# A Paradigmatic Approach to Find Equal Sum Partitions of Zero-Divisors via Complete Graphs

Authors: M. Haris Mateen; M. Khalid Mahmood; Daud Ahmad; Shahbaz Ali; Shajib Ali
Journal: Journal of Chemistry
 (2022)

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## Abstract

In computer science and mathematics, a partition of a set into two or more disjoint subsets with equal sums is a well-known NP-complete problem. This is a hard problem and referred to as the partition problem or number partitioning. In this paper, we solve a particular type of NP-complete problem on the set of all zero-divisors ofZn including zero, where Zn is the ring of residue classes of a positive integer n. In this regard, we introduce and investigate quadratic zero-divisor graph in which we build an edge between zero-divisors zi and zj if and only if zi2≡zj2modn,i≠j. This is denoted as G⏞2,n. We characterize these graphs in term of complete graphs for classes of integers 2α,pα,2αp,2pα and pq, where α is any positive integer and p,q are odd primes.


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## Body

## 1. Introduction

The subject of graph theory plays a vital role in informatics, chemistry, physics, biology, etc. In the field of biology, many applications of graph theory have been elaborated in [1]. In [2], many interesting linkages of graphs have been incorporated in the subject of chemistry. Graphs can also be found in other social and information systems and their related framework (for detail, see [3]). In physics, many circuits can be developed using graphs [4]. Many unknown molecular atomic numbers were computed by using group symmetry of graphs [5]. In computer science, several problems were solved through graphs that were not easily visualized earlier. For combinatorics and discrete mathematics, the relation between number theory and graph theory is of crucial importance. A branch of pure mathematics in the early days, number theory, was basically less applicable in real life. But, in combination with today’s computing technologies, it provides solutions to many current problems. In today’s digital world, cryptography is one of the main fields where cyber security is a major concern. A message sent from a sender to receiver in online communication carries the risk of being seen without proper protection by an unknown person. This problem is solved by the use of the encryption/decryption principle. The message which is sent by the sender is said to be “encrypted” or encoded with the help of a large number, usually prime, which is said to be a “key,” the receiver must have that same key to “decrypt” or decode the message. Through the number theory, such large prime numbers can easily be generated to secure most of the encrypted messages. Maurer [6] derived an efficient algorithm to generate such numbers with the help of number theory. The congruence relations play an essential role in cryptography [7].Rogers discussed the action of a quadratic map on multiplicative groups under modulo a primep using the associated directed graph for which there is an edge from each element to its image [8]. He established a formula to decompose a graph into cyclic components with their attached trees. The necessary and sufficient conditions for the existence of isolated fixed points have also been established. In [9], the structures of graphs of quadratic congruences for composite modulus have been studied by Somer and Krizek. Mahmood and Ahmad proposed many new results of graphs over residues modulo prime powers in [10, 11]. Haris and Khalid investigated the structure of power digraphs associated with the congruence xn≡ymodm in [12–15]. In [16], Yang-Jiang and Gao-Hua introduced the concept of square mapping graphs of the Gaussian ring Zni. In [17], Rezaei et al. discussed the concept of quadratic residues graphs and fully characterized them for each positive integer n over the unit elements of ring of integers. Huo [18] discussed the topological indices of mth chain hex-derived networks of the third type. Zhao et al. [18] investigated the statistics and calculation of entropy of dominating david derived networks. Azeem et al. [19] studied topological aspects of 2D structures of trans-Pd(NH2)S lattice and a metal-organic superlattice. Rashmanlou et al. [20] discussed the cubic graphs with novel application. Amanathulla et al. [21] investigated the distance between the two surjective labellings of paths and interval graphs. To make this paper self-readable, the readers are suggested to read [22–26].Two nonzero integersx and y are called zero-divisors in the ring Zn if and only if xy≡0modn [27]. Let Z be the set of all zero-divisors of Zn. We recall that a graph G⏞2,n=V,E with a vertex set of V=Z∪0 and edges set of E=zizj∣zi2≡zj2modn,i≠j is called a quadratic zero-divisor graph. For n=30, the vertex set is V=0,2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28. By solving the congruences zi2≡zj2mod30 for each zi,zj∈Z, then there are 2, 2, and 6 copies of K1, K4, and K2, respectively, as shown in Figure 1. Note that each copy of K2 and K4 has equal sums, namely, 30 and 60, respectively. Here, Kn is a complete graph (for detail, see [28]).Figure 1
G⏞2,30=2K1⊕6K2⊕2K4.


## 2. Quadratic Zero-Divisor Graphs overZ2α and Zpα

In this section, we characterize quadratic zero-divisor graphs for classes of integers modulo2α and pα for each positive integer α and odd prime p. For the proof of the following theorem, we use the technique of “proof by construction” successively.Theorem 1.
Letn=2αbe an integer. Then(1)G⏞2,2α=Kα,ifα=1,2,2K2,ifα=3,2K4,ifα=4,2K4⊕K8,ifα=5,4K8,ifα=6,6K8⊕K16,ifα=7,8K8⊕4K16,ifα=8,⊕i=1α−7/22α−5−2i−1K23+i−1⊕6K2α−1/2⊕K2α+1/2,ifα≥9,α≡1mod2,⊕i=1α−6/22α−2i−3K22+i⊕22K2α/2,ifα≥10,α≡0mod2.Proof.
