On Some Properties of Addition Signed Cayley Graph ${\mathsf{\Sigma }}_{\mathit{n}}^{\mathbf{\wedge }}$

Abstract

We define an addition signed Cayley graph on a unitary addition Cayley graph $G_n$ represented by ${\mathsf{\Sigma }}_n^{\wedge }$, and study several properties such as balancing, clusterability and sign compatibility of the addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }$. We also study the characterization of canonical consistency of ${\mathsf{\Sigma }}_n^{\wedge }$, for some n.

1. Introduction

We refer to standard books of Harary [1] and West [2] for graph theory. For the signed graphs, we refer to Zaslavsky [3,4]. All the signed graphs considered in this paper are simple, finite and loopless.

For the preliminaries, definition and notation of signed graph S, underlying graph $S^u$, its negation $\eta \left(S\right)$, signed isomorphism and its positive (negative) section, we refer to [5,6].

Some Basic Lemma and Theorems which are used in this paper are stated below as a reference.

Lemma 1

([7]). A signed graph in which every chordless cycle is positive is balanced.

Theorem 1

([8]). A signed graph S is clusterable if—and only if—S does not contains a cycle with exactly one negatively charged edge.

For balancing, clusterability, marking, canonical marking ($\mathcal{C}$-marking), consistency, $\mathcal{C}$-consistency, S consistency, sign compatibility, line signed graph $L\left(S\right)$, line signed root graph, ×-line signed graph, ×-line signed root graph and the common-edge signed graph $C_E\left(S\right)$ of signed graph, S we refer to [6,9,10,11,12,13,14,15,16].

Addition Signed Cayley Graph ${\mathsf{\Sigma }}_n^{\wedge }$

A unitary addition Cayley graph $G_n$, where $n\in I^{+}$, $I^{+}$ is set of positive integers, is a graph in which the vertex set is a ring of integers modulo n, $Z_n$. Any two vertices $x_1$ and $x_2$ are adjacent in $G_n$ if—and only if—$(x_1+x_2)\in U_n$, where $U_n$ denotes the unit set.

Unitary addition Cayley graphs for n = 2, 3, 4, 5, 6 and 7 are shown in Figure 1.

Figure 1. Examples of unitary addition Cayley graphs.

The study of unitary Cayley graphs began in order to gain some insight into the graph representation problem (see [17]), and we can extend it to the signed graphs (see [18]).

Now, we introduce the definition of an addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }$ as follows:

The addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$ is a signed graph whose underlying graph is a unitary addition Cayley graph $G_n$, where $n\in I^{+}$ and for an edge $ab$ of ${\mathsf{\Sigma }}_n^{\wedge }$,

$${\sigma }^{\wedge }\left(ab\right)=\left\{\begin{array}{cc}+\hfill & \mathrm{if}a,b\in U_n,\hfill \\ -\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Examples of addition signed Cayley graph for n = 5, 6 and 10 can be seen in Figure 2a–c. Throughout the paper, we consider $n\ge 2$.

Figure 2. Examples of addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }$.

2. Some Properties of ${\mathsf{\Sigma }}_{\mathit{n}}^{\wedge }$

2.1. Balancing in ${\mathsf{\Sigma }}_n^{\wedge }$

The balancing of some derived signed Cayley graphs has been studied in the literature (see [19]). Here, we find out the property of balancing for the addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }$, for which the following well-known result can be used as a tool.

Theorem 2

([20]). $G_n$, $n\ge 2,$ is bipartite if—and only if—either $n=3$ or n is even.

Lemma 2.

$i\in U_n\Rightarrow (n-i)\in U_n$ and $i\notin U_n\Rightarrow (n-i)\notin U_n$.

Lemma 3.

Addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, for n even, is an all-negative signed graph.

Proof. 

Given an addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, where n is even. Suppose the conclusion is false. Let there be a positive edge, say $ij$, in ${\mathsf{\Sigma }}_n^{\wedge }$. By the definition of ${\mathsf{\Sigma }}_n^{\wedge }$, $i,j\in U_n$. Since n is even, $U_n$ consists only of odd numbers. Thus, i and j are odd numbers and their addition $i+j$ is an even number. This shows that $i+j\notin U_n$, i.e., i and j, are not adjacent in ${\mathsf{\Sigma }}_n^{\wedge }$. Thus, we have a contradiction. Hence, if n is even, then ${\mathsf{\Sigma }}_n^{\wedge }$ is all-negative signed graph. □

Sampathkumar [21] gave the famous characterization to prove the balancing in a signed graph, which is as follows:

Theorem 3

(Marking Criterion [21]). A signed graph $S=(G,\sigma )$ is balanced if—and only if—there exists a marking μ of its vertices such that each edge $uv$ in S satisfies $\sigma \left(uv\right)=\mu \left(u\right)\mu \left(v\right)$.

