Research article Special Issues

A novel approach to find partitions of Zm with equal sum subsets via complete graphs

  • In mathematics and computer sciences, the partitioning of a set into two or more disjoint subsets of equal sums is a well-known NP-complete problem, also referred to as partition problem. There are various approaches to overcome this problem for some particular choice of integers. Here, we use quadratic residue graph to determine the possible partitions of positive integers m=2β,qβ,2βq, 2qβ,qp, where p, q are odd primes and β is any positive integer. The quadratic residue graph is defined on the set Zm={¯0,¯1,,¯m1}, where Zm is the ring of residue classes of m, i.e., there is an edge between ¯x, ¯yZm if and only if ¯x2¯y2(modm). We characterize these graphs in terms of complete graph for some particular classes of m.

    Citation: M. Haris Mateen, Muhammad Khalid Mahmmod, Doha A. Kattan, Shahbaz Ali. A novel approach to find partitions of Zm with equal sum subsets via complete graphs[J]. AIMS Mathematics, 2021, 6(9): 9998-10024. doi: 10.3934/math.2021581

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  • In mathematics and computer sciences, the partitioning of a set into two or more disjoint subsets of equal sums is a well-known NP-complete problem, also referred to as partition problem. There are various approaches to overcome this problem for some particular choice of integers. Here, we use quadratic residue graph to determine the possible partitions of positive integers m=2β,qβ,2βq, 2qβ,qp, where p, q are odd primes and β is any positive integer. The quadratic residue graph is defined on the set Zm={¯0,¯1,,¯m1}, where Zm is the ring of residue classes of m, i.e., there is an edge between ¯x, ¯yZm if and only if ¯x2¯y2(modm). We characterize these graphs in terms of complete graph for some particular classes of m.



    Graph plays a dynamic role in various sciences such as physics, biology, chemistry, and computer science [2,3,4,5,8]. It is used in various frameworks related to social and information systems [4] and also solves many issues related to everyday life. In physics, there are various circuits constructed by considering different graphs [5]. The atomic number of many molecules is evaluated by using group symmetry graphs that are still unknown a few years ago [3]. In computer science, many problems have been discussed using graphs that were not easy to visualize earlier. For discrete mathematics and combinatorics, the applications of number theory and graph theory are of crucial importance. In this work, we employ number theory to investigate the special classes of graphs.

    Rogers [18] discussed the action of a quadratic map on multiplicative groups under modulo a prime p by using the associated directed graph for which there is an edge from each element to its image. 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. Somer and Krizek [19] studied the structures of graphs of quadratic congruences for composite modulus. Mahmood and Ahmad [10,11] proposed many new results of graphs over residues modulo prime powers. Haris and Khalid [12,13,14,15] investigated the structure of power digraphs associated with the congruence xny(modm). Meemark and Wiroonsri [9] discussed the structure of G(R,k) using a quadratic map, where R is the quotient ring of polynomials over finite fields and k is the modulus. Wei and Tang [20] introduced the concept of square mapping graphs of the Gaussian ring Zm[i]. Ali et al. [21,22] introduced new labeling algorithm on various classes of graphs with applications. Some basic and useful result discussed in [1,6,7,16,23,24] as well.

    Let p be a prime and a an integer coprime to p. Then a is called a quadratic residue (modp) if and only if the congruence x2a(modp) has a solution. Otherwise, a is called quadratic non-residue (modp). Two non-zero integers x and y are called zero divisors in the ring Zm if and only if xy0(modm) [17]. Recall that a graph G(2,m), whose vertices are elements of ring Zm and there will be an edge between ¯x and ¯y (¯x¯y) if ¯x2¯y2(modm) then, G(2,m) is termed as a quadratic graph. For m=30 vertex set is

    Zn={¯0,¯1,¯2,¯3,¯4,¯5,¯6,¯7,¯8,¯9,¯10,¯11,¯12,¯13,¯14,¯15,¯16,¯17,¯18,¯19,¯20,¯21,¯22,¯23,¯24,¯25,¯26,¯27, ¯28,¯29}, by solving the congruences ¯x2¯y2(mod30) for each ¯x, ¯yZn then, there are 2, 4, 6 copies of K1, K4 and K2, respectively as shown in Figure 1. We note that each copy of K2 and K4 has equal sum 30 and 60, respectively. Here Kn is a complete graph obtained if each node connected with every other node except itself [4].

    Figure 1.  G(2,30)=2K16K24K4.

    Theorem 1.1. [17] Let p be an odd prime, k be a positive integer, and a an integer such that (a,p)=1. Then

    1. The congruence x2a(modpk) has either no solution or exactly two incongruent solutions modulo pk.

    2. The congruence x2a(modpk) has no solution if a is quadratic non-residue of p and exactly two incongruent solutions modulo p if a is quadratic residue of p.

    Theorem 1.2. [17] Let a be an odd number. Then we have the following:

    1. The congruence equation x2a(mod2) has the unique solution if and only if x1(mod2).

    2. The congruence equation x2a(mod4) either has no solution if a3mod4) or has two solutions x1,3(mod4) if a1(mod4).

    3. When k3, the equation x2a(mod2k) either has no solution if a1(mod8); or has four solutions x1,x1,x1+2k1,(x1+2k1) if a1(mod8).

    In this section, we characterize quadratic residue graphs for some well-known classes of integers 2β and qβ, for each positive integer β and odd prime q.

    Theorem 2.1. Let m=2β be an integer. Then

    G(2,2β)={2Kβ,ifβ=1,2,2K2K4,ifβ=3,4K4,ifβ=4,6K4K8,ifβ=5,4K88K4,ifβ=6,6K8K1616K4,ifβ=7,8K84K1632K4,ifβ=8,β72i=12β52(i1)K23+(i1)6K2(β1)/2K2(β+1)/22β3K4,ifβ9,andβ1(mod2),β62i=12β2i3K22+i22K2β/22β3K4,ifβ10,andβ0(mod2),

    Proof. We discuss two cases to prove this theorem, the zero-divisors and unit elements of the ring Zm. Firstly, we discuss zero-divisors, let S={2m|m=0,1,2,,2β1  1} be the set of all zero-divisors of Z2β with including zero for each positive integer β. To find the number of solutions of η2β2(mod2β), we start from η2β2(mod2), in this case just S={0}. Therefore, η20(mod2) 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 β=2, η2022(mod22) has two solution namely η=0,2, then G(2,22)=K2. If β=3, then there are two congruences

    η20(mod23),andη24(mod23).

