{n}$ of $\mathbb{Z}^{\pi(n)}$, noting that $t\!:K_{n}\to\mathbb{O}_{n}$ is a bijection, and let $T_{n}=\{t(i):i\in[n]\}$. Note that $t$ turns multiplication into vector addition in the sense that $t(kk^{\prime})=t(k)+t(k^{\prime})$ for any $k,k^{\prime}\in K_{n}$. We will find several maps with similar properties in what follows, see for instance definition 4.1, where we call them multiplicative.

Given $A,C\subseteq\mathbb{Z}^{\pi(n)}$, say that $A$ tiles $C$ (or, equivalently, that $C$ can be tiled by $A$) if and only if there is a set $B\subseteq\mathbb{Z}^{\pi(n)}$ such that $C$ is the direct sum of $A$ and $B$, that is,

1. (1)

$C=A+B\coloneqq\{a+b:a\in A,b\in B\}$, and in fact 2. (2)

any $c\in C$ admits a unique decomposition as a sum of a member of $A$ and a member of $B$, that is, there is a unique pair $(a,b)\in A\times B$ 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 $(\alpha_{i}:i\in[\pi(n)])$ in $\mathbb{Z}^{\pi(n)}$ with the corresponding $\pi(n)$-dimensional cubes $\{(x_{i}:i\in[\pi(n)]):\alpha_{i}\leq x_{i}\leq\alpha_{i}+1\}$.

For a fixed value of $n$, consider now the following two statements:

1. (i)

There is an $n$-satisfactory coloring of $K_{n}$. 2. (ii)

$T_{n}$ essentially tiles the orthant $\mathbb{O}_{n}$.

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 $\mathbb{O}_{n}$, say $C=T_{n}+B$, the sum being direct. Note that via this direct sum, each element of $C$, and therefore each $x\in\mathbb{O}_{n}$, belongs to exactly one tile, that is, a unique copy of $T_{n}$. There is a unique $m\in K_{n}$ 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 $a\in T_{n}$. We must argue that this coloring is $n$-satisfactory. Indeed, given $k\in K_{n}$ and $i,j\in[n]$, suppose that $ki$ and $kj$ receive the same color, that is, there are $\alpha\in T_{n}$ and $b_{1},b_{2}\in B$ such that $t(ki)=\alpha+b_{1}$ and $t(kj)=\alpha+b_{2}$. By the multiplicative property of $t$, it follows that

[TABLE]

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 $B^{\prime}$ be the image under the map $t$ of one of the color classes, note that the sum $T_{n}+B^{\prime}$ is direct. Indeed, suppose that $i,j\in[n]$, and $k,k^{\prime}\in K_{n}$ are such that $t(k),t(k^{\prime})\in B^{\prime}$ and $t(i)+t(k)=t(j)+t(k^{\prime})$, that is, $ik=jk^{\prime}$. By removing common factors if necessary, we may further assume that $i,j$ are relatively prime. This means that there is a positive integer $k^{\prime\prime}$ such that $k=jk^{\prime\prime}$ and $k^{\prime}=ik^{\prime\prime}$. Observe that $k^{\prime\prime}\in K_{n}$. The assumption that $B^{\prime}$ is the image of a color class gives us that $c(jk^{\prime\prime})=c(ik^{\prime\prime})$ and, since $c$ is $n$-satisfactory, then $i=j$ and so also $k=k^{\prime}$. This proves that the sum $T_{n}+B^{\prime}$ is indeed direct.

Let now $x\in\mathbb{O}_{n}$ be sufficiently far from the boundary of $\mathbb{O}_{n}$, in the sense that all of $x-t(1),\dots,x-t(n)$ are themselves in $\mathbb{O}_{n}$ (equivalently, if $x=t(m)$, then all of $m,m/2,\dots,m/n$ are positive integers), and fix the image $B^{\prime}$ of a color class. We claim that for some $i\in[n]$, we have that $x-t(i)\in B^{\prime}$. Otherwise, by the pigeonhole principle, for some $i\neq j$, both in $[n]$, it must be that $x-t(i)$ and $x-t(j)$ are in the same image $B^{\prime\prime}$ of a color class. This is impossible, since the decompositions $x=t(i)+(x-t(i))=t(j)+(x-t(j))$ contradict that the sum $T_{n}+B^{\prime\prime}$ is direct, as shown in the previous paragraph.

We have shown that for any image $B^{\prime}$ of a color class, the sum $T_{n}+B^{\prime}$ is direct and contains a translate of $\mathbb{O}_{n}$, for instance $x_{0}+\mathbb{O}_{n}$, where $x_{0}=t(\operatorname{lcm}([n]))$. Setting $B=B^{\prime}-x_{0}$, then $T_{n}+B$ is a direct sum and covers $\mathbb{O}_{n}$, as desired; see figure 2.6. ∎

Now consider the following additional statement:

1. (iii)

$\mathbb{Z}^{\pi(n)}$ 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):

1. (iv)

There is an $n$-satisfactory coloring of $\hat{K}_{n}=\{a/b:a,b\in K_{n}\}$.

Proposition 2.12**.**

With notation as above, (i) implies (iii), and therefore (i)–(iv) are all equivalent.

Proof.

Let $\hat{\mathcal{G}}_{n}$ be the graph with set of vertices $\hat{K}_{n}$ where two points $x,y$ are connected if and only if there is an $m\in\hat{K}_{n}$ such that $x,y\in\{im:i\in[n]\}$ (cf. § 1.3). It is enough to argue that $\chi(\hat{\mathcal{G}}_{n})=n$, 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 $G$ of $\hat{\mathcal{G}}_{n}$, by multiplying all vertices by an appropriate $k\in K_{n}$ we see that $G$ is isomorphic to a finite subgraph of $\mathcal{G}_{n}$ and is therefore $n$-colorable, since (i) is equivalent to the assertion that $\chi(\mathcal{G}_{n})=n$. ∎

Essentially the same argument was also noted in [BDG*+*18], where $\hat{\mathcal{G}}_{n}$ 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

[TABLE]

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 $\mathbb{O}_{n}$, and a cube $D_{m}^{\prime}$ of the same size as $D_{m}$ but completely contained in $\mathbb{O}_{n}$. Now define $d^{m}$ simply by translating $D_{m}^{\prime}$ 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 of $\hat{K}_{n}$, 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 Kőnig’s infinity lemma.

Explicitly: enumerate the points in the grid as $v_{1},v_{2},\dots$. Note that if $m<m^{\prime}$, then $D_{m}\subsetneq D_{m^{\prime}}$, and that the union over all $m$ of the hypercubes $D_{m}$ is the whole space $\mathbb{Z}^{\pi(n)}$ so that, for any $k$, $v_{k}$ is in $D_{m}$ for all sufficiently large $m$. Consider the sequence of colorings $\vec{d}=(d^{m})_{m>0}$. Since only $n$ colors are possible, there is a subsequence of $\vec{d}$ that always assigns to $v_{1}$ the same color. Passing to a subsubsequence, we can also fix the color assigned to $v_{2}$. Going to yet a further subsequence, we can fix the color of $v_{3}$. 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 $i\neq j$ in $[n]$ and some $k\in\hat{K}_{n}$, for $m$ large enough (say, $m\geq m_{0}$) 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 $x=v_{r}$ and $y=v_{s}$ 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 $m$ (using that $r<s$), in particular there is such an $m$ with $m\geq m_{0}$ and therefore $d(x)\neq d(y)$, 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 $k\in K_{n}$, $d$ may not be compatible with any of the translates $c_{k}$ (that is, with the original coloring of any of the translates $\mathbb{O}_{n}+t(k)$) and the question remains whether we can further impose this compatibility requirement. Say that a tiling $T_{n}+B^{\prime}$ of $\mathbb{Z}^{\pi(n)}$ essentially extends a tiling $T_{n}+B$ of a superset of $\mathbb{O}_{n}$ if and only if any tile of $T_{n}+B$ completely contained in $\mathbb{O}_{n}$ is a tile of $T_{n}+B^{\prime}$ (so that, if at all, only partial tiles covering a part of the boundary of the orthant could in principle change).

Question 2.13**.**

Let $n\in\mathbb{Z}^{+}$.

1. (1)

Does any $n$-satisfactory coloring of $K_{n}$ extend to one of $\hat{K}_{n}$? 2. (2)

If $T_{n}$ essentially tiles $\mathbb{O}_{n}$ via a tiling $T_{n}+B$, is there a tiling by $T_{n}$ of all of $\mathbb{Z}^{\pi(n)}$ that essentially extends it?

Question 2.13 has a positive answer for $n=3,4$ but seems delicate in general. It is clear that a multiplicative $n$-satisfactory coloring can be extended as in (1) (see remark 4.5); in particular, (1) has a positive answer if all $n$-satisfactory colorings are multiplicative. In that case, (2) has a positive answer as well. More generally, (2) has a positive answer if the coloring induced by $T_{n}+B$ (as in the proof of proposition 2.11) is multiplicative. Question 2.10 (whether for any $n$-satisfactory $c$ and any $k\in K_{n}$ 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) $\mathbb{O}_{n}$ to one of $\mathbb{O}_{n}-t(k)$. 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 $p_{1}<\dots<p_{\pi(n)}$ 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 $d^{1},d^{2},\dots$ such that $d^{0}={d^{1}}_{p_{1}}$ and, in general,

[TABLE]

for any nonnegative $m$ and any $a\in[\pi(n)]$. If $d^{j}={d^{j+1}}_{p}$, we can think of $d^{j+1}$ as extending the domain of $d^{j}$ to the set $\operatorname{dom}(d^{j+1})=\{a/p:a\in\operatorname{dom}(d^{j})\}$. The colorings $d^{j}$ agree on their common domains as $j$ increases, and any $m\in\hat{K}_{n}$ is eventually included in these domains. This means that there is a unique “limit” $n$-satisfactory coloring $d$ of all of $\hat{K}_{n}$ obtained by this process. ∎

Remark 2.15**.**

The second proof we gave of proposition 2.12 illustrates an application of Kőnig’s lemma that can be interpreted as a construction that the space $C_{K_{n}}$ 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 $\mathbb{O}_{n}$. For each $l\in\mathbb{N}$, let

[TABLE]

be the $\pi(n)$-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 $\mathbf{x}\in\mathbb{O}_{n}$, the coloring $c\upharpoonright(\mathbf{x}+D_{l})$, seen as a coloring of $D_{l}$ in the natural way, coincides with $d$, that is, $d(\mathbf{y})=c(\mathbf{x}+\mathbf{y})$ for any $\mathbf{y}\in D_{l}$.

Define an infinite finitely branching tree $\mathcal{T}$ as follows: start with the empty coloring of the empty set (seen as the only node of $\mathcal{T}$ at level $-1$) and, for each $l\in\mathbb{N}$, use as nodes of the $l^{\mathrm{th}}$ level of $\mathcal{T}$ the colorings $d$ realized by $c$ on infinitely many distinct copies of $D_{l}$ (that is, those $d$ for which there are infinitely many $\mathbf{x}$ as above). Use as immediate successors of such a coloring $d$ the colorings $d^{\prime}$ at the $(l+1)^{\mathrm{st}}$ level that extend $d$ in the sense that $d^{\prime}\upharpoonright D_{l}=d$.

By Kőnig’s lemma, the tree admits an infinite branch, that is, a sequence of colorings $(d^{l}:l\in\mathbb{N})$ such that $\operatorname{dom}(d^{l})=D_{l}$ for each $l$, and the colorings are compatible in the sense that $d^{l}=d^{l+1}\upharpoonright D_{l}$ for each $l$. The union of these colorings is a coloring $d$ of $\mathbb{O}_{n}$ 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 $\mathcal{T}^{\prime}$ 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 $c_{k}$, $k\in K_{n}$, appear as branches of $\mathcal{T}^{\prime}$), while restricting attention to $\mathcal{T}$ may potentially result in different colorings. Moreover, even if, say, $c$ itself appears as a branch through $\mathcal{T}$, 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 $c$ of $\hat{K}_{n}$, 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 $\mathbb{Z}^{\pi(n)}$. From this tiling we can define an $n$-satisfactory coloring $c^{\prime}$ with color classes the preimages under $t$ of the translates $t(i)+B$, $i\in[n]$. Is $c^{\prime}=c$?

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 $\mathbb{O}_{n}$ rather than the whole $\mathbb{Z}^{\pi(n)}$) is somewhat more delicate: now from an essential tiling of $\mathbb{O}_{n}$ 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 5.4). On the other hand, $c$ gives us not just one, but $n$ partial tilings of $\mathbb{O}_{n}$, 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 $c^{\prime}$ of $K_{n}$. Is $c^{\prime}=c$?