Lemma 4.

For the addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, ${\mathsf{\Sigma }}_n^{\wedge }$ is a balanced signed graph, if for any prime p, $n=p^a$.

Proof. 

$n=p^a$, where p is a prime number. Now, we assign a marking $\mu$ to the vertices of ${\mathsf{\Sigma }}_n^{\wedge }$ in such a manner that if $u\in U_n$, then $\mu \left(u\right)=+$ and if $u\notin U_n$, then $\mu \left(u\right)=-$, ∀$u\in V\left({\mathsf{\Sigma }}_n^{\wedge }\right).$ Suppose there is an edge, say $ij$, in ${\mathsf{\Sigma }}_n^{\wedge }$.

Case I: Let ${\sigma }^{\wedge }\left(ij\right)=+.$ Then, $i,j\in U_n$ and according to the marking $\mu \left(i\right)=\mu \left(j\right)=+.$ Thus, ${\sigma }^{\wedge }\left(ij\right)=\mu \left(i\right)\mu \left(j\right)=+.$

Case II: Let ${\sigma }^{\wedge }\left(ij\right)=-.$ Then, there are three possibilities:

$\left(a\right)$

$i\in U_n$, $j\notin U_n$.

$\left(b\right)$

$i\notin U_n$, $j\in U_n$.

$\left(c\right)$

$i,j\notin U_n$.

Now, for (a) and (b), by marking $\mu$, we get $\mu \left(j\right)=-$ and $\mu \left(i\right)=+$ or vice versa. Therefore, ${\sigma }^{\wedge }\left(ij\right)=\mu \left(i\right)\mu \left(j\right)=-.$ Now, if $i,j\notin U_n$. Then, i and j are both multiples of p, and then $i+j=kp$, where k is some positive integer and $i+j\notin U_n$. So $ij\notin E\left({\mathsf{\Sigma }}_n^{\wedge }\right)$. Thus, condition (c) is not possible. So in every condition we get ${\sigma }^{\wedge }\left(ij\right)=\mu \left(i\right)\mu \left(j\right)$. Since $ij$ is an arbitrary edge, using Theorem 3, ${\mathsf{\Sigma }}_n^{\wedge }$ is balanced. □

Theorem 4.

The addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }$ is balanced if—and only if—either n is even or if n has exactly one prime factor, then n is odd.

Proof. 

Necessity: First, suppose ${\sum }_n^{\wedge }$ is balanced. Now, let $n=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_m^{{\alpha }_m}$; $p_1,p_2,\dots ,p_m$ being distinct primes, $p_1\ne 2$, $p_1\le p_2\le \dots \le p_m$.

In the unitary addition Cayley graph $G_n$, $p_1+1\ne k_1{p}_i$ for $i=1,2,\cdots ,m$ and $k_1$ are some positive integers i.e., $p_1+1\in U_n$, so $p_1$ is adjacent with one. Now, we claim that $p_1$ and $p_2$ are adjacent in $G_n$. On the contrary, suppose $p_1{p}_2$ is not an edge in $G_n$. Then, $p_1+p_2\notin U_n$. Thus, $p_1+p_2=k_2{p}_i$ for some $i=1,2,\cdots ,m$ and $k_2$ are some positive integers. Let $p_1+p_2$ be a multiple of $p_1$.

$$\begin{array}{cc}\hfill p_1+p_2& =\alpha p_1\hfill \\ \hfill p_2& =\alpha p_1-p_1\hfill \\ \hfill & =(\alpha -1)p_1\hfill \end{array}$$

for the positive integer $\alpha$, a contradiction. With the same argument, we can show that $p_1+p_2$ is not a multiple of $p_2$. Now, let $p_1+p_2=\alpha p_i$, for $i=3,4,\cdots ,m.$ As we know, the addition of two prime factors is always even; $p_1+p_2$ is even. So, $\alpha$ is even and is at least 2. However, as $p_1<p_2<p_i$, $p_1+p_2$ is always less than any multiple of $p_i$ for $i=3,4,\cdots ,m.$ Thus, $p_1+p_2\in U_n$ and $p_1{p}_2$ is an edge in $G_n$. Next, if $p_2$ is adjacent to 1 in $G_n$, we get a cycle

$$C=(p_1,p_2,1,p_1)$$

in ${\mathsf{\Sigma }}_n^{\wedge }$. Clearly, $p_1$ and $p_2$ are not in $U_n$, then by definition of ${\mathsf{\Sigma }}_n^{\wedge }$, C is a negative cycle. Thus, we have a negative cycle in ${\mathsf{\Sigma }}_n^{\wedge }$, implying that ${\mathsf{\Sigma }}_n^{\wedge }$ is not balanced. Now, suppose $p_2+1\notin E\left(G_n\right)$, since $p_2+1\notin U_n.$ Then, $p_2+1=c{p}_i$; $i=1,2,\cdots ,m$, c are positive integers. Clearly,

$$p_2+1=\alpha p_1$$

(1)

