Divisibility Graphs and Modular Multiplication Tables

Abstract Consider the graph with the residue classes modulo n as vertices, and the following edges: “additive” edges from a to a + 1; “multiplicative” edges from a to ab for some fixed b. This graph illustrates a criterion for divisibility by n for numbers written in base b. By varying n and b, we see a great variety of structures: this topic connects arithmetic to graph theory and has the beauty of string art. For divisibility graphs we investigate the existence of triangles and other cycles, the girth, the minimum/maximum of the vertex degree, the chromatic number, and the planarity.


Introduction
It is natural to display the residue classes modulo n as the vertices of a regular n-gon in cyclic order.Thus adding 1 to a residue lets us move to the next residue along a side of the n-gon, while multiplying by some integer b lets us usually move along a diagonal of the n-gon (occasionally we move along a side or we do not move at all).The divisibility graph D n,b , which is the object of our investigation, is the graph defined as follows: its vertices are the vertices of the regular n-gon; the "additive" edges are the oriented sides from a to a + 1; the "multiplicative" edges go from a to ba and are usually diagonals (occasionally they can be a loop or a side).Our goal is studying the family of divisibility graphs D n,b by varying n and b: as it is to be expected, the graph properties depend on the arithmetic properties of the pair n, b.We also consider the corresponding simplified graphs D ′ n,b , where double edges and loops are removed    and edge orientation is neglected.We investigate for divisibility graphs the existence of cycles (triangles, and rectangles in the canonical representation), the girth, the minimum/maximum vertex degree, the chromatic number, and planarity, see Section 2 for a summary of our results.One motivation for considering the graph D n,b is that, for b ⩾ 2, it illustrates a divisibility criterion for an integer to be divisible by n that makes use of the digits in base b, see Remark 2.2.It seems that divisibility graphs were introduced by David Wilson in 2009 to illustrate the divisibility criterion by n = 7 in base b = 10, according to Tanya Khovanova [1].Another motivation for studying these graphs is that they connect graph theory with arithmetic in a fascinating way, and they are also beautiful objects with a great variety of structures by varying n and b, see for example Figure 1, Figure 2 and Figure 3.
Considering the multiplicative edges of D ′ n,b only, one obtains the graph of modular multiplication by b, which has been considered before to popularize mathematics.These other graphs, although optically similar (the difference being whether to include all sides of the n-gon) have different graph structures though; for example, having the additive edges ensures the existence of cycles.These other graphs are not our object of study, so we refer the reader to the book by Daniel Shanks [2] and to online resources like [3,4].There one finds in particular a geometrical explanation for the (b − 1)-foil pattern that is consequently also to be seen in our divisibility graphs, see Figure 4, Figure 5 and Figure 6.Let us also mention the existence of websites like [5] which output the graph of modular multiplication by b, all based on the parameters n and b entered by the user.
Divisibility graphs are still to be explored, as ours seems to be the first deep investigation on the subject.Beyond our conjectures, some of our results can be improved upon, and there are surely features of divisibility graphs that we have not yet considered.We hope that some of our readers take this challenge.Moreover, we hope that many readers introduce divisibility graphs to a wider audience, because it is an accessible topic (related to known divisibility criteria) that connects arithmetic to graph theory and that possesses the beauty of string art.We canonically represent the vertices of a divisibility graph as the vertices of a regular polygon in cyclic order, see for example Figure 7.
Replacing b by its remainder modulo n does not affect the divisibility graph.Thus, to avoid trivial cases, we will assume that n ⩾ 3 and 1 < b < n.Moreover, we shall denote a vertex by any integer in the given residue class modulo n. . ., 0, take c i additive edges and in case i > 0, further take one multiplicative edge.This assertion can be proven by induction on k: the base case k = 0 is obvious; for the induction step, write the given number as ba+c 0 with a = c k b k−1 +• • •+c 1 b 0 , and complete the known walk for a by taking one multiplicative edge and c 0 additive edges.
For example, 21 = 2 • 3 2 + 1 • 3 1 + 0 • 3 0 is divisible by 7 because in D 7,3 we walk from 0 to 0 by taking two additive edges, one multiplicative edge, one additive edge, and once more a multiplicative edge, see Figure 7. ▲ To ease notation, we call d the greatest common divisor of n and b, we understand that congruences are modulo n (unless specified otherwise), we call an integer a unit if its residue class modulo n is a unit, and we say that two edges overlap if, neglecting their orientation, they coincide.We now recall some notions from graph theory (for references, see for example [6,7]).

