Existentially closed graphs via permutation polynomials over finite fields

Existentially closed graphs via permutation polynomials over finite fields tài liệu, giáo án, bài giảng , luận văn, luận...

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Discrete Applied Mathematics

journal homepage:www.elsevier.com/locate/dam

Existentially closed graphs via permutation polynomials over

finite fields

Nguyen Minh Haia, Tran Dang Phuca, Le Anh Vinhb,∗

aFaculty of Mathematics, Mechanics and Informatics, Hanoi University of Science, Vietnam National University, Hanoi, Viet Nam

bUniversity of Education, Vietnam National University, Hanoi, Viet Nam

a r t i c l e i n f o

Article history:

Received 28 November 2014

Received in revised form 17 June 2015

Accepted 18 May 2016

Available online xxxx

Keywords:

Existentially closed graphs

Permutation polynomial

Distance graphs

a b s t r a c t

For a positive integer n, a graph is n-existentially closed or n-e.c if we can extend all

n-subsets of vertices in all possible ways It is known that almost all finite graphs are n-e.c Despite this result, until recently, only few explicit examples of n-e.c graphs are

known for n>2 In this paper, we construct explicitly a 4-e.c graph via a linear map over finite fields We also study the colored version of existentially closed graphs and construct explicitly many(3,t)-e.c graphs via permutation polynomials and multiplicative groups over finite fields

© 2016 Published by Elsevier B.V

1 Introduction

For a positive integer n, a graph is n-existentially closed or n-e.c if we can extend all n-subsets of vertices in all possible ways Precisely, for every pair of subsets A,B of vertex set V of the graph such that A∩B= ∅and|A| + |B| =n, there is a

vertex z not in A∪B that joined to each vertex of A and no vertex of B From the results of Erdős and Rényi [4], almost all

finite graphs are n-e.c Despite this result, until recently, only few explicit examples of n-e.c graphs are known for n>2 See [2] for a comprehensive survey on the constructions of n-e.c graphs.

In [13], the third listed author studied a multicolor version of this adjacency property Let n,t be positive integers.

A t-edge-colored graph G is(n,t)-e.c or(n,t)-existentially closed if for any t disjoint sets of vertices A1, ,A t with

|A1| + · · · + |A t| =n, there is a vertex x not in A1∪ · · · ∪A t such that all edges from this vertex to the set A iare colored

by the ith color Since the complement of a graph can be viewed as a color class, the usual definition of n-e.c graphs is the special case of t=2

For a positive integer N, the probability space G t(N,1

t)consists of all t-colorings of the complete graph of order N such

that each edge is colored independently by any color with the probability1t The third listed author showed [13, Theorem

1.1] that almost all graphs in G t(N,1

t)have the property(n,t)-e.c as N → ∞ The proof of this theorem is similar to

the proof that almost all finite graphs have n-e.c property (see, for example, [4]) Although this result implies that there are many(n,t)-e.c graphs, it is nontrivial to construct such graphs The third listed author [13, theorem 1.2] constructed

explicitly many graphs satisfying this condition Let q be an odd prime power and F q be the finite field with q elements Let

q be a prime power such that t| (q−1)andνbe a generator of the multiplicative group of the field Fq We identify the color set with the set{0, ,t −1} The graph P q,t is a graph with vertex set Fq, the edge between two distinct vertices being

colored by the ith color if their sum is of the formνj where j≡i mod t One can show that P q,tis an (n,t)-e.c graph when

∗Corresponding author.

E-mail addresses:nguyenminhhai06@gmail.com (N.M Hai), trandangphuc234@gmail.com (T.D Phuc), vinhla@vnu.edu.vn (L.A Vinh).

http://dx.doi.org/10.1016/j.dam.2016.05.017

0166-218X/ © 2016 Published by Elsevier B.V.

q is large enough More precise, if q is a prime power such that

then P q,thas the(n,t)-e.c property (Note that, from the probabilistic argument, the upper bound for the smallest order of

an(n,t)-e.c is better than the bound in(1.1) The probabilistic bound, however, is not explicit.)