LetS=2n∣n=0,1,2,⋯,2α−1−1 be the set of all zero-divisors of Z2α including zero for each positive integer α. To find the number of solutions of α2≡β2mod2α, we start from γ2≡β2mod2; in this case just, S=0. Therefore, γ2≡0mod2 has only one solution which is zero, but by the definition of quadratic zero-divisors graph, there will be a no loop, so G⏞2,2=K1. For α=2, γ2≡0≡22mod22 has two solutions, namely, γ=0,2, so G⏞2,22=K2. If α=3, then there are two congruences:
(2)γ2≡0mod23,γ2≡4mod23.The roots of these congruences areγ=0,4 and γ=2,6, respectively. Thus, there are two copies of K2.Forα=4, we have
(3)γ2≡0mod24,γ2≡4mod24.The roots of these congruences are0,4,8,12 and 2,6,10,14, respectively. Therefore, G⏞2,24=2K4. There are three congruences for α=5:
(4)γ2≡0mod25,γ2≡4mod25,γ2≡16mod25.The solution sets of these congruences are0,8,16,24, 2,6,10,14,18,22,26,30, and 4,12,20,28, respectively. Therefore, there are 2 copies of K4 and one copy of K8. For α=6, we have
(5)γ2≡0mod26,γ2≡4mod26,γ2≡16mod26,γ2≡36mod26.The zeroes of these congruence are0,8,16,24,32,40,48,54, 2,18,34,50∪62,46,30,14,4,12,20,28,36,44,52,60, and 6,22,38,54∪58,42,26,10, respectively. When α=7, then there are 7 different congruences:
(6)γ2≡2k2mod27,k=0,1,3,4,5,7,γ2≡16mod27.The roots of these congruences are2k+16n∣n=0,1,2⋯23−1, 2k+32n∣n=0,1,2⋯22−1∪27−32n−2k∣n=0,1,2⋯22−1, k=1,3,4,5,7, and 4+8n∣n=0,1,2⋯24−1. Hence, there are six copies of K8 and one copy of K16. There are 12 number of congruences when α=8:
(7)γ2≡2+4k2mod28,k=0,1,⋯,23−1,γ2≡4k2mod28,k=0,1,⋯,22−1.The zeroes of these congruences are2+4k+64n∣n=0,1,2⋯22−1∪28−64n−4k−2∣n=0,1,2⋯22−1, k=0,1,⋯,23−1, 4k+32n∣n=0,1,2⋯24−1, 4k+32nn=0,1,2⋯23−1∪28−32n−4kn=0,1,2⋯23−1, and k=1,⋯,22−1. Thus, we have G⏞2,28=8K4⊕4K16. For α=9, there are 23 different congruences:
(8)γ2≡2+4k2mod29,k=0,1,⋯,24−1,(9)γ2≡4k2mod29,k=0,1,3,4,5,7,(10)γ2≡64mod29.The solutions of Congruences (8)–(10) are 2+4k+128n∣n=0,1,2⋯22−1∪29−128n−4k−2∣n=0,1,2⋯22−1, k=0,1,⋯,26−1, 4k+32n∣n=0,1,2⋯24−1, 4k+64n∣n=0,1,2⋯23−1∪29−64n−4k∣n=0,1,2⋯23−1, k=1,3,4,5,7, and 8+16n∣n=0,1,2⋯25−1. That is, G⏞2,29=16K8⊕6K16⊕K32. When α=10, then there are 44 different congruences:
(11)γ2≡2+4k2mod210,k=0,1,⋯,25−1,(12)γ2≡4+8k2mod210,k=0,1,⋯,23−1,(13)γ2≡8k2mod210,k=0,1,⋯,22−1.Sequences of roots of Congruences (11)–(13) are
(14)2+4k+256n∣n=0,1,2⋯22−1∪210−256n−4k−2∣n=0,1,2⋯22−1,k=0,1,⋯,25−1,4+8k+128n∣n=0,1,2⋯23−1∪210−128n−8k−4∣n=0,1,2⋯23−1,k=0,1,⋯,23−1,8k+64n∣n=0,1,2⋯25−1,8k+64n∣n=0,1,2⋯24−1∪210−64n−8k∣n=0,1,2⋯24−1,k=1,⋯,22−1.That is,G⏞2,210=32K8⊕8K16⊕4K32. For α=11:
(15)γ2≡2+4k2mod211,k=0,1,⋯,26−1,(16)γ2≡4+8k2mod211,k=0,1,⋯,24−1,(17)γ2≡8k2mod211,k=0,1,3,4,5,7,(18)γ2≡256mod211.The zeroes of Congruences (15)–(18) are 2+4k+512n∣n=0,1,2⋯22−1∪211−512n−4k−2∣n=0,1,2⋯22−1, k=0,1,⋯,26−1, 4+8k+256n∣n=0,1,2⋯23−1∪211−256n−8k−4∣n=0,1,2⋯23−1, k=0,1,⋯,24−1, 8k+64n∣n=0,1,2⋯25−1, 8k+128n∣n=0,1,2⋯24−1∪211−128n−8k∣n=0,1,2⋯24−1, k=1,3,4,5,7, and16+32n∣n=0,1,2⋯26−1. Therefore, we have G⏞2,211=64K8⊕16K16⊕6K32⊕K64. There are 172 congruences when α=12:
(19)γ2≡2+4k2mod212,k=0,1,⋯,27−1,(20)γ2≡4+8k2mod212,k=0,1,⋯,25−1,(21)γ2≡8+16k2mod212,k=0,1,⋯,23−1,(22)γ2≡16k2mod212,k=0,1,⋯,22−1.The sets of roots of Congruences (19)–(22) are 2+4k+1024n∣n=0,1,2⋯22−1∪212−1024n−4k−2∣n=0,1,2⋯22−1, k=0,1,⋯,27−1, 4+8k+512n∣n=0,1,2⋯23−1∪212−512n−8k−4∣n=0,1,2⋯23−1, k=0,1,⋯,25−1, 8k+256n∣n=0,1,2⋯24−1∪212−256n−8k∣n=0,1,2⋯24−1, k=0,1,⋯,23−1, 16k+128n∣n=0,1,2⋯26−1, 16k+128n∣n=0,1,2⋯25−1∪212−128n−8k∣n=0,1,2⋯25−1, and k=1,⋯,22−1. That is, G⏞2,212=128K8⊕32K16⊕8K32⊕4K64. Generalize the sequence for α≥9 with α an odd; then, there are 242α−7−1+21/3 number of congruences. These are
(23)γ2≡2i+2i+1k2mod2α,k=0,1,⋯,2α−2i−3−1,i=1,2,3,⋯,α−72,(24)γ2≡2α−5/2k2mod2α,k=0,1,3,4,5,7,(25)γ2≡2α−3mod2α.Sequences of roots of Congruences (23)–(25) are 2i+2i+1k+2α−i−1n∣n=0,1,2⋯2i+1−1∪2α−2α−i−1n−2i+1k−2i∣n=0,1,2⋯2i+1−1, k=0,1,⋯,2α−2i−3−1, i=1,2,3,⋯,α−7/2, 2α−5/2k+2α+1/2n∣n=0,1,2,⋯2α−1/2−1,2α−5/2k+2α+3/2n∣n=0,1,2,⋯2α−3/2−1∪2α−2α+3/2n−2α−5/2k∣n=0,1,2,⋯2α−3/2−1, k=1,3,4,5,7, and 2α−3/2+2α−1/2n∣n=0,1,2⋯2α+1/2−1. That is, G⏞2,2α=⊕i=1α−7/22α−5−2i−1K23+i−1⊕6K2α−1/2⊕K2α+1/2. For the second case, when α is even and α≥10, then we have 252α−8−1+36/3 number of congruences:
(26)γ2≡2i+2i+1k2mod2α,k=0,1,⋯,2α−2i−3−1,i=1,2,3,⋯,α−62,(27)γ2≡2α−4/2k2mod2α,k=0,1,⋯,22−1.The sequences of zeroes of Congruences (26)–(27) are 2i+2i+1k+2α−i−1n∣n=0,1,2⋯2i+1−1∪2α−2α−i−1n−2i+1k−2i∣n=0,1,2⋯2i+1−1, k=0,1,⋯,2α−2i−3−1, i=1,2,3,⋯,α−6/2, 2α−4/2k+2α+2/2n∣n=0,1,2⋯,2α/2−1, 2α−4/2k+2α+2/2n∣n=0,1,2⋯,2α−2/2−1∪2α−2α+2/2n−2α−4/2k∣n=0,1,2α−2/2−1, and k=1,⋯,22−1. Hence, G⏞2,2α=⊕i=1α−6/22α−2i−3K22+i⊕22K2α/2.The quadratic zero-divisor graph forn=128 is shown in Figure 2.Theorem 2.
Letp be an odd prime. Then,
(28)G⏞2,pα=K1,ifα=1,Kp,ifα=2,p−12K2p⊕Kp,ifα=3,p−12pK2p⊕Kp2,ifα=4,⊕i=1α−1/2pα−2i−pα−2i−12K2pi⊕Kpα−1/2,ifα≥5,α≡1mod2,⊕i=1α−2/2pα−2i−pα−2i−12K2pi⊕Kpα/2,ifα≥6,α≡0mod2.Proof.