    The roots of these congruences are η=0,4, and η=2,6, respectively. Thus, there exist two copies of K2. For β=4, we have

    η20(mod24),andη24(mod24).

    The corresponding roots of congruence are {0,4,8,12} and {2,6,10,14}, respectively. Therefore, G(2,24)=2K4. There are three congruences for β=5

    η20(mod25),η24(mod25),andη216(mod25).

    The solution sets of these congruences are {0,8,16,24}, {2,6,10,14,18,22,26,30}, and {4,12,20,28}, respectively. Therefore, there are two copies of K4 and one copy of K8. For β=6, we have

    η20(mod26),η24(mod26),η216(mod26),andη236(mod26).

    The zeros of these congruences are {0,8,16,24,32,40,48,54}, {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. For β=7, there are 7 different congruences. We have

    η2(2t)2(mod27),t=0,1,3,4,5,7, (2.1)
    η216(mod27). (2.2)

    Roots of these congruences are {2t+16m|m=0,1,2231}, {2t+32m|m=0,1,2221}{2732m2t|m=0,1,2221}, t=1,3,4,5,7, {4+8m|m=0,1,2241}. Hence, there are six copies of K8 and one copy of K16. For β=8, there are 12 congruences. We have

    η2(2+4t)2(mod28),t=0,1,,231, (2.3)
    η2(4t)2(mod28),t=0,1,,221. (2.4)

    Zeroes of these congruences are {2+4t+64m|m=0,1,2221}{2864m4t2|m=0,1,2221}, t=0,1,,231, {4t+32m|m=0,1,2241}, {4t+32m|m=0,1,2231}{2832m4t|m=0,1,2231}, t=1,,221. Thus, we have G(2,28)=8K44K16. For β=9, there are 23 different congruences. We have

    η2(2+4t)2(mod29),t=0,1,,241, (2.5)
    η2(4t)2(mod29),t=0,1,3,4,5,7, (2.6)
    η264(mod29). (2.7)

    Solutions of congruences (2.5)(2.7) are {2+4t+128m|m=0,1,2221}{29128m4t2|m=0,1,2221}, t=0,1,,261, {4t+32m|m=0,1,2241}, {4t+64m|m=0,1,2231}{2964m4t|m=0,1,2231}, t=1,3,4,5,7, {8+16m|m=0,1,2251}. That is, G(2,29)=16K86K16K32. For β=10, there are 44 different congruences. We have

    η2(2+4t)2(mod210),t=0,1,,251, (2.8)
    η2(4+8t)2(mod210),t=0,1,,231, (2.9)
    η2(8t)2(mod210),t=0,1,,221, (2.10)

    Sequences of roots of congruences (2.8)(2.10) are {2+4t+256m|m=0,1,2221}{210256m4t2|m=0,1,2221} t=0,1,,251, {4+8t+128m|m=0,1,2231}{210128n8t4|m=0,1,2231}, t=0,1,,231, {8t+64m|m=0,1,2251}, {8t+64m|m=0,1,2241}{21064m8t|m=0,1,2241}, t=1,,221. That is, G(2,210)=32K88K164K32. For β=11,

    η2(2+4t)2(mod211),t=0,1,,261, (2.11)
    η2(4+8t)2(mod211),t=0,1,,241, (2.12)
    η2(8t)2(mod211),t=0,1,3,4,5,7, (2.13)
    η2256(mod211). (2.14)

    Zeroes of congruences (2.11)(2.14) are {2+4t+512m|m=0,1,2221}{211512m4t2|m=0,1,2221}, t=0,1,,261, {4+8t+256m|m=0,1,2231}{211256m8t4|m=0,1,2231}, t=0,1,,241, {8t+64m|m=0,1,2251}, {8t+128m|m=0,1,2241}{211128m8t|m=0,1,2241}, t=1,3,4,5,7, {16+32m|m=0,1,2261}.

    Therefore, we have G(2,211)=64K816K166K32K64. For β=12, there are 172 congruences. We have

    η2(2+4t)2(mod212),t=0,1,,271, (2.15)
    η2(4+8t)2(mod212),t=0,1,,251, (2.16)
    η2(8+16t)2(mod212),t=0,1,,231, (2.17)
    η2(16t)2(mod212),t=0,1,,221. (2.18)

    Sets of roots of congruences (2.15)(2.18) are {2+4t+1024m|m=0,1,2221}{2121024m4t2|m=0,1,2221}, t=0,1,,271, {4+8t+512m|m=0,1,2231}{212512m8t4|m=0,1,2231}, t=0,1,,251, {8t+256m|m=0,1,2241}{212256m8t|m=0,1,2241}, t=0,1,,231, {16t+128m|m=0,1,2261}, {16t+128m|m=0,1,2251}{212128m8t|m=0,1,2251}, t=1,,221. That is, G(2,212)=128K832K168K324K64.

    The generalize sequence for β9, where β is an odd number, there are 24(2β71)+213 number of congruences. we have

    η2(2i+2i+1t)2(mod2β),t=0,1,,2β2i31, (2.19)
    i=1,2,3,,β72,η2(2β52t)2(mod2β),t=0,1,3,4,5,7, (2.20)
    η22β3(mod2β). (2.21)

    Sequences of roots of congruences (2.19)(2.21) are {2i+2i+1t+2βi1m|m=0,1,22i+11}{2β2βi1m2i+1t2i|m=0,1,22i+11}, t=0,1,,2β2i31, i=1,2,3,,β72, {2β52t+2β+12m|m=0,1,2,2β121}, {2β52t+2β+32m|m=0,1,2,2β321}{2β2β+32m2β52k|m=0,1,2,2β321}, t=1,3,4,5,7, {2β32+2β12m|m=0,1,22β+121}. That is, G(2,2β)=β72i=12β52(i1)K23+(i1)   6K2(β1)/2  K2(β+1)/2.