Definition.
Let G be an undirected graph.A triangle in G is a cycle of length 3. The girth of G is the length of a smallest cycle.The vertex degree for a vertex of G is the amount of incident edges.We denote by δ(G) (respectively,  ∆(G)) the minimum (respectively, maximum) degree, namely the minimum (respectively, maximum) among all vertex degrees.The chromatic number of G, denoted by χ(G), is the smallest positive integer k such that G admits a kcoloring, namely a map from the vertex set to {0, . . ., k − 1} such that any two adjacent vertices have distinct images.The graph G is called planar if it can be drawn on the plane without crossing edges.▲ Our main results are the following assertions: 2.4 Theorem.For simplified divisibility graphs, the following hold: (1) The graph D ′ n,b contains a triangle for every b if and only if n = 4, n is prime, or n is a power of 3.
(2) There is a rectangle in the canonical representation of D ′ n,b for some 1 < b < n if and only if n = 4 or n ≡ 0, ±2, ±6 mod 16.The outline of the paper is as follows: after some preliminary results in Section 3, we consider cycles in D ′ n,b in Section 4 and the vertex degree in Section 5. Finally, in Section 6 and Section 7, we respectively consider the chromatic number of D ′ n,b and investigate whether D ′ n,b is planar.In particular, we provide evidence for the following two conjectures:

Preliminaries
We begin with some general observations: 3.1 Remark.The simplified divisibility graph D ′ n,b (with the additive edges forming a regular n-gon) is symmetric with respect to the line connecting 0 to the center of the n-gon.Indeed, the symmetry maps the n-gon to itself, and it maps the multiplicative edge (a, ba) to the edge (n − a, b(n − a)).
Moreover, D ′ n,b passes through the center of the n-gon if and only if for the 2-adic valuation we have v 2 (b − 1) < v 2 (n).Indeed, we precisely need the congruence bx ≡ x + n 2 to be solvable.Notice that in D ′ 125,51 (see Figure 1) the multiplicative edges do not connect vertices that are almost opposite.▲

Remark.
Consider the divisibility graph D n,b and some vertex a.The outgoing multiplicative edge at a is a loop if and only if (b − 1)a ≡ 0, thus we have a loop at 0, and this is the only loop if and only if b − 1 is a unit.The outgoing multiplicative edge at a can overlap an additive edge: this happens if and only if (b−1)a ≡ ±1, so it happens only if b−1 is a unit and in this case there is precisely one multiplicative edge that equals an additive edge (respectively, the reverse of an additive edge).▲ Notice that all vertices are preperiodic under iterates of this map.Moreover, a vertex a is periodic if and only if gcd(ba, n) = gcd(a, n), hence all vertices are periodic if and only if b is a unit.

Remark. A vertex in
To see this, call g the largest integer of the form gcd(b x , n) for some positive integer x, so that for every prime number p, the p-adic valuation v p (g) equals v p (n) if p divides b, and it is 0 otherwise.Then the periodic vertices are precisely the multiples of g.For example, if n is even, then n 2 is periodic if and only if b is odd.If a is periodic, then the period of its orbit is the smallest positive integer x such that (b x − 1)a ≡ 0 mod n g .Considering Euler's totient function ϕ, the period of the orbit divides ϕ(n); it equals ϕ(n) if and only if all units lie in one same orbit.▲ If G is a graph, then a minor of G is a graph that can be obtained from G by deletion and contraction of edges, as well as deletion of some isolated vertices.Given a subset of the vertices of G, the induced subgraph has precisely those vertices, and it has as edges all edges of G between them.).An example for G d can be found in Figure 12. ▲