We can also ask whether the partial tilings can be extended to essential tilings in compatible ways. We address question 2.17 in remark 5.5 where we show that, perhaps surprisingly, there are instances where the answer is negative.

2.4. Satisfactory colorings with $n\leq 5$

For $n\leq 5$ 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 list is no longer possible even for $n=6$. We use $\mathbb{N}$ for the set of natural numbers, including 0.

• •

$n=1$.

Trivially, there is only one 1-satisfactory coloring of $K_{1}=\{1\}$.

• •

$n=2$.

Similarly, there is only one 2-satisfactory coloring $c$ of $K_{2}=\{2^{a}:a\in\mathbb{N}\}$: presented as an equivalence relation with 2 classes $c(1)\neq c(2)$, we have

[TABLE]

for all $\alpha\in\mathbb{N}$.

This introduces a recurring theme: we could describe the coloring simply as $c(2^{\alpha})=(\alpha\bmod 2)$, 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 $c(m)=(m\bmod 3)$ 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 $K_{3}=\{2^{\alpha}3^{\beta}:\alpha,\beta\in\mathbb{N}\}$. For $a$ a positive integer we have that $c(a)$, $c(2a)$ and $c(3a)$ are all different.

It follows that $c(2a),c(4a),c(6a)$ are different, and so are $c(3a),c(6a),c(9a)$. In particular, $c(6a)\neq c(2a),c(3a)$, so $c(6a)=c(a)$ and therefore $c(4a)=c(3a)$, from which we conclude that, in fact, $c(2^{\alpha}3^{\beta})=c(2^{\alpha+2\beta})$. Also, since $c(a),c(2a),c(4a)$ are different, we have that $c(2a),c(4a),c(8a)$ are different as well, and it follows that $c(8a)=c(a)$. This means that

[TABLE]

for all $\alpha,\beta\in\mathbb{N}$. 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 $\alpha,\beta$, the unit square with bottom left corner at $(\alpha,\beta)$ has color $(\alpha+2\beta\bmod 3)$, so that as before, the coloring of a number $m=2^{\alpha}3^{\beta}\in K_{3}$ is given as a linear equation in the exponents of the prime factorization of $m$.

Note that the permutation $(23)$ is an automorphism of $([3],|)$, see figure 2.7. By lemma 2.9, the coloring $\tilde{c}$ is also 3-satisfactory, where

[TABLE]

and, again, we require $\tilde{c}(1),\tilde{c}(2),\tilde{c}(3)$ to be different. Uniqueness, of course, simply means that $c$ is invariant under this permutation, i.e., $c=\tilde{c}$.

• •

$n=4$.

The argument is similar to the previous case: if $c$ is a 4-satisfactory coloring of $K_{4}=K_{3}$, then for any positive integer $a$, the colors $c(a),c(2a),c(3a),c(4a)$ are different, and $c(6a)\neq c(2a),c(4a),c(8a)$ and $c(6a)\neq c(3a),c(9a),c(12a)$, so $c(6a)=c(a)$ and $c(8a)=c(3a)$.

Also, since $c(a),c(2a),c(4a),c(8a)$ are different, we see that $c(16a)=c(a)$, see figure 2.8. Thus,

[TABLE]

for all $\alpha,\beta\in\mathbb{N}$. Conversely, equation (2.4) and the requirement that $c(1),c(2)$, $c(3),c(4)$ are distinct describes a 4-satisfactory coloring of $K_{4}$. This is the coloring indicated in figure 2.3. Again the coloring can be succinctly described by a linear equation as $c(2^{\alpha}3^{\beta})=(\alpha+3\beta\bmod 4)$.

Again by uniqueness and the result of § 1.1 (since $4+1=5$ is prime), the coloring can also be described by $c(m)=(m\bmod 5)$.

• •

$n=5$.

Note first that if $c$ is a 5-satisfactory coloring of $K_{5}$, then $c(8a)\neq c(a)$ for any $a\in K_{5}$. Indeed, otherwise $c(10a)=c(3a)$ and $c(6a)=c(5a)$, which further forces $c(12a)=c(2a)$ and no color can be assigned to $20a$. This means that either $c(6a)=c(a)$ or $c(10a)=c(a)$. In particular, either $c(6)=c(1)$ or $c(10)=c(1)$.

Consider first the case where $c(6)=c(1)$, and let

[TABLE]

so that $1\in K^{1}$. Suppose that $a\in K^{1}$, that is, $c(6a)=c(a)$. We have that $c(10a)=c(3a)$ and $c(8a)=c(5a)$, thus $c(12a)=c(2a)$ or $2a\in K^{1}$. It follows that $c(15a)=c(4a)$ and $c(9a)=c(5a)$.

Now: we just proved that $a\in K^{1}$ implies $c(9a)=c(5a)$; since it also implies that $2a\in K^{1}$, it follows that $c(6\cdot 3a)=c(18a)=c(9\cdot 2a)=c(5\cdot 2a)=c(10a)=c(3a)$ and, similarly, $c(6\cdot 5a)=c(30a)=c(15\cdot 2a)=c(4\cdot 2a)=c(8a)=c(5a)$. That is, $3a,5a\in K^{1}$ as well.

This means that $K^{1}=K_{5}$ and

[TABLE]

for all $a\in K_{5}$. Now we can proceed as in the previous cases: note that $c(5a)=c(8a)$ and $c(3a)=c(10a)=c(5\cdot 2a)=c(8\cdot 2a)=c(16a)$, so

[TABLE]

for all $\alpha,\beta,\gamma\in\mathbb{N}$. Conversely, equation (2.5) and the requirement that $c(1),\dots,c(5)$ are distinct describes a 5-satisfactory coloring of $K_{5}$; moreover, this is the only such coloring with $c(6)=c(1)$.

A similar analysis shows that

[TABLE]

for all $\alpha,\beta,\gamma\in\mathbb{N}$, and the requirement that $c(1),\dots,c(5)$ are distinct describes the unique 5-satisfactory coloring of $K_{5}$ with $c(10)=c(1)$.

Actually, the latter analysis can be avoided by noting that 3 and 5 are indiscernible in $K_{5}$ in the sense that the transposition $(35)$ is an automorphism of the Hasse diagram for divisibility in $[5]$, see figure 2.9. By lemma 2.9, there is a one-to-one correspondence between 5-satisfactory colorings with $c(2\cdot 5)=c(1)$ and those with $c(2\cdot 3)=c(1)$, so in particular there is a unique 5-satisfactory coloring with $c(10)=c(1)$, and the lemma allows us to recover the precise form of equation (2.6).

In terms of linear equations, the two colorings we obtained are

[TABLE]

where the superindices in $c^{1},c^{5}$ refer to whether $c(6)=c(1)$ or $c(6)=c(5)$, respectively.

(Note that typically the same coloring can be described by several linear equations. For example, the 5-coloring that to $2^{\alpha}3^{\beta}5^{\gamma}$ assigns $(2\alpha+\beta+3\gamma\bmod 5)$ coincides with $c^{1}$.)

Finally, we can now illustrate why lemma 2.9 cannot be strengthened by allowing $\rho$ to be any permutation of the set of primes in $[n]$. Indeed, consider the transposition $\rho=(23)$, extend it to a permutation of $K_{5}$ as in equation (2.1), and note that the coloring $\tilde{c^{1}}$, given by $\tilde{c^{1}}(a)=\tilde{c^{1}}(b)$ if and only if $c^{1}(\rho(a))=c^{1}(\rho(b))$ is not a 5-satisfactory coloring, since $\tilde{c^{1}}(4)=c^{1}(9)=c^{1}(5)=\tilde{c^{1}}(5)$.

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 $n\leq 31$: given such an $n$, let $k=\pi(n)$ and let $p_{1}<\dots<p_{k}$ be the prime numbers less than or equal to $n$. We exhibit coefficients $a_{1},\dots,a_{k}$ such that the $n$-coloring $c$ of $K_{n}$ given by

[TABLE]

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 $\prod_{j\leq i}p_{i}^{\beta_{i}}$. 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

[TABLE]

so that if $c(m)=(k\bmod 7)$, say, then

[TABLE]

and $c$ is indeed 7-satisfactory.

We revisit this approach and provide additional context through the notion of partial isomorphism in § 4.2.

3. Generalizing the approach for $p$ prime

3.1. Strong representatives

In this section, we present a condition on $n$ that, if satisfied, ensures the existence of $n$-satisfactory colorings. The construction below was first noticed by the third-named author in 2009. It was suggested independently in MathOverflow by Victor Protsak888See https://mathoverflow.net/q/26358/. It has also been considered before in connection with Graham’s conjecture discussed in § 1.3, see for instance [FP90, § 2] and references therein.

Theorem 3.1**.**

If $n,k$ are positive integers such that $p=kn+1$ is prime and $1^{k},2^{k},\dots,n^{k}$ are distinct modulo $p$, then $c(m)=({m^{k}}\bmod p)$, $m\in K_{n}$, is an $n$-satisfactory coloring of $K_{n}$.

Proof.

We begin noting that there are exactly $n$ pairwise incongruent nonzero $k^{\mathrm{th}}$ power residues modulo $kn+1$.

For $i\neq j\in[n]$ and $a\in K_{n}$, we note that $c(ia)\neq c(ja)$ since the hypothesis implies that $a^{k}i^{k}\not\equiv a^{k}j^{k}\pmod{p}$. ∎

In particular, we recover the proof of problem A.506 from KöMaL given in the introduction since the assumption that $1,2,\dots,n$ 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).

Corollary 3.2**.**

If $p=2n+1$ is prime, then $c(m)=({m^{2}}\bmod p)$ is an $n$-satisfactory coloring of $K_{n}$.

Proof.

It is enough to verify that $1^{2},\dots,n^{2}$ are pairwise incongruent modulo $p$. This is immediate since $i^{2}\equiv j^{2}\pmod{p}$ if and only if either $i\equiv j\pmod{p}$ or $i\equiv-j\pmod{p}$, but the latter is impossible if $i,j\in[n]$. ∎

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 $1^{k},\dots,n^{k}$ are pairwise distinct modulo $p$, and $c(m)=(m^{k}\bmod p)$ for all $m\in K_{n}$. 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 $n\leq 33$ the smallest strong representative of order $n$.

As table 3.1 suggests, unlike the cases $k=1,2$, the primality of $kn+1$ for $k>2$ does not automatically ensure that the hypothesis of theorem 3.1 is satisfied. For example, if $n=3$, then $p=4n+1=13$ is prime. However, $2^{4}=16$, $3^{4}=81$, and $16\equiv 81\pmod{13}$. This is further discussed in § 3.2.

Remark 3.4**.**

In terms of notions introduced below, all colorings obtained through strong representatives are multiplicative, and in fact are $\mathbb{Z}/n\mathbb{Z}$-colorings. However, there are satisfactory colorings that are nonmultiplicative (see theorem 5.6), multiplicative colorings that are not $\mathbb{Z}/n\mathbb{Z}$-colorings (see table 4.20 for $a=1$), and $\mathbb{Z}/n\mathbb{Z}$-colorings that do not admit a strong representative (see table 4.12).

3.2. $k$-representatives

Definition 3.5**.**

Let $k\in\mathbb{Z}^{+}$. 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 $1^{k},\dots,n^{k}$ 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 2011,999See https://www.komal.hu/feladat?a=honap&h=201111&t=mat&l=en. proposed by the third-named author.

Theorem 3.6**.**

If $p=3n+1$ is prime, then $p$ is not a 3-representative.

In particular, if $n>2$, then there is an $i\in[n]$, $i>2$, such that $i^{3}\equiv 1\pmod{p}$ or $i^{3}\equiv 8\pmod{p}$.

Proof.

For $n=2$ we have that $1^{3}\equiv 2^{3}\pmod{7}$. Suppose now that $n>2$ and $p=3n+1$ is prime. Work in $\mathbb{Z}/p\mathbb{Z}$. Note that $x^{3}=1$ and $x\neq 1$ if and only if $x^{2}+x+1=0$ if and only if $4x^{2}+4x+4=0$, or $(2x+1)^{2}=-3$. Also, $x^{3}=8$ and $x\neq 2$ if and only if $x^{2}+2x+4=0$, or $(x+1)^{2}=-3$.

We claim that at least one of these two situations must happen for some $x\in[n]$. Note first that $-3$ is a quadratic residue modulo $p$:

[TABLE]

where $\left(\frac{q}{p}\right)$ 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 $x=y-1\in[n]$. Suppose otherwise. Note that either $y$ or $p-y$ is odd. Call it $z$, and note that $z\leq 2p/3$, and therefore $x=(z-1)/2$ is at most $(p-1)/3$, so it is in $[n]$. ∎

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 $p\equiv 1\pmod{4}$.

