A CLASSIFICATION OF RAMANUJAN COMPLEMENTS OF UNITARY CAYLEY GRAPHS

The unitary Cayley graph on n vertices, Xn, has vertex set Zn, where two vertices a and b are connected by an edge if and only if they differ by a multiplicative unit modulo n, i.e. gcd(ab, n) = 1. A k-regular graph X is Ramanujan if and only if λ(X) ≤ 2 √ k − 1 where λ(X) is the second largest absolute value of the eigenvalues of the adjacency matrix of X. We obtain a complete characterization of the cases in which the complements of unitary Cayley graph X̄n is a Ramanujan graph. AMS Subject Classification: 05C25


Introduction
The eigenvalues of a graph are eigenvalues of its adjacency matrix, and the spectrum of a graph is the collection of its eigenvalues together with multiplicities.If λ 1 , λ 2 , . . ., λ k are distinct eigenvalues of a graph X and m 1 , m 2 , . . ., m k the corresponding multiplicities, then we denote the spectrum of X by Given a finite group G and a symmetric subset S of G, we define the Cayley graph X = Cay(G, S) to be the graph whose vertex set is G, in which two vertices v and u in G are connected by an edge if and only if vu −1 is in S. A Cayley graph of the form Cay(G, S) with G = Z n is called a circulant graph.
It is easy to see that Cay(G, S) is a simple, |S|-regular graph.
The unitary Cayley graph on n vertices, X n , is defined to be the undirected graph whose vertex set is Z n , in which two vertices a and b are connected by an edge if and only if gcd(a − b, n) = 1.This can also be stated as X n = Cay(Z n , U n ), where Z n is the additive group of integers modulo n and U n = Z * n is the set of multiplicative units modulo n.X n is a simple, ϕ(n)-regular graph, where ϕ is the Euler totient function.When discussing X n , we always assume n > 3 Lemma 1.The eigenvalues of any adjacency matrix of X n are where µ is the Mobius function.
The complement X of a graph X is the graph with the same vertex set as X such that two vertices are adjacent in X if and only if they are not adjacent in X.

Lemma 3 ([2]
).Let X be an k-regular graph with n vertices.Then X and X have the same eigenvectors, and their largest eigenvalues are k, and Lemma 1 and Corollary2 together imply the following result.

Ramanujan Unitary Cayley Graphs
Recall that the adjacency matrix of any k-regular graph X has eigenvalues between k and k, where k is an eigenvalue with multiplicity precisely equal to the number of connected components of X.Furthermore, if λ(X) denotes the largest absolute value of the eigenvalues of the adjacency matrix of X, smaller than k, then the graph X is called Ramanujan if and only if Note that λ(X) is only defined for regular graphs X with 3 or more vertices.Writing n in the form and positive integers α 1 , . . ., α k , we can determine λ(X n ) as follows.Since X n is ϕ(n)-regular, we find the maximum absolute value of an eigenvalue λ m (n) of the adjacency matrix of X n , smaller than ϕ(n).This can be accomplished by looking at 1. Indeed, we see that if n = 2 α ; then the eigenvalues have absolute value of either 0 or ϕ(n) (since, the only values of m, 0 ≤ m ≤ n − 1, which make n (n,m) square-free are m = 0 and m = 2 α−1 ), resulting in eigenvalues ϕ(n) and −ϕ(n).Thus λ(X 2 α ) = 0 and so X 2 α satisfies 2 and thus is Ramanujan.It is known from [1] that the graph X n is Ramanujan if and only if n satisfies one of the following conditions for some distinct odd primes p < q and natural α.

Ramanujan Complements of Unitary Cayley Graphs
k , be the canonical factorization of an integer n into prime powers, where p 1 < p 2 < • • • < p k are primes and each Theorem 6.Let n ≥ 2 be a intager.Then Xn is Ramanujan if and only if n is one of the following forms: 1. n = p a with p a prime and a ≥ 1 3. n = pq with p and q are primes and 3 ≤ p ≤ 5, 5 ≤ q ≤ 7.
Case 1.1: k = 3 and n = 2 a p b q c .Since (p − 1) and so Xn is not Ramanujan.It is easy to see that if b ≥ 2 or c ≥ 2, then ϕ(n) 2 > 4n, and so Xn is not Ramanujan.It remains to consider the case where n = 2pq or 4pq.If n = 2pq, then If p = 3 or p = 3 and q ≥ 7, then ϕ(n) 2 > 4n, and so Xn is not Ramanujan.It is easy to see that n = 2 Xn is not Ramanujan.If n = 4pq, similar to earlier state if p = 3, or p = 3 and q = 5, then Xn is not Ramanujan.For n = 4 • 3 • 5, the condition (6) is not established and so Xn is not Ramanujan.Case 2: p 1 ≥ 3.In this case λ( Xn ) = ϕ(n) p 1 −1 − 1 and so Xn is Ramanujan if and only if this inequality is equivalent to and this condition is not satisfied unless If k ≥ 4, then by Lemma 5, ϕ(n) 2 /n > ϕ(n)/2 k−1 > 4(p − 1) and so Xn is not Ramanujan.Assume k ≤ 3 and n = p a q b r c .
Case 2.2: k = 2 and n = p a q b .In this case by Lemma 5, and so if a ≥ 3, b ≥ 2 or a = b = 2, then (9) is violated and Xn is not Ramanujan.If n = pq, then from (8), we have (q − 2) 2 ≤ 4p and so if q > 7, then Xn is not Ramanujan.It is easy to see that n and so Xn is not Ramanujan.As in previous cases, if n = p 2 q, then Xn is not Ramanujan.