Cycles
To study cycles, we first classify triangles in D ′ n,b according to the number of additive edges: Proof of Theorem 2.4 (1).For the "if" implication, we can check by hand the case n = 4 and the above classification allows us to deal with n ⩾ 5 prime.Now suppose that n is a power of 3 and write k := v 3 (n) and v := v 3 (b − 1).
If v = 0, we may resort to the above classification, so suppose v > 0. Notice that a := 3 k−v−1 is an integer and a(b 3 − 1) ≡ 0. So the vertices a, ab, ab 2 form a triangle, as they are pairwise distinct: n,b with additive edges because, for instance, ab ≡ a + 2 implies apx ≡ 2; this is impossible because 2 is a unit and p divides n.We also exclude triangles with zero additive edges because ab 3 ≡ a means apx(p 2 x 2 + 3px + 3) ≡ 0, implying apx ≡ 0 and hence ab ≡ a(px + 1) ≡ a. ■ 4.1 Lemma.Let n be a positive, composite, odd integer and let p > 3 be a prime factor of n.Then there exists a positive integer x < n p such that p 2 x 2 + 3px + 3 is coprime to n.
Proof.If n is a power of p we can take x := 1, while if n is a multiple of 3 we can take x := n 3 v 3 (n) p because n has no prime factor in common with Finally, if n is coprime to 3 and it has a prime factor q > 3 different from p, we may take x := tn pq vq (n) for some integer 0 < t < q.Indeed, n and p 2 x 2 + 3px + 3 can have at most the prime factor q in common and, for some choice of t, the latter integer is not divisible by q: that polynomial in x has at most two roots modulo q, and different choices for t give different values of x modulo q.
For b 2 ≡ b, we have the cycle (0, 1, b, b − 1).For b not a unit, there is an integer a ̸ ≡ 0, 1, −1 such that we have the cycle (0, 1, b, a + 1, a) (if b ≡ a + 1, we take a cycle of length 4 instead).For b a unit such that b 2 ̸ ≡ 1, calling a its inverse, we consider the following cycle: It is of length 5. if n is odd hence we will suppose that n is even.

Remark.
In [8] Neeyanth Kopparapu gives a criterion to have a rectangle of the form (a, ba, b 2 a, b 3 a) for some vertex a.In short, we must have b 4 − 1 ≡ 0 and v 2 (b 2 − 1) < v 2 (n) (in particular, n ≡ 0 mod 8 and ϕ(n) ≡ 0 mod 4).▲ Having all additive edges in a rectangle is only possible for n = 4.If the edge (a, a ± 1) is multiplicative, then b is odd and the edge (a + n 2 , a + n 2 ± 1) is also multiplicative, so the number of edges in a rectangle that are not multiplicative is even.The remaining case is described below: 4.4 Remark.We classify the rectangles with precisely two edges that are not multiplicative (they must be opposite).If the two multiplicative edges have the same orientation, then the rectangle is without loss of generality of the form (bx, x, x + 1, bx − 1) with b(x + 1) ≡ bx − 1 and x − bx + 1 ≡ n 2 , which means b = n − 1 and 2x + 1 ≡ n 2 .In particular, n ≡ 2 mod 4 and we may take x = n−2 4 .Now suppose that the two parallel edges have opposite orientation, and let (x, x + 1) be an additive edge.If the multiplicative edge is incoming at x, the vertices are (bx, x, x + 1, bx − 1) and we have b(bx − 1) ≡ x + 1 and bx − 1 − x ≡ n 2 .Else, the vertices are (x, x + 1, bx + b, bx + b + 1) and we have b(bx + b + 1) ≡ x and bx + b − x ≡ n 2 .Notice that D ′ n,b either contains rectangles of both cases, or of none: given a rectangle of one case, one obtains a rectangle of the other case by replacing its vertices by their opposites modulo n. ▲

Proof of Theorem 2.4 (2).
Clearly there is no rectangle for n = 3, while the additive edges form a square for n = 4. Now suppose that n ⩾ 5 is even.By Remark 4.4 there is a rectangle in D n,n−1 if n 2 is odd.Moreover, if n ≡ 0 mod 16, then D ′ n,3 contains the rectangle (x, 3x, 9x, 27x) where x = n 16 .To conclude we have to prove that there is no rectangle if n ≡ ±4, 8 mod 16.
whose left member is odd, a contradiction.