    For second case when β is an even number and β10, we have 25(2β81)+363 congruences as follows

    η2(2i+2i+1t)2(mod2β),t=0,1,,2β2i31, (2.22)
    i=1,2,3,,β62,η2(2β42t)2(mod2β),t=0,1,,221. (2.23)

    Sequences of zeroes of congruences (2.22)(2.23) are {2i+2i+1t+2βi1m|m=0,1,22i+11}{2β2βi1m2i+1t2i|m=0,1,22i+11}, t=0,1,,2β2i31, i=1,2,3,,β62, {2β42t+2β+22m|m=0,1,2,2β21}, {2β42t+2β+22m|m=0,1,2,2β221}{2β2β+22m2β42t|m=0,1,2β221}, t=1,,221.

    Thus, G(2,2β)=β62i=12β2i3K22+i22K2β/2.

    Now, we discuss the unit elements of Zm. For m=2,4, the result is straightforward. For β3, the graph G(2,m) contains ϕ(2β)=2β1   number of vertices. We calculate the least positive residues of the square of the integers, which are smaller than and relatively prime with m. Hence, there are ϕ(2β)=2β1   obtained. By Theorem 1.2, the congruence x2a(mod2β) has either no solution or exactly four incongruent solutions. This implies that, there are always ϕ(m)4=ϕ(2β1  )4=2β3 quadratic residues among all the vertices. Thus G(2,m)=2β3K4. By combining both cases, we get the desired result. The quadratic residues graph for n=128 is shown in Figure 2.

    Figure 2.  G(2,27)=6K8K1616K4.

    Theorem 2.2. Let q be an odd prime. Then G(2,qβ)

    ={K1q12K2,ifβ=1,Kqq(q1)2K2,ifβ=2,(q12)K2pKqq2(q1)2K2,ifβ=3,(q12)qK2qKq2q3(q1)2K2,ifβ=4,β12i=1(qβ2iqβ2i12)K2qiKq(β1)/2qβ1  (q1)2K2,ifβ5,andβ1(mod2),β22i=1(qβ2iqβ2i12)K2qiKqβ/2qβ1  (q1)2K2,ifβ6,andβ0(mod2).

    Proof. To prove this theorem first we assume zero-divisors of the ring Zm. Let q is an odd prime and S={tqβ|t=0,1,2,,qβ1  1} be zero-divisors including zero of qβ for each positive integer β. To solve congruence η2β2(modqβ) for each β1, we start with η2β2(modq). In this case, 0 is only root of this congruence, but there is no edge between two vertices when they are same, so G(2,q)=K1. For β=2, there is only one congruence namely η20(modq2), roots of this congruence are η=0,q,2q,,(q1)q. There are q solutions and will be complete graph of order q. For β=3, there are q+12 congruences. We have

    η20(modq3), (2.24)
    η2(qt)2(modq3),t=1,2,,q12. (2.25)

    Zeros of these congruences are {η=q2(m1)|m=1,2,,q}, and {η=q2(m1)+qt|t=1,2,,q12,m=1,2,,q}{η=q3q2(m1)qt|t=1,2,,q12,m=1,2,,q}, respectively. For β=4, the number of distinct congruences is (q2q+1)/2. These are

    η20(modq4), (2.26)
    η2(qt)2(modq4),t=1,2,,q212,buttql,l=1,2,,q12. (2.27)

    Sequences of roots of these congruences are {η=q2(m1)|m=1,2,,q2}, and {η=q3(m1)+qt|t=1,2,,q212,buttql,l=1,2,,q12,m=1,2,,q}{η=q4q3(1)pt|t=1,2,,p212,buttpl,l=1,2,,q12,m=1,2,,q}, respectively.

    For β=5, we have

    η20(modq5), (2.28)
    η2(q2t)2(modq5),t=1,2,,q12 (2.29)
    η2(qt)2(modq5),t=1,2,,q312,buttpl,l=1,2,,q212. (2.30)

    Zeroes of these congruences are {η=q3(m1)|m=1,2,,q2}, {η=q3(m1)+q2t|t=1,2,,q12,m=1,2,,q2}{η=q5q3(m1)q2t|t=1,2,,q12,m=1,2,,q2}, and {η=q4(m1)+qt|t=1,2,,q312,buttql,l=1,2,,q212,m=1,2,,q}{η=q5q4(m1)qt|t=1,2,,q312,buttql,l=1,2,,q212,m=1,2,,q}, respectively.

    If β=6, then we have

    η20(modq6), (2.31)
    η2(q2t)2(modq6),t=1,2,,q212, (2.32)
    buttql,l=1,2,,q12η2(qt)2(modq6),t=1,2,,q412,buttql,l=1,2,,q312. (2.33)

    Sequences of roots of congruences (2.31)(2.33) are {η=q3(m1)|m=1,2,,q3}, {η=q4(m1)+q2t|t=1,2,,q212,buttql,l=1,2,,q12,m=1,2,,q2}{η=q6q3(m1)q2t|t=1,2,,q212,buttql,l=1,2,,q12,m=1,2,,q2}, and {η=q5(m1)+qt|t=1,2,,q412,buttql,l=1,2,,q312,m=1,2,,q}{η=q6q5(m1)qt|t=1,2,,q412,buttql,l=1,2,,q312,m=1,2,,q}, respectively.