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 $1\leq x<y$ such that $p=x^{2}+y^{2}$. Note that if $p\geq k^{2}$, then $p^{2}/k^{2}=p\cdot\frac{p}{k^{2}}\geq p>y^{2}$, so $x<y\leq p/k=n+1/k$ and therefore in fact $x<y\in[n]$, but $x^{2}\equiv-y^{2}\pmod{p}$, so $x^{k}\equiv y^{k}\pmod{p}$. ∎

The bound on $p$ found in the theorem allows us to identify by a quick exhaustive search all the possible values of $p$ that are $k$-representatives, for any given value of $k$ that is a multiple of 4. Table 3.2 lists these values $p=4mn+1$ for all $k=4m\leq 100$.

We now proceed to the general case. The key observation is that if $p$ is prime and $G\leq(\mathbb{Z}/p\mathbb{Z})^{*}$ is nontrivial, then $\sum_{g\in G}g=0$. Indeed, let $S=\sum_{g\in G}g$ and let $h\in G$ be different from the identity. The map $g\mapsto hg$ is a permutation of $G$, and we have $S=\sum_{g\in G}hg=hS$.

Suppose now that $(n,k)\neq(1,1)$ and $p=kn+1$ is prime. The observation, applied to the case where $G$ is the group of $k^{\mathrm{th}}$ powers of nonzero elements of $(\mathbb{Z}/p\mathbb{Z})^{*}$, gives us that if $p$ is a $k$-representative, then $\sum_{i=1}^{n}i^{k}\equiv 0\pmod{p}$. 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 $c\neq 0$, such that $aP_{k}(x)$ has integer coefficients and

[TABLE]

For $x=n$ we have that $p=kn+1$ divides $P_{k}(n)$ (by the observation) and therefore $p\mid c$, and there are only finitely many possibilities for $p$, all of which lie among the 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 $\sum_{i=1}^{n}i^{2k}\equiv 0\pmod{p}$, and the polynomial $P_{2k}(n)$ is now $B_{2k+1}(n+1)$ for $B_{2k+1}(x)$ the $(2k+1)^{\mathrm{st}}$ 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 Harcos101010See https://mathoverflow.net/q/78270/.

Note also that for $k=1,2$ we have that

[TABLE]

so the argument above fails (as it should) since $kn+1$ is in both cases a factor of the corresponding polynomial.

Proof.

Let $B(t,x)=\frac{te^{tx}}{e^{t}-1}$. The Bernoulli polynomials $B_{m}(x)$ are defined as follows using the power series expansion in terms of $t$ of $B(t,x)$:

[TABLE]

It is well known that each $B_{m}(x)$ is a polynomial in $x$ of degree $m$ with rational coefficients, and

[TABLE]

for all positive integers $n$, see for instance [Was97, chapter 4].

Writing

[TABLE]

the numbers $b_{k}=B_{k}(0)$ are usually called the Bernoulli numbers; they satisfy $b_{2k+1}=0$ for all $k\geq 1$. As indicated above, it will be important for us to know all the rational linear factors of the polynomial $B_{m}(x)-B_{m}(0)$; when $m$ is odd this reduces to determining the rational linear factors of $B_{m}(x)$. The following result of K. Inkeri [Ink59, theorem 3] solves this problem.

Theorem 3.9** (Inkeri).**

The rational roots of a Bernoulli polynomial $B_{m}(x)$ can be only 0, $1/2$, and 1. Moreover, all these are roots when $m>1$ is odd.

Suppose $p=kn+1$ is a $k$-representative. We claim that

[TABLE]

To see this, notice that there are precisely $\frac{p-1}{d}=n/\gcd(2,n)$ incongruent $(2k)^{\mathrm{th}}$ 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 $1^{2k},\dots,n^{2k}$ are all distinct and are precisely all the nonzero $(2k)^{\mathrm{th}}$ powers. If $n$ is even, this means that each nonzero $(2k)^{\mathrm{th}}$ power appears exactly twice among these numbers. In either case, it follows that the sum is zero by the same argument as above.

Since

[TABLE]

it must be the case that $(kn+1)\mid B_{2k+1}(n+1)$. By Inkeri’s theorem 3.9, since $k>2$, the polynomial $kx+1$ is relatively prime to the polynomial $B_{2k+1}(x+1)$. Thus, there must be polynomials $u,v\in\mathbb{Q}[x]$ such that

[TABLE]

(In fact, $v$ is a constant.)

Multiplying this identity by an appropriate integer constant $L=L_{1}L_{2}$, it follows that there are polynomials $\check{u}=Lu$, $\check{B}_{2k+1}=L_{1}B_{2k+1}$, and $\check{v}=L_{2}v$, all in $\mathbb{Z}[x]$ such that

[TABLE]

Since $B_{2k+1}(n+1)\equiv 0\pmod{kn+1}$, evaluating the last displayed equation at $x=n$ gives us that $p=kn+1\mid L$. 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 $B_{2k+1}(x+1)$.

3.3. Examples

Let us illustrate theorem 3.8 with some examples, for which it suffices to consider $\sum_{i=1}^{n}i^{k}$ rather than the sum of $(2k)^{\mathrm{th}}$ powers.

• •

$k=3$.

Recall that

[TABLE]

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.

• •

$k=4$.

We have that

[TABLE]

If $4n+1$ is a 4-representative, then it must divide $3n^{2}+3n-1$. But

[TABLE]

so $4n+1$ must divide $25$. Hence $n=1$ and $p=5$ is the only 4-representative.

• •

$k=5$.

We have that

[TABLE]

If $5n+1$ is a 5-representative, then it must divide $2n^{2}+2n-1$. But

[TABLE]

so $5n+1$ must divide $33$. Hence $n=2$. Since $2^{5}=32\equiv-1\not\equiv 1\pmod{11}$, it follows that $p=11$ is indeed the only 5-representative.

• •

$k=6$.

We have that

[TABLE]

If $6n+1$ is a 6-representative, then it must divide $3n^{4}+6n^{3}-3n+1$. But

[TABLE]

so $6n+1$ must divide $637=7^{2}\cdot 13$, and $n=1$ or $n=2$.

Since $2^{6}=64\equiv-1\not\equiv 1\pmod{13}$, it follows that $p=7$ and $p=13$ are the only 6-representatives.

• •

$k=7$.

We have

[TABLE]

If $7n+1$ is a 7-representative, then it must divide $3n^{4}+6n^{3}-n^{2}-4n+2$. But

[TABLE]

so $7n+1$ must divide $6086=2\cdot 17\cdot 179$. However, since none of these prime factors is congruent to 1 modulo 7, it follows that there are no 7-representatives.

• •

$k=8$.

We have that

[TABLE]

If $8n+1$ is an 8-representative, it must divide $5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}+9n-3$. But

[TABLE]

so $8n+1$ must divide $1077555=3\cdot 5\cdot 71837$. However, since none of these prime factors is congruent to 1 modulo 8, it follows that there are no 8-representatives.

Although it ends up not making a significant difference here, using theorem 5.21, we only had to consider primes not exceeding 64.

• •

$k=9$.

We have that

[TABLE]

If $9n+1$ is a 9-representative, then it must divide $n^{2}+n-1$ or $2n^{4}+4n^{3}-n^{2}-3n+3$. The first case is impossible since

[TABLE]

so $9n+1$ would have to divide $89\not\equiv 1\pmod{9}$. Now, since

[TABLE]

then in the second case $9n+1$ must divide $21755=5\cdot 19\cdot 229$, so the only possibility is $n=2$. Since $2^{9}=512\equiv-1\not\equiv 1\pmod{19}$, it follows that $p=19$ is indeed the only 9-representative.

• •

$k=10$.

We have that

[TABLE]

If $10n+1$ is a $10$-representative, it must divide one of the two factors $n^{2}+n-1$ or $3n^{6}+9n^{5}+2n^{4}-11n^{3}+3n^{2}+10n-5$. But

[TABLE]

and

[TABLE]

so $10n+1$ must divide $109$ or $5958887=11^{5}\cdot 37$, so $n=1$. It follows that $p=11$ is the only $10$-representative.

3.4. Density of strong representatives

All satisfactory $n$-colorings with $n\leq 5$ admit strong representations: first, for each $n\leq 4$ 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 $6^{2}=36\equiv 25=5^{2}\pmod{11}$.

Similarly, $421=84\cdot 5+1$ is prime, and is a strong representative of order 5 for $c^{1}$ since

[TABLE]

Strong representatives are hardly unique. For instance, any prime is a strong representative of order 1. The case $n=2$ is more interesting.

Fact 3.10**.**

A prime $p$ is a strong representative of order 2 if and only if

[TABLE]

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 $2^{k}=2^{\frac{p-1}{2}}\not\equiv 1\pmod{p}$, which is equivalent to asserting that $2$ is not a square modulo $p$. The supplementary law tells us that

[TABLE]

which equals $-1$ if and only if $p\equiv\pm 3\pmod{8}$. By the prime number theorem for arithmetic progressions, half of all primes are of this form in the sense of natural density: simply note that $\phi(8)=4$, where $\phi(\cdot)$ is Euler’s totient function, and that, asymptotically, $1/4$ of all primes have the form $8k+a$ for any given $a\in[8]$ relatively prime with 8, see [Dav00, chapter 22]. ∎

Some natural questions occur at this point.

Question 3.11**.**

Let $n\in\mathbb{Z}^{+}$.

1. (1)

If $n$ admits a strong representative $p$, does it admit infinitely many? 2. (2)

If the answer to item (1) is positive, is the set of such primes of positive natural density among all primes? 3. (3)

Further, suppose $n$ admits a strong representative. For a satisfactory $n$-coloring $c$, 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 $n\leq 10$. 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.

Example 3.12**.**

The prime $p=701=140\cdot 5+1$ is a strong representative of order 5 for $c^{1}$. In effect,

[TABLE]

Similarly, one can check that $p=2311=462\cdot 5+1$ is a strong representative of order 5 for $c^{5}$.

Given a real $x$, denote by $\mathcal{C}^{1}(x)$ and $\mathcal{C}^{5}(x)$ the sets of primes $p\leq x$ that are strong representatives of order 5 for $c^{1}$ and $c^{5}$, respectively, and let $\mathcal{C}(x)=\mathcal{C}^{1}(x)\cup\mathcal{C}^{5}(x)$. Also, write $\mathcal{C}_{T}(x)$ for the set of all primes $p\leq x$ of the form $5n+1$. Table 3.3 provides some numerical evidence suggesting a positive answer to item (3) of question 3.11 for $n=5$.

For $x\geq 0$, denote by $\pi_{n}(x)$ 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 $n\leq 10$.

We now mention some remarks explaining that items (1) and (2) of question 3.11 admit a positive answer. First, we recall a well-known observation.

Lemma 3.13**.**

Suppose that $n$ and $p$ are prime. If not all numbers are $n^{\mathrm{th}}$ powers modulo $p$, then $p\equiv 1\pmod{n}$.

Proof.

Indeed, all numbers are $n^{\mathrm{th}}$ powers modulo $n$. For $p$ of the form $nk+a$ with $1<a<n$, let $\alpha\in[n]$ be the multiplicative inverse of $1-a$ modulo $n$, and note that $x^{\alpha(p-1)+1}\equiv x\pmod{p}$ for any $x$ and that $\alpha(p-1)+1\equiv 0\pmod{n}$. ∎

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 3.14**.**

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

[TABLE]

and this is equivalent to asserting that 2, 3, and $12=2^{3}\cdot 3/2$ are not cubes modulo $p$ (note that the fact that $p\equiv 1\pmod{3}$ follows from the assertion that 2 is not a cube modulo $p$, by lemma 3.13). This indicates that the key technical result needed to determine whether question 3.11 holds is Chebotarëv’s theorem, see [Lan94, Theorem VIII.10]111111Lang states the result in terms of Dirichlet density, but the same conclusion holds for natural density, see [Lan94, § XV.5].. We recall the theorem and some basic facts from algebraic number theory that should allow us to apply it in the case at hand.

Denote by $\mathbb{P}$ the set of integral primes. Recall that a set $A\subseteq\mathbb{P}$ has natural density $\delta$ in $\mathbb{P}$ if and only if $\lim_{n\to\infty}|A\cap[n]|/\pi(n)$ exists and equals $\delta$.