■
We conclude this section with two remarks on k-cycles:

Vertex degree
An oriented graph is Eulerian if and only if at each vertex there is the same amount of oncoming and incoming edges.4) the latter condition means that, for every integer a, at least one between a and a − 1 is a unit (if n had distinct prime factors p, q, then the congruences a ≡ 0 mod p and a ≡ 1 mod q have a common solution).▲ Proof.By Remark 5.2 and Theorem 2.4 5, to determine ∆(D ′ n,b ) we may neglect the vertex 0.Moreover, if d > 1, or if there is a vertex of degree d + 2, then we may neglect the vertices which are not a multiple of d.
If bd ≡ d (hence d > 1), then for every k we have bkd ≡ kd hence the vertex degree of kd is d + 1.Now suppose that bd ̸ ≡ d, and notice that b 2 d ≡ d holds if and only if for every integer k we have b 2 kd ≡ kd.The latter condition means that the outgoing multiplicative edge at kd overlaps an incoming multiplicative edge.We conclude because the additive edges at d do not overlap multiplicative edges by our assumption on b − 1.Finally, let d ′ > 1 and set k := n dd ′ hence bkd ≡ 0. In particular, the outgoing multiplicative edge at kd does not overlap an additive edge, and neither does an incoming multiplicative edge, else we would have b(kd ± 1) ≡ kd hence b ≡ ±kd, thus bd ≡ 0, contradiction.In the second case, we get a 3-coloring by assigning to each multiple of d the color 0 and by alternating between 1 and 2 on each interval between two multiples of d.In the third case, we obtain a 3-coloring as follows: assign the color 0 to the Above, the polygon consists of the additive edges, with 0 at the bottom.▲

Example.
If n ≡ 2 mod 3 (respectively, n ≡ 1 mod 6), then a 3-coloring for D ′ n,n−2 can be defined by mapping a vertex to its remainder modulo 3 (respectively, additionally replacing the color of 0 to be 2).Indeed, for all a the vertices a − 1, a, a + 1 have distinct colors, so we are left to show that a and ba have distinct colors for every a ̸ = 0: the remainder of ba modulo n is n − 2a or 2n − 2a, thus not congruent to a modulo 3.
Similarly, if k ⩾ 2 is not divisible by 3, then D ′ 3k,k+1 and D ′ 3k,2k+1 have a 3-coloring by mapping a vertex to its remainder modulo 3 (as b ̸ ≡ 1 mod 3 and n ≡ 0 mod 3, the remainder of ba modulo n is not congruent to a modulo 3).▲

Planarity
To investigate planarity for divisibility graphs, we may clearly consider simplified divisibility graphs instead. 2 −1 ) can be drawn as in Figure 19 (respectively, Figure 20), where the dashed edge ( n 2 , 0) is not present if n 2 is odd (respectively, even).▲ Wagner's theorem states that a graph is planar if and only if it does not contain as a minor K 5 nor K 3,3 , which are the following graphs: For example, D 6,15 contains a minor isomorphic to K 5 .We will often exhibit a minor isomorphic to K 3,3 : the vertices will be listed in cyclic order; the vertices are {x 1 , x 2 , x 3 , bx 1 , bx 2 , bx 3 }, and x 1 , x 2 , x 3 will be marked in bold.Proof.An odd number ⩾ 13 is an odd multiple of a number ⩾ 13 which is prime or ⩽ 121, so we conclude by Lemma 7.4.An even number ⩾ 16 is a number for which the conjecture has been verified or it is a power of 2 times an odd number ⩾ 13, so we conclude by Lemma 7.4 and the previous case.