Note that the main motivation of that work is to construct new classes of n-e.c graphs From any(n,t)-e.c graph, we can

obtain an n-e.c graph by dividing the color set into two sets For a positive integer N and 0< ρ <1, the probability space

G(N, ρ)consists of graphs with vertex set of size N so that two distinct vertices are joined independently with probability

ρ It is known that almost all graphs in G(N, ρ)have the n-e.c graphs The above construction supports this statement by constructing explicitly n-e.c graphs with edge density p for any 0< ρ <1

For any positive integers n and t, let f(n,t)be the order of the smallest(n,t)-e.c graph It follows from(1.1)that

f(n,t) ≤9(t− 1 )n+n2(t− 1 )n.

In particular, if n = 3 then f(3,t) = O(93t), which is of exponential order We recall that the expressions A ≪ B and

A=O(B)are each equivalent to the statement that|A| ≤cB for some constant c>0 In [15], the second listed author gave new explicit constructions of(3,t)-graphs of polynomial order Let p be a prime such that t| (p−1), Fp be the finite field of p

elements, andνbe a generator of the multiplicative group of the field We identify the color set with the set{0, ,t−1}

For any d≥2, the graph Q p d,tis the complete graph with the vertex set Fd , the edge between two distinct vertices x,y being

colored by the ith color if their distance

∥x−y∥ = (x1−y1)2+ · · · + (x d−y d)2

is of the formνj where j≡i mod t The third listed author [15, Theorem 1.1] showed that Q p d,tis an(3,t)-e.c graph when

p≥t6and d≥5 As an immediate corollary, f(3,t) =O(t30), which is of polynomial order

1.1 Permutation polynomials

The main purpose of this paper is to give other explicit constructions of(3,t)-graphs via permutation polynomials with

two advantages over previous known results First, we can relax the condition t| (p−1) Second, we can construct explicitly (3,t)-e.c graphs with arbitrarily color density Let p be a prime and F p be the finite field of p elements Suppose that f(x)is

a polynomial over Fp of degree smaller than p A basic question in the theory of finite fields is to estimate the size V f of the value set{f(a) |a∈Fq} Because a polynomial f(x)cannot assume a given value of more than deg(f)times over a field, one has the trivial bound



p−1

deg(f)



If the lower bound in(1.2)is attained, then f(x)is called a minimal value set polynomial The classification of minimal value set polynomials is the subject of several papers; see [3,5,6,10] The results in these papers assume that p is large compared to the degree of f(x)

If the upper bound in(1.2)is attained, then f(x)is called a permutation polynomial The classification of permutation polynomials has received considerable attention See the book of Lidl and Niederreiter [9] and the survey article by Mullen [11] We identify Fpwith the set{0,1, ,p−1} LetA = A1∪ · · · ∪A t be a partition of Fp such that each A i

is a block of consecutive numbers in Fp, that is for any 1≤i≤t, there exist t i,s i such that A i = {t i+1, ,t i+s i} Let

f ∈Fp[x]be a permutation polynomial of degree at least 2 We also need to assume that p is large compared to the degree

of f(x) For any 1≤i≤t, set V i = {f(a) :a∈A i} For any d≥2, the graph G d

f, Ais the complete graph with the vertex set

Fd ; the edge between two distinct vertices x,y being colored by the ith color if their distance∥x−y∥ ∈V i We claim that

G d f, Ais an(3,t)-e.c graph when d≥5 and|A i| ≫deg(f)p5/ 6log p for all 1≤i≤t.

Theorem 1.1 Let f be a nonlinear permutation polynomial over F p and letA=A1∪ · · · ∪A t be a partition of F p such that each

A i is a block of consecutive numbers of cardinality|A i| ≫deg(f)p5/ 6log p For any d≥5, the graph G d f, Ahas(3,t)-e.c property.

Note that these graphs are just Cayley graphs of Fd To construct non-Cayley(3,t)-e.c graphs, we need to adjust the

definition of G d f, Aslightly using the following notion of mixed distance between two points x,y∈Fd p:

∥x−y∥m=2x1y1+ (x2−y2)2+ · · · + (x d−y d)2.

Theorem 1.2 Let f be a nonlinear permutation polynomial over F p and letA= A1∪ · · · ∪A t be a partition of F p such that each A i is a block of consecutive numbers of cardinality|A i| ≫deg(f)p5/ 6log p For any d≥6, the graph H f d, Ais the complete graph with the vertex set F d ; the edge between two distinct vertices x,y being colored by the ith color if their mixed distance

∥x−y∥m∈ {f(a) :a∈A i}

Then H d

f, Ais a non-Cayley(3,t)-e.c graphs.