Theorem 3.15** (Chebotarëv).**

Let $L/k$ be a Galois extension with Galois group $G=\operatorname{Gal}(L/k)$, and let $C$ be a conjugacy class of $G$. The set of primes $\mathfrak{p}$ of $k$ that are unramified in $L$ and for which the Frobenius symbol $\sigma_{\mathfrak{p}}$ of $\mathfrak{p}$ 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

[TABLE]

has no roots modulo $p$. This suggests to consider $L=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\zeta_{3})$, the splitting field of $f$ over $k=\mathbb{Q}$, where $\zeta_{3}$ denotes a primitive cubic root of unity. Note that $[L:\mathbb{Q}]=18$. In fact, we can quickly check that $G$ is the generalized dihedral group for the elementary abelian group of order 9, that is,

[TABLE]

any automorphism in $G$ is determined by its action on $\sqrt[3]{2}$, $\sqrt[3]{3}$, and $\zeta_{3}$; the former two correspond to independent copies of $\mathbb{Z}/3\mathbb{Z}$, while the latter corresponds to the abelian group of order 2, which acts on $(\mathbb{Z}/3\mathbb{Z})^{2}$ via the inverse map. We can list $G$ as

[TABLE]

where $\pi_{a,b,c}$ is the field automorphism of $L$ that maps $\sqrt[3]{2}$ to $\zeta_{3}^{a}\sqrt[3]{2}$, $\sqrt[3]{3}$ to $\zeta_{3}^{b}\sqrt[3]{3}$, and $\zeta_{3}$ to $\zeta_{3}^{c}$.

We concentrate on those primes $p$ that do not ramify over $L$. For this, note that $L$ is the compositum of $\mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}(\sqrt[3]{3}),\mathbb{Q}(\zeta_{3})$, 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 3.15, namely the Frobenius, or Artin symbol. Suppose an integral prime $p$ is unramified in $L$ and $\mathfrak{q}$ is a prime of $L$ lying over $p$. The Frobenius is the unique $\sigma\in G$ such that

[TABLE]

for all $\alpha\in L$, where ${\mathrm{N}}_{L/\mathbb{Q}}(\cdot)$ is the norm. The choice of $\sigma=\sigma_{p}$ depends on $\mathfrak{q}$, 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 $g\in\mathbb{Q}[x]$ is monic and $K$ is its splitting field. Given an integral prime $p$, denote by $\mathbb{F}_{p}$ the field of $p$ elements. In $\mathbb{F}_{p}[x]$, $g$ factors into irreducible polynomials, say $g=g_{1}\cdots g_{m}$. Letting $n_{i}$ denote the degree of $g_{i}$ for $i\in[m]$, we can associate to $g$ the partition

[TABLE]

of $\deg(g)$. Letting $\mathcal{G}=\operatorname{Gal}(K/\mathbb{Q})$, we can identify $\mathcal{G}$ with a group of permutations of the roots of $g$ in $K$. Fixing an ordering of the roots, we can write each $\sigma\in\mathcal{G}$ as a product of disjoint cycles, say $\sigma=\tau_{1}\cdots\tau_{l}$. Letting $t_{i}$ denote the length of $\tau_{i}$, we can associate to $\sigma$ its cycle pattern

[TABLE]

Frobenius’s theorem states that the set of $p$ unramified in $K$ with associated partition $\Pi_{p}$ has natural density in the set of primes equal to the fraction of $\sigma\in\mathcal{G}$ whose associated cycle pattern $\Lambda_{\sigma}$ coincides with $\Pi_{p}$.

In the case under consideration, the condition on $2,3,12$ means that we are looking at those integral primes $p$ with $\Pi_{p}(f)=(3,3,3)$. Since each $\pi_{a,b,c}$ fixes (setwise) the sets $\{\zeta_{3}^{s}\sqrt[3]{r}:s\in[3]\}$ for $r=2,3,12$, what we need to count is those automorphisms that do not fix any of the $\zeta_{3}^{s}\sqrt[3]{r}$.

Very explicitly: fix $a,b\in\{-1,0,1\}$ and $c\in\{-1,1\}$. We see that $\pi_{a,b,c}$ maps each $\zeta_{3}^{j}\sqrt[3]{2}$ to $\zeta_{3}^{jc+a}\sqrt[3]{2}$, each $\zeta_{3}^{k}\sqrt[3]{3}$ to $\zeta_{3}^{kc+b}\sqrt[3]{3}$, and each $\zeta_{3}^{l}\sqrt[3]{12}$ to $\zeta_{3}^{lc+2a+b}\sqrt[3]{12}$, and we need that $jc+a\neq j$, $kc+b\neq k$ and $lc+2a+b\neq l$ 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 $a-j\neq j$, or $a\neq 2j$, but this is impossible to satisfy simultaneously for all $j$. Second, if $c=1$, what we need is that $a+j\neq j$, $k+b\neq k$ and $l+2a+b\neq l$, that is, $a$, $b$ and $2a+b$ should all be different from 0, and the last requirement is equivalent to asking that $a\neq b$. There are precisely two members of $G$ that satisfy all these conditions, namely $\pi_{1,-1,1}$ and $\pi_{-1,1,1}$.

This means that the set of strong representatives of order 3 has natural density $2/18=1/9$. ∎

Remark 3.16**.**

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 $a\in[18]$, and table 3.4 strongly suggests that, indeed, $r=1/9=0.\overline{1}$.

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 3.11 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 $i^{k}\not\equiv j^{k}\pmod{p}$ whenever $i<j$ are in $[n]$, that is, $(j/i)^{k}\not\equiv 1\pmod{p}$, which means that $j\cdot i^{n-1}$ is not an $n^{\mathrm{th}}$ power modulo $p$, or, what is the same, that the polynomial $x^{n}-j\cdot i^{n-1}$ 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 $\mathbb{Q}$, namely $L=\mathbb{Q}(\zeta_{n},\root n \of{j}:j\in[n]\cap\mathbb{P})$, the splitting field over $\mathbb{Q}$ of the polynomial

[TABLE]

where $\zeta_{n}$ denotes a primitive $n^{\mathrm{th}}$ 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 $\mathbb{F}_{p}[x]$ that are compatible with $f$ not having roots in $\mathbb{F}_{p}$.

As before, $L$ is the compositum of the fields $\mathbb{Q}(\sqrt[n]{j})$, $\mathbb{Q}(\zeta_{n})$, for $j\in[n]\cap\mathbb{P}$, 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 3.13 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

[TABLE]

where $(\mathbb{Z}/n\mathbb{Z})^{*}$ denotes the group of units modulo $n$, a group of order $\phi(n)=n-1$. 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 $\sqrt[n]{j}$ and $\zeta_{n}$. For instance, for $n=8$ and $\zeta_{8}=e^{2\pi\sqrt{-1}/8}$, we have $\zeta_{8}+\zeta_{8}^{-1}=\sqrt{2}$ or, if $n=2p$ where $p$ is prime and $p\equiv 1\pmod{4}$, then $\sqrt{p}\in\mathbb{Q}(\zeta_{n})$. 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 $\operatorname{Gal}(\mathbb{Q}(\zeta_{n})/\mathbb{Q})$, 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 3.17**.**

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 $(c(ij):i\leq j\in[n])$, 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 $[L:\mathbb{Q}]=n^{\pi(n)}\cdot\phi(n)$. 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 3.10. Table 3.5 shows the degree of the corresponding extension for each prime $n$ with $2\leq n\leq 10$ and the density that table 3.4 suggests accordingly. For $n=4$ no additional relations occur either. In that case, the extension has degree $4^{2}\cdot 2=32$, and the expected density is $1/16=0.0625$.121212The discussion here incorporates suggestions of Felipe Voloch to the first-named author at https://mathoverflow.net/q/141993 and through private communication. Thanks are also due to David E Speyer.

Remark 3.18**.**

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.

3.5. Asymptotics of coincidences

Fix $k>2$. For primes $p=kn+1$ sufficiently large, theorem 3.8 shows the existence of coincidences

[TABLE]

with $1\leq a<b\leq n$. 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. Elkies131313See https://mathoverflow.net/q/78270/, and what follows is closely based on his argument.

Theorem 3.19** (Elkies).**

For $k>2$, the number of coincidences $a^{k}\equiv b^{k}\pmod{p}$ for $p$ of the form $kn+1$ and sufficiently large, and distinct $a,b\in[n]$ is

[TABLE]

where

[TABLE]

and $\epsilon(k)=1/\phi(k)$.

Proof.

First, for $a,b$ nonzero and distinct modulo $p$, that

[TABLE]

is equivalent to saying that $b\equiv ma\pmod{p}$ where $m\neq 1$ is a $k^{\mathrm{th}}$ root of unity: $m^{k}\equiv 1\pmod{p}$. Since we are only interested in the case where $a,b\in[n]$, for $k$ even we further exclude $m=-1$. Fix $m$, and consider the nonzero vectors $(a,b)$ in $\mathbb{Z}^{2}$ defined by the relation $b\equiv ma\pmod{p}$, $a,b\in[n]$. Note that for any such vector, $p\mid a^{k}-b^{k}$, and the latter factors into homogeneous polynomials in $a,b$ of degree at most $\phi(k)$, none of which is zero, and therefore the length of the vector is $\Omega(p^{\epsilon(k)})$.

This means that the solutions to the equation $b\equiv ma\pmod{p}$ with $a,b\in[n]$ are the lattice points in the square with sides parallel to the axis of side length $n\approx p/k$ 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

[TABLE]

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 $(b,a)$). ∎

The argument can be strengthened to estimate for $k,n,p$ as before the proportion of distinct $k^{\mathrm{th}}$ powers of members of $n$. A quick computation verifies that the fractions

[TABLE]

stay rather close to $2/3$ for $k=3$, and to $84/125$ for $k=5$. For instance, for $k=3$, $n=387,642$, and $p=1,162,927$, the fraction is

[TABLE]

while for $k=5$, $n=35,804$ and $p=179,021$, the fraction is

[TABLE]

The result, also due to Elkies, shows that these values are to be expected.

Theorem 3.20** (Elkies).**

For $k>2$ and $p$ of the form $kn+1$ and sufficiently large, the fraction $|\{(i^{k}\bmod p):i\in[n]\}|/n$ of distinct $k^{\mathrm{th}}$ powers of members of $[n]$ is asymptotic to $1-((k-1)^{k}+1)/k^{k}$.

In particular, for $k=3$ the fraction approaches $1-\frac{2^{3}+1}{3^{3}}=2/3$ and for $k=5$ it approaches $1-\frac{4^{5}+1}{5^{5}}=84/125$, as expected, and, as $k\to\infty$, the proportion of $k^{\mathrm{th}}$ powers with small $k^{\mathrm{th}}$ roots approaches $1-(1/e)$.

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 $k^{\mathrm{th}}$ 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 $k^{\mathrm{th}}$ power is asymptotic to

[TABLE]

while there are no such subsets with $j=k$ because the sum of all $k$ solutions of $a^{k}\equiv c\pmod{p}$ vanishes (for any $c$). By an inclusion-exclusion argument one then obtains the estimate indicated in theorem 3.20.

4. Multiplicative colorings

4.1. Multiplicativity

As shown in the previous section, any $n$-satisfactory coloring for $n\leq 5$ 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

[TABLE]

The group $G$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. The map $h\!:K_{n}\to G$ given by

[TABLE]

satisfies

[TABLE]

for any $a,b\in K_{n}$, where $ab$ is the usual product of $a$ and $b$ and $h(a)\cdot h(b)$ is the product in $G$. We generalize this setting in the following definition.

Definition 4.1**.**

A satisfactory coloring $c$ of $K_{n}$ is multiplicative if and only if there exists a group $(G,\cdot)$ of order $n$ and a bijection $\varphi\!:[n]\to G$ such that, thinking of $c$ as a map $c\!:K_{n}\to[n]$ with $c(i)=i$ for all $i\in[n]$, and letting $h=\varphi\circ c$, we have that

[TABLE]

for all $a,b\in K_{n}$. In this case, we say that $c$ is a $G$-coloring.

Multiplicative colorings of $\mathbb{Z}^{+}$ are defined the same way, only requiring that the domain of $c$ be $\mathbb{Z}^{+}$ and that equation 4.1 holds for all positive integers.

The usefulness of the notion is stated explicitly in theorem 4.10 below, the point is that to describe a multiplicative coloring it is enough to describe what we call a partial isomorphism, see § 4.2, which reduces the problem of searching for a multiplicative coloring to a finite question.

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 $G_{1}\cong G_{2}$.

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.

Definition 4.3**.**

If $(G,+)$ is an abelian group (of order $n$) and the map $h\!:K_{n}\to G$ satisfies that $h(ab)=h(a)+h(b)$ for any $a,b\in K_{n}$, we say that $h$ is multiplicative.

Corollary 4.4**.**

For any $n$, there are only finitely many multiplicative colorings of $K_{n}$.