■
Notice that, in case the lower bound for n in Conjecture 2.6 gets updated, one could still reduce to prime numbers and an explicit finite number set.Above, a denotes a multiple of k such that kn − ℓn < a < ℓn. ■

2. 2
Remark.The divisibility graph D n,b provides a divisibility criterion by n considering the digits of a number expressed in base b.This is because the number c k b k + • • • + c 1 b 1 + c 0 b 0 (with k, c 0 , . . ., c k ⩾ 0) corresponds to the end vertex of the following walk in D n,b : start at 0; for i = k, .

Figure 10 :
Figure 10: D ′ 21,8 has girth 7 Figure 11: D ′ 56,43 has girth 7 D n,b has no incoming multiplicative edge unless it is a multiple of d (as n and b are multiples of d), and in this case it has precisely d incoming multiplicative edges (a solution for bx ≡ d being given by Bézout's identity).Consider the multiplication by b on the residue classes modulo n, represented by the multiplicative edges of D n,b .

Figure 12 :
Figure 12: G2 for D52,10 ,b is a minor of D n,b .Indeed, by contracting the edges between multiples of m we obtain a graph with n/m vertices such that D n/m,b is a subgraph (notice that bk ≡ k ′ mod n m implies bkm ≡ k ′ m for all integers k, k ′ ).▲ 3.5 Remark.The subgraph D × n,b of D n,b obtained by removing the additive edges is the union of orbits for the multiplication by b.A periodic orbit is a cycle (possibly consisting of one or two vertices).Two orbits are either disjoint or have their periodic cycle in common.A connected component of D × n,b is as follows: there is one periodic cycle and attached to each vertex of the cycle there are d − 1 vertices outside the cycle (one incoming multiplicative edge belongs to the cycle).Each of the above vertices outside the cycle is the root of a d-ary tree: these trees are pairwise disjoint and do not intersect the cycle.▲ 3.6 Remark.If m > 1 is a divisor of n, then we define G m as the induced subgraph on the vertices of D ′ n,b that are multiples of m.Thus G m can be obtained from D n/m,b (identifying multiples of m modulo n and residue classes modulo n/m) by removing additive edges and loops, and neglecting edge directions.Moreover, G m can be obtained from D × n,b as the induced subgraph on the multiples of m, removing loops and neglecting edge directions (for m = d, the induced subgraph only looses leaves with respect to D × n,b

Figure 10 and
Figure 10 and Figure 11 show that the girth of D ′ n,b can be as large as 7.It can, however, not be greater: Proof of Theorem 2.4 (3).By the above classification, D ′ n,b contains a triangle whenever b − 1 is a unit, so we may exclude this case.For b = n − 1 or b 2 ≡ 0

Figure 13 :Figure 14 :
Figure 13: A rectangle with four multiplicative edges

■ 4 . 2
Remark.Consider those triangles inside D ′ n,b that are equilateral triangles in the canonical representation, as in Figure 9.For n > 3 there is such a triangle based at some vertex a if and only if the three numbers (b − 1)a, (b − 1)ba, (b − 1)b 2 a are congruent to n 3 .So we must have (b − 1)a ≡ n 3 mod n and b ≡ 1 mod 3. We deduce that there is an equilateral triangle if and only if n = 3 or 1 ⩽ v 3 (b − 1) ⩽ v 3 (n) − 1.Similarly, to find a regular k-gon inside the canonical representation of D ′ n,b (beyond n = k) it is necessary and sufficient that for any prime divisor p of k we have v p (k) ⩽ v p (b − 1) ⩽ v p (n) − v p (k). ▲ Now we turn our attention to rectangles in the canonical representation of simplified divisibility graphs.For example D ′ 64,3 contains a rectangle build with sides and diagonals of a regular octagon.Observe that a 4-cycle (x, y, z, w) in D ′ n,b forms a rectangle if and only if x − z ≡ y − w ≡ n 2 (i.e.there are two pairs of diametrically opposite vertices).In particular, there is no rectangle in D ′ n,b If D ′ n,b contains a rectangle of the form (x, bx, b 2 x, b 3 x), then x(b 2 − 1) ≡ n 2 ≡ x(b 3 − b), so that b n 2 ≡ n 2 , hence n 2 (b − 1) ≡ 0 and b must be odd.Since x(b 2 − 1) ≡ n 2 and b 2 − 1 is a multiple of 8, also n 2 is a multiple of 8 and hence n ≡ 0 mod 16.If D ′ n,b contains a rectangle with two multiplicative edges of the same direction, it holds that n ≡ 2 mod 4 and thus n ̸ ≡ ±4, 8 mod 16.By Remark 4.4, the only remaining case to consider is D ′ n,b containing a rectangle of the form (bx, x, x + 1, bx − 1).We prove that n ̸ ≡ 0 mod 4 by contradiction.Write n = 4k.Now b(bx − 1) ≡ x + 1 and bx