$\alpha$ is a positive integer.

Since $p_2\notin U_n$, according to Lemma 2, $n-p_2\notin U_n$. Next, we claim that $n-p_2$ is adjacent to 1 or $n-p_2+1=n-(p_2-1)\in U_n$. If $p_2-1\in U_n,$ then according to Lemma 2, $n-p_2+1=n-(p_2-1)\in U_n.$ Suppose $p_2-1\notin U_n.$ Then, $p_2-1=\beta p_i$; $i=1,2,\cdots ,m$, $\beta$ are positive integers. Let $p_2-1=\beta p_1.$ However, from Equation (1), $p_2=\alpha p_1-1.$ This implies

$$\begin{array}{cc}\hfill p_2-1& =\beta p_1\hfill \\ \hfill \alpha p_1-1-1& =\beta p_1\hfill \\ \hfill \alpha p_1-2& =\beta p_1\hfill \\ \hfill \alpha p_1-\beta p_1& =2\hfill \\ \hfill (\alpha -\beta )p_1& =2.\hfill \end{array}$$

This is not possible, as $p_1$ is at least $3.$ Thus, $p_2-1$ is not a multiple of any of the $p_i$s, whence $p_2-1\in U_n$. Hence, $n-p_2+1=n-(p_2-1)\in U_n,$ whence $n-p_2$ is adjacent to 1 in $G_n$. Now, $n-p_2+p_1=n-(p_2-p_1).$ Since $p_1<p_2<\cdots p_m$, $p_2-p_1\ne k{p}_i$; $i=2,3,\cdots m.$, k is a positive integer. Additionally, $p_2-p_1$ is not a multiple of $p_1.$ This shows that $p_2-p_1\in U_n$ and by Lemma 2, $n-(p_2-p_1)\in U_n.$ This shows that $n-p_2$ is adjacent to $p_1$ in ${\mathsf{\Sigma }}_n.$ Thus, we get a cycle

$$C^{\prime }=(p_1,n-p_2,1,p_1)$$

in ${\mathsf{\Sigma }}_n.$ Clearly, $p_1$ and $n-p_2$ do not belong to $U_n$ and $1\in U_n.$ Then, by definition ${\mathsf{\Sigma }}_n^{\wedge }$, we have a cycle $C^{\prime }$ with three negative edges. Thus, a contradiction. So, by contraposition, necessity is true.

Sufficiency: Let n be even. Then, according to Lemma 3, ${\mathsf{\Sigma }}_n^{\wedge }$ is an all-negative signed graph. Additionally, according to Theorem 2, $G_n$ is a bipartite graph. Hence, ${\mathsf{\Sigma }}_n^{\wedge }$, by Lemma 3 and Theorem 2, is balanced.

Now, let n be odd, with exactly one prime factor. Then, according to Lemma 4, ${\mathsf{\Sigma }}_n^{\wedge }$ is balanced, hence the theorem. □

2.2. Clusterability in ${\mathsf{\Sigma }}_n^{\wedge }$

Theorem 5.

The addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$ is clusterable.

Proof. 

Given an addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$. Suppose $v\in V\left({\mathsf{\Sigma }}_n^{\wedge }\right)$. Define $V^{*}\subseteq V\left({\mathsf{\Sigma }}_n^{\wedge }\right)$, such that $V^{*}=\left\{u_i:u_i\in V\left({\mathsf{\Sigma }}_n^{\wedge }\right)\mathrm{and}{\sigma }^{\wedge }\left(v{u}_i\right)=+\right\}$. By the definition of ${\mathsf{\Sigma }}_n^{\wedge }$, clearly $u_i$ and v are in $U_n$.