Proof.

Suppose $c$ is multiplicative as witnessed by $(G,+),\varphi$. Let $h=\varphi\circ c$, where as before, $c$ is interpreted as a map $c\!:K_{n}\to[n]$ with $c(i)=i$ for $i\in[n]$, so $h(ab)=h(a)+h(b)$ for all $a,b\in K_{n}$. Note that this induces a group structure $\oplus$ on $[n]$ isomorphic to $G$ because $c$ is the identity on $[n]$, so if $a\in[n]$, then

[TABLE]

and we are setting $a\oplus b=d$ for $a,b,d\in[n]$ if and only if $\varphi(d)=\varphi(a)+\varphi(b)$. By identifying $(G,+)$ with $([n],\oplus)$, it follows that we may assume that $\varphi$ is the identity so $h=c$. But now we see that $([n],\oplus)$ completely determines $c$. In effect, if $p_{1}<\dots<p_{\pi(n)}$ are the primes less than or equal to $n$ and $s=\pi(n)$, then the multiplicity requirement gives us

[TABLE]

where $\alpha_{i}c(p_{i})$ is the result of adding $c(p_{i})$ to itself $\alpha_{i}$ times in $([n],\oplus)$.

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 $(G,\cdot)$, the corresponding tiling of $\mathbb{O}_{n}$ 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 $x\in G$. In turn, this also gives periodicity of the tiling by $T_{n}$, see § 4.3.

In what follows, given an abelian group $(G,\oplus)$, we will denote $\alpha$-fold sums of the form \underbrace{g\oplus\cdots\oplus g}_{\text{\alpha times}} by $g^{\oplus\alpha}$ rather than $\alpha g$ as above. We extend the notation to include negative values of $\alpha$ (including $\alpha=-1$).

Remark 4.6**.**

Note that not every abelian group structure on $[n]$ gives rise to a multiplicative coloring. For example, if $n=4$, then $\oplus$ is given by

[TABLE]

and, in particular, $([4],\oplus)$ is isomorphic to $\mathbb{Z}/4\mathbb{Z}$ and not to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

The case when $G\cong\mathbb{Z}/n\mathbb{Z}$, as in the case of a strong representation, deserves special attention.

Question 4.7**.**

Does every $\mathbb{Z}/n\mathbb{Z}$-coloring admit a strong representation?

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, see the analysis of multiplicative 8-colorings in § 4.5 and in particular the coloring described in table 4.12. 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 and § 5.3 for dramatic examples. Similarly, not every multiplicative coloring is a $\mathbb{Z}/n\mathbb{Z}$-coloring. Examples are presented in § 4.5, in particular see table 4.20 for $a=1$. 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 except for § 5.2 and § 5.3.

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 Delanoy141414See https://mathoverflow.net/q/26358/ considered the case $G=\mathbb{Z}/n\mathbb{Z}$. Though not identical, it is closely related to the concept of Freiman homomorphism in additive combinatorics, see [TV06, definition 5.21].

Definition 4.8**.**

Let $(G,+)$ be an abelian group of order $n$. A map $h\!:[n]\to G$ is a partial $G$-isomorphism if and only if $h$ is a bijection and, whenever $a,b\in[n]$, if $ab\in[n]$, then $h(ab)=h(a)+h(b)$. If $G=\mathbb{Z}/n\mathbb{Z}$, we simply call $h$ a partial isomorphism.

Remark 4.9**.**

We require $G$ to be abelian as our goal is to relate partial $G$-isomorphisms to satisfactory colorings. This is done via an explicit construction in theorem 4.10 below, and although the coloring we describe is perhaps the “natural” one, our formula requires that $G$ is abelian, and we do not see a way to proceed otherwise. 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 $\oplus$ on $[n]$ such that

[TABLE]

and $\oplus$ extends the partial graph of multiplication on $[n]$. Our use of the term isomorphism here is perhaps further justified by noting that if $h$ is a partial $G$-isomorphism, then $h(1)=h(1\cdot 1)=h(1\oplus 1)=h(1)+h(1)$, and it follows that $h(1)=0_{G}$.

Theorem 4.10**.**

If $h\!:[n]\to G$ is a partial $G$-isomorphism, then $h$ can be uniquely extended to a multiplicative map $\hat{h}\!:K_{n}\to G$. Moreover, $h^{-1}\circ\hat{h}\!:K_{n}\to[n]$ is a $G$-coloring of $K_{n}$.

Proof.

Let $h\!:[n]\to G$ be a partial $G$-isomorphism. Letting $p_{1},\dots,p_{s}$ be the primes less than or equal to $n$, a map $\hat{h}\!:K_{n}\to G$ extends $h$ and is multiplicative if and only if for any $a_{1},\dots,a_{s}\in\mathbb{N}$, we have

[TABLE]

This proves the existence and uniqueness of the extension $\hat{h}$.

Moreover, if $1\leq i<j\leq n$ and $a\in K_{n}$, then

[TABLE]

because $\hat{h}\upharpoonright[n]=h$ is a bijection.

Letting $c=h^{-1}\circ\hat{h}$, this gives us that $c\!:K_{n}\to[n]$ is a $G$-coloring. ∎

Remark 4.11**.**

Note the similarity between this argument and the proof of corollary 4.4.

Obviously, if $\hat{h}\!:K_{n}\to G$ is multiplicative and $h=\hat{h}\upharpoonright[n]$ is a bijection, then $h$ is a partial $G$-isomorphism. Therefore, if $c\!:K_{n}\to[n]$ is a $G$-coloring as witnessed by the bijection $\varphi\!:[n]\to G$, then $\varphi$ is a partial $G$-isomorphism as, by definition, $h=\varphi\circ c$ is multiplicative, and $\varphi=h\upharpoonright[n]$.

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:

Definition 4.12**.**

Given an abelian group $(G,+)$ of order $n$, we say that an abelian group structure on $[n]$, $([n],\oplus)$, is a $G$-satisfactory group if and only if

[TABLE]

and $a\oplus b=ab$ whenever $a,b,ab\in[n]$.

We say that the $G$-coloring resulting from extending $\oplus$ as in theorem 4.10 is associated to $([n],\oplus)$.

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 $G$.

For example, if $h_{1}\!:[6]\to\mathbb{Z}/6\mathbb{Z}$ and $h_{2}\!:[6]\to\mathbb{Z}/6\mathbb{Z}$ are the maps

[TABLE]

and

[TABLE]

then both give rise to the $\mathbb{Z}/6\mathbb{Z}$-coloring strongly represented by $7=1\cdot 6+1$, and this coloring is associated to the $G$-satisfactory group shown in table 4.1.

In § 4.5, we use systematically the notation of $G$-satisfactory groups to identify all multiplicative colorings with at most eight colors.

Remark 4.13**.**

We are now in a position to explain how $\mathbb{Z}/n\mathbb{Z}$-colorings or, equivalently, partial isomorphisms are closely related to strong representations. In fact, we can prove that any $\mathbb{Z}/n\mathbb{Z}$-coloring admits a “weak” representation. Let $h\!:[n]\to\mathbb{Z}/n\mathbb{Z}$ be a partial isomorphism. As before, let $p_{1},\dots,p_{s}$ be the primes less than or equal to $n$. Extend $h$ to a map from $K_{n}$ to $\mathbb{Z}/n\mathbb{Z}$ 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 $(\mathbb{Z}/P\mathbb{Z})^{*}$. In other words, the powers $g^{ki}$ are precisely the $k^{\mathrm{th}}$ power residues modulo $P$. Invoking again Dirichlet’s theorem, for each $p_{i}$ we can find a prime $q_{i}$ such that

[TABLE]

Now for $x\in K_{n}$ define $d\!:K_{n}\to(\mathbb{Z}/P\mathbb{Z})^{*}$ by $d(x)=g^{kh(x)}$.

If $x=\prod_{i=1}^{s}p_{i}^{a_{i}}$, then $h(x)=\sum_{i}a_{i}h(p_{i})$ and

[TABLE]

where of course the products are computed modulo $P$.

The point is that if $i,j\in[n]$, then $d(i)\neq d(j)$ because $h(i)\neq h(j)$, $h$ being a bijection. If $0\leq h(i)<h(j)<n$, then $0\leq kh(i)<kh(j)<kn$, and $g^{kh(i)}\neq g^{kh(j)}$, since $g$ is a primitive root. It follows that $d(ix)=d(i)d(x)\neq d(j)d(x)=d(jx)$, and $d$ defines a $\mathbb{Z}/n\mathbb{Z}$-coloring.

Note how close the coloring given by equation (4.3) 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 $n\leq 31$ are given in table 2.1. In appendix B of the second-named author’s master’s thesis151515See http://scholarworks.boisestate.edu/td/231/, this is extended to all $n\leq 54$. The authors of [FP90] have verified their existence for all $n<195$.

Given $n$, define $M$ and $M_{K_{n}}$ as the sets of multiplicative colorings of $\mathbb{Z}^{+}$ and of $K_{n}$, respectively. In corollary 4.4 we showed that $M_{K_{n}}$ 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).

Theorem 4.14**.**

If $n>1$ and $M_{K_{n}}\neq\emptyset$, then $|M|=\mathfrak{c}$.

Proof.

As in corollary 2.7, it is enough to show that $n^{\aleph_{0}}\leq|M|$. Let $c\!:K_{n}\to[n]$ be a multiplicative coloring associated to the $G$-satisfactory group $([n],\oplus)$. To each prime $p$ assign a number $a_{p}\in[n]$ with the only restriction that $a_{p}=p$ if $p\in n$. Now define $c^{\prime}\!:\mathbb{Z}^{+}\to[n]$ as follows: if $m\in\mathbb{Z}^{+}$, let $\prod_{i}p_{i}^{b_{i}}$ be its prime factorization, and set

[TABLE]

It is immediate that any $c^{\prime}$ defined this way is multiplicative and extends $c$, and that different sequences $(a_{p}:p$ prime$)$ give rise to different $c^{\prime}$, and therefore we have associated $n^{\aleph_{0}}$ colorings in $M$ to each $c\in M_{K_{n}}$. ∎

4.3. 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 $k\in K_{n}$, then $c_{k}$ is the coloring where two numbers $m,m^{\prime}\in K_{n}$ receive the same color precisely when $c(km)=c(km^{\prime})$.

Definition 4.15**.**

A coloring $c$ of $K_{n}$ is translation invariant if and only if $c_{k}=c$ for all $k\in K_{n}$.

For any $k\in K_{n}$, we can naturally identify $\mathbb{O}_{n}$ and $\mathbb{O}_{n}+t(k)$. This induces a coloring of $\mathbb{O}_{n}$ from one of $\mathbb{O}_{n}+t(k)$. If we start with $c$, the resulting coloring is precisely $c_{k}$. That $c$ is translation invariant means that, for any $k\in K_{n}$, 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 figure 4.1, showing the four color classes of the 4-satisfactory coloring depicted in figure 2.3. In the figure, we have also indicated the axes of an orthant $\mathbb{O}_{4}+t(k)$, and the reader can see that the four color classes of the induced coloring of this orthant look precisely like the original ones.

Fact 4.16**.**

Let $c$ be a translation invariant finite coloring of $K_{n}$. For any $k\in K_{n}$ there is a least positive integer $o(k)$ such that $c(k^{o(k)}m)=c(m)$ for all $m\in K_{n}$.

If $N$ is the number of colors used by $c$, then $o(k)\mid N$.

Proof.

We simply use the standard argument for Lagrange’s theorem: the list of colors $c(1),c(k),c(k^{2}),\dots$ must eventually have repetitions. If $i<j$ and

[TABLE]

then, since $c_{k^{i}}=c$, we see that $c(1)=c(k^{j-i})$, and it follows that there is a least $o(k)>0$ with $c(k^{o(k)})=c(1)$. By translation invariance, in fact $c(k^{o(k)}m)=c(m)$ for all $m\in K_{n}$.

For any $m\in K_{n}$, the colors $c(m),c(km),\dots,c(k^{o(k)-1}m)$ are all distinct, since a coincidence $c(k^{a}m)=c(k^{b}m)$ with $a<b$ implies $c(1)=c(k^{b-a})$ by translation invariance. Let $L_{m}=\{c(m),c(km),\dots,c(k^{o(k)-1}m)\}$, and note that any two such sets $L_{m},L_{m^{\prime}}$ are either disjoint or coincide, since if $c(mk^{a})=c(m^{\prime}k^{b})$, then, letting $l=b-a$ if $b\geq a$ or $b+o(k)-a$ otherwise, we see that $c(m)=c(m^{\prime}k^{l})$, from which $L_{m}=L_{m^{\prime}}$ follows. This shows that the sets $L_{m}$ partition the set of colors into classes of the same size, which completes the proof. ∎

Arguably, a coloring $c$ of $K_{n}$ deserves to be called periodic if there is a $k\in K_{n}$ larger than 1 such that $c(km)=c(m)$ for all $m\in K_{n}$. Our actual definition is somewhat more stringent.