4. 5 4 . 6
Remark.Let k ⩾ 3.In [9] Hans-Peter Stricker gives a criterion to have a k-cycle of the form (a, ba, b 2 a, b 3 a, . . ., b k−1 a) for some vertex a. Replace n by m := n gcd(n,a) and hence suppose that a is a unit.Then we precisely need that b is a unit with multiplicative order k.So the general criterion becomes that there exists a divisor m of n such that (b mod m) is a unit with multiplicative order k.▲ Remark.Let k ⩾ 3. Assume that n ⩾ k and that b − 1 is a unit.Then there exists a k-cycle: for example, the vertices (a, a + 1, . . ., a + k − 1) where a := (b − 1) −1 (k − 1) form a k-cycle because we have ba ≡ a + k − 1. ▲

2 .■ 5 . 3
If b is a unit (respectively, b is not a unit and b−1 is a unit), then 0 (respectively, the multiplicative inverse of b−1) has vertex degree 2. Now suppose that b and b−1 are not units: no multiplicative edge overlaps an additive edge; we have δ(D ′ n,b ) ⩾ 3 because, if the outgoing multiplication edge is a loop, then there is an incoming multiplication edge that is not a loop (ba ≡ a implies b(a + c) ≡ a for any c ̸ ≡ 0 such that bc ≡ 0).We have ∆(D ′ n,b ) ⩽ d + 3 because at every vertex there are at most d incoming multiplicative edges; we show that the vertex degree of b is at least d + 1.For 0 ⩽ i < d, the d incoming multiplicative edges (1 + i n d , b) are distinct: if one of them is a loop, they don't overlap additive edges (b 2 ≡ b implies b ̸ ≡ b(b ± 1)); else, not both additive edges at b overlap incoming multiplicative edges (as b 3 ≡ b(b + 1) ≡ b(b − 1) ≡ b implies b = n 2 , so b 2 ≡ 0 gives b even while n 2 (b 2 − 1) ≡ 0 gives b odd).Remark.An integer n ⩾ 3 is a prime power if and only if for all b we have δ(D ′ n,b ) = 2. Indeed, by Theorem 2.4 (

Figure 15 ,
Figure 15, Figure 16, and Figure 17 are examples for the three cases in the following result.

