**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 *Su*, its negation *η*(*S*), 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 (C-*marking*), consistency, C-*consistency*, *S* consistency, sign compatibility, line signed graph *L*(*S*), line signed root graph, ×-line signed graph, ×-line signed root graph and the common-edge signed graph *CE*(*S*) of signed graph, *S* we refer to [6,9–16].

*Addition Signed Cayley Graph* Σ<sup>∧</sup> *n*

A *unitary addition Cayley graph Gn*, where *n* ∈ *I*+, *I*<sup>+</sup> is set of positive integers, is a graph in which the vertex set is a ring of integers modulo *n*, *Zn*. Any two vertices *x*<sup>1</sup> and *x*<sup>2</sup> are adjacent in *Gn* if—and only if—(*x*<sup>1</sup> + *x*2) ∈ *Un*, where *Un* denotes the unit set.

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

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* Σ<sup>∧</sup>

*<sup>n</sup>* as follows: The *addition signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧) is a signed graph whose underlying graph is a unitary addition Cayley graph *Gn*, where *n* ∈ *I*<sup>+</sup> and for an edge *ab* of Σ<sup>∧</sup> *n* ,

$$
\sigma^\wedge(ab) = \begin{cases} + & \text{if } a, b \in \mathcal{U}\_{n\prime} \\ - & \text{otherwise}. \end{cases}
$$

**Citation:** Wardak, O.; Dhama, A.; Sinha, D. On Some Properties of Addition Signed Cayley Graph Σ<sup>∧</sup> *n* . *Mathematics* **2022**, *10*, 3492. https:// doi.org/10.3390/math10193492

Academic Editor: Emeritus Mario Gionfriddo

Received: 24 July 2022 Accepted: 8 September 2022 Published: 25 September 2022

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**Figure 1.** Examples of unitary addition Cayley graphs.

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* ≥ 2.

**Figure 2.** Examples of addition signed Cayley graph Σ<sup>∧</sup> *n* .

#### **2. Some Properties of Σ***∧ n*

*2.1. Balancing in* Σ<sup>∧</sup> *n*

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 Σ<sup>∧</sup> *<sup>n</sup>* , for which the following well-known result can be used as a tool.

**Theorem 2** ([20])**.** *Gn*, *n* ≥ 2, *is bipartite if—and only if—either n* = 3 *or n is even.*

**Lemma 2.** *i* ∈ *Un* ⇒ (*n* − *i*) ∈ *Un and i* ∈ *Un* ⇒ (*n* − *i*) ∈ *Un.*

**Lemma 3.** *Addition signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧)*, for n even, is an all-negative signed graph.*

**Proof.** Given an addition signed Cayley graph Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧), where *n* is even. Suppose the conclusion is false. Let there be a positive edge, say *ij*, in Σ<sup>∧</sup> *<sup>n</sup>* . By the definition of Σ<sup>∧</sup> *<sup>n</sup>* , *i*, *j* ∈ *Un*. Since *n* is even, *Un* 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* ∈/ *Un*, i.e., *i* and *j*, are not adjacent in Σ<sup>∧</sup> *<sup>n</sup>* . Thus, we have a contradiction. Hence, if *n* is even, then Σ<sup>∧</sup> *<sup>n</sup>* 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*, *σ*) *is balanced if—and only if—there exists a marking μ of its vertices such that each edge uv in S satisfies σ*(*uv*) = *μ*(*u*)*μ*(*v*)*.*

**Lemma 4.** *For the addition signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧)*,* Σ<sup>∧</sup> *<sup>n</sup> is a balanced signed graph, if for any prime p, n* = *pa.*

**Proof.** *n* = *pa*, where *p* is a prime number. Now, we assign a marking *μ* to the vertices of Σ<sup>∧</sup> *<sup>n</sup>* in such a manner that if *u* ∈ *Un*, then *μ*(*u*)=+ and if *u* ∈/ *Un*, then *μ*(*u*) = −, ∀ *u* ∈ *V*(Σ<sup>∧</sup> *<sup>n</sup>* ). Suppose there is an edge, say *ij*, in Σ<sup>∧</sup> *n* .

**Case I:** Let *σ*∧(*ij*)=+. Then, *i*, *j* ∈ *Un* and according to the marking *μ*(*i*) = *μ*(*j*)=+. Thus, *σ*∧(*ij*) = *μ*(*i*)*μ*(*j*)=+.