    Now we are going to derive generalize sequence for both odd and even distinct congruences for β5 and β1(mod2). These are

    η2(qit)2(modqβ),t=1,2,,qβ2i12, (2.34)
    buttql,l=1,2,,qβ2i112,i=1,2,,β12,η20(modqβ). (2.35)

    Sequences of roots are β12i=1{{η=qβi(m1)+qit|t=1,2,,qβ2i12,butlql,l=1,2,3,,qβ2i112,m=1,2,,qi}, {η=qβqβi(m1)qit|t=1,2,,qβ2i12,buttql,l=1,2,,qβ2i112,m=1,2,,qi}} and {η=qβ+12(m1)|m=1,2,,qβ12}, respectively. Therefore, for every positive integer β5 with β1(mod2), G(2,qβ)=β12i=1(qβ2iqβ2i12)K2qiKq(β1)/2. In second case, when β6 and β1(mod2), the number of distinct congruences is qβ1  +q+22(q+1). We have

    η2(qit)2(modqβ),t=1,2,,qβ2i12, (2.36)
    buttql,l=1,2,,qβ2i112,i=1,2,,β22,η20(modqβ). (2.37)

    Zeroes of congruences (2.36) and (2.37) are β22i=1{{η=qβi(m1)+qit|t=1,2,,qβ2i12,buttql,l=1,2,3,,qβ2i112,m=1,2,,qi}{η=qβqβi(m1)qit|t=1,2,,qβ2i12,buttql,l=1,2,,qβ2i112,m=1,2,,qi}}, and {η=qβ2(m1)|m=1,2,,qβ2}, respectively. Thus, for β6 with β0(mod2), G(2,qβ)=β22i=1(qβ2iqβ2i12)K2qiKqβ/2.

    Now, we assume the set of unit elements of Zm. The graph G(2,m) contains ϕ(qβ)=qβ1  (q1), vertices, where β1. We determine the least positive residue of the square of the integers which are less than and relatively prime with m. Because, there are ϕ(qβ)=qβ1  (q1) squares to be found. By Theorem 1.1, the congruence x2a(modqβ) has either no solution or exactly two incongruent solutions. This implies, there are always ϕ(qβ)2=(qβ1  (q1))2 quadratic residues among all the vertices. Thus, G(2,qβ)=(qβ1  (q1))2K2. By combining both cases, we get the desired result. Quadratic graph for m=162 is shown in Figure 3.

    Figure 3.  G(2,243)=9K6K9K1881K2.

    In this section, we characterize quadratic residues graphs for n=2βq,2qβ,qp.

    Theorem 3.1. Let q be an odd prime. Then G(2,2βq)

    ={2K1(q1)K2,ifβ=1,2K2(q1)K4,ifβ=2,2K2qK4q12K8,ifβ=3,4K42(q1)K8,ifβ=4,6K4(3q2)K8q12K16,ifβ=5,8K44qK82(q1)K16,ifβ=6,16K42(4q1)K8(3q2)K16q12K32,ifβ=7,32K48(2q1)K84qK162(q1)K32,ifβ=8,64K416(2q1)K8(8q2)K16(3q2)K32q12K64,ifβ=9,2i=12β2i1K2i+1β82i=1(2β42iq2β52i)K23+i4qK2β22(q1)K2β+222β1  (q1)8K8,ifβ10,andβ0(mod2),2i=12β2i1K2i+1β92i=1(2β42iq2β52i)K23+i(8q2)K2β12(3q2)K2β+12q12K2β+322β1  (q1)8K8,ifβ11,andβ1(mod2).

    Proof. Let n=2βq be an integer, where q is an odd prime. For β=1, S={2m|m=0,1,2,q1}{q} is set of zero-divisors of Z2q including 0. There are q+12 distinct congruences. We have

    η20(mod2q),η2q(mod2q),andη2(2t)2(mod2q),t=1,2,,q12,

    γ=0, γ=q, and γ=2t,2q2t,t=1,2,3,,q12 zeroes of congruences, respectively. When β=2, there exit following congruences given as

    η20(mod22q),η2q(mod22q),andη2(2t)2(mod22q),t=1,2,,q12.

    Sets of roots of these congruence are {0,2q}, {q,3q}, {2t,2q2t}{22q2t,22q2q+2t},t=1,2,,q12, respectively. There are q+2 congruences for β=3. They are η20(mod23q),η2(2q)2(mod23q),η2q2(mod23q),andη2(2t)2(mod23q),t=1,2,,q1. The zeroes of these congruences are {0,22q}, {2q,23q2q}, {q,3q,5q,7q} and {2t,22q2t}{23q2t,23q22q+2t},t=1,2,,q1, respectively. For β=4, there are q+3 congruences. We have

    η20(mod24q),η2q2(mod24q),η2(2q)2(mod24q),η2(3q)2(mod24q),
    andη2(2t)2(mod24q),t=1,2,,q1,

    zeroes are {0,22q,23q,22q+23q}, {q,7p,9p,15q}, {2q,6q,10q,14q}, {3q,5q,11q,13q} and {2t,22q2t,22q+2t,23q2t}{24q2t,24q22q+2t,24q22q2t,24q23q+2t},t=1,2,,q1, respectively.

    For β=5, there are (3q+11)/2 congruences

    η20(mod25q), (3.1)
    η2(qt)2(mod25q),t=1,3,5,7, (3.2)
    η2(4q)2(mod25q), (3.3)
    η2(2q)2(mod25q), (3.4)
    η2(2t)2(mod25q),t=2i,i=1,2,3,,q1, (3.5)
    η2(2t)2(mod25q),t=2i1,i=1,2,3,,q12. (3.6)

    Zeroes of congruences (3.1)(3.6) are {0,23q,24q,23q+24q}, t{1,3,5,7}{qt,24q+qt}{25qqt,25q24qqt}, {4q,12q,20q,28q}, {2q,6q,10q,14q,18q,22q,26q,30q}, t=2l,1lq1{{23qj+2t,j=0,1,2,,221}{25q23qj2t,j=0,1,2,,221}}, q12l=i{{22qi+4l2,i=0,1,2,,231}{25q22qi4l+2,i=0,1,2,,231}}.