The proof ofTheorem 1.2is exactly the same as the proof ofTheorem 1.1and is left to the interested reader

1.2 Linear maps

We also study the case of linear maps over finite fields Let f(x) = αx+ β, whereα ∈F∗p,β ∈Fp, be a linear map over the finite field Fp We identify Fpwith the residue set{0,1, ,p−1} We partition Fpinto three equitable disjoint sets as

follows For p=3k+1, set

A1= {0,1, ,k} ,

A2= {k+1,k+2, ,2k} ,

A3= {2k+1,2k+2, ,3k} ,

and for p=3k+2, set

A1= {0,1, ,k} ,

A2= {k+1,k+2, ,2k+1} ,

A3= {2k+2,2k+3, ,3k+1}

For any 1≤i≤3, set V i = {f(a) :a∈A i} For any d≥2, the graph G d

f, 3is the complete graph with the vertex set Fd; the

edge between two distinct vertices x,y being colored by the ith color if their distance∥x−y∥ ∈V i Our next result is the following theorem

Theorem 1.3 Let f(x) = αx+ β, whereα ∈F∗p ,β ∈Fp , be a linear map over the finite field F p Suppose that p≥20 is an odd

prime and d≥5 be an integer Then the graph G d

f, 3has(3,3)-e.c property.

Note that we can partition the finite field Fp into t equitable disjoint sets of consecutive numbers and define the graph

G d

f,t similarly The case t=2 has been studied in [14] More precisely, let A1= {0,1, , (p−1)/2}, the graph G dis defined

as follows: the vertices of the graph G dare the points of Fd ; and there is an edge between two vertices x and y if and only if

∥x−y∥ = (x1−y1)2+ · · · + (x d−y d)2∈A1 The third listed author [14] proved that G d2is 3-e.c for p≥7 and d≥5

In this paper, we will show further that G dis indeed a 4-e.c graph More precisely, we have the following theorem

Theorem 1.4 Let p≥11 be an odd prime and d≥6 be an integer, then the graph G d is 4-e.c.

On the other hand, one can prove that the graph G d is not a 5-e.c graph for every odd prime p and integer d Furthermore,

in contrary to the case of higher degree polynomials, the graph G d f,t, however, does not have(3,t)-e.c property when f is a linear map and t≥4 The main reason these constructions fail is that the image set of a linear map of consecutive numbers over Fpis an arithmetic progression This motivates our next construction from the multiplicative group of Fp

Letνbe a generator of the multiplicative group of the finite field Fp, then multiplicative group F∗

p ≡ { ν, ν2, , νp− 1} LetA=A1∪ · · · ∪A tbe a partition of{1, ,p−1}such that each A iis a block of consecutive numbers in Z/(p−1)Z, that

is for any 1≤i≤t, there exist t i,s i such that A i = {t i+1, ,t i+s i}mod(p−1) For any 1≤i≤t, set V i= { νa:a∈A i}

For any d≥2, the graph H d

Ais the graph with the vertex set Fd ; the edge between two distinct vertices x,y being colored

by the ith color if their distance∥x−y∥ ∈V i We claim that HAd is an(3,t)-e.c graph when d≥5 and|A i| ≫p5/ 6log p for

all 1≤i≤t.

Theorem 1.5 LetA =A1∪ · · · ∪A t be a partition of {1, ,p−1}such that each A i is a block of consecutive numbers of cardinality|A i| ≫p5/ 6log p For any d≥5, the graph HAd has(3,t)-e.c property.

2 The(3,t)-e.c property of the graphs G d

f, Aand H d

A

We first give a proof ofTheorem 1.1in this section Our proof is similar to that of Theorem 1 in [15] It suffices to show

that for any three distinct points a= (a1, ,a d), b= (b1, ,b d), c = (c1, ,c d)in Fd and i,j,k∈ {1,2,3}, there is a

point x= (x1, ,x d) ∈Fd , x̸=a,b,c such that∥x−a∥ ∈V i,∥x−b∥ ∈V jand∥x−c∥ ∈V k Therefore, we only need to

show that there exist u∈V i, v ∈V j, andw ∈V ksuch that the following system has at least four solutions (in this case, one

of these solutions is different from a, b, and c),

By eliminating x2

i’s from(2.2)and(2.3), we get an equivalent system of equations

where x·y is the usual dot product between two vectors x and y We first show that the system of two equations(2.5)and (2.6)has a solution x0for some choices of u∈V i,v ∈V j, andw ∈V k We consider two cases