Definition 4.17**.**

Let $c$ be a coloring of $K_{n}$, $k\in K_{n}$ be larger than 1, and $l\in\mathbb{Z}^{+}$. 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 $m\in K_{n}$.

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 $o(k)$, $k\in K_{n}$, 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 of $\hat{K}_{n}$; moreover, the extension $c$ is translation invariant in the strong sense that $c_{k}=c$ for all $k\in\hat{K}_{n}$. 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 $m,m^{\prime}\in K_{n}$, let

[TABLE]

This is well-defined, in the sense that if $m^{\prime}/m=s^{\prime}/s$ for $s,s^{\prime}\in K_{n}$, then

[TABLE]

because

[TABLE]

and therefore $c(m^{\prime}m^{o(m)-1})=c(s^{\prime}s^{o(s)-1})$, by translation invariance.

Similarly, if $a,m_{1},m_{2}\in K_{n}$, then

[TABLE]

because

[TABLE]

We argue that the extension is translation invariant in the strong sense indicated above: let $k\in\hat{K}_{n}$. We must show that $c_{k}=c$, that is, that if $a,b\in\hat{K}_{n}$, then $c_{k}(a)=c_{k}(b)$ if and only if $c(a)=c(b)$. For this, let $m_{1},\dots,m_{6}\in K_{n}$ be such that $a=m_{1}/m_{2}$, $b=m_{3}/m_{4}$ and $k=m_{5}/m_{6}$, and note that $c_{k}(a)=c_{k}(b)$ if and only if

[TABLE]

or, equivalently,

[TABLE]

which, in turn, is equivalent to

[TABLE]

that is, to $c(a)=c(b)$, as wanted.

Suppose now that $c$ is in addition $n$-satisfactoty. To see that the extension is again $n$-satisfactory, note that if $i,j\in[n]$ and $c(im^{\prime}/m)=c(jm^{\prime}/m)$, then $c(im^{\prime}m^{o(s)-1})=c(jm^{\prime}m^{o(s)-1})$, and it follows that $i=j$.

To see that the extension we defined is the only possible translation invariant extension using the same colors, suppose $c^{\prime}$ is such an extension of $c$, and that $a,b,i\in K_{n}$ are such that $c^{\prime}(a/b)=c^{\prime}(i)=c(i)$. By translation invariance of $c^{\prime}$, this is equivalent to asserting that $c^{\prime}(bi)=c^{\prime}(a)$, that is $c(bi)=c(a)=c(ab^{o(b)})$ which, again by translation invariance, is in turn equivalent to $c^{\prime}(i)=c(i)=c(ab^{o(b)-1})$. 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 to $\hat{K}_{n}$ 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 $\mathbb{O}_{n}$ given by the coordinate axes. In the absence of such a frame of reference, the color classes are entirely indistinguishable from one another.

Fact 4.20**.**

The canonical extension of a multiplicative coloring is again multiplicative.

Proof.

Suppose $c$ is the multiplicative $n$-satisfactory coloring determined by the abelian group $([n],\oplus)$. For $a,b\in\hat{K}_{n}$, let $m_{1},m_{2},m_{3},m_{4}\in K_{n}$ be such that $a=m_{1}/m_{2}$ and $b=m_{3}/m_{4}$. We have that

[TABLE]

by equation (4.4). Since $c$ is multiplicative on $K_{n}$, the last expression equals $c(m_{1}m_{2}^{o(m_{2})-1})\oplus c(m_{3}m_{4}^{o(m_{4})-1})=c(a)\oplus c(b)$, as wanted. ∎

Remark 4.21**.**

With notation as in the proof of fact 4.20, let $k\in\hat{K}_{n}$, and write its prime factorization as

[TABLE]

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 $\alpha_{i}$, 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 $i\in[n]$, we have that

[TABLE]

that is, equation (4.2) still holds, regardless of the sign of the exponents.

For $l$ finite, a tiling of $\mathbb{Z}^{l}$ by $T$, say $T+B$ where the sum is direct, is periodic if and only if there is a finite index subgroup $\Lambda$ of $\mathbb{Z}^{l}$ such that $B+\Lambda=B$.

In § 2.3 we defined the natural bijective map $t\!:K_{n}\to\mathbb{O}_{n}$ associating to a point $k$ in $K_{n}$ the point in $\mathbb{O}_{n}$ whose coordinates are the exponents of the prime factorization of $k$. The same definition gives us an extension of this map to $\hat{K}_{n}$ that we again denote by $t$ and is now a bijection with $\mathbb{Z}^{\pi(n)}$. 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 $\mathbb{Z}^{\pi(n)}$.

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 $\mathbb{Z}^{\pi(n)}$ by $T_{n}$ is periodic. In particular, this holds for multiplicative colorings.

Proof.

Let $p_{1}<\dots<p_{\pi(n)}$ be the primes less than or equal to $n$. For $i\in[\pi(n)]$ let $\mathbf{x}_{i}=t(p_{i}^{o(p_{i})})$, and let $\Lambda=\langle x_{i}:i\in[\pi(n)]\rangle$, so that $\Lambda$ has finite index in $\mathbb{Z}^{\pi(n)}$. We claim that $B+\Lambda=B$. The point is that if $k\in t^{-1}(\Lambda)$, then $c(k)=c(1)$ and if $k^{\prime}\in t^{-1}(B)$, then $c(kk^{\prime})=c(k^{\prime})$, so that $t(k)+t(k^{\prime})\in B$, that is, $B+\Lambda\subseteq B$. But, since $\mathbf{0}\in\Lambda$, clearly $B\subseteq B+\Lambda$. ∎

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 $(G,\oplus)$, let $B$ be the image under $t$ of a color class of the canonical extension of $c$, and let $\Lambda$ be the image under $t$ of the color class of 1. We have that $\Lambda$ is a finite index subgroup of $\mathbb{Z}^{\pi(n)}$ and that $B+\Lambda=\Lambda$. In fact, $\mathbb{Z}^{\pi(n)}/\Lambda\cong G$ and the images of the color classes are the cosets of $\Lambda$ in $\mathbb{Z}^{\pi(n)}$.

Note that, in particular, $\mathbb{Z}^{\pi(n)}=T_{n}+\Lambda$, and the sum is direct.

Proof.

That $\Lambda$ is a group is clear from the fact that $c$ is multiplicative: first,

[TABLE]

if $c(k)=c(1)$, then $c(k^{-1})=c(k)^{\oplus(-1)}=c(1)$, so $\Lambda$ is closed under inverses; and since $c(k_{1}k_{2})=c(k_{1})\oplus c(k_{2})$, then $\Lambda$ is closed under products. That it has finite index in $\mathbb{Z}^{\pi(n)}$ follows from the fact that it contains the group we denoted by $\Lambda$ in the proof of lemma 4.22. That $B+\Lambda=B$ is exactly as in that proof.

In fact, let $B$ be the image under $t$ of the color class of $i\in[n]$, and note that $T_{n}\cap B=\{t(i)\}$. We claim that $B=t(i)+\Lambda$, from which it follows that $\mathbb{Z}^{\pi(n)}/\Lambda\cong G$ and that the images of the color classes are the cosets of $\Lambda$. Clearly $B\supseteq t(i)+\Lambda$, and if $t(k)\in B$, then

[TABLE]

so that

[TABLE]

This completes the proof. ∎

Figure 4.2 illustrates the group $\Lambda$ for the unique 4-satisfactory coloring.

Remark 4.24**.**

We can easily address question 2.16 for multiplicative colorings via lemma 4.23. Suppose $c$ is a multiplicative $n$-satisfactory coloring. Let $\Lambda$ be the image under $t$ of the color class of 1, so any color class $B$ is a coset $t(i_{0})+\Lambda$. The coloring $c^{\prime}$ derived from the tiling $T_{n}+B$ has color classes $t(i)+(t(i_{0})+\Lambda)$ for $i\in[n]$, but these are precisely the color classes of $c$, as $(c(ii_{0}):i\in[n])$ is just a permutation of $[n]$.

We now establish the main result of this subsection.

Theorem 4.25**.**

A translation invariant $n$-satisfactory coloring is multiplicative.

Proof.

Let $c$ be translation invariant and $n$-satisfactory. Using the convention that $c(i)=i$ for $i\in[n]$, we need to verify that the map $\oplus\!:[n]\times[n]\to[n]$ given by $i\oplus j=c(ij)$ defines a group operation on $[n]$. Clearly, $\oplus$ is commutative and, for any $j$, the sequence $(i\oplus j:i\in[n])$ is a permutation of $[n]$. The issue is whether $\oplus$ is associative, that is, whether for $i,j,k\in[n]$ we have

[TABLE]

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 $a,b\in K_{n}$, $c(ab)=c(a)\oplus c(b)$. For this, note that $c(a)=c(c(a))$ and $c(b)=c(c(b))$ so, by translation invariance,

[TABLE]

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 $T\subseteq\mathbb{Z}^{l}$, if $T$ tiles $\mathbb{Z}^{l}$, then it also does so periodically. With $l=\pi(n)$ 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 5.6 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

[TABLE]

The group $G$ in the theorem is isomorphic to $\mathbb{Z}/(p^{2}-p)\mathbb{Z}$, but visualizing it as indicated was essential to identifying this coloring.

Proof.

For $i\in[p^{2}-p]$ such that $p\nmid i$, let $g_{i}\equiv i\pmod{p^{2}}$, and for $j\in[p-1]$, let $g_{pj}\equiv p^{2}-j\pmod{p^{2}}$, so that $g_{1},g_{2},\dots,g_{p^{2}-p}$ lists the elements of a reduced residue system modulo $p^{2}$.

We claim that $g_{ij}=g_{i}g_{j}$, whenever $ij\leq n$. If $p\nmid i,j$, then the statement is clear. If $p\nmid i$, but $j=pj^{\prime}$, then

[TABLE]

Finally, if $p\mid i,j$, then $ij$ is too large. For $G=(\mathbb{Z}/p^{2}\mathbb{Z})^{*}$, this shows that the map $g\!:[p^{2}-p]\to G$ given by $i\mapsto g_{i}$ 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 $n\leq 8$

In this subsection we list all $G$-satisfactory groups for $6\leq|G|\leq 8$, thus determining all multiplicative colorings with at most eight colors. The construction is relatively simple in each case and proceeds by explicitly exhibiting the multiplication table of all possible $G$-satisfactory groups $([n],\oplus)$ for $6\leq n\leq 8$. For instance, we use the fact that if $([n],\oplus)$ is $G$-satisfactory for some $G$, then

[TABLE]

coincides with the set of odd integers in $[n]$, and repeatedly make use of the fact that $\oplus$ must be associative and commutative, that each row and column of the multiplication table must be a permutation of $[n]$, and that the order in $([n],\oplus)$ of any element $a$ must divide $n$. In a sense, identifying these colorings is akin to solving a Sudoku puzzle. The complete analysis is perhaps a bit too tedious to present in full. We provide essentially all details for $n=8$, the most involved case, and sketch the cases $n=6,7$; these sketches can be fleshed out along the same lines as for $n=8$ but more straightforwardly.

• •

To begin with, the only abelian $G$ of size 6 is $\mathbb{Z}/6\mathbb{Z}$, and there are precisely five $\mathbb{Z}/6\mathbb{Z}$-colorings of $K_{6}$.

To see this, begin by building the partial multiplication table of a putative $G$-satisfactory group $([6],\oplus)$. The only entries we know originally are those of the form $a\oplus b$ with $ab\leq 6$ (in which case $a\oplus b=ab$). Note that $\{2\oplus a:a>3\}=\{1,3,5\}$. We consider three cases, according to the value of $a$ with $2\oplus a=1$.

If $2\oplus 5=1$, then $4\oplus 5=2$ and all values of the table are completely determined from the elementary observations two paragraphs above; the result is table 4.5 below.

If $2\oplus 6=1$, similarly all values are completely determined; the result is table 4.2.

If $2\oplus 4=1$, then $2\oplus 5=3$ (it cannot be 5 since already $1\oplus 5=5$) and $2\oplus 6=5$. Thus, $4\oplus 2=2\oplus 4=1$, $4\oplus 3=2\oplus 6=5$, $4\oplus 4=2\oplus(2\oplus 4)=2\oplus 1=2$, and $4\oplus 6\neq 6$, so $4\oplus 5=6$ and $4\oplus 6=3$. We cannot complete the table just yet, but we do as soon as we choose the value of $3\oplus 3$, which must be one of 1, 2, or 4; the results are shown in tables 4.3, 4.1 and 4.4, respectively.