■ 5 . 5
Theorem.Suppose that b − 1 is a unit.Then ∆(D ′ n,b ) = d + 1 holds if and only if either n = 2d or we have n = 3d and bd ̸ ≡ 0.Moreover, for n ⩾ 6 we have ∆(D ′ n,b ) = d + 3 if and only if n ⩾ 4d and b 2 d ̸ ≡ d.Proof.If n = 2d, then d = b and b is even, and we only need to remark that the vertex degree for d is d + 1 (the additive edges at d overlap multiplicative ones).Now suppose n = 3d hence d = b.If bd ̸ ≡ 0 (hence 2bd ̸ ≡ 0), the vertex degree of d and of 2d is d + 1: we have bd ≡ 2d and 2bd ≡ d hence the outgoing multiplicative edge overlaps an incoming one; at least one additive edge overlaps an incoming multiplicative edge.If bd ≡ 0, then the vertex degree of d or of 2d is d + 2: there are d distinct incoming multiplicative edges and a non-overlapping outgoing multiplicative edge (towards 0); the additive edges at d and 2d cannot all overlap multiplicative edges; at least one additive edge at d (respectively, 2d) overlaps a multiplicative edge, as d = b.Finally, suppose n ⩾ 4d.By Remark 3.2 there exists k such that the additive edges at kd do not overlap multiplicative ones.If b 2 d ≡ d, then for every k the outgoing multiplicative edge at kd overlaps an incoming multiplicative edge.From now on, supposing b 2 d ̸ ≡ d, we find a vertex with vertex degree d + 3.If d = 1 (thus b is a unit), then there are three non-consecutive vertices that are units, namely (1, b, n − 1), or (1, 4, n − 1) for b = 2, n − 2 hence n ⩾ 7: some of them is suitable, as at most two additive edges overlap a multiplicative edge.Now suppose d > 1.If bd ≡ 0, there exists k such that the additive edges at kd do not overlap multiplicative ones.Now suppose bd ̸ ≡ 0, and call d ′ := gcd( n d , d).If d ′ = 1, then for 1 ⩽ k ⩽ k ′ ⩽ n d − 1 we have bkd ̸ ≡ 0 and hence bkd ̸ ≡ bk ′ d unless k = k ′ .In particular, (b − 1)b 4 d ̸ ≡ 0 (as b − 1 is a unit).Thus the vertices d, bd, b 2 d are distinct and their outgoing multiplicative edges do not overlap incoming multiplicative edges: they are not connected by additive edges (as d > 1), and for at most two of them an additive edge overlaps a multiplicative edge.

■ 6
Chromatic numberFor n ⩾ 4 the simplified divisibility graph D ′ n,b is connected, and neither an odd cycle nor a complete graph, so Brooks' Theorem givesχ(D ′ n,b ) ⩽ ∆ n,b .Proof of Theorem 2.4(5).For the first assertion we may suppose that n ⩾ 5 and also by Theorem 2.4 (4) that d > 1.The subgraph G d induced by the multiples of d admits a 3-coloring by Remark 3.6, and we can extend it to a 4-coloring of D ′ n,b .Indeed, there are no incoming multiplicative edges for the vertices outside G d , so for 0 ⩽ k ⩽ n d − 1 we can iteratively color the vertices kd + i for 1 ⩽ i ⩽ d − 1: the vertex kd + i sees at most two (respectively, three) previously chosen colors for i ̸ = d − 1 (respectively, i = d − 1).If n ≡ 2 mod 4, then D ′ n, n 2 +1 has a 2-coloring given by mapping a vertex to its remainder modulo 2 (the multiplication by b maps odd vertices to even vertices, and it maps any even vertex to itself).If χ(D ′ n,b ) = 2, then (considering the additive edges) n is even and w.l.o.g. the 2-coloring maps a vertex to its remainder modulo 2: if b ̸ = n 2 + 1 (respectively, b = n 2 + 1 and n ≡ 0 mod 4), the vertices 2 and 2b (respectively, 1 and b) are distinct, adjacent, and have the same color.■ Conjecture 2.5 has been verified with SageMath [10] for n ⩽ 291.Moreover, it has been proven in the following cases: 6.1 Proposition.If b 2 ≡ 1, bd ≡ d, or bd ≡ 0, then χ(D ′ n,b ) ⩽ 3. Proof.In first case, we can check by hand D ′ 3,2 , D ′ 4,3 , D ′ 5,4 , and for n ⩾ 6 we have ∆(D ′ n,b ) ⩽ d + 2 = 3 by Theorem 5.4 and Theorem 5.5.

7. 1 , n 2 (
Example.The graph D ′ n,n−1 , and for n even the graphs D ′ Indeed, D ′ n,n−1 can be drawn as in Figure 18, while the graph D ′ nrespectively, D ′ n, n
5.1 Remark.▲ 5.2 Remark.The vertex degree is at most 3 for vertices that are not multiples of d by Remark 3.3.Moreover, by Remark 3.2 for d > 1 no additive edge overlaps a multiplicative edge, and for d ⩾ 1 no additive edge at 0 overlaps a multiplicative edge.Thus the vertex degree at 0 is d + 1 (the outgoing multiplicative edge is a loop).▲ Proof of Theorem 2.4 (4).Recall Remark 3.