Case 1 Suppose that b−a and c−a are linearly independent For any u∈V i,v ∈V j, andw ∈V k, it is clear that there is

a solution x0to the system of two equations(2.5)and(2.6)

Case 2 Suppose that b−a and c−a are linearly dependent Since b−a̸=c−a̸=0, c−a=l(b−a)for some l̸=0,1 The two equations(2.5)and(2.6)have a common solution if we can choose u∈V i,v ∈V j, andw ∈V ksuch that

u− w + ∥c∥ − ∥a∥ =l(u− v + ∥b∥ − ∥a∥ ),

or equivalently,

whereγ = ∥c∥ + (l−1)∥a∥ −l∥b∥ ∈Fp

For anyγ ∈Fq, let

Nδ= |{ (u, v, w) ∈V i×V j×V k: (l−1)u−lv + w = γ }|.

For any x∈Fp , let e p(x) =e2 πix/p From the orthogonality property of the exponential sum [9]

p− 1



s= 0

e p(st) =p if t =0,and 0 if t̸=0,

we have

Nγ = 1

p



(u,v,w)∈Vi×Vj×Vk

p− 1



s= 0

e p(s((l−1)u−lv + w − γ )),

where the inner sum is p if(l−1)u−lv + w = γ and zero otherwise This implies that

Nγ = |V i| |V j| |V k|

1

p



(u,v,w)∈Vi×Vj×Vk

p− 1



s= 1

e p(s((l−1)u−lv + w − γ ))

= |V i| |V j| |V k|

1

p

p− 1



s= 1

e p(−sγ )





u∈Ai

e p(s(l−1)u)

 





v∈Aj

e p(−slv)









w∈Ak

e p(sw)



Note that deg(f) ≥2 It is well-known (see, for example, [1, Lemma 2] and [16]) that









m≤x≤m+h

e p(sf(x))







for any s̸=0 and 0≤m,h≤p−1

Putting(2.8)and(2.9)together, we have

Nγ ≥ |V i| |V j| |V k|

p −O

 (p−1)p1/ 2(log p)3

>0 for everyγ ∈Fq This implies that Eq.(2.7)has a solution(u, v, w) ∈V i×V j×V k Hence we always can choose u ∈V i,

v ∈V j, andw ∈V ksuch that the two equations(2.5)and(2.6)have a common solution x0

Take a basis of solutions of the system

x· (b−a) =0

x· (c−a) =0,

and the solution x0 Substitute them into(2.4), we get a single quadratic equation of d−2 variables Since d−2>3, this

quadratic equation has at least q(≥4)solutions.Theorem 1.1follows immediately

Note that, proof ofTheorem 1.5is similar to the proof ofTheorem 1.1 One only need to replace the bound(2.9)in the above proof by the following well known inequality, which can be found in [7,8,12] Letνbe a generator of F∗

p , for any a∈F∗

p

and any 0≤S,T ≤p−1, the bound









S≤u≤S+T

e p(aνu)







3 The(3,3)-e.c property of the graph G f d,3

We give a proof ofTheorem 1.3in this section It suffices to show that for any three distinct points a = (a1, ,a d),

b= (b1, ,b d), c = (c1, ,c d)in Fd and i,j,k∈ {1,2,3}, there is a point x= (x1, ,x d) ∈Fd , x̸=a,b,c such that

∥x−a∥ ∈V i,∥x−b∥ ∈V jand∥x−c∥ ∈V k Follows the proof ofTheorem 1.1in Section2, we only need to show that for any 2≤l≤p−1 andδ ∈Fp, the equation

has a solution u∈V i, v ∈V j, w ∈V k Let u1= (u− β)/α,v1= (v − β)/α, andw1= (w − β)/α, then(3.1)becomes

w1+ (l−1)u1−lv1= γ /α,

where u1∈A i, v1∈A j, w1∈A k Hence, we only need to show that for any 2≤l≤p−1 and for anyδ ∈Fp, the equation

has a solution u∈A i, v ∈A j, w ∈A k

Let A0= {0,1, ,k−1}, then

A0⊂A1,

A0+ (k+1) ≡ {k+1, ,2k} ⊂A2,

A0+ (2k+1) ≡ {2k+1, ,3k} ⊂A3.