    For β=6, we have

    η2(qt)2(mod26q),t=2i1,i=1,2,3,,23, (3.7)
    η2(2t)2(mod26q),t=qi,i=0,1,2,3, (3.8)
    η2(2t)2(mod26q),t=4l,l=1,2q12, (3.9)
    η2(2)2(mod26q),t=3q4iqj,i=1,2,3,,3q4,j=0,1,2,,3q4iq. (3.10)

    Solution sets of congruences (3.7)(3.10) are

    23t=2i1,i=1{qt,25q+qt}{26qqt,26q25qqt},{0,23q,24q,25q,23q+24q,23q+25q,24q+25q,23q+24q+25q},
    3t=qi,i=1{2t,24t2t,24t+2t,25t2t}{26q2t,26q24t+2t,26q24t2t,26q25t+2t},
    q12t=4l,l=1{23qj+2t,j=0,1,2,,231}{26q23qj2t,j=0,1,2,,231},

    3q4iq1i3q4,j=0{24ql+6p8i2qj,l=0,1,,221}{26q24ql6q+8i+2qj,l=0,1,,221}{24ql+24+6q8i2qj,l=0,1,,221}{26q24ql246q+8i+2qj,l=0,1,,221}.

    For β=7, we obtain

    η2(qt)2(mod27q),t=2i1,i=1,2,3,,24, (3.11)
    η2(2t)2(mod27q),t=qi,i=0,1,3,4,5,7, (3.12)
    η2(2t)2(mod27q),t=2q, (3.13)
    η2(2t)2(mod27q),t=4l,l=1,2,,q1, (3.14)
    η2(2t)2(mod27q),t=7q8i2qj,i=1,2,3,,7q8,j=0,1,2,,7q8i2q, (3.15)
    η2(2t)2(mod27q),t=4l2,l=1,2q12. (3.16)

    Zeroes of congruences (3.11)(3.16) are

    24t=2i1,i=1{qt,26q+qt}{27qqt,27q26qqt},{0,24q,25q,26q,24q+25q,24q+26q,25q+26q,24q+25q+26q},
    t=qi,i{1,3,4,5,7}{{2t,25t2t,25t+2t,26t2t}{27q2t,27q25t+2t,27q25t2t,27q26t+2t}},{23qj+2t,t=2q,j=0,1,2,,241},
    t=4l,1lq1{{24qj+2t,j=0,1,2,,231}{27q24qj2t,j=0,1,2,,231}},
    7q8i2q1i7q8,j=0{{25ql+14q16i4qj,l=0,1,2,,221}{27q25ql14q+16i+4qj,l=0,1,2,,221}{25ql+23q+22q+14q16i4qj,l=0,1,2,,221}{27q25ql23q22q14q+16i+4qj,l=0,1,2,,221}},
    q12l=i{{23qi+8l4,i=0,1,2,,241}{27q23qi8l+4,i=0,1,2,,241}}.

    For β=8, the following congruence equations turning out to be

    η2(qt)2(mod28q),t=2i1,i=1,2,3,,25, (3.17)
    η2(2t)2(mod28q),t=pl,l=2i1i=1,2,3,,23, (3.18)
    η2(2t)2(mod28q),t=2qi,i=0,1,2,,,221, (3.19)
    η2(2t)2(mod28q),t=15q16i2qj,i=1,2,3,,15q16,j=0,1,2,,15q16i2q, (3.20)
    η2(2t)2(mod28q),t=8l,l=1,2q12, (3.21)
    η2(2t)2(mod28q),t=6q8i2qj,i=1,2,3,,6q8,j=0,1,2,,6q8i2q. (3.22)

    Zeroes of congruences (3.17)(3.22) are

    25t=2i1,i=1{qt,27q+qt}{28qqt,28q27qqt},
    23t=q(2i1),i=1{26qj+2t,j=0,1,2,,221}{28q26qj2t,j=0,1,2,,221},
    3k=2qi,i=1{25qj+2t,j=0,1,2,,231}{28q25qj2t,j=0,1,2,,231},{24qj,j=0,1,2,,241},
    (15q16i)/2q1i15q/16,j=0{{26ql+30q32i4qj,l=0,1,,221}{28q26ql30q+32i+4qj,l=0,1,,221}{26ql+23q+22q+30q32i4qj,l=0,1,,221}{28q26ql23q22q30q+32i+4qj,l=0,1,,221}},
    (15q16i)/2q1i15q/16,j=0{{26ql+30q32i4qj,l=0,1,,221}{28q26ql30q+32i+4qj,l=0,1,,221}{26ql+23q+22q+30q32i4qj,l=0,1,,221}{28q26ql23q22q30q+32i+4qj,l=0,1,,221}},
    (q1)/2t=8l,l=1{24qj+2t,j=0,1,2,,241}{28q24qj2t,j=0,1,2,,241},
    (6q8i)/2q1i6q/8,j=0{{24ql+12q16i4qj,l=0,1,2,,241}{28q24ql12q+16i+4qj,l=0,1,2,,241}}.

    For β=9, we have

    η2(qt)2(mod29q),t=2i1,i=1,2,3,,26, (3.23)
    η2(2t)2(mod29q),t=ql,l=2i1i=1,2,3,,24, (3.24)
    η2(2t)2(mod29q),t=2qi,i=0,1,3,4,5,7, (3.25)
    η2(2t)2(mod29q),t=31q32i2qj,i=1,2,3,,31p32,j=0,1,3,,31q32i2q, (3.26)
    η2(2t)2(mod29q),t=4q, (3.27)
    η2(2t)2(mod29q),t=8l,l=1,2,,q1, (3.28)
    η2(2t)2(mod29q),t=14q16i4qj,i=1,2,3,,14q16,j=0,1,3,,14q16i4q, (3.29)
    η2(2t)2(mod29q),t=8l4,l=1,2q12. (3.30)