The resulting five colorings are all strongly representable, namely, by $7=1\cdot 6+1$, $13=2\cdot 6+1$, $103=17\cdot 6+1$, $487=81\cdot 6+1$, and $547=91\cdot 6+1$, and the tables are listed in the increasing order of these primes. Note that the construction given in the proof of theorem 4.26 for $p=3$ results in table 4.5.

• •

The only abelian $G$ of size 7 is $\mathbb{Z}/7\mathbb{Z}$, and there are precisely six $\mathbb{Z}/7\mathbb{Z}$-colorings of $K_{7}$.

We argue as before, by starting with the partial multiplication table of a putative $G$-satisfactory group $([7],\oplus)$ and analyzing cases. Note that $2\oplus 4\neq 1$, since 1 is the identity of $([7],\oplus)$ and every member of $\mathbb{Z}/7\mathbb{Z}$ other than the identity has order 7. Also, $2\oplus 6=4\oplus 3\neq 3$. Similarly, $2\oplus a\neq a$ for $a=5,7$. Hence, the sequence $(2\oplus a:4\leq a\leq 7)$ is a permutation of the numbers $1,3,5,7$ that does not begin with 1, does not have 5 as its second element, does not have 3 as its third element, and does not end with 7.

Although nine permutations satisfy these requirements, $(2\oplus a:4\leq a\leq 7)$ cannot be any of the sequences $(3,7,1,5)$, $(5,1,7,3)$, and $(7,3,5,1)$: the first would imply that 2 has order 5, while the other two would imply that it has order 4. For instance, if $(2\oplus a:4\leq a\leq 7)=(3,7,1,5)$, then $4\oplus 2=3$ and $6\oplus 2=1$, so $4\oplus 4=6$ and $6\oplus 2=2^{\oplus 5}$.

The remaining six sequences lead indeed to the multiplication table of a $\mathbb{Z}/7\mathbb{Z}$-satisfactory group.

They are all strongly representable, by $659=94\cdot 7+1$, $1429=204\cdot 7+1$, $2087=298\cdot 7+1$, $3557=508\cdot 7+1$, $17431=2490\cdot 7+1$, and $21911=3130\cdot 7+1$, and are described by tables 4.6, 4.7, 4.8, 4.9, 4.10, and 4.11, respectively.

• •

There are three abelian groups of order 8, namely $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$, and $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. There are no $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$-satisfactory groups, for the simple reason that $2\oplus 2=4\neq 1$. There are precisely four $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$-colorings of $K_{8}$, and four $\mathbb{Z}/8\mathbb{Z}$-colorings admitting strong representatives. There are also six additional $\mathbb{Z}/8\mathbb{Z}$-colorings that do not admit strong representatives.

We provide some details. Consider first the sequence $(2\oplus a:5\leq a\leq 8)$, noting that it must be a permutation of the numbers $1,3,5,7$ that does not begin with 5 and does not have 7 as a third element. Moreover, 3 cannot be the second element, since $2\oplus 6=3$ would imply that $4\oplus 6=6$. This means that the sequence must be one of the following: $(1,5,3,7)$, $(1,7,3,5)$, $(1,7,5,3)$, $(3,1,5,7)$, $(3,5,1,7)$, $(3,7,1,5)$, $(3,7,5,1)$, $(7,1,3,5)$, $(7,1,5,3)$, $(7,5,1,3)$, or $(7,5,3,1)$.

However, $(1,7,3,5)$, $(3,5,1,7)$, and $(7,1,5,3)$ are not possible.

1. a.

Consider $(1,7,3,5)$: if $2\oplus 6=7$ and $2\oplus 7=3$, then $8\oplus 3=3$. 2. b.

Consider $(3,5,1,7)$: if $2\oplus 5=3$ and $2\oplus 6=5$, then again $8\oplus 3=3$. 3. c.

Consider $(7,1,5,3)$: if $2\oplus 6=1$ and $2\oplus 8=3=4\oplus 4$, then $4^{\oplus 3}=3\oplus 4=1$, against Lagrange’s theorem.

Of the remaining eight sequences, six of them determine $\oplus$ uniquely as shown below. In all cases, the resulting group is $\mathbb{Z}/8\mathbb{Z}$-satisfactory and 2 is a generator. As we will see below, none of the associated colorings is strongly representable.

For $(1,5,3,7)$, see table 4.12; for $(1,7,5,3)$, see table 4.13; for $(3,1,5,7)$, see table 4.14; for $(3,7,1,5)$, see table 4.15; for $(7,1,3,5)$, see table 4.16; and for $(7,5,1,3)$, see table 4.17.

The remaining two sequences do not contain sufficient information to determine $\oplus$. 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 $2^{\oplus 4}=1$. We conclude by observing that the value of $3\oplus 3=a$ 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.

In both cases, we obtain $\mathbb{Z}/8\mathbb{Z}$-satisfactory groups if and only if $a=2$ or 8. The associated colorings admit strong representatives, as follows: for the sequence $(3,7,5,1)$, if $a=2$, take $5417=677\cdot 8+1$, and if $a=8$, take $117017=14627\cdot 8+1$. For the sequence $(7,5,3,1)$, if $a=2$, take $3617=452\cdot 8+1$, and if $a=8$, take $17=2\cdot 8+1$.

If instead we let $a=1$ or 4, we obtain $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$-satisfactory groups. If $a=1$, in both cases, the unique group homomorphism that maps 2 to $(0,1)$ and 3 to $(1,0)$ is an isomorphism between $([8],\oplus)$ and $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$. If $a=4$ and the sequence is $(3,7,5,1)$, the corresponding isomorphism is obtained by considering the homomorphism that maps 2 to $(0,1)$ and 5 to $(1,0)$. If $a=4$ and the sequence is $(7,5,3,1)$, consider instead the homomorphism that maps 2 to $(0,1)$ and 7 to $(1,0)$.

Finally, we argue that the colorings associated with the first six $\mathbb{Z}/8\mathbb{Z}$-satisfactory groups we listed are not strongly representable. For this, simply note that if they were, any strong representative must be of the form $p=8k+1$, so $2^{4k}\equiv 1\pmod{p}$. But $2^{4k}=(2^{4})^{k}$, so the corresponding coloring $c$ must satisfy

[TABLE]

that is, we must have $2\oplus 8=1$.

5. 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.

5.1. Groupless numbers

Theorem 5.1** (Forcade-Pollington [FP90]).**

There are positive integers $n$ for which no multiplicative colorings exist. The smallest such $n$ is $n=195$.

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 § 3.4 would prove the existence of multiplicative colorings for all $n$.161616See for instance https://mathoverflow.net/q/26358/

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 $n\leq 500$. The data for the table was supplied by Rodney Forcade. This is sequence OEIS A204811 in the Online Encyclopedia of Integer Sequences171717See http://oeis.org/A204811.

In [BM12], S. R. Blackburn and J. F. McKee study partial $\mathbb{Z}/n\mathbb{Z}$-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 3.11.

In [BM12, section 5.2], question 1.2 is considered (independently), in the language of tilings of powers of $\mathbb{Z}$. In [BM12, section 9.2], Blackburn and McKee discuss the number of partial isomorphisms for a given $n$ and present a table listing those $n\leq 300$ that do not admit partial $\mathbb{Z}/n\mathbb{Z}$-isomorphisms (their table coincides with the beginning of our table 5.1). They further ask, motivated by numerical evidence, whether it is the case that if $n$ is large enough, then there is a partial $\mathbb{Z}/n\mathbb{Z}$-isomorphism if and only if either $n+1$ or $2n+1$ is prime. Note, however, that this turns out to be false, by theorem 4.26.

Nevertheless, the suggestion from the computations of [BM12] is that the set of groupless $n$ is large. It is thus natural to ask the following, as suggested by the referee.

Question 5.3**.**

Is the set of groupless $n$ infinite, or even of natural density 1?

5.2. Nonmultiplicative 6-satisfactory colorings

We finish 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 $k\in K_{n}$, to find all $n$-satisfactory colorings $d$ with $d_{k}=c$ (meaning that two numbers $m,m^{\prime}$ receive the same color under $c$ if and only if $km$ and $km^{\prime}$ 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 $\mathbb{O}_{n}$ to one of $\mathbb{O}_{n}-t(k)$.

Theorem 5.4**.**

Let $c$ be the 6-satisfactory coloring with strong representative 7, that is, $c(m)=(m\bmod 7)$ for $m\in K_{6}$. 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 $\mathbb{Z}/6\mathbb{Z}$-coloring, it is determined by table 4.1. The reader may consider just as well the 6-satisfactory coloring $c^{\prime}$ with strong representative 487, that is, $c^{\prime}(m)=(m^{81}\bmod{487})$, 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 $\mathbb{Z}/6\mathbb{Z}$-coloring, $c^{\prime}$ is determined by table 4.4.

Proof.

Note that $c$ is multiplicative, $c(9)=2$ and $c(3^{5})=5$, so that

[TABLE]

and, in particular, for fixed $\beta,\gamma$, $\{c(2^{\alpha}3^{\beta}5^{\gamma}):\alpha\in\mathbb{N}\}$ is either $\{1,2,4\}$ or $\{3,5,6\}$, the values alternating depending on the parity of $\beta+\gamma$. Indeed, $\beta+\gamma$ and $\beta+5\gamma$ have the same parity, and the powers of 3 are given modulo 7 by $1,3,2,6,4,5,1,3,\dots$. Moreover, since $c(2^{3})=1$ while $c(i)=i$ for $i=1,2,4$, the value of $c(2^{\alpha}3^{\beta}5^{\gamma})$ is periodic in $\alpha$ with period 3, see figure 5.1.

We proceed to verify the claim that there are precisely six 6-satisfactory colorings $d$ with $d_{5}=c$. Note that if $d$ is multiplicative, then $d_{k}=d$ for all $k$. In particular, five of these colorings are nonmultiplicative.

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 $\operatorname{dom}(d)=\frac{1}{5}\cdot K_{6}$ and also the points in the 2-dimensional grid corresponding to $\gamma=-1$ 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 $i\in[6]$ we show in the grid for $\gamma=-1$ 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 $\gamma=0$), 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 $\gamma=-1$, all colors are determined except for those in the bottom row, corresponding to $\beta=0$, see figure 5.4.

As for what colors $d$ must assign to points in that row, what we see is that if $d(5^{-1})=u,d(2\cdot 5^{-1})=v,d(4\cdot 5^{-1})=w$, then $\{u,v,w\}=\{3,5,6\}$, and $d$ still satisfies that $d(8\cdot x)=d(x)$ 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 5.4 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 $\frac{1}{5}\cdot K_{6}$, all of them are given by equation (5.1) except for $\beta=0,\gamma=-1$, where they are given as described in the last paragraph of the proof.

Remark 5.5**.**

We can extend the argument of theorem 5.4 in a few ways. For instance, their periodicity in the direction of 2 allows us to consider these colorings as defined on $\{a/2^{n}:a\in K_{6},n\in\mathbb{N}\}$. And we can iterate the construction: assign to each permutation of $\{3,5,6\}$ a number in $[6]$, and do the same to the permutations of $\{1,2,4\}$. Denote by $[6]^{<\mathbb{N}}$ the set of finite strings of members of $[6]$, that is,

[TABLE]

For $\sigma\in[6]^{<\mathbb{N}}$ denote its length by $|\sigma|$. Starting with $d^{\emptyset}=c$, we can associate to each finite string $\sigma\in[6]^{<\mathbb{N}}$ a coloring $d^{\sigma}$ with the following properties:

1. (1)

\operatorname{dom}(d^{\sigma})=\bigl{\{}\frac{a}{2^{n}5^{m}}:a\in K_{6},n\in\mathbb{N},0\leq m\leq|\sigma|\bigr{\}}* and $d^{\sigma}$ is 6-satisfactory on its domain.* 2. (2)

The functions $d^{\sigma{}^{\frown}\langle i\rangle}$ for $i\in[6]$ are all the 6-satisfactory maps $d$ on their domain such that $d_{5}=d^{\sigma}$ (under the convention of theorem 5.4, where we think of the equation $d_{5}=d^{\prime}$ for a given $d^{\prime}$ as seeking a map $d$ with domain $\frac{1}{5}\cdot\operatorname{dom}(d^{\prime})$). 3. (3)

Given $d^{\sigma}$, $d^{\sigma{}^{\frown}\langle i\rangle}$ is completely determined by its values on numbers of the form $2^{\alpha}5^{-|\sigma|-1}$, $\alpha\in\mathbb{Z}$, and in turn these values are given by the permutation associated to $i$ corresponding to $\{3,5,6\}$ if $\sigma$ is even and to $\{1,2,4\}$ if $\sigma$ is odd, as follows: if the permutation is $(u_{0},u_{1},u_{2})$, then $d^{\sigma{}^{\frown}\langle i\rangle}(2^{\alpha}5^{-|\sigma|-1})=u_{j}$, where $\alpha\equiv j\pmod{3}$.