**Case II:** Let *σ*∧(*ij*) = −. Then, there are three possibilities:


Now, for (*a*) and (*b*), by marking *μ*, we get *μ*(*j*) = − and *μ*(*i*)=+ or vice versa. Therefore, *σ*∧(*ij*) = *μ*(*i*)*μ*(*j*) = −. Now, if *i*, *j* ∈/ *Un*. Then, *i* and *j* are both multiples of *p*, and then *i* + *j* = *kp*, where *k* is some positive integer and *i* + *j* ∈/ *Un*. So *ij* ∈/ *E*(Σ<sup>∧</sup> *<sup>n</sup>* ). Thus, condition (*c*) is not possible. So in every condition we get *σ*∧(*ij*) = *μ*(*i*)*μ*(*j*). Since *ij* is an arbitrary edge, using Theorem 3, Σ<sup>∧</sup> *<sup>n</sup>* is balanced.

**Theorem 4.** *The addition signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup> 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 ∑<sup>∧</sup> *<sup>n</sup>* is balanced. Now, let *<sup>n</sup>* <sup>=</sup> *<sup>p</sup>α*<sup>1</sup> <sup>1</sup> *<sup>p</sup>α*<sup>2</sup> <sup>2</sup> ... *<sup>p</sup>α<sup>m</sup> <sup>m</sup>* ; *p*1, *p*2, ... , *pm* being distinct primes, *p*<sup>1</sup> = 2, *p*<sup>1</sup> ≤ *p*<sup>2</sup> ≤ ... ≤ *pm*.

In the unitary addition Cayley graph *Gn*, *p*<sup>1</sup> + 1 = *k*<sup>1</sup> *pi* for *i* = 1, 2, ... , *m* and *k*<sup>1</sup> are some positive integers i.e., *p*<sup>1</sup> + 1 ∈ *Un*, so *p*<sup>1</sup> is adjacent with one. Now, we claim that *p*<sup>1</sup> and *p*<sup>2</sup> are adjacent in *Gn*. On the contrary, suppose *p*<sup>1</sup> *p*<sup>2</sup> is not an edge in *Gn*. Then, *p*<sup>1</sup> + *p*<sup>2</sup> ∈/ *Un*. Thus, *p*<sup>1</sup> + *p*<sup>2</sup> = *k*<sup>2</sup> *pi* for some *i* = 1, 2, ... , *m* and *k*<sup>2</sup> are some positive integers. Let *p*<sup>1</sup> + *p*<sup>2</sup> be a multiple of *p*1.

$$\begin{aligned} p\_1 + p\_2 &= \mathfrak{a}p\_1\\ p\_2 &= \mathfrak{a}p\_1 - p\_1\\ &= (\mathfrak{a} - 1)p\_1 \end{aligned}$$

for the positive integer *α*, a contradiction. With the same argument, we can show that *p*<sup>1</sup> + *p*<sup>2</sup> is not a multiple of *p*2. Now, let *p*<sup>1</sup> + *p*<sup>2</sup> = *αpi*, for *i* = 3, 4, ... , *m*. As we know, the addition of two prime factors is always even; *p*<sup>1</sup> + *p*<sup>2</sup> is even. So, *α* is even and is at least 2. However, as *p*<sup>1</sup> < *p*<sup>2</sup> < *pi*, *p*<sup>1</sup> + *p*<sup>2</sup> is always less than any multiple of *pi* for *i* = 3, 4, ... , *m*. Thus, *p*<sup>1</sup> + *p*<sup>2</sup> ∈ *Un* and *p*<sup>1</sup> *p*<sup>2</sup> is an edge in *Gn*. Next, if *p*<sup>2</sup> is adjacent to 1 in *Gn*, we get a cycle

$$\mathbb{C} = (p\_{1\prime}, p\_{2\prime}, 1, p\_1)$$

in Σ<sup>∧</sup> *<sup>n</sup>* . Clearly, *p*<sup>1</sup> and *p*<sup>2</sup> are not in *Un*, then by definition of Σ<sup>∧</sup> *<sup>n</sup>* , *C* is a negative cycle. Thus, we have a negative cycle in Σ<sup>∧</sup> *<sup>n</sup>* , implying that Σ<sup>∧</sup> *<sup>n</sup>* is not balanced. Now, suppose *p*<sup>2</sup> + 1 ∈/ *E*(*Gn*), since *p*<sup>2</sup> + 1 ∈ *Un*. Then, *p*<sup>2</sup> + 1 = *cpi*; *i* = 1, 2, ... , *m*, *c* are positive integers. Clearly,

$$p\_2 + 1 = \mathfrak{a}p\_1\tag{1}$$

*α* is a positive integer.