    Sets of solution of congruences (3.23)(3.30) are

    26t=2i1,i=1{qt,28q+qt}{29qqt,29q28qqt},
    24t=q(2i1),i=1{27qj+2t,j=0,1,2,,221}{29q27qj2t,j=0,1,2,,221},
    t=2qi,i{1,3,4,5,7}{26qj+2t,j=0,1,2,,231}{29q26qj2t,j=0,1,2,,231},{25qj,j=0,1,2,,241},
    (31q32i)/2q1i31p/32,j=0{{27ql+62q64i4qj,l=0,1,2,,221}{29q (3.31)
    27ql62q+64i+4qj,l=0,1,2,,221}{27ql+26q+ (3.32)
    23q+22q+62q64i4qj,l=0,1,2,,221}{29q (3.33)
    27ql26q23q22q62q+64i+4qj,l=0,1,2,,221}}, (3.34)
    t=8l,1lq1{{25qj+2t,j=0,1,2,,241}{29q25qj2t,j=0, (3.35)
    1,2,,241}},{24qj+2t,t=4q,j=0,1,2,,251}, (3.36)
    (14q16i)/4q1i14q/16,j=0{{26ql+28q32i8qj,l=0,1,2,,231}{29q26ql28q+32i+8qj,l=0,1,2,,231}{26ql+24q+23q+28q32i8qj,l=0,1,2,,231}{29q26ql24q23q28q+32i+8qj,l=0,1,2,,231}},
    (q1)/2l=1{{24qi+16l8,i=0,1,2,,251}{29q24qi16l+8,i=0,1,2,,251}}.

    For β=10, we get

    η2(qt)2(mod210q),t=2i1,i=1,2,3,,27, (3.37)
    η2(2t)2(mod210q),t=ql,l=2i1i=1,2,3,,25, (3.38)
    η2(2t)2(mod210q),t=2qi,i=2j1,j=1,3,,23, (3.39)
    η2(2t)2(mod210q),t=63q64i2qj,i=1,2,3,,63q64,j=0,1,3,,63q64i2q, (3.40)
    η2(2t)2(mod210q),t=4qi,i=0,1,2,,,221, (3.41)
    η2(2t)2(mod210q),t=30q32i4qj,i=1,2,3,,30q32,j=0,1,3,,30q32i4q, (3.42)
    η2(2t)2(mod210q),t=16l,l=1,2q12, (3.43)
    η2(2t)2(mod210q),t=12q16i4qj,i=1,2,3,,12q16,j=0,1,3,,12q16i4q. (3.44)

    Zeroes of congruences (3.37)(3.44) are

    27t=2i1,i=1{qt,29+qt}{210qqt,210q29qqt},
    25t=q(2i1),i=1{28qj+2t,j=0,1,2,,221}{210q28qj2t,j=0,1,2,,221},
    23t=2q(2i1),i=1{27qj+2t,j=0,1,2,,231}{210q27qj2t,j=0,1,2,,231},
    3t=4qi,i=1{26qj+2t,j=0,1,2,,241}{210q26qj2t,j=0,1,2,,241},{25qj,j=0,1,2,,251},
    (30q32i)/4q1i30q/32,j=0{{27qt+60q64i8qj,l=0,1,2,,231}{210q27ql60q+64i+8qj,l=0,1,2,,231}{27ql+24q+23q+60q64i8qj,l=0,1,2,,231}{210q27ql24q23q60q+64i+8qj,l=0,1,2,,231}},
    (q1)/2t=16l,l=1{25qj+2t,j=0,1,2,,251}{210q25qj2t,j=0,1,2,,251},
    (12q16i)/4q1i12q/16,j=0{{25ql+24q32i8qj,l=0,1,2,,251}{210q25ql24q+32i+8qj,l=0,1,2,,251}}.

    For β=11, we obtain

    η2(qt)2(mod211q),t=2i1,i=1,2,3,,28, (3.45)
    η2(2t)2(mod211q),t=ql,l=2i1i=1,2,3,,26, (3.46)
    η2(2t)2(mod211q),t=2qi,i=2j1,j=1,3,,24, (3.47)
    η2(2t)2(mod211q),t=127q128i2qj,i=1,2,3,,127q128,j=0,1,3,,127q128i2q, (3.48)
    η2(2t)2(mod211q),t=4qi,i=0,1,3,4,5,7, (3.49)
    η2(2t)2(mod211q),t=62q64i4qj,i=1,2,3,,62q64,j=0,1,3,,62q64i4q, (3.50)
    η2(2t)2(mod211q),t=8q, (3.51)
    η2(2t)2(mod211q),t=16l,l=1,2,,q1, (3.52)
    η2(2t)2(mod211q),t=28q32i8qj,i=1,2,3,,28q32,j=0,1,3,,28q32i8q, (3.53)
    η2(2t)2(mod211q),t=16l8,l=1,2q12. (3.54)

    Sets of solution of congruences (3.45)(3.54) are

    24t=2q(2i1),i=1{28qj+2t,j=0,1,2,,231}{211q28qj2t,j=0,1,2,,231},
    (127q128i)/2q1i127q/128,j=0{{29qt+254q256i4qj,t=0,1,,221}{211q29ql254q+256i+4qj,l=0,1,,221}{29ql+27q+26q+23q+22q+254q256i4qj,l=0,1,,221}{211q29ql27q26q23q22q254q+256i+4qj,l=0,1,,221}},
    t=2qi,i{1,3,4,5,7}{27qj+2t,j=0,1,2,,241}{211q27qj2t,j=0,1,2,,241},{26qj,j=0,1,2,,251},
    (62q64i)/4q1i62q/64,j=0{{28ql+124q128i8qj,l=0,1,2,,231}{211q28ql124q+128i+8qj,l=0,1,2,,231}{28ql+27q+24q+23q+124q128i8qj,l=0,1,2,,231}{211q28ql27q24q23q124q+128i+8qj,l=0,1,2,,231}},
    t=16l,1lq1{{26qj+2t,j=0,1,2,,251}{211q26qj2t,j=0,1,2,,251}},{25qj+2t,t=8q,j=0,1,2,,261},
    (14q16i)/8q1i28q/32,j=0{{27ql+56q64i16qj,l=0,1,2,,241}{211q27ql56q+64i+16qj,l=0,1,2,,241}{27ql+25q+24q+56q64i16qj,l=0,1,2,,241}{211q27ql25q24q56q+64i+16qj,l=0,1,2,,241}},
    (q1)/2l=1{{25qi+32l16,i=0,1,2,,261}{211q25qi32l+16,i=0,1,2,,261}}.