If, for i and j, ($i\ne j$), $u_i$ and $u_j$ are adjacent, then ${\sigma }^{\wedge }\left(u_i{u}_j\right)=+$. Thus, $U_n\subseteq V^{*}$. Since $\left|U_n\right|=\varphi \left(n\right)$, $n-\varphi \left(n\right)=k$ (say) vertices are not in $U_n$. Thus, only negative edges are incident on these k vertices. Put all these vertices in the k partition $V_1,V_2,\cdots ,V_k$, such that each partition contains exactly only one vertex. The clearly induced subgraph $<V^{*}>$ is all positive. Additionally, no positive edge joins the vertex of $V^{*}$ with the vertex of any of $V_i$, for $i=1,2,\cdots ,k$, and there is no edge $xy$, such that ${\sigma }^{\wedge }\left(xy\right)=-$ and $x,y\in V^{*}$. Thus, there exists a partition of the $V\left({\mathsf{\Sigma }}_n^{\wedge }\right)$, such that every positive edge has end vertices within the same subset and every negative edge has end vertices in a different subset. Hence, the proof. □

2.3. Sign-Compatibility in ${\mathsf{\Sigma }}_n^{\wedge }$

Theorem 6

([22]). A signed graph S is sign compatible if—and only if—S does not contain a sub signed graph isomorphic to either of the two signed graphs. $S_1$ formed by taking the path $P_4=(x,u,v,y)$ with both edges $xu$ and $vy$ negative and edge $uv$ positive, and $S_2$ formed by taking $S_1$ and identifying the vertices x and y (Figure 3).

Figure 3. Two forbidden sub signed graphs for a sign-compatible signed graph [13].

Theorem 7.

Addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }$ is sign compatible if—and only if—n is 3 or even.

Proof. 

Let addition signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }$ be sign compatible. If possible, suppose the conclusion is not true. Let n be odd but not 3. Now, $01\in E\left({\mathsf{\Sigma }}_n^{\wedge }\right)$. As, $n-2+1=n-1\in U_n$, $1(n-2)\in E\left({\mathsf{\Sigma }}_n^{\wedge }\right)$. Additionally, $n-2+0=n-2\in U_n$. Thus, we have a triangle $(0,1,n-2,0)$ with one positive edge $1(n-2)$ and two negative edges 01 and $(n-2)0$, which again contradict Theorem 6. Hence, the condition is necessary.

Next, let n be even. Thus, according to Lemma 3, ${\mathsf{\Sigma }}_n^{\wedge }$, which is all-negative, is trivially sign compatible. If $n=3$, then ${\mathsf{\Sigma }}_n^{\wedge }$ is $P_3$, which is trivially sign compatible. □

Acharya and Sinha [23] showed that every line signed graph is sign compatible. Next, we discuss the value of n for which ${\mathsf{\Sigma }}_n^{\wedge }$ is a line signed graph.

Theorem 8.

$G_n$ is a line graph if—and only if—n is equal to 2 or 3 or 4 or 6.

Proof. 

Necessity: Let $G_n$ be a line graph. Meanwhile, n is not equal to 2, 3, 4 and 6.

Case I: n is prime. It is clear that $n\ge 5.$ Here, n is prime, so by the definition of $U_n$, there are numbers from 1 to $(n-1)$ in $U_n.$ 0 is connected to every vertex of $G_n.$ The other vertex, $i\ne 0$, in $G_n$ is not connected to only $(n-i)$ by definition. For any $i,j\in V\left(G_n\right)$; $i\ne 0$, $j\ne 0$ there is an induced subgraph in $G_n$ (see Figure 4).

Figure 4. A forbidden subgraph for a line graph in $G_n$.

Thus, $G_n$ contains forbidden subgraph for a line graph. Thus, $G_n$ is not a line graph.

Case II: n is not prime. 1 is connected to 0 in $G_n.$ Next, 1 is connected to $p_1,$ as $p_1+1\in U_n,$ where $p_1$ is the smallest factor of $n.$ Let $\alpha p_1=n$, for a positive integer $\alpha$. Now,

$$\begin{array}{cc}\hfill 1+(\alpha -1)p_1& =1+\alpha p_1-p_1\hfill \\ \hfill & =1+n-p_1\hfill \\ \hfill & =n-(p_1-1).\hfill \end{array}$$

Since $p_1-1\in U_n,$ by Lemma 2, $n-(p_1-1)\in U_n.$ Thus, 1 and $(a-1)p_1$ are adjacent in $G_n.$ Additionally, 0 is not adjacent to $p_1$ and $(a-1)p_1$, because their sum is a multiple of $p_1.$ In the same way, $p_1$ and $(a-1)p_1$ are not connected in $G_n$ because their sum is a multiple of $p_1.$ So, we have an induced subgraph in $G_n$ (see Figure 5). Thus, there is a forbidden subgraph $K_{1,3}$ of a line graph. Additionally, $G_n$ is not a line graph.