This implies that it suffices to show that for any 2≤l ≤ p−1 and for anyδ ∈ Fp, Eq.(3.2)has a solution u, v, w ∈ A0

For any u, w, w, set f l(u, v, w) = w + (l−1)u−lv For anyδ ∈ Fp, we will show an explicit solution(u, v, w) ∈ A30of

f l(u, v, w) = δ

First, let u= v =0 and letwrun from 0 to k−1, we have

f l(0,0,0) =0,

f l(0,0,1) =1,

.

f l(0,0,k−1) =k−1.

Next, letw =0 and u= v ∈ {1, ,k−1}, we have

f l(1,1,0) =p−1,

f l(2,2,0) =p−2,

.

f l(k−1,k−1,0) =p− (k−1).

We now show explicit solutions for the remaining casesδ ∈ {k,k+1, ,p−k} We consider four cases

Case 1 Suppose that l=2 Then f2(u, v, w) = (w +u) −2v Letv =0, w =k−1, and u runs from 1 to k−1, then we have

f2(1,0,k−1) =k,

f2(2,0,k−1) =k+1,

.

f2(k−1,0,k−1) =2k−2.

For remaining values ofδ ∈ {p−k−3,p−k−2,p−k−1,p−k}, let(u, v, w) ∈ {(k−5,k−1,0), (k−4,k−1,0), (k−3,

k−1,0), (k−2,k−1,0)}, respectively This implies that f2(A3) ≡Fp

Case 2 Suppose that 3≤l≤k+1 Then l−2∈A0, and we have

f l(1,0,k+1−l) =k,

f l(2,0,k+1−l) =k+ (l−1),

.

f l(k−1,0,k+1−l) =k+ (k−2)(l−1).

For any 1≤s≤k−1, letwrun from k+1−l to k−1, we have

f l(s,0,k+1−l) =k+ (s−1)(l−1),

f l(s,0,k+2−l) =k+ (s−1)(l−1) +1,

.

f(s,0,k−1) =k+ (s−1)(l−1) + (l−2).

Hence, we obtained explicit solutions forδ ∈ {k, ,k+ (k−2)(l−1)} On the other hand, l≥3 so k+ (k−2)(l−1) ≥

k+2(k−2) ≥p−k This implies that f l(A30) ≡Fpfor any 3≤l≤k+1

Case 3 Suppose that k+2≤l≤2k−1 Set a= (l−1) −k, then 1≤a≤k−2 We have

f l(a+1,a,0) = (l−1)(a+1) −la=l−a−1=k,

f l(a+1,a,1) =k+1,

.

f l(a+1,a,k−1) =2k−1,

and

f l(0,1,a−1) =a−1−l=p−k−2,

f l(0,1,a) =p−k−1,

f l(0,1,a+1) =p−k.

This implies that f l(A3) ≡Fp for any k+2≤l≤2k−1

Case 4 Suppose that 2k≤l≤p−1 Set l′=p−l+1 then 2≤l′<2k−1 and

f l(u, v, w) =f l′(v,u, w).

It follows immediately from Case 2.3 that f l(A3) =f l′(A3) ≡Fp

Therefore, we showed that for any 2≤l≤p−1 andδ ∈Fq, Eq.(3.1)has a solution u∈V i, v ∈V j, w ∈V k.Theorem 1.3 now follows as in the proof ofTheorem 1.1 

4 The 4-e.c property of the graph G2d

We give a proof ofTheorem 1.4in this section Let A2=Fp\A1 We will show that for every four distinct vertices a,b,c,d

of the graph G d2, andα, β, γ , δ ∈ {1,2}, there exists a vertex x̸=a,b,c,d such that

By eliminating x i’s from(4.2),(4.3), and(4.4), we have the following equivalent system of equations

x· (b−a) =u− v + ∥b∥ − ∥a∥

x· (c−a) = w − v + ∥c∥ − ∥a∥

x· (d−a) =t− v + ∥d∥ − ∥a∥

We first show that the system of three equations(4.6),(4.7),(4.8)has a solution x0for some choices of u∈Aα,v ∈Aβ,

w ∈Aγ, and t∈Aδ We consider three cases

Case 1 Suppose that three vectors b−a, c−a, and d−a are linear independent For any u∈Aα, v ∈Aβ, w ∈Aγ, and

t ∈Aδ, it is clear that there is a solution x0to the system of three equations(4.6),(4.7), and(4.8)

If three vectors b−a, c−a, and d−a are not linear independent, then we have two cases.