    For β10 with β0(mod2), there are 2β55+(2β81)2β532β8(2q1)+6q2 congruences. We have

    η2(pt)2(mod2βq),t=2i1,i=1,2,3,,2β3, (3.55)
    η2(2t)2(mod2βq),t=ql,l=2i1i=1,2,3,,2β5. (3.56)

    For ν=1,2,3,,β82,

    η2(2t)2(mod2βq),t=2νqi,i=2j1,j=1,3,,2β7, (3.57)
    η2(2t)2(mod2βq),t=(2β5+ν  2ν1  )q(2β5+ν  )i2νqj,i=1,2,,2β5+ν  2ν1  q2β5+ν  ,j=0,1,,(2β5+ν  2ν1  )q(2β5+ν  )i2νq, (3.58)
    η2(2t)2(mod2βq),t=2β62qi,i=0,1,2,,,221, (3.59)
    η2(2t)2(mod2βq),t=(2β22β82)q2β2i2β62qj,i=1,2,3,,(2β22β82)q2β2,j=0,1,2,,(2β22β82)q2β2i2β62q, (3.60)
    η2(2t)2(mod2βq),t=2β22l,l=1,2q12, (3.61)
    η2(2t)2(mod2βq),t=(2β222β62)q2β22i2β62qj,i=1,2,3,,(2β222β62)q2β22,j=0,1,2,,(2β222β62)q2β22i2β62q. (3.62)

    Zeroes of congruences (3.55)(3.62) are

    2β3t=2i1,i=1{qt,2β1  q+qt}{2βqqt,2β1  q2β1  qqt},2β5t=q(2i1),i=1{2β2qj+2t,j=0,1,2,,221}{2βq2β2qj2t,j=0,1,2,,221},

    For ν=1,2,3,,β82,

    2β7t=2νq(2i1),i=1(β8)/2ν=1{2β2νqj+2t,j=0,1,2,,231}{2βq2β2νqj2t,j=0,1,2,,231},
    B1iA,j=0(β8)/2ν=1{{2β1νql+2t,l=0,1,,221}{2βq2β1νql2t,l=0,1,,221}{2β1νqk+2ν1  C+2t,l=0,1,,221}{2βq2β1νql2ν1  C2t,l=0,1,,221}},

    where A=2β5+ν  2ν1  q2β5+ν  , B=(2β5+ν  2ν1  )q(2β5+ν  )i2νq, C=27q+26q+23q+22q and t=(2β5+ν  2ν1  )q(2β5+ν  )i2νqj.

    3t=2β62qi,i=1{2β+22qj+2t,j=0,1,2,,2β221}{2βq2β+22qj2t,j=0,1,2,,2β221},{2β2qj,j=0,1,2,,2β21},E1iD,j=0{{2β+42ql+2t,l=0,1,2,,2β421}{2βq2β+42ql2t,l=0,1,2,,2β421}{2β+42ql+2(β10)/2F+2t,l=0,1,2,,2β421}{2βq2(β10)/2F2t,l=0,1,2,,2β421}},

    where D=(2β22β82)q2β2, E=(2β22β82)q2β2i2β62q, F=24q+23q and t=(2β22β82)q2β2i2β62qj.

    (q1)/2t=2(β2)/2l,l=1{2β/2qj+2t,j=0,1,2,,2β/21}{2βq2β/2qj2t,j=0,1,2,,2β/21},H1iG,j=0{{2β/2ql+2t,l=0,1,2,,2β/21}{2βq2βql2t,l=0,1,2,,2β/21}},

    where G=(2β222β62)q2β22, H=(2β222β62)q2β22i2β62q and t=(2β222β62)q2β22i2β62qj.

    For β11 with β1(mod2), there are 2β55+(2β81)2β532β8(2q1)+23q92 congruences. We have

    η2(qt)2(mod2βq),t=2i1,i=1,2,3,,2β3, (3.63)
    η2(2t)2(mod2βq),t=ql,l=2i1i=1,2,3,,2β5. (3.64)

    For ν=1,2,3,,β92,

    η2(2t)2(mod2βq),t=2νqi,i=2j1,j=1,3,,2β7, (3.65)
    η2(2t)2(mod2βq),t=(2β5+ν  2ν1  )q(2β5+ν  )i2νqj,i=1,2,,2β5+ν  2ν1  q2β5+ν  ,j=0,1,,(2β5+ν  2ν1  )q(2β5+ν  )i2νq, (3.66)
    η2(2t)2(mod2βq),t=2β+12qi,i=0,1,3,4,5,7, (3.67)
    η2(2t)2(mod2βq),t=(2β+122β92)q2β+12i2β72qj,i=1,2,3,,2β+122β92)q2β+12,j=0,1,3,,(2β+122β92)q2β+12i2β72q, (3.68)
    η2(2t)2(mod2βq),t=2β52q, (3.69)
    η2(2t)2(mod2βq),t=2β32l,l=1,2,,q1, (3.70)
    η2(2t)2(mod2βq),t=(2β122β72)q2β12i2β52qj,i=1,2,3,,(2β122β72)q2β12,j=0,1,3,,(2β122β72)q2β12i2β52q, (3.71)
    η2(2t)2(mod2βq),t=2β32l2β52,l=1,2q12. (3.72)

    Zeroes of congruences (3.63)(3.72) are

    2β3t=2i1,i=1{qt,2β1  q+qt}{2βqqt,2β1  q2β1  qqt},2β5t=q(2i1),i=1{2β2qj+2t,j=0,1,2,,221}{2βq2β2qj2t,j=0,1,2,,221},

    For ν=1,2,3,,β92,

    2β7t=2νq(2i1),i=1(β9)/2ν=1{2β2νqj+2t,j=0,1,2,,231}{2βq2β2νqj2t,j=0,1,2,,231},B1iA,j=0(β9)/2ν=1{{2β1νql+2t,l=0,1,,221}{2βq2β1νql2t,l=0,1,,221}{2β1νqt+2ν1  C+2t,l=0,1,,221}{2βq2β1νql2ν1  C2t,l=0,1,,221}},

    where A=2β5+ν  2ν1  q2β5+ν  , B=(2β5+ν  2ν1  )q(2β5+ν  )i2νq, C=27q+26q+23q+22q and t=(2β5+ν  2ν1  )q(2β5+ν  )i2νqj.