Since *p*<sup>2</sup> ∈ *Un*, according to Lemma 2, *n* − *p*<sup>2</sup> ∈ *Un*. Next, we claim that *n* − *p*<sup>2</sup> is adjacent to 1 or *n* − *p*<sup>2</sup> + 1 = *n* − (*p*<sup>2</sup> − 1) ∈ *Un*. If *p*<sup>2</sup> − 1 ∈ *Un*, then according to Lemma 2, *n* − *p*<sup>2</sup> + 1 = *n* − (*p*<sup>2</sup> − 1) ∈ *Un*. Suppose *p*<sup>2</sup> − 1 ∈ *Un*. Then, *p*<sup>2</sup> − 1 = *βpi*; *i* = 1, 2, ... , *m*,

*β* are positive integers. Let *p*<sup>2</sup> − 1 = *βp*1. However, from Equation (1), *p*<sup>2</sup> = *αp*<sup>1</sup> − 1. This implies

$$\begin{aligned} p\_2 - 1 &= \beta p\_1 \\ \alpha p\_1 - 1 - 1 &= \beta p\_1 \\ \alpha p\_1 - 2 &= \beta p\_1 \\ \alpha p\_1 - \beta p\_1 &= 2 \\ (\alpha - \beta) p\_1 &= 2 \end{aligned}$$

This is not possible, as *p*<sup>1</sup> is at least 3. Thus, *p*<sup>2</sup> − 1 is not a multiple of any of the *pi*s, whence *p*<sup>2</sup> − 1 ∈ *Un*. Hence, *n* − *p*<sup>2</sup> + 1 = *n* − (*p*<sup>2</sup> − 1) ∈ *Un*, whence *n* − *p*<sup>2</sup> is adjacent to 1 in *Gn*. Now, *n* − *p*<sup>2</sup> + *p*<sup>1</sup> = *n* − (*p*<sup>2</sup> − *p*1). Since *p*<sup>1</sup> < *p*<sup>2</sup> < ··· *pm*, *p*<sup>2</sup> − *p*<sup>1</sup> = *kpi*; *i* = 2, 3, ... *m*., *k* is a positive integer. Additionally, *p*<sup>2</sup> − *p*<sup>1</sup> is not a multiple of *p*1. This shows that *p*<sup>2</sup> − *p*<sup>1</sup> ∈ *Un* and by Lemma 2, *n* − (*p*<sup>2</sup> − *p*1) ∈ *Un*. This shows that *n* − *p*<sup>2</sup> is adjacent to *p*<sup>1</sup> in Σ*n*. Thus, we get a cycle

$$\mathbb{C}' = (p\_1, n - p\_2, 1, p\_1).$$

in Σ*n*. Clearly, *p*<sup>1</sup> and *n* − *p*<sup>2</sup> do not belong to *Un* and 1 ∈ *Un*. Then, by definition Σ<sup>∧</sup> *<sup>n</sup>* , we have a cycle *C* with three negative edges. Thus, a contradiction. So, by contraposition, necessity is true.

*Sufficiency*: Let *n* be even. Then, according to Lemma 3, Σ<sup>∧</sup> *<sup>n</sup>* is an all-negative signed graph. Additionally, according to Theorem 2, *Gn* is a bipartite graph. Hence, Σ<sup>∧</sup> *<sup>n</sup>* , by Lemma 3 and Theorem 2, is balanced.

Now, let *n* be odd, with exactly one prime factor. Then, according to Lemma 4, Σ<sup>∧</sup> *<sup>n</sup>* is balanced, hence the theorem.

#### *2.2. Clusterability in* Σ<sup>∧</sup> *n*

**Theorem 5.** *The addition signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧) *is clusterable.*

**Proof.** Given an addition signed Cayley graph Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧). Suppose *v* ∈ *V*(Σ<sup>∧</sup> *<sup>n</sup>* ). Define *V*<sup>∗</sup> ⊆ *V*(Σ<sup>∧</sup> *<sup>n</sup>* ), such that *V*<sup>∗</sup> = {*ui* : *ui* ∈ *V*(Σ<sup>∧</sup> *<sup>n</sup>* ) and *σ*∧(*vui*)=+}. By the definition of Σ<sup>∧</sup> *<sup>n</sup>* , clearly *ui* and *v* are in *Un*.