Case 2 Suppose that d−a=t1(b−a) +t2(c−a), with(t1,t2) ̸∈ {(0,0), (0,1), (1,0)}and b−a, c−a are linearly

independent In this case, the system of equations(4.6),(4.7),(4.8)has a solution if we can choose u∈Aα, v ∈Aβ, w ∈Aγ,

and t∈Aδsuch that

u−t+ ∥d∥ − ∥a∥ =t1(u− v + ∥b∥ − ∥a∥ ) +t2(u− w + ∥c∥ − ∥a∥ ),

or

where a= ∥d∥ − ∥a∥ −t1(∥b∥ − ∥a∥ ) +t2(∥c∥ − ∥a∥ )

Let A0= {0,1, ,k−1}, then

A0⊂A1,

A0+ (k+1) =A2,

hence, it suffices to consider the case(u, v, w,t) ∈A4 We have the following subcases

Case 2.1 At least one of 1−t1−t2,t1,t2is different from 0,1,p−1 (Note that, we identify Fpwith the set{0,1, ,p−1}) Without loss of generality, we can assume that 1−t1−t2̸=0,1,p−1 as other cases are similar We label Fparound the circle Fixvandw, then Eq.(4.9)becomes t= (1−t1−t2)u+b (where b=t1v +t2w +a) Going|A0| =k steps of length

(1−t1−t2)around the circle, we cannot avoid any block of k consecutive points Hence, we can choose u∈A0such that (1−t1−t2)u+b∈Aδ This implies that the system of equations(4.6),(4.7), and(4.8)has a solution

Case 2.2 Suppose that 1−t1−t2,t1,t2∈ {0,1,p−1} Since

(t1,t2) ̸∈ {(0,0), (0,1), (1,0)},

we only need to consider three cases: t1=1,t2=1; t1=1,t2=p−1; and t1=p−1,t2=1

If t1= 1 and t2= 1, then t = −u+ v + w +a Fix u and letv = w, we are back to Case 2.1 We can do similarly for

t1=1,t2=p−1 and t1=p−1,t2=1

Case 3 Suppose that c−a=t1(b−a)and d−a=t2(b−a)with t1̸=t2̸=0,1 Similar to Case 2, we will show that

for any t1,t2,c1,c2∈Fp, the following system of equations has a solution(u, v, w,t) ∈A40

Without loss of generality, we assume that t1 < t2 Let k′ = k−1, then p = 2k′+3 and A0 = {0,1, ,k′} Let

g t2(u, v, w,t) =t+ (t2−1)u−t2vand N= { (u, v, w,t) ∈A4: w + (t1−1)u−t1v =c1} We will show that g t2(N) ≡Fp Note that if(u, v, w,t)is a solution for the case(t1,t2,c1,c2)then(k′−u,k′− v,k′− w,k′−t)is a solution for the case(t1,t2,p−c1,p−c2)and(v,u, w,t)is a solution for the case(p−t1+1,p−t2+1,c1,c2) Therefore, we only need

to consider the following sub cases

Case 3.1 Suppose that 2 ≤ t1 < t2 ≤ k′and 0 ≤ c1 ≤ k′ We choose i,m such that k′−c1 = it1+m and 0 ≤ i,

0≤m≤t1−1 Let s be the smallest natural number such that c1+s=j(t1−1) We have j(t1−1) ≤k′+ (t1−1) −1

We start with some specific values of(u, v, w,t) ∈N:

g t2(0,0,c1,t) =t, 0≤t≤k′

g t2(1,1,c1+1,0) =p−1,

g t2(2,2,c1+2,0) =p−2,

· · ·

g t2(m,m,c1+m,0) =p−m,

g t2(m,m+1,c1+m+t1,0) =p−m−t2,

.

g t2(m,m+i,c1+m+it1,0) =p−m−it2.

Since t2≤k′, for any 0≤i′≤i, we start with(u, v, w,t) = (m,m+i′,c1+m+i′t1,0)and let t run from 0 to k′then

g t2obtains all possible values between−m−i′t2and−m− (i′−1)t2 Next, we consider the following values of g t2:

g t2(0,0,c1,k′) =k′,

g t2(s+1,s,c1+s− (t1−1),k′) =t2−1+k′−s,

g t2(s+2,s,c1+s−2(t1−1),k′) =2(t2−1) +k′−s,

· · ·

g t2(s+j,s,c1+s−j(t1−1),k′) =j(t2−1) +k′−s.