    t=2(β+1)/2qi,i{1,3,4,5,7}{2(β+3)/2qj+2t,j=0,1,2,,2(β3)/21}{2βq2(β+3)/2qj2t,j=0,1,2,,2(β3)/21},{2(β+1)/2qj,j=0,1,2,,2(β1)/21},
    M1iL,j=0{{2(β+5)/2ql+2t,l=0,1,2,,2(β5)/21}{2βq2(β+5)/2ql2t,l=0,1,2,,2(β5)/21}{2(β+5)/2ql+2(β11)/2N+2t,l=0,1,2,,2(β5)/21}{2βq2(β+5)/2ql2(β11)/2N2t,l=0,1,2,,2(β5)/21}},

    where L=2β+122β92)q2β+12, M=(2β+122β92)q2β+12i2β72q, N=27q+24q+23q and t=(2β+122β92)q2β+12i2β72qj.

    t=2(β3)/2l,1lq1{{2(β+1)/2qj+2t,j=0,1,,2(β1)/21}{2βq2(β+1)/2qj2t,j=0,1,,2(β1)/21}},{2(β1)/2qj+2t,t=(β5)/2q,j=0,1,2,,2(β+1)/21},
    R1iQ,j=0{{2(β+3)/2ql+2t,l=0,1,2,,2(β3)/21}{2βq2(β+3)/2ql2t,l=0,1,2,,2(β3)/21}{2(β+3)/2ql+2(β11)/2S+2t,l=0,1,2,,2(β3)/21}{2βq2(β+3)/2ql2(β11)/2S2t,l=0,1,2,,2(β3)/21}},

    where Q=(2β122β72)q2β12, R=(2β122β72)q2β12i2β52q, S=25q+24q and t=(2β122β72)q2β12i2β52qj.

    (q1)/2l=1{{2(β1)/2qi+2(β1)/2l2(β3)/2,i=0,1,2,,2(β+1)/21}{2βq2(β1)/2qi2(β1)/2l+2(β3)/2,i=0,1,2,,2(β+1)/21}}.

    Now, we consider the set of unit elements of Zm. The x2a (mod2) has one solution and x2a(mod22) has two solution. By Theorem 1.2, if β3, the congruence x2a(mod2β) has either no solution or exactly 4 incongruent solutions. Furthermore, again by Theorem 1.1, for an odd prime the congruence x2a(modq) has either no solution or exactly 2 incongruent solutions. By using Chinese remainder theorem, if β=0or1 then, x2a(mod2β) has either no solution or exactly 2 incongruent solutions. If β=2 then, x2a(mod2β) has either no solution or exactly 4 incongruent solutions. Lastly, if β3, then x2a(mod2β) has either no solution or exactly 8 incongruent solutions. Hence, G(2,2βq)=ϕ(n)8K8,β3. We achieve the desired outcome by combining both cases.

    Proposition 3.2. If q and p are odd primes, then G(2,qp)=K1q+p22K2(q1)(p1)4K4.

    Proof. To prove this first we discuss the zero-divisors elements of ring Zm. Let {0,nq,yp|n=1,2,3,,p1,y=1,2,3,,q1} be a set of zero-divisors of Zqp with zero. There are q+p2 distinct congruences. We have

    η20(modqp), (3.73)
    η2(qt)2(modqp),t=1,2,3,,p12, (3.74)
    η2(pt)2(modqp),t=1,2,3,,q12. (3.75)

    Zeroes of these congruences are η=0, η=qt,qpqt,t=1,2,3,,p12, and η=pt,qppt,t=1,2,3,,q12. Thus, G(2,qp)=K1q+p22K2. Now we discuss the unit elements of Zm. By Theorem 1.1, for distinct odd primes q and p, the congruence x2a(modq) has either no solution or exactly 2 incongruent solutions. Similarly, the congruence x2a(modp) has either no solution or exactly 2 incongruent solutions. By using Chines remainder theorem we have, x2a(modqβ) has either no solution or exactly 4 solutions. Thus, G(2,qp)=ϕ(qp)4K4=ϕ(q1)(p1))4K4. We get the desired outcome by combining both cases.

    Theorem 3.3. Let q be an odd prime. Then, G(2,2qβ)

    ={2K1(q1)K2,ifβ=1,q(q1)K22Kq,ifβ=2,(q1)q2K22Kp(q1)K2q,ifβ=3,(q1)qβ1  K2β22i=1qβ12i(q1)K2qi2Kqβ/2,ifβ4,andβ0(mod2),(q1)qβ1  K2β12i=1qβ12i(q1)K2qi2Kq(β1)/2,ifβ5andβ1(mod2).

    Proof. The proof is on similar lines as illustrated in the proof of Theorem 3.1.

    Figures 4 and 5 reflect Theorems 3.1 and 3.3, respectively.

    Figure 4.  ˜G(2,96)=6K47K8K16.
    Figure 5.  ˜G(2,486)=2K182K918K6162K2.

    In this article, we investigated the mapping xαyα,(modm) for α=2 over the ring of integers. A problem of partitions of a given set into the form of subsets with equal sums is NP problem. A paradigmatic approach was introduced to find equal sum partitions of quadratic maps via complete graphs. Moreover, we characterized quadratic graphs associated with the mapping xαyα,(modm), α=2 for well-known classes m=2β,qβ,2βq,2qβ,qp, in terms of complete graphs, where q,p is an odd prime. Later on, we intend to extend our research to higher values of α over various rings. We hope that this work will open new inquiry opportunities in various fields for other researchers and knowledge seekers.

    This work was funded by the Deanship of Scientific Research (DSR) at King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSR for technical and financial support.

    The authors declare that they have no conflict of interest regarding the publication of the research article.



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