(The proof of this is a straightforward extension of that of theorem 5.4, we omit the details.) In turn, this implies that if $\tilde{K}_{6}=\{2^{\alpha}3^{\beta}5^{\gamma}:\alpha,\gamma\in\mathbb{Z},\beta\in\mathbb{N}\}$, then $|C_{\tilde{K}_{6}}|=\mathfrak{c}$, since to each infinite sequence $x\in[6]^{\mathbb{Z}^{+}}$ we can associate the coloring

[TABLE]

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 $\hat{K}_{6}$.

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 $t$ of $\tilde{K}_{6}$, note that this can be extended to an essential tiling, and let $(d^{x})^{\prime}$ be the coloring coming from this tiling. We see that $(d^{x})^{\prime}\neq d^{x}$. 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 $\gamma=-1$ we use the permutation $(3,5,6)$ and for any other $\gamma$ we use either $(3,6,5)$ or $(1,2,4)$. The point is that tiles contained in the orthant (so $\gamma\geq 0$) 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 5.4 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 $\beta,\gamma$ in $\mathbb{N}$, the set $\{c(2^{\alpha}3^{\beta}5^{\gamma}):\alpha\in\mathbb{N}\}$ is either $\{1,2,4\}$ or $\{3,5,6\}$, and this only depends on the parity of $\beta+\gamma$. This is most readily apparent geometrically: consider an essential tiling of $\mathbb{O}_{6}$ by $T_{6}$ induced by $c$. For each fixed $\gamma$, the trace on this tiling on the plane grid corresponding to $\gamma$ 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.

Theorem 5.6**.**

$|C_{K_{6}}|=\mathfrak{c}$.

Proof.

Write $\mathbb{O}_{6}=\{(\alpha,\beta,\gamma):\alpha,\beta,\gamma\in\mathbb{N}\}$, identifying each point $(\alpha,\beta,\gamma)$ with the number $2^{\alpha}3^{\beta}5^{\gamma}\in K_{6}$. We will define a family of 6-satisfactory colorings $d$ by specifying certain essential tilings of $\mathbb{O}_{6}$. For each of them, as in the example just discussed, see also figure 5.5, the trace of the tiling on any plane $\gamma=\gamma_{0}$ 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.

Fixing $\gamma=\gamma_{0}$, each of the horizontal strips we consider has the form $H_{k}(\gamma_{0})$ for some $k\in\mathbb{N}$, where

[TABLE]

We follow our usual convention that $d(i)=i$ for $i\in[6]$, meaning that $T_{6}$ itself is one of the tiles we use (equivalently, if the tiling is $T_{6}+B\supset\mathbb{O}_{6}$, the sum being direct, then $\mathbf{0}\in B$). 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 $H_{k}(\gamma)$ with $k+\gamma$ even is of type $j\in[3]$, then $H_{k-1}(\gamma+1)$ is of type $j+1$ (using cyclic notation, so that 4 is identified with 1). This includes the case $k=0$ in cases where $\gamma$ is even.

We are ready to describe the colorings, they are of the form $d^{x}$ for $x\in[3]^{\mathbb{Z}^{+}}$, where $d^{x}$ is defined as follows: the tiling induced by the coloring $d^{x}$ has the following trace on the plane $\gamma=0$: $H_{0}(0)$ has type 1 (as required by our convention). For each $k>0$, $H_{2k}(0)$ has type $x(k)$. For $\gamma>0$, the traces are recursively defined according to the last sentence of the previous paragraph.

It is immediate from the construction that if $x\neq x^{\prime}$, then $d^{x}\neq d^{x^{\prime}}$, so we have defined $3^{\aleph_{0}}=\mathfrak{c}$ 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 $\mathbb{O}_{6}$. ∎

Note that, for $x^{1}$ the constant function taking the value 1, the coloring $d^{x^{1}}$ just described is the $\mathbb{Z}_{6}$-coloring with strong representative 103 and determined by table 4.3. Also, note that the set $\{d^{x}:x\in[3]^{\mathbb{Z}^{+}}\}$ not only has the same size as the reals but is in fact a perfect subset of $C_{K_{6}}$.

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 5.7**.**

We previously obtained a different proof of theorem 5.6 that used the fact that $C_{K_{6}}$ is closed in its natural topology. The argument proceeded in two stages. First, theorem 5.4 was extended using a variant of the construction in remark 5.5 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 $m\in K_{6}$. 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 5.6 to show that the result applies to $C_{K_{8}}$ as well.

Theorem 5.8**.**

$|C_{K_{8}}|=\mathfrak{c}$.

Proof.

We construct a perfect set of 8-satisfactory colorings of $K_{8}$ by describing the associated essential tilings of $\mathbb{O}_{8}=\{(\alpha,\beta,\gamma,\delta):\alpha,\beta,\gamma,\delta\in\mathbb{N}\}$. As before, the tilings have the form $T_{8}+B$ with $\mathbf{0}\in B$ and are given by tiling strips, which are now of the form $H_{k}(\gamma_{0},\delta_{0})$ for some fixed $\gamma_{0},\delta_{0}$ and some $k\in\mathbb{N}$, where

[TABLE]

We begin by describing the four possible types a strip $H_{k}(\gamma_{0},\delta_{0})$ 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 $m\in K_{8}$.

We say the tiling is of type

• •

1 if and only if it is (precisely) the depicted tiling with $a=0$,

• •

2 if and only if it is the depicted tiling with $a=-1$,

• •

3 if and only if it is the depicted tiling with $a=-2$,

• •

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 $k+\gamma_{0}+\delta_{0}$ is even and $H_{k}(\gamma_{0},\delta_{0})$ is of type $i$, then $H_{k-1}(\gamma_{0}+1,\delta_{0})$ is of type $i+2$ and $H_{k-1}(\gamma_{0},\delta_{0}+1)$ 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 $\mathbb{O}_{8}$ once the colors of points in the (intersection of the orthant with the) plane $\gamma=\delta=0$ are specified. For this, first note that the recursion describes how to assign colors to points in $H_{k}(\gamma,\delta)$ for $k$ of the same parity as $\gamma+\delta$ so, in particular, fixing the values of $\gamma$ and $\delta$, for any point $x$ in the resulting plane there is a unique $k\geq-1$ with $x\in H_{k}(\gamma,\delta)$ and $k$ of the relevant parity. Now, by induction on $\gamma+\delta$, let $i$ be the type of $H_{k+\gamma+\delta}(0,0)$, and check that the rules specify that the type of $H_{k}(\gamma,\delta)$ is precisely $i+2\gamma+3\delta$ (using the cyclic convention), so the color assigned to $x$ is indeed unambiguous.

Finally, we define the colorings $d^{x}$ for $x\in[4]^{\mathbb{Z}^{+}}$ 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 $\{d^{x}:x\in[4]^{\mathbb{Z}^{+}}\}$ is a perfect subset of $C_{K_{8}}$. ∎

Note that for $x^{1}$ the constant function taking the value 1, the coloring $d^{x^{1}}$ is the $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$-coloring determined by table 4.21 for $a=1$.

Remark 5.9**.**

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.

6. Open questions

For the reader’s convenience, we close the paper by listing the questions we have mentioned throughout the paper. We omit those that, like question 4.7, were asked as rhetorical devices and are answered in the text.

Question 1.2.

Given any positive integer $n$, is there a coloring of the positive integers using $n$ colors such that for any positive integer $a$, the numbers $a,2a,\dots,na$ all have different colors?

Question 1.3.

Assuming that question 1.2 has a negative answer for $n$, can we find a better bound than the smallest prime larger than $n$ on the number of colors required to ensure a positive answer?

We refine the original formulation of question 2.2 as follows:

Question 2.2.

Given $n>1$, how many $n$-satisfactory colorings of $K_{n}$ are there, if any at all? Is the map $n\mapsto|C_{K_{n}}|$ that assigns to each $n$ the number of $n$-satisfactory colorings of the core a recursive function (taking values in $\mathbb{N}\cup\{\aleph_{0},\mathfrak{c}\}$)?

Question 2.8.

Given $n\in\mathbb{Z}^{+}$, suppose that $C_{K_{n}}$ is nonempty. Should it have isolated points?

For the operation $d\mapsto d_{k}$ on colorings, see item (3) in § 2.2.

Question 2.10.

Given an $n$-satisfactory coloring $c$ and $k\in K_{n}$, is there an $n$-satisfactory coloring $d$ such that $d_{k}=c$? In that case, how many such colorings $d$ are there?

Question 2.13.

Let $n\in\mathbb{Z}^{+}$.

1. (1)

Does any $n$-satisfactory coloring of $K_{n}$ extend to one of $\hat{K}_{n}$? 2. (2)

If $T_{n}$ essentially tiles $\mathbb{O}_{n}$ via a tiling $T_{n}+B$, is there a tiling by $T_{n}$ of all of $\mathbb{Z}^{\pi(n)}$ that essentially extends it?

Question 2.16.

Given an $n$-satisfactory coloring $c$ of $\hat{K}_{n}$, 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 $\mathbb{Z}^{\pi(n)}$. From this tiling we can define an $n$-satisfactory coloring $c^{\prime}$ with color classes the preimages under $t$ of the translates $t(i)+B$, $i\in[n]$. Is $c^{\prime}=c$?

See the discussion surrounding the original presentation of question 2.17 for additional details of the setting it references. Briefly, given $n$, from an $n$-satisfactory coloring $c$ of $K_{n}$, we obtain $n$ partial tilings of the orthant $\mathbb{O}_{n}$, and any point in $\mathbb{O}_{n}$ belongs to at least one of the resulting direct sums $T_{n}+B$. Any of these sums in turn defines a partial $n$-satisfactory coloring of $K_{n}$.

Question 2.17.

Are the resulting partial colorings compatible? If they are, their union gives us a coloring $c^{\prime}$ of $K_{n}$. Is $c^{\prime}=c$?

Originally, question 3.11 was listed with three parts, but we proceeded to solve positively the first two. The following remains, though we expect the answer to be negative and easily accessible from the techniques we discuss in § 3.4.

Question 3.11.

Let $n\in\mathbb{Z}^{+}$. Suppose $n$ admits a strong representative. For a satisfactory $n$-coloring $c$, is the natural density of the set of strong representatives of order $n$ for $c$ independent of $c$?

Question 5.3.

Is the set of groupless $n$ infinite, or even of natural density 1?

Many combinatorial questions remain besides those just listed. They appear intractable with current methods.

Question 6.1**.**

Is there an $n$ admitting precisely a countable infinity of $n$-satisfactory colorings of $K_{n}$? Which finite $m$ are precisely the number of $n$-satisfactory colorings of $K_{n}$ for some $n$?

Acknowledgements

We thank Amanda Francis for creating figure 5.2. Thanks are also due to Zach Teitler for alerting us of [FLLP86], and to Rodney Forcade for providing us with the data for table 5.1. Special thanks to Dömötör Pálvölgyi for promoting the question we study in this paper (and for making the first-named author aware of it!) by posting it as question 26358 in MathOverflow. Thanks are also due to the MathOverflow community for their ideas and suggestions, and in particular to Darij Grinberg, Gergely Harcos and Noam D. Elkies for allowing us to include their results. Thanks to Ben Barber and Péter Csikvári for their interest on the topic of this paper and their suggestions. Thanks to Felipe Voloch and David E Speyer for their illuminating suggestions and assistance regarding the subject of § 3.4, and to Lon Mitchell for his interest on the topic of § 5.2. We also want to thank the anonymous referee for their careful reading of the manuscript and valuable suggestions.

The first-named author gave talks on this topic at Albion College, Albion, MI, and at Miami University, Oxford, OH, and wants to thank his respective hosts for the invitations and support. The visit to Miami University was partially supported by NSF grant DMS-1201494. The third-named author was supported by the Lendület program of the Hungarian Academy of Sciences (MTA), the National Research, Development and Innovation Office NKFIH (Grant Nr. PD115978, K129335 and BME NC TKP2020), the New National Excellence Program of the Ministry of Human Capacities (UNKP-18-4) and the János Bolyai Research Scholarship of the Hungarian Academy of Sciences.