Supposep is an odd prime and S=kpαk=0,1,2,⋯,pα−1−1 be the set of zero-divisors including zero of pα for each positive integer α. To solve congruence γ2≡β2modpα for each α≥1, we start with γ2≡β2modp. In this case, 0 is the only root of this congruence, but there is no edge between two vertices when they are same, so G⏞2,p=K1. When α=2, then there are only one congruence, namely, γ2≡0modp2; the roots of this congruence are γ=0,p,2p,⋯,p−1p. There arepnumber of solution that is graph will be a complete graph of orderp. For α=3, there are p+1/2 number of distinct congruences:
(29)γ2≡0modp3,γ2≡pk2modp3,k=1,2,⋯,p−12.Figure 2
G⏞2,27=6K8⊕K16.The zeroes of these congruences areγ=p2n−1n=1,2,⋯,p and γ=p2n−1+pkk=1,2,⋯,p−1/2,n=1,2,⋯,p∪γ=p3−p2n−1−pkk=1,2,⋯,p−1/2,n=1,2,⋯,p, respectively. When α=4 then the number of distinct congruences are p2−p+2/2. These are
(30)γ2≡0modp4,γ2≡pk2modp4,k=1,2,⋯,p2−12,butk≠pt,t=1,2,⋯,p−12.The sequences of the roots of these congruences areγ=p2n−1n=1,2,⋯,p2, and γ=p3n−1+pkk=1,2,⋯,p2−1/2,butk≠pt,t=1,2,⋯,p−1/2,n=1,2,⋯,p∪γ=p4−p3n−1−pkk=1,2,⋯,p2−1/2,butk≠pt,t=1,2,⋯,p−1/2,n=1,2,⋯,p,, respectively. For α=5, we have
(31)γ2≡0modp5,γ2≡p2k2modp5,k=1,2,⋯,p−12,γ2≡pk2modp5,k=1,2,⋯,p3−12,butk≠pt,t=1,2,⋯,p2−12.The zeroes of these congruences areγ=p3n−1∣n=1,2,⋯,p2, γ=p3n−1+p2kk=1,2,⋯,p−1/2,n=1,2,⋯,p2∪γ=p5−p3n−1−p2kk=1,2,⋯,p−1/2,n=1,2,⋯,p2 and γ=p4n−1+pkk=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2,n=1,2,⋯,p∪γ=p5−p4n−1−pkk=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2,n=1,2,⋯,p, respectively. If α=6, then we have
(32)γ2≡0modp6,(33)γ2≡p2k2modp6,k=1,2,⋯,p2−12,(34)butk≠pt,t=1,2,⋯,p−12(35)γ2≡pk2modp6,k=1,2,⋯,p4−12,(36)butk≠pt,t=1,2,⋯,p3−12.The sequences of roots of Congruences (32)–(35) are γ=p3n−1∣n=1,2,⋯,p3,γ=p4n−1+p2kk=1,2,⋯,p2−1/2,butk≠pt,t=1,2,⋯,p−1/2,n=1,2,⋯,p2∪γ=p6−p3n−1−p2kk=1,2,⋯,p2−1/2,butk≠pt,t=1,2,⋯,p−1/2,n=1,2,⋯,p2, and γ=p5n−1+pkk=1,2,⋯,p4−1/2,butk≠pt,t=1,2,⋯,p3−1/2,n=1,2,⋯,p∪γ=p6−p5n−1−pkk=1,2,⋯,p4−1/2,butk≠pt,t=1,2,⋯,p3−1/2,n=1,2,⋯,p, respectively. Now, we are going to derive generalize sequence for both odd and even α≥5. There are pα−1+2p+1/2p+1 number of distinct congruences, when α≥5 and α≡1mod2. These are
(37)γ2≡pik2modpα,k=1,2,⋯,pα−2i−12,but,k≠pt,t=1,2,⋯,pα−2i−1−12,i=1,2,⋯,α−12,γ2≡0modpα.The sequences of roots are∪i=1α−1/2γ=pα−in−1+pikk=1,2,⋯,pα−2i−1/2,butk≠pt,t=1,2,3,⋯,pα−2i−1−1/2,n=1,2,⋯,pi∪γ=pα−pα−in−1−pikk=1,2,⋯,pα−2i−1/2,butk≠pt,t=1,2,⋯,pα−2i−1−1/2,n=1,2,⋯,pi and γ=pα+1/2n−1∣n=1,2,⋯,pα−1/2, respectively. Therefore, for every positive integer α grater than 5 with α≡1mod2, G⏞2,pα=⊕i=1α−1/2pα−2i−pα−2i−1/2K2pi⊕Kpα−1/2. In the second case, when α≥6 and α≡1mod2, then a number of distinct congruences are pα−1+p+2/2p+1. We have
(38)γ2≡pik2modpα,k=1,2,⋯,pα−2i−12,(39)but,k≠pt,t=1,2,⋯,pα−2i−1−12,i=1,2,⋯,α−22,(40)γ2≡0modpα.The zeroes of Congruences (38) and (40) are ∪i=1α−2/2γ=pα−in−1+pikk=1,2,⋯,pα−2i−1/2,butk≠pt,t=1,2,3,⋯,pα−2i−1−1/2,n=1,2,⋯,pi∪γ=pα−pα−in−1−pikk=1,2,⋯,pα−2i−1/2,butk≠pt,t=1,2,⋯,pα−2i−1−1/2,n=1,2,⋯,pi and γ=pα/2n−1∣n=1,2,⋯,pα/2, respectively. Hence, for α=6 with α≡0mod2, G⏞2,pα=⊕i=1α−2/2pα−2i−pα−2i−1/2K2pi⊕Kpα/2.The quadratic zero-divisor graph forn=243 is shown in Figure 3.Figure 3
G⏞2,243=9K6⊕K9⊕K18.


## 3. Quadratic Zero-Divisor Graphs overZ2αp,Zpq, and Z2pα

In this section, we characterize quadratic zero-divisor graphs forn=2αp,2pα,pq, where p,q are odd primes.Theorem 3.
Letp be an odd prime. Then,
G⏞2,2αp=2K1⊕p−1/2K2,ifα=1,2K2⊕p−1/2K4,ifα=2,2K2⊕pK4,ifα=3,4K4⊕p−1K8,ifα=4,6K4⊕pK8⊕p−1/2K16,ifα=5,8K4⊕4K8⊕2p−1K16,ifα=6,16K4⊕6K8⊕3p−2K16⊕p−1/2K32,ifα=7,32K4⊕8K8⊕4pK16⊕2p−1K32,ifα=8,64K4⊕16K8⊕8p−2K16⊕3p−2K32⊕p−1/2K64,ifα=9,⊕i=122α−2i−1K2i+1⊕i=1α−8/22α−4−2ip−2α−5−2iK23+i⊕4pK2α/2⊕2p−1K2α+2/2,ifα≥10,α≡0mod2,⊕i=122α−2i−1K2i+1⊕i=1α−9/22α−4−2ip−2α−5−2iK23+i⊕8p−2K2α−1/2⊕3p−2K2α+1/2⊕p−1/2K2α+3/2,ifα≥11,α≡1mod2.Proof.
Letn=2αp be an integer, where p is an odd prime. For α=1, S=2m∣m=0,1,2,⋯p−1∪p is set of zero-divisors of Z2p including 0. There are p+1/2 number of distinct congruences:
(41)γ2≡0mod2p,γ2≡pmod2p,γ2≡2k2mod2p,k=1,2,⋯,p−12.