Figure 5. A forbidden subgraph for a line graph in $G_n$.

Sufficiency: Let $n=2$ or $n=3$ or $n=4$ or $n=6$. Then, $G_2\cong L\left(P_3\right)$, $G_3\cong L\left(P_4\right)$, $G_4\cong L\left(C_4\right)$ and $G_6\cong L\left(C_6\right)$ (see Figure 6). Hence, the result. □

Figure 6. Showing $G_2,$ $G_3,$ $G_4$ and $G_6$.

Theorem 9.

${\mathsf{\Sigma }}_n^{\wedge }$ is a line signed graph if—and only if—$n=2$ or $n=3$ or $n=4$ or $n=6$.

Proof. 

Necessity: Let, if possible, n be unequal to 2, 3, 4 and 6. Theorem 8 shows that $G_n\neg \cong L\left(G\right)$, for any graph G. Thus, a contradiction and the condition are necessary.

Sufficiency: Now, suppose $n=2$ or $n=3$ or $n=4$ or $n=6$. Line signed graphs of an addition signed Cayley graph, for these values of n, are displayed in Figure 7, hence the sufficiency. □

Figure 7. Showing ${\mathsf{\Sigma }}_2^{\wedge }$, ${\mathsf{\Sigma }}_3^{\wedge }$, ${\mathsf{\Sigma }}_4^{\wedge }$ and ${\mathsf{\Sigma }}_6^{\wedge }$ and its line signed root graphs.

Remark 1.

${\mathsf{\Sigma }}_n^{\wedge }$ is a ×-line signed graph if—and only if—$n=2$ or $n=3$ or $n=4$ or $n=6$.

Proof. 

Let ${\mathsf{\Sigma }}_n^{\wedge }$ be a $\times -$line signed graph. We know that the underlying structure for line signed graphs and $\times -$line signed graphs is the same. Thus, the condition comes from Theorem 8.

Next, let $n\in \{2,3,4,6\}$. ${\mathsf{\Sigma }}_2^{\wedge }$, ${\mathsf{\Sigma }}_3^{\wedge }$, ${\mathsf{\Sigma }}_4^{\wedge }$ and ${\mathsf{\Sigma }}_6^{\wedge }$ and its $\times -$line signed root graphs are displayed in Figure 8. From Theorem 4, it is clear that for these values of $n,$ an addition signed Cayley graph is balanced. Additionally, $L_{\times }\left(S\right)$ of any signed graph is always balanced, and its underlying graph is a line graph (see [24]). This result comes from Theorems 4 and 8. □

Figure 8. Showing ${\mathsf{\Sigma }}_2^{\wedge }$, ${\mathsf{\Sigma }}_3^{\wedge }$, ${\mathsf{\Sigma }}_4^{\wedge }$ and ${\mathsf{\Sigma }}_6^{\wedge }$ and its $\times -$line signed root graphs.

2.4. $\mathcal{C}$-Consistency of ${\mathsf{\Sigma }}_n^{\wedge }$

Lemma 5.

For any prime p, $p\ne 2$ and $n=p^{\alpha }$, the $d^{-}\left(2\right)$ and $d^{-}\left(4\right)$ in ${\mathsf{\Sigma }}_n^{\wedge }$ is odd.

Proof. 

Given a ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, where $n=p^{\alpha }$ and p is an odd prime. Since n is odd; 2, $4\in U_n$. It is obvious that $d^{-}\left(2\right)$ and $d^{-}\left(4\right)$ in ${\mathsf{\Sigma }}_n^{\wedge }$ appear only when 2 and 4 are adjacent to $kp$, where k is some positive integer. Now, $(2+4)+cp\ne kp$; positive integers c and k. Additionally, 2 and 4 are connected to all the multiples of p, which are $p^{\alpha -1}$. Therefore $d^{-}\left(2\right)$$\left(d^{-}\left(4\right)\right)=p^{\alpha -1}$ is odd, hence the lemma. □

Theorem 10

([25]). Let a, b and m be integers with m positive. The linear congruence $ax\equiv b(modm)$ is soluble if and only if $(a,m)|b$. If $x_0$ is a solution, there are exactly $(a,m)$ incongruent solutions given by $\{x_0+tm/(a,m)\}$, where $t=0,1,\dots ,(a,m)-1$.

Corollary 1.

If $(a,m)=1$ then the congruence $ax\equiv b(modm)$ has exactly one incongruent solution.

Lemma 6.