If, for *i* and *j*, (*i* = *j*), *ui* and *uj* are adjacent, then *σ*∧(*uiuj*)=+. Thus, *Un* ⊆ *V*∗. Since |*Un*| = *φ*(*n*), *n* − *φ*(*n*) = *k* (say) vertices are not in *Un*. Thus, only negative edges are incident on these *k* vertices. Put all these vertices in the *k* partition *V*1, *V*2, ... , *Vk*, such that each partition contains exactly only one vertex. The clearly induced subgraph < *V*<sup>∗</sup> > is all positive. Additionally, no positive edge joins the vertex of *V*<sup>∗</sup> with the vertex of any of *Vi*, for *i* = 1, 2, ... , *k*, and there is no edge *xy*, such that *σ*∧(*xy*) = − and *x*, *y* ∈ *V*∗. Thus, there exists a partition of the *V*(Σ<sup>∧</sup> *<sup>n</sup>* ), 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* Σ<sup>∧</sup> *n*

**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*<sup>1</sup> *formed by taking the path P*<sup>4</sup> = (*x*, *u*, *v*, *y*) *with both edges xu and vy negative and edge uv positive, and S*<sup>2</sup> *formed by taking S*<sup>1</sup> *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* Σ<sup>∧</sup> *<sup>n</sup> is sign compatible if—and only if—n is* 3 *or even.*

**Proof.** Let addition signed Cayley graph Σ<sup>∧</sup> *<sup>n</sup>* be sign compatible. If possible, suppose the conclusion is not true. Let *n* be odd but not 3. Now, 01 ∈ *E*(Σ<sup>∧</sup> *<sup>n</sup>* ). As, *n* − 2 + 1 = *n* − 1 ∈ *Un*, 1(*n* − 2) ∈ *E*(Σ<sup>∧</sup> *<sup>n</sup>* ). Additionally, *n* − 2 + 0 = *n* − 2 ∈ *Un*. 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, Σ<sup>∧</sup> *<sup>n</sup>* , which is all-negative, is trivially sign compatible. If *n* = 3, then Σ<sup>∧</sup> *<sup>n</sup>* 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 Σ<sup>∧</sup> *<sup>n</sup>* is a line signed graph.

**Theorem 8.** *Gn is a line graph if—and only if—n is equal to* 2 *or* 3 *or* 4 *or* 6*.*

**Proof.** *Necessity*: Let *Gn* 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* ≥ 5. Here, *n* is prime, so by the definition of *Un*, there are numbers from 1 to (*n* − 1) in *Un*. 0 is connected to every vertex of *Gn*. The other vertex, *i* = 0, in *Gn* is not connected to only (*n* − *i*) by definition. For any *i*, *j* ∈ *V*(*Gn*); *i* = 0, *j* = 0 there is an induced subgraph in *Gn* (see Figure 4).

**Figure 4.** A forbidden subgraph for a line graph in *Gn*.

Thus, *Gn* contains forbidden subgraph for a line graph. Thus, *Gn* is not a line graph. **Case II:** *n* is not prime. 1 is connected to 0 in *Gn*. Next, 1 is connected to *p*1, as *p*<sup>1</sup> + 1 ∈ *Un*, where *p*<sup>1</sup> is the smallest factor of *n*. Let *αp*<sup>1</sup> = *n*, for a positive integer *α*. Now,

$$\begin{aligned} 1 + (a - 1)p\_1 &= 1 + \alpha p\_1 - p\_1 \\ &= 1 + n - p\_1 \\ &= n - (p\_1 - 1) . \end{aligned}$$

Since *p*<sup>1</sup> − 1 ∈ *Un*, by Lemma 2, *n* − (*p*<sup>1</sup> − 1) ∈ *Un*. Thus, 1 and (*a* − 1)*p*<sup>1</sup> are adjacent in *Gn*. Additionally, 0 is not adjacent to *p*<sup>1</sup> and (*a* − 1)*p*1, because their sum is a multiple of *p*1. In the same way, *p*<sup>1</sup> and (*a* − 1)*p*<sup>1</sup> are not connected in *Gn* because their sum is a multiple of *p*1. So, we have an induced subgraph in *Gn* (see Figure 5). Thus, there is a forbidden subgraph *K*1,3 of a line graph. Additionally, *Gn* is not a line graph.