γ=0, γ=p, and γ=2k,2p−2k,k=1,2,3,⋯,p−1/2 are zeroes of congruences, respectively. When α=2, then there are congruences are
(42)γ2≡0mod22p,γ2≡pmod22p,γ2≡2k2mod22p,k=1,2,⋯,p−12.The set of roots of these congruences are0,2p, p,3p, 2k,2p−2k∪22p−2k,22p−2p+2k,k=1,2,⋯,p−1/2, respectively. There are p+2 number of congruences for α=3; these are γ2≡0mod23p,γ2≡2p2mod23p,γ2≡p2mod23p,γ2≡2k2mod23p,k=1,2,⋯,p−1. The zeroes of these congruences are 0,22p, 2p,23p−2p, p,3p,5p,7p, and 2k,22p−2k∪23p−2k,23p−22p+2k,k=1,2,⋯,p−1, respectively. When α=4, then there are p+3 number of congruences:
(43)γ2≡0mod24p,γ2≡p2mod24p,γ2≡2p2mod24p,γ2≡3p2mod24p,γ2≡2k2mod24p,k=1,2,⋯,p−1,where the zeroes are 0,22p,23p,22p+23p, p,7p,9p,15p, 2p,6p,10p,14p, 3p,5p,11p,13p, and 2k,22p−2k,22p+2k,23p−2k∪24p−2k,24p−22p+2k,24p−22p−2k,24p−23p+2k,k=1,2,⋯,p−1, respectively. If α=5, then there are 3p+11/2 distinct congruences:
(44)γ2≡0mod25p,(45)γ2≡pk2mod25p,k=1,3,5,7,(46)γ2≡4p2mod25p,(47)γ2≡2p2mod25p,(48)γ2≡2k2mod25p,k=2i,i=1,2,3,⋯,p−1,(49)γ2≡2k2mod25p,k=2i−1,i=1,2,3,⋯,p−12.The roots of the Congruences (44)–(49) are 0,23p,24p,23p+24p,∪k∈1,3,5,7pk,24p+pk∪25p−pk,25p−24p−pk,4p,12p,20p,28p,2p,6p,10p,14p,18p,22p,26p,30p,∪k=2t,1≤t≤p−123pj+2k,j=0,1,2,⋯,22−1∪25p−23pj−2k,j=0,1,2,⋯,22−1, and ∪t=ip−1/222pi+4t−2,i=0,1,2,⋯,23−1∪25p−22pi−4t+2,i=0,1,2,⋯,23−1.Forα=6, we have
(50)γ2≡pk2mod26p,k=2i−1,i=1,2,3,⋯,23,(51)γ2≡2k2mod26p,k=pi,i=0,1,2,3,(52)γ2≡2k2mod26p,k=4t,t=1,2⋯p−12,(53)γ2≡2k2mod26p,k=3p−4i−pj,i=1,2,3,⋯,3p4,(54)j=0,1,2,⋯,3p−4ip.The solutions of the Congruences (50)–(54) are ∪k=2i−1,i=123pk,25p+pk∪26p−pk,26p−25p−pk,0,23p,24p,25p,23p+24p,23p+25p,24p+25p,23p+24p+25p,∪k=pi,i=132k,24k−2k,24k+2k,25k−2k∪26p−2k,26p−24k+2k,26p−24k−2k,26p−25k+2k,∪k=4t,t=1p−1/223pj+2k,j=0,1,2,⋯,23−1∪26p−23pj−2k,j=0,1,2,⋯,23−1,∪1≤i≤3p/4,j=03p−4i/p24pt+6p−8i−2pj,t=0,1,⋯,22−1∪26p−24pt−6p+8i+2pj,t=0,1,⋯,22−1∪24pt+24+6p−8i−2pj,t=0,1,⋯,22−1∪26p−24pt−24−6p+8i+, and 2pj,t=0,1,⋯,22−1.Forα=7, we obtain
(55)γ2≡pk2mod27p,k=2i−1,i=1,2,3,⋯,24,(56)γ2≡2k2mod27p,k=pi,i=0,1,3,4,5,7,(57)γ2≡2k2mod27p,k=2p,(58)γ2≡2k2mod27p,k=4t,t=1,2,⋯,p−1,(59)γ2≡2k2mod27p,k=7p−8i−2pj,i=1,2,3,⋯,7p8,(60)j=0,1,2,⋯,7p−8i2p,(61)γ2≡2k2mod27p,k=4t−2,t=1,2⋯p−12.The roots of the Congruences (55)–(61) are
(62)∪k=2i−1,i=124pk,26p+pk∪27p−pk,27p−26p−pk,0,24p,25p,26p,24p+25p,24p+26p,25p+26p,24p+25p+26p,∪k=pi,i∈1,3,4,5,72k,25k−2k,25k+2k,26k−2k∪27p−2k,27p−25k+2k,27p−25k−2k,27p−26k+2k,23pj+2k,k=2p,j=0,1,2,⋯,24−1,∪k=4t,1≤t≤p−124pj+2k,j=0,1,2,⋯,23−1∪27p−24pj−2k,j=0,1,2,⋯,23−1,∪1≤i≤7p8,j=07p−8i2p25pt+14p−16i−4pj,t=0,1,2,⋯,22−1∪27p−25pt−14p+16i+4pj,t=0,1,2,⋯,22−1∪25pt+23p+22p+14p−16i−4pj,t=0,1,2,⋯,22−1∪27p−25pt−23p−22p−14p+16i+4pj,t=0,1,2,⋯,22−1,∪t=ip−1/223pi+8t−4,i=0,1,2,⋯,24−1∪27p−23pi−8t+4,i=0,1,2,⋯,24−1.Forα=8, we get
(63)γ2≡pk2mod28p,k=2i−1,i=1,2,3,⋯,25,(64)γ2≡2k2mod28p,k=pt,t=2i−1i=1,2,3,⋯,23,(65)γ2≡2k2mod28p,k=2pi,i=0,1,2,,⋯,22−1,(66)γ2≡2k2mod28p,k=15p−16i−2pj,i=1,2,3,⋯,15p16,(67)j=0,1,2,⋯,15p−16i2p,(68)γ2≡2k2mod28p,k=8t,t=1,2⋯p−12,(69)γ2≡2k2mod28p,k=6p−8i−2pj,i=1,2,3,⋯,6p8,(70)j=0,1,2,⋯,6p−8i2p.The zeroes of the Congruences (63)–(70) are