In addition, signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, if $n=p_1^{a_1}{p}_2^{a_2}$, where $p_1$ and $p_2$ are two distinct odd primes, then $d^{-}\left(2\right)\left(d^{-}\left(4\right)\right)$ = odd.

Proof. 

Given that $n=p_1^{a_1}{p}_2^{a_2}$ in ${\mathsf{\Sigma }}_n^{\wedge }$, $p_1$ and $p_2$ are distinct odd primes. As n is odd, $2\in U_n$. Now, the negative degree of 2 of ${\mathsf{\Sigma }}_n^{\wedge }$ appears only when 2 is adjacent with the multiples of $p_1$ and $p_2$. Let $A_i=\left\{c{p}_i;\mathrm{c}\mathrm{certain}\mathrm{positive}\mathrm{integers},i=1,2\right\}$. Then,

$$\begin{array}{c}\hfill \left|A_1\right|=p_1^{a_1-1}{p}_2^{a_2}\\ \hfill \left|A_2\right|=p_1^{a_1}{p}_2^{a_2-1}\end{array}$$

and

$$\begin{array}{c}\hfill \left|A_1\cap A_2\right|=p_1^{a_1-1}{p}_2^{a_2-1}\end{array}$$

Thus, using the inclusion–exclusion principle

$$\begin{array}{c}\hfill \left|A_1\cup A_2\right|=p_1^{a_1-1}{p}_2^{a_2}+p_1^{a_1}{p}_2^{a_2-1}-p_1^{a_1-1}{p}_2^{a_2-1}\end{array}$$

Since $c{p}_1\left(p_2\right)+2=p_2\left(p_1\right)$, for certain positive integers c and so, $c{p}_1\left(p_2\right)2\notin E\left({\mathsf{\Sigma }}_n^{\wedge }\right)$ for those c. Thus, according to Theorem 10, we have

$$\begin{array}{c}\hfill p_{1}x\equiv -2\left(\mathrm{mod}{p}_2\right)\end{array}$$

(2)

and

$$\begin{array}{c}\hfill p_{2}y\equiv -2\left(\mathrm{mod}{p}_1\right)\end{array}$$

(3)

Due to Corollary 1, we have an incongruent solution $x_0$ (say), which is unique for Equation (2). So, for Equation (2) where $p_{1}x+2<n$, we have:

$$\begin{array}{c}\hfill x_0+0\left(p_2\right),x_0+1\left(p_2\right),x_0+2\left(p_2\right),\cdots ,x_0+(p_1^{a_1-1}{p}_2^{a_2-1}-1)\left(p_2\right)\end{array}$$

(4)

Thus, Equation (2) has $p_1^{a_1-1}{p}_2^{a_2-1}$ total solutions. Similarly, the total solutions of Equation (3) are $p_1^{a_1-1}{p}_2^{a_2-1}$. Hence,

$$\begin{array}{cc}\hfill d^{-}\left(2\right)& =p_1^{a_1-1}{p}_2^{a_2}+p_1^{a_1}{p}_2^{a_2-1}-p_1^{a_1-1}{p}_2^{a_2-1}-p_1^{a_1-1}{p}_2^{a_2-1}-p_1^{a_1-1}{p}_2^{a_2-1}\hfill \\ \hfill & =p_1^{a_1-1}{p}_2^{a_2-1}(p_1+p_2-3)\hfill \end{array}$$

$p_1$ and $p_2$ are odd primes, which implies $d^{-}\left(2\right)$ is odd. The proof for $d^{-}\left(4\right)$ is analogous. □

Lemma 7.

In ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, if $n=3^{a_1}{5}^{a_2}$, then $d^{-}\left(7\right)$ = odd.

Proof. 

This is easy to prove using the same logic as mentioned in Lemma 6. □

Theorem 11.

Let n have at most two distinct odd prime factors, then ${\mathsf{\Sigma }}_n^{\wedge }$ is $\mathcal{C}$ consistent if—and only if—n is even or 3.

Proof. 

Necessity: Let n have, at most, two distinct prime factors and let ${\mathsf{\Sigma }}_n^{\wedge }$ be $\mathcal{C}$ consistent. If possible, let n be odd but not 3.