*Sufficiency*: Let *<sup>n</sup>* = 2 or *<sup>n</sup>* = 3 or *<sup>n</sup>* = 4 or *<sup>n</sup>* = 6. Then, *<sup>G</sup>*<sup>2</sup> ∼= *<sup>L</sup>*(*P*3), *<sup>G</sup>*<sup>3</sup> ∼= *<sup>L</sup>*(*P*4), *<sup>G</sup>*<sup>4</sup> ∼= *<sup>L</sup>*(*C*4) and *<sup>G</sup>*<sup>6</sup> ∼= *<sup>L</sup>*(*C*6) (see Figure 6). Hence, the result.

**Figure 5.** A forbidden subgraph for a line graph in *Gn*.

**Figure 6.** Showing *G*2, *G*3, *G*<sup>4</sup> and *G*6.

**Theorem 9.** Σ<sup>∧</sup> *<sup>n</sup> 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 *Gn* ∼= *L*(*G*), 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 Σ<sup>∧</sup> <sup>2</sup> , <sup>Σ</sup><sup>∧</sup> <sup>3</sup> , <sup>Σ</sup><sup>∧</sup> <sup>4</sup> and <sup>Σ</sup><sup>∧</sup> <sup>6</sup> and its line signed root graphs.

**Remark 1.** Σ<sup>∧</sup> *<sup>n</sup> is a* ×*-line signed graph if—and only if—n* = 2 *or n* = 3 *or n* = 4 *or n* = 6*.*

**Proof.** Let Σ<sup>∧</sup> *<sup>n</sup>* be a ×−line signed graph. We know that the underlying structure for line signed graphs and ×−line signed graphs is the same. Thus, the condition comes from Theorem 8.

Next, let *n* ∈ {2, 3, 4, 6}. Σ<sup>∧</sup> <sup>2</sup> , <sup>Σ</sup><sup>∧</sup> <sup>3</sup> , <sup>Σ</sup><sup>∧</sup> <sup>4</sup> and <sup>Σ</sup><sup>∧</sup> <sup>6</sup> and its ×−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*×(*S*) 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 Σ<sup>∧</sup> <sup>2</sup> , <sup>Σ</sup><sup>∧</sup> <sup>3</sup> , <sup>Σ</sup><sup>∧</sup> <sup>4</sup> and <sup>Σ</sup><sup>∧</sup> <sup>6</sup> and its ×−line signed root graphs.

#### *2.4.* C*-Consistency of* Σ<sup>∧</sup> *n*

**Lemma 5.** *For any prime p, p* = 2 *and n* = *pα, the d*−(2) *and d*−(4) *in* Σ<sup>∧</sup> *<sup>n</sup> is odd.*

**Proof.** Given a Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧), where *n* = *p<sup>α</sup>* and *p* is an odd prime. Since *n* is odd; 2, 4 ∈ *Un*. It is obvious that *d*−(2) and *d*−(4) in Σ<sup>∧</sup> *<sup>n</sup>* appear only when 2 and 4 are adjacent to *kp*, where *k* is some positive integer. Now, (2 + 4) + *cp* = *kp*; positive integers *c* and *k*. Additionally, 2 and 4 are connected to all the multiples of *p*, which are *pα*<sup>−</sup>1. Therefore *d*−(2) (*d*−(4)) = *pα*−<sup>1</sup> is odd, hence the lemma.

**Theorem 10** ([25])**.** *Let a, b and m be integers with m positive. The linear congruence ax* ≡ *b* (mod *m*) *is soluble if and only if* (*a*, *m*)|*b. If x*<sup>0</sup> *is a solution, there are exactly* (*a*, *m*) *incongruent solutions given by* {*x*<sup>0</sup> + *tm*/(*a*, *m*)}*, where t* = 0, 1, . . . ,(*a*, *m*) − 1*.*