(71)∪k=2i−1,i=125pk,27p+pk∪28p−pk,28p−27p−pk,∪k=p2i−1,i=12326pj+2k,j=0,1,2,⋯,22−1∪28p−26pj−2k,j=0,1,2,⋯,22−1,∪k=2pi,i=1325pj+2k,j=0,1,2,⋯,23−1∪28p−25pj−2k,j=0,1,2,⋯,23−1,24pj,j=0,1,2,⋯,24−1,∪1≤i≤15p/16,j=015p−16i/2p26pt+30p−32i−4pj,t=0,1,⋯,22−1∪28p−26pt−30p+32i+4pj,t=0,1,⋯,22−1∪26pt+23p+22p+30p−32i−4pj,t=0,1,⋯,22−1∪28p−26pt−23p−22p−30p+32i+4pj,t=0,1,⋯,22−1,∪1≤i≤15p/16,j=015p−16i/2p26pt+30p−32i−4pj,t=0,1,⋯,22−1∪28p−26pt−30p+32i+4pj,t=0,1,⋯,22−1∪26pt+23p+22p+30p−32i−4pj,t=0,1,⋯,22−1∪28p−26pt−23p−22p−30p+32i+4pj,t=0,1,⋯,22−1,∪k=8t,t=1p−1/224pj+2k,j=0,1,2,⋯,24−1∪28p−24pj−2k,j=0,1,2,⋯,24−1,∪1≤i≤6p/8,j=06p−8i/2p24pt+12p−16i−4pj,t=0,1,2,⋯,24−1∪28p−24pt−12p+16i+4pj,t=0,1,2,⋯,24−1.Forα=9, we have
(72)γ2≡pk2mod29p,k=2i−1,i=1,2,3,⋯,26,(73)γ2≡2k2mod29p,k=pt,t=2i−1i=1,2,3,⋯,24,(74)γ2≡2k2mod29p,k=2pi,i=0,1,3,4,5,7,(75)γ2≡2k2mod29p,k=31p−32i−2pj,i=1,2,3,⋯,31p32,(76)j=0,1,3,⋯,31p−32i2p,(77)γ2≡2k2mod29p,k=4p,(78)γ2≡2k2mod29p,k=8t,t=1,2,⋯,p−1,(79)γ2≡2k2mod29p,k=14p−16i−4pj,i=1,2,3,⋯,14p16,(80)j=0,1,3,⋯,14p−16i4p,(81)γ2≡2k2mod29p,k=8t−4,t=1,2⋯p−12.The roots of the Congruences (72)–(81) are
(82)∪k=2i−1,i=126pk,28p+pk∪29p−pk,29p−28p−pk,∪k=p2i−1,i=12427pj+2k,j=0,1,2,⋯,22−1∪29p−27pj−2k,j=0,1,2,⋯,22−1,∪k=2pi,i∈1,3,4,5,726pj+2k,j=0,1,2,⋯,23−1∪29p−26pj−2k,j=0,1,2,⋯,23−1,25pj,j=0,1,2,⋯,24−1,∪1≤i≤31p/32,j=031p−32i/2p27pt+62p−64i−4pj,t=0,1,2,⋯,22−1∪29p−27pt−62p+64i+4pj,t=0,1,2,⋯,22−1∪27pt+26p+23p+22p+62p−64i−4pj,t=0,1,2,⋯,22−1∪,29p−27pt−26p−23p−22p−62p+64i+4pj,t=0,1,2,⋯,22−1,∪k=8t,1≤t≤p−125pj+2k,j=0,1,2,⋯,24−1∪29p−25pj−2k,j=0,1,2,⋯,24−1,24pj+2k,k=4p,j=0,1,2,⋯,25−1,∪1≤i≤14p/16,j=014p−16i/4p26pt+28p−32i−8pj,t=0,1,2,⋯,23−1∪29p−26pt−28p+32i+8pj,t=0,1,2,⋯,23−1∪26pt+24p+23p+28p−32i−8pj,t=0,1,2,⋯,23−1∪,29p−26pt−24p−23p−28p+32i+8pj,t=0,1,2,⋯,23−1,∪t=1p−1/224pi+16t−8,i=0,1,2,⋯,25−1∪29p−24pi−16t+8,i=0,1,2,⋯,25−1.Ifα=10, then we have
(83)γ2≡pk2mod210p,k=2i−1,i=1,2,3,⋯,27,(84)γ2≡2k2mod210p,k=pt,t=2i−1i=1,2,3,⋯,25,(85)γ2≡2k2mod210p,k=2pi,i=2j−1,j=1,3,⋯,23,(86)γ2≡2k2mod210p,k=63p−64i−2pj,i=1,2,3,⋯,63p64,(87)j=0,1,3,⋯,63p−64i2p,(88)γ2≡2k2mod210p,k=4pi,i=0,1,2,,⋯,22−1,(89)γ2≡2k2mod210p,k=30p−32i−4pj,i=1,2,3,⋯,30p32,(90)j=0,1,3,⋯,30p−32i4p,(91)γ2≡2k2mod210p,k=16t,t=1,2⋯p−12,(92)γ2≡2k2mod210p,k=12p−16i−4pj,i=1,2,3,⋯,12p16,(93)j=0,1,3,⋯,12p−16i4p.The zeroes of the Congruences (83)–(93) are
(94)∪k=2i−1,i=127pk,29+pk∪210p−pk,210p−29p−pk,∪k=p2i−1,i=12528pj+2k,j=0,1,2,⋯,22−1∪210p−28pj−2k,j=0,1,2,⋯,22−1,∪k=2p2i−1,i=12327pj+2k,j=0,1,2,⋯,23−1∪210p−27pj−2k,j=0,1,2,⋯,23−1,∪1≤i≤63p/64,j=063p−64i/2p28pt+126p−128i−4pj,t=0,1,⋯,22−1∪210p−28pt−126p+128i+4pj,t=0,1,⋯,22−1∪28pt+27p+26p+23p+22p+126p−128i−4pj,t=0,1,⋯,22−1∪210p−28pt−27p−26p−23p−22p−126p+128i+4pj,t=0,1,⋯,22−1,∪k=4pi,i=1326pj+2k,j=0,1,2,⋯,24−1∪210p−26pj−2k,j=0,1,2,⋯,24−1,25pj,j=0,1,2,⋯,25−1,∪1≤i≤30p/32,j=030p−32i/4p27pt+60p−64i−8pj,t=0,1,2,⋯,23−1∪210p−27pt−60p+64i+8pj,t=0,1,2,⋯,23−1∪27pt+24p+23p+60p−64i−8pj,t=0,1,2,⋯,23−1∪210p−27pt−24p−23p−60p+64i+8pj,t=0,1,2,⋯,23−1,∪k=16t,t=1p−1/225pj+2k,j=0,1,2,⋯,25−1∪210p−25pj−2k,j=0,1,2,⋯,25−1,∪1≤i≤12p/16,j=012p−16i/4p25pt+24p−32i−8pj,t=0,1,2,⋯,25−1∪210p−25pt−24p+32i+8pj,t=0,1,2,⋯,25−1.Whenα=11, we have