Case (a): Let $n\equiv 1(mod3)$ or $n\equiv 2(mod3)$. As n is odd, $2\in U_n$. Clearly, 0 is adjacent with 1, 2, $n-1$ in ${\mathsf{\Sigma }}_n^{\wedge }.$ Since, $n-1+2=1\in U_n$, $n-1$ and 2 are connected in ${\mathsf{\Sigma }}_n^{\wedge }.$ Since, 3 is not a factor of n, $3\in U_n.$ Now, $2+1=3\in U_n$. Hence, 2 and 1 are adjacent in ${\mathsf{\Sigma }}_n^{\wedge }.$ Now, the cycles $Z_1=(0,1,2,0)$, $Z_2=(0,2,n-1,0)$ have a common chord with end vertices 0 and 2. By Lemma 6,

$${\mu }_{\sigma }\left(2\right)=-$$

Since the vertex $0\notin U_n$, $d\left(0\right)=d^{-}\left(0\right)=\varphi \left(n\right)=$ even. It follows,

$${\mu }_{\sigma }\left(0\right)=+.$$

Now, if either $Z_1$ or $Z_2$ is not a $\mathcal{C}$-consistent cycle, a contradiction. Thus, $Z_1$ and $Z_2$ both cycle are $\mathcal{C}$-consistent. The common chord with end vertices zero and two are oppositely marked, in contradiction with (Theorem 2, [26]).

Case (b): Let $n\equiv 0(mod3).$ Then, either $n=3^{a_1}$ or $n=3^{a_1}\times p_2^{a_2}.$ First, suppose $p_2\ne 5.$ Since, n is odd, $2,4\in U_n.$ According to Lemma 2, $n-2\in U_n.$ Clearly, 0 is adjacent to 1, 4 and $n-2$ in ${\mathsf{\Sigma }}_n^{\wedge }.$ Since, $n-2+4=n+2=2\in U_n$, $n-2$ is adjacent to 4 in ${\mathsf{\Sigma }}_n^{\wedge }.$ Now, for cycle $Z_1=(0,1,4,0)$, $Z_2=(0,4,n-2,0)$; $Z_1$, $Z_2$ have a common chord with end vertices 0 and 4. According to Lemma 6,

$${\mu }_{\sigma }\left(4\right)=-$$

Since the vertex $0\notin U_n,$$d\left(0\right)=d^{-}\left(0\right)=\varphi \left(n\right)=$ even. It follows,

$${\mu }_{\sigma }\left(0\right)=+.$$

Now, if either $Z_1$ or $Z_2$ is a cycle which is not $\mathcal{C}$ consistent, a contradiction. Therefore, $Z_1$ and $Z_2$ are the cycles which are $\mathcal{C}$-consistent. However, there is a chord whose end vertices 0 and 4 have opposite marking. Here again, we find a contradiction to the (Theorem 2, [26]).

Now, suppose $p_2=5.$ In this case, we consider two cycles $Z_1=(0,1,7,0)$ and $Z_2=(0,7,10,13,0)$ in ${\mathsf{\Sigma }}_n^{\wedge }.$ For cycles $Z_1$, $Z_2$ have a common chord with end vertices 0 and 7, according to Lemma 7,

$${\mu }_{\sigma }\left(7\right)=-$$

Since the vertex $0\notin U_n,$$d\left(0\right)=d^{-}\left(0\right)=\varphi \left(n\right)=$ even. It follows that

$${\mu }_{\sigma }\left(0\right)=+.$$

Now, if either $Z_1$ or $Z_2$ is a cycle which is not $\mathcal{C}$ consistent, this is a contradiction. Therefore, $Z_1$ and $Z_2$ are the cycles which are $\mathcal{C}$ consistent. However, the end vertices 0 and 7 have the opposite marking. Here, we have a contradiction to the (Theorem 2, [26]). Hence, n is either even or $n=3$.

Sufficiency: Let n be even. According to Lemma 3, ${\mathsf{\Sigma }}_n^{\wedge }$ is all negative. Additionally, according to Theorem 13, $d\left(v\right)=d^{-}\left(v\right)$ = even ∀$v\in V\left({\mathsf{\Sigma }}_n^{\wedge }\right).$ So, according to canonical marking ${\mu }_{\sigma }\left(v\right)=+$∀$v\in V\left({\mathsf{\Sigma }}_n^{\wedge }\right).$ So when n is even, ${\mathsf{\Sigma }}_n^{\wedge }$ is trivially $\mathcal{C}$ consistent. If $n=3$, then $G_3$ is a path, which is trivially $\mathcal{C}$- consistent, hence the result. □

3. Balance in Certain Derived Signed Graphs of ${\mathsf{\Sigma }}_{\mathit{n}}^{\wedge }$

Theorem 12.

$\eta \left({\mathsf{\Sigma }}_n^{\wedge }\right)$ is balanced if—and only if—n is 3 or even.

Proof. 