**Corollary 1.** *If* (*a*, *m*) = 1 *then the congruence ax* ≡ *b* (mod *m*) *has exactly one incongruent solution.*

**Lemma 6.** *In addition, signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *<sup>σ</sup>*∧)*, if <sup>n</sup>* <sup>=</sup> *<sup>p</sup>a*<sup>1</sup> <sup>1</sup> *<sup>p</sup>a*<sup>2</sup> <sup>2</sup> *, where p*<sup>1</sup> *and p*<sup>2</sup> *are two distinct odd primes, then d*−(2)(*d*−(4)) *= odd.*

**Proof.** Given that *n* = *pa*<sup>1</sup> <sup>1</sup> *<sup>p</sup>a*<sup>2</sup> <sup>2</sup> in <sup>Σ</sup><sup>∧</sup> *<sup>n</sup>* , *p*<sup>1</sup> and *p*<sup>2</sup> are distinct odd primes. As *n* is odd, 2 ∈ *Un*. Now, the negative degree of 2 of Σ<sup>∧</sup> *<sup>n</sup>* appears only when 2 is adjacent with the multiples of *p*<sup>1</sup> and *p*2. Let *Ai* = {*cpi*; *c* certain positive integers, *i* = 1, 2}. Then,

$$\begin{aligned} |A\_1| &= p\_1^{a\_1 - 1} p\_2^{a\_2} \\ |A\_2| &= p\_1^{a\_1} p\_2^{a\_2 - 1} \end{aligned}$$

and

$$|A\_1 \cap A\_2| = p\_1^{a\_1 - 1} p\_2^{a\_2 - 1}$$

Thus, using the inclusion–exclusion principle

$$|A\_1 \cup A\_2| = p\_1^{a\_1 - 1} p\_2^{a\_2} + p\_1^{a\_1} p\_2^{a\_2 - 1} - p\_1^{a\_1 - 1} p\_2^{a\_2 - 1}$$

Since *cp*1(*p*2) + 2 = *p*2(*p*1), for certain positive integers *c* and so, *cp*1(*p*2)2 ∈/ *E*(Σ<sup>∧</sup> *n* ) for those *c*. Thus, according to Theorem 10, we have

$$p\_1 \underline{x} \equiv -2 \pmod{p\_2} \tag{2}$$

and

$$p\_2 y \equiv -2 \pmod{p\_1} \tag{3}$$

Due to Corollary 1, we have an incongruent solution *x*<sup>0</sup> (say), which is unique for Equation (2). So, for Equation (2) where *p*1*x* + 2 < *n*, we have:

$$(\mathbf{x}\_0 + \mathbf{0}(p\_2), \mathbf{x}\_0 + \mathbf{1}(p\_2), \mathbf{x}\_0 + \mathbf{2}(p\_2), \dots, \mathbf{x}\_0 + (p\_1^{a\_1 - 1} p\_2^{a\_2 - 1} - 1)(p\_2) \tag{4}$$

Thus, Equation (2) has *pa*1−<sup>1</sup> <sup>1</sup> *<sup>p</sup>a*2−<sup>1</sup> <sup>2</sup> total solutions. Similarly, the total solutions of Equation (3) are *pa*1−<sup>1</sup> <sup>1</sup> *<sup>p</sup>a*2−<sup>1</sup> <sup>2</sup> . Hence,

$$\begin{aligned} d^-(2) &= p\_1^{a\_1 - 1} p\_2^{a\_2} + p\_1^{a\_1} p\_2^{a\_2 - 1} - p\_1^{a\_1 - 1} p\_2^{a\_2 - 1} - p\_1^{a\_1 - 1} p\_2^{a\_2 - 1} - p\_1^{a\_1 - 1} p\_2^{a\_2 - 1} \\ &= p\_1^{a\_1 - 1} p\_2^{a\_2 - 1} (p\_1 + p\_2 - 3) \end{aligned}$$

*p*<sup>1</sup> and *p*<sup>2</sup> are odd primes, which implies *d*−(2) is odd. The proof for *d*−(4) is analogous.

**Lemma 7.** *In* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧)*, if n* = 3*a*<sup>1</sup> 5*a*<sup>2</sup> *, then d*−(7) *= 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* Σ<sup>∧</sup> *<sup>n</sup> is* C *consistent if—and only if—n is even or* 3*.*

**Proof.** *Necessity*: Let *n* have, at most, two distinct prime factors and let Σ<sup>∧</sup> *<sup>n</sup>* be C consistent. If possible, let *n* be odd but not 3.