(95)γ2≡pk2mod211p,k=2i−1,i=1,2,3,⋯,28,(96)γ2≡2k2mod211p,k=pt,t=2i−1i=1,2,3,⋯,26,(97)γ2≡2k2mod211p,k=2pi,i=2j−1,j=1,3,⋯,24,(98)γ2≡2k2mod211p,k=127p−128i−2pj,i=1,2,3,⋯,127p128,(99)j=0,1,3,⋯,127p−128i2p,(100)γ2≡2k2mod211p,k=4pi,i=0,1,3,4,5,7,(101)γ2≡2k2mod211p,k=62p−64i−4pj,i=1,2,3,⋯,62p64,(102)j=0,1,3,⋯,62p−64i4p,(103)γ2≡2k2mod211p,k=8p,(104)γ2≡2k2mod211p,k=16t,t=1,2,⋯,p−1,(105)γ2≡2k2mod211p,k=28p−32i−8pj,i=1,2,3,⋯,28p32,(106)j=0,1,3,⋯,28p−32i8p,(107)γ2≡2k2mod211p,k=16t−8,t=1,2⋯p−12.The zeroes of the Congruences (95)–(107) are
(108)∪k=2i−1,i=128pk,210p+pk∪211p−pk,211p−210p−pk,∪k=p2i−1,i=12629pj+2k,j=0,1,2,⋯,22−1∪211p−29pj−2k,j=0,1,2,⋯,22−1,∪k=2p2i−1,i=12428pj+2k,j=0,1,2,⋯,23−1∪211p−28pj−2k,j=0,1,2,⋯,23−1,∪1≤i≤127p/128,j=0127p−128i/2p29pt+254p−256i−4pj,t=0,1,⋯,22−1∪211p−29pt−254p+256i+4pj,t=0,1,⋯,22−1∪29pt+27p+26p+23p+22p+254p−256i−4pj,t=0,1,⋯,22−1∪211p−29pt−27p−26p−23p−22p−254p+256i+4pj,t=0,1,⋯,22−1,∪k=2pi,i∈1,3,4,5,727pj+2k,j=0,1,2,⋯,24−1∪211p−27pj−2k,j=0,1,2,⋯,24−1,26pj,j=0,1,2,⋯,25−1,∪1≤i≤62p/64,j=062p−64i/4p28pt+124p−128i−8pj,t=0,1,2,⋯,23−1∪211p−28pt−124p+128i+8pj,t=0,1,2,⋯,23−1∪28pt+27p+24p+23p+124p−128i−8pj,t=0,1,2,⋯,23−1∪211p−28pt−27p−24p−23p−124p+128i+8pj,t=0,1,2,⋯,23−1,∪k=16t,1≤t≤p−126pj+2k,j=0,1,2,⋯,25−1∪211p−26pj−2k,j=0,1,2,⋯,25−1,25pj+2k,k=8p,j=0,1,2,⋯,26−1,∪1≤i≤28p/32,j=014p−16i/8p27pt+56p−64i−16pj,t=0,1,2,⋯,24−1∪211p−27pt−56p+64i+16pj,t=0,1,2,⋯,24−1∪27pt+25p+24p+56p−64i−16pj,t=0,1,2,⋯,24−1∪211p−27pt−25p−24p−56p+64i+16pj,t=0,1,2,⋯,24−1,∪t=1p−1/225pi+32t−16,i=0,1,2,⋯,26−1∪211p−25pi−32t+16,i=0,1,2,⋯,26−1.Ifα≥10 with α≡0mod2, then there are 2α−55+2α−8−12α−5/3⋅2α−82p−1+6P−2 number of congruences:
(109)γ2≡pk2mod2αp,k=2i−1,i=1,2,3,⋯,2α−3,(110)γ2≡2k2mod2αp,k=pt,t=2i−1i=1,2,3,⋯,2α−5.Forν=1,2,3,⋯,α−8/2,(111)γ2≡2k2mod2αp,k=2νpi,i=2j−1,j=1,3,⋯,2α−7,(112)γ2≡2k2mod2αp,k=2α−5+ν−2ν−1p−2α−5+νi−2νpj,(113)i=1,2,⋯,2α−5+ν−2ν−1p2α−5+ν,(114)j=0,1,⋯,2α−5+ν−2ν−1p−2α−5+νi2νp,(115)γ2≡2k2mod2αp,k=2α−6/2pi,i=0,1,2,,⋯,22−1,(116)γ2≡2k2mod2αp,k=2α/2−2α−8/2p−2α/2i−2α−6/2pj,(117)i=1,2,3,⋯,2α/2−2α−8/2p2α/2,(118)j=0,1,2,⋯,2α/2−2α−8/2p−2α/2i2α−6/2p,(119)γ2≡2k2mod2αp,k=2α−2/2t,t=1,2⋯p−12,(120)γ2≡2k2mod2αp,k=2α−2/2−2α−6/2p−2α−2/2i−2α−6/2pj,(121)i=1,2,3,⋯,2α−2/2−2α−6/2p2α−2/2,(122)j=0,1,2,⋯,2α−2/2−2α−6/2p−2α−2/2i2α−6/2p.The roots of the Congruences (109)–(122) are
(123)∪k=2i−1,i=12α−3pk,2α−1p+pk∪2αp−pk,2α−1p−2α−1p−pk,∪k=p2i−1,i=12α−52α−2pj+2k,j=0,1,2,⋯,22−1∪2αp−2α−2pj−2k,j=0,1,2,⋯,22−1,Forν=1,2,3,⋯,α−8/2,(124)∪k=2νp2i−1,i=12α−7∪ν=1α−8/22α−2−νpj+2k,j=0,1,2,⋯,23−1∪2αp−2α−2−νpj−2k,j=0,1,2,⋯,23−1,∪1≤i≤A,j=0B∪ν=1α−8/22α−1−νpt+2k,t=0,1,⋯,22−1∪2αp−2α−1−νpt−2k,t=0,1,⋯,22−1∪2α−1−νpt+2ν−1C+2kt=0,1,⋯,22−1∪2αp−2α−1−νpt−2ν−1C−2k,t=0,1,⋯,22−1,where A=2α−5+ν−2ν−1p/2α−5+ν, B=2α−5+ν−2ν−1p−2α−5+νi/2νp, C=27p+26p+23p+22p, and k=2α−5+ν−2ν−1p−2α−5+νi−2νpj.
(125)∪k=2α−6/2pi,i=132α+2/2pj+2k,j=0,1,2,⋯,2α−2/2−1∪2αp−2α+2/2pj−2k,j=0,1,2,⋯,2α−2/2−1,2α/2pj,j=0,1,2,⋯,2α/2−1,∪1≤i≤D,j=0E2α+4/2pt+2k,t=0,1,2,⋯,2α−4/2−1∪2αp−2α+4/2pt−2k,t=0,1,2,⋯,2α−4/2−1∪2α+4/2pt+2α−10/2F+2k,t=0,1,2,⋯,2α−4/2−1∪2αp−−2α−10/2F−2k,t=0,1,2,⋯,2α−4/2−1,where D=2α/2−2α−8/2p/2α/2,E=2α/2−2α−8/2p−2α/2i/2α−6/2p,F=24p+23p, and k=2α/2−2α−8/2p−2α/2i−2α−6/2pj.(126)∪k=2α−2/2t,t=1p−1/22α/2pj+2k,j=0,1,2,⋯,2α/2−1∪2αp−2α/2pj−2k,j=0,1,2,⋯,2α/2−1,∪1≤i≤G,j=0H2α/2pt+2k,t=0,1,2,⋯,2α/2−1∪2αp−2αpt−2k,t=0,1,2,⋯,2α/2−1,where G=2α−2/2−2α−6/2p/2α−2/2,H=2α−2/2−2α−6/2p−2α−2/2i/2α−6/2p, and k=2α−2/2−2α−6/2p−2α−2/2i−2α−6/2pj.Forα≥11 with α≡1mod2, there are 2α−55+2α−8−12α−5/3⋅2α−82p−1+23p−9/2 number of congruences:
(127)γ2≡pk2mod2αp,k=2i−1,i=1,2,3,⋯,2α−3,(128)γ2≡2k2mod2αp,k=pt,t=2i−1i=1,2,3,⋯,2α−5.Forν=1,2,3,⋯,α−9/2,(129)γ2≡2k2mod2αp,k=2νpi,i=2j−1,j=1,3,⋯,2α−7,(130)γ2≡2k2mod2αp,k=2α−5+ν−2ν−1p−2α−5+νi−2νpj,(131)i=1,2,⋯,2α−5+ν−2ν−1p2α−5+ν,(132)j=0,1,⋯,2α−5+ν−2ν−1p−2α−5+νi2νp,(133)γ2≡2k2mod2αp,k=2α+1/2pi,i=0,1,3,4,5,7,(134)γ2≡2k2mod2αp,k=2α+1/2−2α−9/2p−2α+1/2i−2α−7/2pj,(135)i=1,2,3,⋯,2α+1/2−2α−9/2p2α+1/2,(136)j=0,1,3,⋯,2α+1/2−2α−9/2p−2α+1/2i2α−7/2p,(137)γ2≡2k2mod2αp,k=2α−5/2p,(138)γ2≡2k2mod2αp,k=2α−3/2t,t=1,2,⋯,p−1,(139)γ2≡2k2mod2αp,k=2α−1/2−2α−7/2p−2α−1/2i−2α−5/2pj,(140)i=1,2,3,⋯,2α−1/2−2α−7/2p2α−1/2,(141)j=0,1,3,⋯,2α−1/2−2α−7/2p−2α−1/2i2α−5/2p,(142)γ2≡2k2mod2αp,k=2α−3/2t−2α−5/2,t=1,2⋯p−12.The zeroes of Congruences (127)–(142) are
(143)∪k=2i−1,i=12α−3pk,2α−1p+pk∪2αp−pk,2α−1p−2α−1p−pkand∪k=p2i−1,i=12α−52α−2pj+2k,j=0,1,2,⋯,22−1∪2αp−2α−2pj−2k,j=0,1,2,⋯,22−1.Forν=1,2,3,⋯,α−9/2,(144)∪k=2νp2i−1,i=12α−7∪ν=1α−9/22α−2−νpj+2k,j=0,1,2,⋯,23−1∪2αp−2α−2−νpj−2k,j=0,1,2,⋯,23−1,∪1≤i≤A,j=0B∪ν=1α−9/22α−1−νpt+2k,t=0,1,⋯,22−1∪2αp−2α−1−νpt−2k,t=0,1,⋯,22−1∪2α−1−νpt+2ν−1C+2kt=0,1,⋯,22−1∪2αp−2α−1−νpt−2ν−1C−2k,t=0,1,⋯,22−1,where A=2α−5+ν−2ν−1p/2α−5+ν,B=2α−5+ν−2ν−1p−2α−5+νi/2νp,C=27p+26p+23p+22p, and k=2α−5+ν−2ν−1p−2α−5+νi−2νpj.(145)∪k=2α+1/2pi,i∈1,3,4,5,72α+3/2pj+2k,j=0,1,2,⋯,2α−3/2−1∪2αp−2α+3/2pj−2k,j=0,1,2,⋯,2α−3/2−1,2α+1/2pj,j=0,1,2,⋯,2α−1/2−1,∪1≤i≤L,j=0M2α+5/2pt+2k,t=0,1,2,⋯,2α−5/2−1∪2αp−2α+5/2pt−2k,t=0,1,2,⋯,2α−5/2−1∪2α+5/2pt+2α−11/2N+2k,t=0,1,2,⋯,2α−5/2−1∪2αp−2α+5/2pt−2α−11/2N−2k,t=0,1,2,⋯,2α−5/2−1,where L=2α+1/2−2α−9/2p/2α+1/2,M=2α+1/2−2α−9/2p−2α+1/2i/2α−7/2p,N=27p+24p+23p, and k=2α+1/2−2α−9/2p−2α+1/2i−2α−7/2pj.(146)∪k=2α−3/2t,1≤t≤p−12α+1/2pj+2k,j=0,1,⋯,2α−1/2−1∪2αp−2α+1/2pj−2k,j=0,1,⋯,2α−1/2−1,2α−1/2pj+2k,k=α−5/2p,j=0,1,2,⋯,2α+1/2−1,∪1≤i≤Q,j=0R2α+3/2pt+2k,t=0,1,2,⋯,2α−3/2−1∪2αp−2α+3/2pt−2k,t=0,1,2,⋯,2α−3/2−1∪2α+3/2pt+2α−11/2S+2k,t=0,1,2,⋯,2α−3/2−1∪2αp−2α+3/2pt−2α−11/2S−2k,t=0,1,2,⋯,2α−3/2−1,where Q=2α−1/2−2α−7/2p/2α−1/2,R=2α−1/2−2α−7/2p−2α−1/2i/2α−5/2p,S=25p+24p, and k=2α−1/2−2α−7/2p−2α−1/2i−2α−5/2pj.(147)∪t=1p−1/22α−1/2pi+2α−1/2t−2α−3/2,i=0,1,2,⋯,2α+1/2−1∪2αp−2α−1/2pi−2α−1/2t+2α−3/2,i=0,1,2,⋯,2α+1/2−1.The quadratic zero-divisor graph forn=96 is shown in Figure 4.Proposition 4.
Ifp and q are odd primes, then G⏞2,pq=K1⊕p+q−2/2K2.Proof.
Let0,mp,zqm=1,2,3,⋯,q−1,z=1,2,3,⋯,p−1 be a set of zero-divisors of Zpq with zero. There are p+q/2 number of distinct congruences:
(148)γ2≡0modpq,γ2≡pk2modpq,k=1,2,3,⋯,q−12,γ2≡qk2modpq,k=1,2,3,⋯,p−12.Figure 4
G~2,96=6K4⊕3K8⊕K16.The zeroes of these congruences areγ=0,γ=pk,pq−pk,k=1,2,3,⋯,q−1/2, and γ=qk,pq−qk,k=1,2,3,⋯,p−1/2. Thus, G⏞2,pq=K1⊕p+q−2/2K2..Theorem 5.
Letp be an odd prime. Then,
(149)G⏞2,2pα=2K1⊕p−12K2,ifα=1,p−12pK2⊕2Kp,ifα=2,p−12p2K2⊕2Kp⊕p−1K2p,ifα=3,p−12pα−1K2⊕i=1α−2/2pα−1−2ip−1K2pi⊕2Kpα/2,ifα≥4,α≡0mod2,p−12pα−1K2⊕i=1α−1/2pα−1−2ip−1K2pi⊕2Kpα−1/2,ifα≥5,α≡1mod2.Proof.