To obtain all values between k′,t2−1+k′−s,2(t2−1) +k′−s, ,j(t2−1) +k′−s, we let t run backward from k′

to 0 We, however, need to check that s+j≤k′ We have

s+j≤ (t1−2) +k

′+t1−2

t1−1 = (t1−1) +k

′−1

t1−1.

Let k′=r(t1−1) +s′with 0≤r,0≤s′≤t1−2 Then

(t1−1) +k′−1

t1−1 = (t1−1) +r(t1−1) +s′−1

t1−1

= (t1−1) +r+ s′−1

t1−1 ≤r(t1−1) +1+ (s′−1) =k′. Therefore, we have a solution(u, v, w,t) ∈A40of the system of equations(4.10)and(4.11) Note that the above listed

values of g t2 cover Fp unless p−m−it2 >j(t2−1) +k′−s+1 Since m = k′−c1−it1,s =c1−j(t1−1), we have

2k′+3>2k′+ (i+j)(t −t ) +1, or 1≥ (i+j)(t −t )

We consider two sub cases.

Subcase 3.1.1 Suppose that i=j=0 This implies that c1=0,s=0 and k′=m≤t1−1, which is a contradiction

Subcase 3.1.2 Suppose that i =1,j= 0 (the case i =0,j=1 is similar) This implies that c1+s= 0, and c1 = 0 If

c2 ≤k′, then(u, v, w,t) = (0,0,0,c2)is a solution of the system of equations(4.10)and(4.11) If c2 ≥k′+3, then we have a solution(u, v, w,t) = (p−c2,p−c2,p−c2,0) If c2=k′+1 or k′+2, then(t1,t1−1,0,k′+1− (t2−t1))and (k′−t1,k′− (t1−1),k′, (t2−t1) −1)are solutions of the system, respectively

Case 3.2 Suppose that 2≤t1<t2≤k′and c1=k′+1 Let

h t2(u, v, w,t) =g t2(u, v, w,t) − (t2−1).

Choose i,m such that k′+ (t1−1) −c1=it1+m and 0≤i, 0≤m≤t1−1 Let s be the smallest natural number such that

c1− (t1−1)+s=j(t1−1) Hence, j(t1−1) ≤k′+ (t1−1)−1 We consider the following specific values of(u, v, w,t) ∈N

h t2(1,0,c1−t1+1,t) =t, 0≤t≤k′,

h t2(2,1,c1−t1+2,0) =p−1,

h t2(3,2,c1−t1+3,0) =p−2,

· · ·

h t2(m,m−1,c1−t1+m,0) =p−m+1,

h t2(m,m,c1−t1+m+t1,0) =p−m−t2+1,

· · ·

h t2(m,m+i−1,c1−t1+m+it1,0) =p−m−it2+1,

and

h t2(1,0,c1−t1+1,k′) =k′,

h t2(s+1,s,c1−t1+1+s− (t1−1),k′) =k′−s,

h t2(s+2,s,c1−t1+1+s−2(t1−1),k′) = (t2−1) +k′−s,

· · ·

h t2(s+j,s,c1−t1+1+s−j(t1−1),k′) = (j−1)(t2−1) +k′−s.

The remaining of this case can be done similarly as in the proof of Case 3.1

Case 3.3 Suppose that 2≤ t1 <t2 ≤ k′and c1 ≥ k′+2 Note that for some c1,c2such that the system of equations (4.10)and(4.11)has a solution then the system also has a solution for−c1, −c2 Since 0≤ −c1≤k′+1, we come back to Cases 3.1 and 3.2

Case 3.4 Suppose that 2≤t1≤k′<t2 We can rewrite the system of equations(4.10)and(4.11)as following

where t′

2=p−t2+1 One can show explicitly solutions for the cases t2=k′+1,k′+2,k′+3 When k′+4≤t2, we can repeat the proof of the above cases

Case 3.5 Suppose that k′ < t1 < t2 Similarly, one can show explicitly solutions for the case k′ < t2 ≤k′+3 When

k′+4≤t2, we consider(p−t1+1,p−t2+1)instead of(t1,t2)and go back to the above cases

Hence we always can choose u∈ Aα,v ∈Aβ,w ∈ Aγ, and t ∈Aδsuch that the system of three equations(4.6),(4.7), (4.8)has a solution x0