**Case (a)**: Let *n* ≡ 1 (mod 3) or *n* ≡ 2 (mod 3). As *n* is odd, 2 ∈ *Un*. Clearly, 0 is adjacent with 1, 2, *n* − 1 in Σ<sup>∧</sup> *<sup>n</sup>* . Since, *n* − 1 + 2 = 1 ∈ *Un*, *n* − 1 and 2 are connected in Σ<sup>∧</sup> *n* . Since, 3 is not a factor of *n*, 3 ∈ *Un*. Now, 2 + 1 = 3 ∈ *Un*. Hence, 2 and 1 are adjacent in Σ<sup>∧</sup> *<sup>n</sup>* . Now, the cycles *Z*<sup>1</sup> = (0, 1, 2, 0), *Z*<sup>2</sup> = (0, 2, *n* − 1, 0) have a common chord with end vertices 0 and 2. By Lemma 6,

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

Since the vertex 0 /∈ *Un*, *d*(0) = *d*−(0) = *φ*(*n*) = even. It follows,

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

Now, if either *Z*<sup>1</sup> or *Z*<sup>2</sup> is not a C-consistent cycle, a contradiction. Thus, *Z*<sup>1</sup> and *Z*<sup>2</sup> both cycle are C-consistent. The common chord with end vertices zero and two are oppositely marked, in contradiction with (Theorem 2, [26]).

**Case (b)**: Let *n* ≡ 0 (mod 3). Then, either *n* = 3*a*<sup>1</sup> or *n* = 3*a*<sup>1</sup> × *pa*<sup>2</sup> <sup>2</sup> . First, suppose *p*<sup>2</sup> = 5. Since, *n* is odd, 2, 4 ∈ *Un*. According to Lemma 2, *n* − 2 ∈ *Un*. Clearly, 0 is adjacent to 1, 4 and *n* − 2 in Σ<sup>∧</sup> *<sup>n</sup>* . Since, *n* − 2 + 4 = *n* + 2 = 2 ∈ *Un*, *n* − 2 is adjacent to 4 in Σ<sup>∧</sup> *n* . Now, for cycle *Z*<sup>1</sup> = (0, 1, 4, 0), *Z*<sup>2</sup> = (0, 4, *n* − 2, 0); *Z*1, *Z*<sup>2</sup> have a common chord with end vertices 0 and 4. According to Lemma 6,

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

Since the vertex 0 /∈ *Un*, *d*(0) = *d*−(0) = *φ*(*n*) = even. It follows,

$$
\mu\_{\mathcal{F}}(0) = +.
$$

Now, if either *Z*<sup>1</sup> or *Z*<sup>2</sup> is a cycle which is not C consistent, a contradiction. Therefore, *Z*<sup>1</sup> and *Z*<sup>2</sup> are the cycles which are 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*<sup>2</sup> = 5. In this case, we consider two cycles *Z*<sup>1</sup> = (0, 1, 7, 0) and *Z*<sup>2</sup> = (0, 7, 10, 13, 0) in Σ<sup>∧</sup> *<sup>n</sup>* . For cycles *Z*1, *Z*<sup>2</sup> have a common chord with end vertices 0 and 7, according to Lemma 7,

*μσ*(7) = −

Since the vertex 0 /∈ *Un*, *d*(0) = *d*−(0) = *φ*(*n*) = even. It follows that

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

Now, if either *Z*<sup>1</sup> or *Z*<sup>2</sup> is a cycle which is not C consistent, this is a contradiction. Therefore, *Z*<sup>1</sup> and *Z*<sup>2</sup> are the cycles which are 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, Σ<sup>∧</sup> *<sup>n</sup>* is all negative. Additionally, according to Theorem 13, *d*(*v*) = *d*−(*v*) = even ∀ *v* ∈ *V*(Σ<sup>∧</sup> *<sup>n</sup>* ). So, according to canonical marking *μσ*(*v*)=+ ∀ *v* ∈ *V*(Σ<sup>∧</sup> *<sup>n</sup>* ). So when *n* is even, Σ<sup>∧</sup> *<sup>n</sup>* is trivially C consistent. If *n* = 3, then *G*<sup>3</sup> is a path, which is trivially C- consistent, hence the result.