Letn=2pα be an integer, where p is an odd prime. For α=1, S=2m∣m=0,1,2,⋯p−1∪p is a set of zero-divisors of Z2p including 0. There are p+1/2 number of distinct congruences:
(150)γ2≡0mod2p,γ2≡pmod2p,γ2≡2k2mod2p,k=1,2,⋯,p−12.γ=0, γ=p, and γ=2k,2p−2k,k=1,2,3,⋯,p−1/2 are the zeroes of congruences, respectively. When α=2, then there are congruences are
(151)γ2≡0mod2p2,(152)γ2≡p2mod2p2,(153)γ2≡2k2mod2p2,k=1,2,3,⋯,p2−12,k≠pt,t=1,2,⋯,p−12.The roots of Congruences (151)–(153) are 2kp∣k=0,1,2,⋯,p−1,2k−1p∣k=1,2,⋯,p, and 2k,2p2−2kk=1,2,3,⋯,p2−1/2,k≠pt,t=1,2,⋯,p−1/2. For α=3,
(154)γ2≡0mod2p3,(155)γ2≡p3mod2p3,(156)γ2≡2kp2mod2p3,k=1,2,⋯,p−12,(157)γ2≡2k−1p2mod2p3,k=1,2,⋯,p−12,(158)γ2≡2k2mod2p2,k=1,2,3,⋯,p3−12,k≠pt,t=1,2,⋯,p2−12.The roots of Congruences (154)–(158) are 2kp2∣k=0,1,2,⋯,p−1,2k−1p2∣k=1,2,⋯,p, 2kp∪2p3−2kp∪2kppi−1i,k=1,2,⋯,p−1/2∪2p3−2kppi−1i,k=1,2,⋯,p−1/2∪2kppi+1i,k=1,2,⋯,p−1/2∪2p3−2kppi+1i,k=1,2,⋯,p−1/2, 2k−1p∪2p3−2k−1p∪2k−1ppi−1i,k=1,2,⋯,p−1/2∪2p3−2k−1ppi−1i,k=1,2,⋯,p−1/2∪2k−1ppi+1i,k=1,2,⋯,p−1/2∪2p3−2k−1ppi+1i,k=1,2,⋯,p−1/2, and 2k,2p3−2kk=1,2,3,⋯,p3−1/2,k≠pt,t=1,2,⋯,p2−1/2. If α=4, we have
(159)γ2≡0mod2p4(160)γ2≡p4mod2p4(161)γ2≡2kp2mod2p4,k=1,2,⋯,p2−12,butk≠pt,t=1,2,⋯,p−12(162)γ2≡2k−1p2mod2p4,k=1,2,⋯,p2−12,butk≠pt,t=1,2,⋯,p−12(163)γ2≡2k2mod2p4,k=1,2,3,⋯,p4−12,k≠pt,t=1,2,⋯,p3−12.The roots of Congruences (159)–(163) are 2kp3∣k=0,1,2,⋯,p−1,2k−1p3∣k=1,2,⋯,p, 2kp∪2p4−2kp∪2kpp2i−1∣i,k=1,2,⋯,p−1/2∪2p4−2kpp2i−1∣i,k=1,2,⋯,p−1/2∪2kpp2i+1∣i,k=1,2,⋯,p−1/2∪2p4−2kpp2i+1∣i,k=1,2,⋯,p−1/2, 2k−1p∪2p4−2k−1p∪2k−1pp2i−1∣i,k=1,2,⋯,p−1/2∪2p3−2k−1pp2i−1∣i,k=1,2,⋯,p−1/2∪2k−1pp2i+1∣i,k=1,2,⋯,p−1/2∪2p4−2k−1pp2i+1∣i,k=1,2,⋯,p−1/2, and 2k,2p4−2k∣k=1,2,3,⋯,p4−1/2,k≠pt,t=1,2,⋯,p3−1/2. For α=5, we have
(164)γ2≡0mod2p5,(165)γ2≡p5mod2p5,(166)γ2≡2kp2mod2p5,k=1,2,⋯,p3−12,butk≠pt,t=1,2,⋯,p2−12,(167)γ2≡2k−1p2mod2p5,k=1,2,⋯,p2−12,butk≠pt,t=1,2,⋯,p2−12,(168)γ2≡2kp22mod2p5,k=1,2,⋯,p−12,(169)γ2≡2k−1p22mod2p5,k=1,2,⋯,p−12,(170)γ2≡2k2mod2p5,k=1,2,3,⋯,p5−12,k≠pt,t=1,2,⋯,p4−12.The zeroes of Congruences (164)–(170) are 2kp4∣k=0,1,2,⋯,p−1,2k−1p4∣k=1,2,⋯,p, 2kp∪2p5−2kp∪2kpp3i−1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2∪2p5−2kpp3i−1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2∪2kpp3i+1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2∪2p5−2kpp3i+1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2,, 2k−1p∪2p5−2k−1p∪2k−1pp3i−1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2∪2p5−2k−1pp3i−1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2∪2k−1pp3i+1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2∪2p5−2k−1pp3i+1∣i=1,2,⋯,p−1/2,k=1,2,⋯,p3−1/2,butk≠pt,t=1,2,⋯,p2−1/2,, 2kp2∪2p5−2kp2∪2kpp2i−1∣i=1,2,⋯,p2−1/2,k=1,2,⋯,p−1/2∪2p5−2kpp2i−1∣i=1,2,⋯,p2−1/2,k=1,2,⋯,p−1/2∪2kpp2i+1∣i=1,2,⋯,p2−1/2,k=1,2,⋯,p−1/2∪2p5−2kpp2i+1∣i=1,2,⋯,p2−1/2,k=1,2,⋯,p−1/2, and 2k,2p4−2k∣k=1,2,3,⋯,p4−1/2,k≠pt,t=1,2,⋯,p3−1/2. When α≥6, then we obtain
(171)γ2≡0mod2pα,(172)γ2≡pαmod2pα,(173)γ2≡2kpi2mod2pα,k=1,2,⋯,pα−2i−12,butk≠pt,t=1,2,⋯,pα−2i−1−12,(174)i=1,2,⋯,α−22,(175)γ2≡2k−1pi2mod2pα,k=1,2,⋯,pα−2i−12,butk≠pt,t=1,2,⋯,pα−2i−1−12,(176)i=1,2,⋯,α−22,(177)γ2≡2k2mod2pα,k=1,2,3,⋯,pα−12,k≠pt,t=1,2,⋯,pα−1−12.The zeroes of the Congruences (171)–(172) are 2kpα−1∣k=0,1,2,⋯,p−1 and 2k−1kpα−1∣k=1,2,⋯,p. Substituting k=k=1,2,⋯,pα−2i−1/2,butk≠pt,t=1,2,⋯,pα−2i−1−1/2,i=1,2,⋯,α−2/2, and r=1,2,⋯,pi−1/2, then the solutions of the Congruences (174)–(176) are 2kpi∪2pα−2kpi∪2kppα−2r−1∪2pα−2kppα−2r−1∪2kppα−2r+1∪2pα−2−2kppα−2r+1, and 2k−1pi∪2pα−2k−1pi∪2k−1ppα−2r−1∪2pα−2k−1ppα−2r−1∪2k−1ppα−2r+1∪2pα−2−2k−1ppα−2r+1. The roots of the Congruence (177) are 2k,2pα−1−2kk=1,2,3,⋯,p4−1/2,k=1,2,3,⋯,pα−1/2,k≠pt,t=1,2,⋯,pα−1−1/2.If α≥7, then we have
(178)γ2≡0mod2pα,(179)γ2≡pαmod2pα,(180)γ2≡2kpi2mod2pα,k=1,2,⋯,pα−2i−12,butk≠pt,t=1,2,⋯,pα−2i−1−12,(181)i=1,2,⋯,α−12,(182)γ2≡2k−1pi2mod2pα,k=1,2,⋯,pα−2i−12,butk≠pt,t=1,2,⋯,pα−2i−1−12,(183)i=1,2,⋯,α−12,(184)γ2≡2k2mod2pα,k=1,2,3,⋯,pα−12,k≠pt,t=1,2,⋯,pα−1−12,where the solution sets of the Congruences (178)–(184) are same as Congruences (171)–(177); just the variation of i will be change as i=1,2,⋯,α−1/2.The quadratic zero-divisor graph forn=486 is shown in Figure 5.Figure 5
G⏞2,486=81K2⊕18K6⊕2K9⊕2K18.


## 4. Conclusion

In this article, we investigated the mappingum≡vmmodn for m=2 over the zero-divisors including zero of the ring of integers. A problem of partitions of a given set into subsets of equal sums is a particular NP-complete problem. A novel approach is introduced to find equal sum partitions of zero-divisors via complete graphs. Later on, we intended to extend our research to higher values of m over various rings. We hope that this work will open new inquiry and opportunities in various fields for the other researchers and knowledge seekers.



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