Let $\eta \left({\mathsf{\Sigma }}_n^{\wedge }\right)$ be balanced. If possible, n is odd but not 3, and p is the smallest prime factor of $n$. Since $n-2+1=n-1\in U_n$, $n-2$ and 1 are connected in ${\mathsf{\Sigma }}_n^{\wedge }$. $p+1\in U_n$ implies that p and 1 are connected in ${\mathsf{\Sigma }}_n^{\wedge }.$ Additionally, as n is odd, $2\in U_n$ and $n-2\in U_n.$ according to Lemma 2. Since, $n-2+p=n+(p-2)=p-2\in U_n$, $(n-2)p\in E\left({\mathsf{\Sigma }}_n^{\wedge }\right).$ Now, for the cycle $Z=(1,p,n-2,1)$ in ${\mathsf{\Sigma }}_n^{\wedge }$ we have a one positive edge $1(n-2)$ and two negative edges $1p$ and $p(n-2)$ in $Z.$ However, in $\eta \left({\mathsf{\Sigma }}_n^{\wedge }\right)$, there is a cycle $Z^{\prime }=(1,p,n-2,1)$ with one negative edge $1(n-2)$ and two positive edges $1p$ and $p(n-2).$ Thus, we have a negative cycle that contradicts the given condition. Therefore, the only possibility is that n is 3 or even.

Conversely, let n be even. ${\mathsf{\Sigma }}_n^{\wedge }$, according to Lemma 3 is an all-negative signed graph. So $\eta \left({\mathsf{\Sigma }}_n^{\wedge }\right)$ is balanced and is all positive. $\eta \left({\mathsf{\Sigma }}_n^{\wedge }\right)$ for $n=3$ is a tree which is trivially balanced, hence the converse. □

We present the following theorem for the degree of the vertices of $G_n$ (see [20]).

Theorem 13

([20]). Let m be any vertex of the unitary addition Cayley graph $G_n$. Then,

$$d\left(m\right)=\left\{\begin{array}{cc}\varphi \left(n\right)\hfill & \mathit{if}n\mathit{is}\mathit{even},\hfill \\ \varphi \left(n\right)\hfill & \mathit{if}n\mathit{is}\mathit{odd}\mathit{and}(m,n)\ne 1,\hfill \\ \varphi \left(n\right)-1\hfill & \mathit{if}n\mathit{is}\mathit{odd}\mathit{and}(m,n)=1.\hfill \end{array}\right.$$

Additionally, for a signed graph S, the balance property of $L\left(S\right)$ is discussed in ([27], Theorem 4).

Theorem 14.

For an additional signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, its line signed graph $L\left({\mathsf{\Sigma }}_n^{\wedge }\right)$ is balanced if—and only if—$n\in \left\{2,3,4,6\right\}.$

Proof. 

Let $L\left({\mathsf{\Sigma }}_n^{\wedge }\right)$ be balanced and $n\ne 2,3,4$ and 6. Now, according to Theorem 13, $d\left(0\right)=d^{-}\left(0\right)=\varphi \left(n\right)$ = even, which implies $d\left(0\right)=d^{-}\left(0\right)=\varphi \left(n\right)\ge 4.$ This shows that condition $ii$ (of Theorem 4, [27]) is not satisfied for ${\mathsf{\Sigma }}_n^{\wedge }.$ This is a contradiction. Hence, $n\in \left\{2,3,4,6\right\}.$ The converse part is easy to prove. □

For a signed graph S, the balance property of $C_E\left(S\right)$ is discussed in ([9], Theorem 13).

Theorem 15.

For an additional signed Cayley graph ${\mathsf{\Sigma }}_n^{\wedge }=(G_n,{\sigma }^{\wedge })$, its common-edge signed graph $C_E\left({\mathsf{\Sigma }}_n^{\wedge }\right)$ is balanced if—and only if—$n\in \left\{3,4,6\right\}.$

Proof. 

Let $n\notin \left\{3,4,6\right\}.$ It is clear that $0\notin U_n.$ Now, by Theorem 13, $d\left(0\right)=d^{-}\left(0\right)=\varphi \left(n\right)$ = even, which implies $d\left(0\right)=d^{-}\left(0\right)=\varphi \left(n\right)\ge 4.$ This shows that condition $ii$ (of Theorem 13, [9]) is not satisfied for ${\mathsf{\Sigma }}_n^{\wedge }.$ Thus, $C_E\left({\mathsf{\Sigma }}_n^{\wedge }\right)$ is not balanced, which is a contradiction. Hence, $n\in \left\{3,4,6\right\}$. The converse part is easy to prove. □