#### **3. Balance in Certain Derived Signed Graphs of Σ***∧ n*

**Theorem 12.** *η*(Σ<sup>∧</sup> *<sup>n</sup>* ) *is balanced if—and only if—n is* 3 *or even.*

**Proof.** Let *η*(Σ<sup>∧</sup> *<sup>n</sup>* ) 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 ∈ *Un*, *n* − 2 and 1 are connected in Σ<sup>∧</sup> *<sup>n</sup>* . *p* + 1 ∈ *Un* implies that *p* and 1 are connected in Σ<sup>∧</sup> *<sup>n</sup>* . Additionally, as *n* is odd, 2 ∈ *Un* and *n* − 2 ∈ *Un*. according to Lemma 2. Since, *n* − 2 + *p* = *n* + (*p* − 2) = *p* − 2 ∈ *Un*, (*n* − 2)*p* ∈ *E*(Σ<sup>∧</sup> *<sup>n</sup>* ). Now, for the cycle *Z* = (1, *p*, *n* − 2, 1) in Σ<sup>∧</sup> *<sup>n</sup>* we have a one positive edge 1(*n* − 2) and two negative edges 1*p* and *p*(*n* − 2) in *Z*. However, in *η*(Σ<sup>∧</sup> *<sup>n</sup>* ), there is a cycle *Z* = (1, *p*, *n* − 2, 1) with one negative edge 1(*n* − 2) and two positive edges 1*p* 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. Σ<sup>∧</sup> *<sup>n</sup>* , according to Lemma 3 is an all-negative signed graph. So *η*(Σ<sup>∧</sup> *<sup>n</sup>* ) is balanced and is all positive. *η*(Σ<sup>∧</sup> *<sup>n</sup>* ) 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 *Gn* (see [20]).

**Theorem 13** ([20])**.** *Let m be any vertex of the unitary addition Cayley graph Gn. Then,*

$$d(m) = \begin{cases} \phi(n) & \text{if } n \text{ is even,} \\ \phi(n) & \text{if } n \text{ is odd and } (m, n) \neq 1, \\ \phi(n) - 1 & \text{if } n \text{ is odd and } (m, n) = 1. \end{cases}$$

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

**Theorem 14.** *For an additional signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧)*, its line signed graph L*(Σ<sup>∧</sup> *n* ) *is balanced if—and only if—n* ∈ {2, 3, 4, 6}.

**Proof.** Let *L*(Σ<sup>∧</sup> *<sup>n</sup>* ) be balanced and *n* = 2, 3, 4 and 6. Now, according to Theorem 13, *d*(0) = *d*−(0) = *φ*(*n*) = even, which implies *d*(0) = *d*−(0) = *φ*(*n*) ≥ 4. This shows that condition *ii* (of Theorem 4, [27]) is not satisfied for Σ<sup>∧</sup> *<sup>n</sup>* . This is a contradiction. Hence, *n* ∈ {2, 3, 4, 6}. The converse part is easy to prove.

For a signed graph *S*, the balance property of *CE*(*S*) is discussed in ([9], Theorem 13).

**Theorem 15.** *For an additional signed Cayley graph* Σ<sup>∧</sup> *<sup>n</sup>* = (*Gn*, *σ*∧)*, its common-edge signed graph CE*(Σ<sup>∧</sup> *<sup>n</sup>* ) *is balanced if—and only if—n* ∈ {3, 4, 6}.

**Proof.** Let *n* ∈/ {3, 4, 6}. It is clear that 0 ∈/ *Un*. Now, by Theorem 13, *d*(0) = *d*−(0) = *φ*(*n*) = even, which implies *d*(0) = *d*−(0) = *φ*(*n*) ≥ 4. This shows that condition *ii* (of Theorem 13, [9]) is not satisfied for Σ<sup>∧</sup> *<sup>n</sup>* . Thus, *CE*(Σ<sup>∧</sup> *<sup>n</sup>* )is not balanced, which is a contradiction. Hence, *n* ∈ {3, 4, 6}. The converse part is easy to prove.

**Author Contributions:** Conceptualization, D.S.; Formal analysis, O.W., D.S. and A.D.; Methodology, O.W.; Supervision, D.S.; Writing—review & editing, O.W. and D.S. All authors have read and agreed to the published version of the manuscript.

**Funding:** The first author thanks the South Asian University for research grant support. The third author is grateful to DST [MTR/2018/000607] for the support under the Mathematical Research Impact Centric Support (MATRICS).

**Data Availability Statement:** No data were used to support the findings of the study.

**Conflicts of Interest:** All the authors declare that they have no conflict of interest regarding the publication of this paper.