Take a basis of solutions of the system

x· (b−a) =0

x· (c−a) =0

x· (d−a) =0,

and the solution x0 Substitute them into(4.5), we get a single quadratic equation of d−3 variables Since d−3>3, this

quadratic equation has at least q(≥5)solutions.Theorem 1.1follows immediately 

5 Remarks

It is remarkable that the graph G d2is not n-e.c for any n≥5 It is clear that we only need to show that for n=5

Suppose for the contrary that the graph G d is 5-e.c then for every five distinct vertices a,b,b,c,d,e of the graph G d, and

α, β, γ , δ, ϵ ∈ {1,2}, there exists a vertex x̸=a,b,b,c,d,e such that

∥x−b∥ = v ∈Aβ (5.2)

By eliminating x i’s from(5.2),(5.3),(5.4), and(5.5)we have the following equivalent system of equations

x· (b−a) =u− v + ∥b∥ − ∥a∥

x· (c−a) =u− w + ∥c∥ − ∥a∥

x· (d−a) =u−t+ ∥d∥ − ∥a∥

x· (e−a) =u−r+ ∥e∥ − ∥a∥

We will consider the case c−a, d−a, and e−a are linearly dependent on b−a and show that the above system of

equations has no solution Similarly as in the previous section, it is suffice to consider the case(u, v, w,t,r) ∈A5 By the

linearly dependent of the above vectors, there exist t1,t2,t3such that c−a=t1(b−a), d−a=t2(b−a)v e−a=t3(b−a)

Since a,b,b,c,d,e are distinct, we have t1,t2,t3̸=0,1 If the system of equations(5.7)–(5.10)has a solution, then there exit(u, v, w,t,r) ∈A5such that

u− w + ∥c∥ − ∥a∥ =t1(u− v + ∥b∥ − ∥a∥ )

u−r+ ∥d∥ − ∥a∥ =t2(u− v + ∥b∥ − ∥a∥ )

u−t+ ∥e∥ − ∥a∥ =t3(u− v + ∥b∥ − ∥a∥ ),

or

where

c1= ∥c∥ − ∥a∥ −t1(∥b∥ − ∥a∥ ),

c2= ∥d∥ − ∥a∥ −t2(∥b∥ − ∥a∥ ),

c3= ∥e∥ − ∥a∥ −t3(∥b∥ − ∥a∥ ).

We will show a specific case such that the system of equations(5.11),(5.12)v(5.13)has no solution Let t1 = 2,t2 =

3,t3 = 4,c1 = 0,c2 = p−5

2 ,c3 = p− 2, we will show that the following system of equations has no solution (u, v, w,t,r) ∈A50

t =3v −2u+p−5

Consider Eq.(5.14) Suppose that 2v −u < 0 Since 2v −u ≥ −u ≥ −p−3

2 , we have−p−3

2 ≤ w < 0, which is a contradiction tow ∈A0 Hence, 2v −u≥0

On the other hand, 2v −u≤2v ≤p−3 Hence, it follows fromw ∈A0that

0≤2v −u≤ p−3

We have the following condition ofvand u

u

2 ≤ v ≤ u

p−3

Suppose that 4v −3u<0 From(5.16)and(5.17), we have

p>r=p−2+4v −3u≥p−2+2u−3u=p−2−u≥p−2− p−3

2 > p−3

which is a contradiction to 0≤r≤p−3 Hence, 4v −3u≥0

Since we need r∈A0, it follows that

ip≤4v −3u+p−2≤ip+p−3

for a positive integer i.

Suppose that i≥2 From(5.18), we have 4v −3u+p−2≥2p, hencev ≥ 3

4u+p+2

4 , which is contradiction to(5.17)

Therefore, i=1, substituting into(5.18), we have

p≤4v −3u+p−2≤p+p−3

which is equivalent to

3

4u+

2

4 ≤ v ≤ 3

4u+

p+1

From(5.15)and(5.19), we have

t =3v −2u+p−5

9

4u+

6

4 −2u+

p−5

u

2p−4

4 > p−3

Since we need t ∈A0, it follows that 3v −2u+p−5

2 ≥p, or 3v −2u≥ p+5

2 Hence,

v ≥ 2

3u+

p+5

From(5.19)and(5.20), we have23u+p+5

6 ≤3

4u+p+1

8 orp+217 ≤u, which is a contradiction It concludes that the graph

G d is not n-e.c for any n≥5

Acknowledgment

The third author’s research was supported by Vietnam National University Hanoi Project QG.13.02

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