Some New Families of Integral Trees of Diameter Four


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1 2011 ¼ 6 «Æ Æ Æ 15 ² Æ 2 June, 2011 Operations Research Transactions Vol.15 No.2 Some New Families of Integral Trees of Diameter Four WANG Ligong 1 ZHANG Zheng 1 Abstract An integral graph is a graph of which all the eigenvalues of its adjacency matrix are integers. This paper investigates integral trees of diameter 4. Many new classes of such integral trees are constructed infinitely by solving some certain Diophantine equations. These results generalize some results of Wang, Li and Zhang (see Families of integral trees with diameters 4, 6 and 8, Discrete Applied Mathematics, 2004, 136: ). Keywords Operations research, integral tree, characteristic polynomial, diophantine equation, graph spectrum Subject Classification (GB/T ) Ú Ô 4 ÙÓÕÒ ÝÜÛ 1 Þ ß 1 Ø ¹ È 4»Ä Å ±ÎÀ É Å ³ µæ ( Families of integral trees with diameters 4, 6 and 8, Discrete Applied Mathematics, 2004, 136: ) ËÄ º «Æ Ã Å ¾ ÖÐ Ñ (GB/T ) Introduction Throughout this paper we shall consider only simple graphs (i.e. undirected graphs without loops and/or multiple edges). Let G be a simple graph with vertex set V (G) = {v 1,v 2,,v n } and edge set E(G). The adjacency matrix A(G) = [a ij ] of G is an n n symmetric matrix of 0 s and 1 s with a ij = 1 if and only if v i and v j are joined by an edge. The characteristic polynomial of G is the polynomial P(G,x) = det(xi n A(G)), where I n always denotes the n n identity matrix. The 2009 ½ 1 6 Supported by the NNSF of China (No ), the Natural Science Foundation of Shaanxi Province (No. SJ08A01) and SRF for ROCS, SEM. 1. Department of Applied Mathematics, School of Science, Northwestern Polytechnical University, Xi an , China; Á Ê Ç ÇÏÌÍÇÂ Á
2 2 WANG Ligong, ZHANG Zheng 15 ² spectrum of A(G) is also called the spectrum of G. The research on integral graphs was initiated by Harary and Schwenk in 1974 [1]. A graph G is called integral if all eigenvalues of its characteristic polynomial P(G,x) are integers. In general, the problem of finding (describing) all graphs with integral spectrum seems to be very difficult. Thus, it makes sense to restrict our investigations to interesting families of graphs. Particular results on integral graphs were obtained only for some certain classes of graphs [2]. Trees present another important family of graphs for which the problem has been considered in [113]. We know that trees of diameter 4 can be formed by joining the centers of r stars K 1,m1,K 1,m2,,K 1,mr to a new vertex v. The tree is denoted by S(r;m 1,,m r ) or simply S(r;m i ). Assume that the number of distinct integers of m 1,m 2,,m r is s. Without loss of generality, assume that the first s ones are the distinct integers such that 0 m 1 < m 2 < < m s. Suppose that a i is the multiplicity of m i for each i = 1,2,,s. The tree S(r;m i ) is also denoted by S(a 1 + a a s ;m 1,m 2,,m s ), where r = s a i and V = 1+ s a i (m i +1). For all other facts on graph spectra (or terminology), see [4]. In this paper, we investigate integral trees S(r;m i ) = S(a 1 +a 2 +a 3 ;m 1,m 2,m 3 ) of diameter 4. We shall construct infinitely many new classes of such integral trees by solving some certain Diophantine equations. These results are different from those in the existing literature. These results generalize some results of [10]. These integral graphs constructed in this paper are new members of integral graphs. We believe that it is useful for constructing other integral trees. 1 Preliminaries In this section, we state some known results on integral trees of diameter 4 and also obtain a new result on integral trees of diameter 4. Lemma 1.1 [7,10] For the tree S(r;m i ) = S(a 1 + +a s ;m 1,,m s ) of diameter 4, then we have P[S(r;m i ),x] =P[S(a a s ;m 1,,m s ),x] =x 1+ s a i (m i 1) [ (x 2 m i ) a i 1 (x 2 m i ) s a i j=1,j i ] (x 2 m j ).
3 2 Some New Families of Integral Trees of Diameter Four 3 Lemma 1.2 [10,13] The tree S(r;m i ) = S(a a s ;m 1,,m s ) of diameter 4 is integral if and only if (i) a i = 1 must hold if m i is not a perfect square, (ii) all solutions of the following equation are integers: (x 2 m i ) s a i j=1,j i (x 2 m j ) = 0. (1.1) We now discuss (1.1) to get more information. First, we divide both sides of (1.1) by s (x 2 m i ), and obtain F(x) = s a i x 2 m i 1 = 0. (1.2) Clearly, ± m i are not roots of (1.1) for 1 i s. Hence, all solutions of (1.1) are the same as those of (1.2). Because F(x) is an even function, we need only to consider the roots of F(x) on the real interval [0,+ ). Note that F(x) is discontinuous at each point m i. For 1 i s, when m i 0, we have that F( m i 0) =, F( m i + 0) = +, F(+ ) = 1. When m i = 0, we have that F(0 + 0) = +, and F (x) = 2x s 1 (x 2 m i. We deduce that F(x) is strictly monotone decreasing ) 2 on each of the continuous intervals over the real interval [0,+ ). Using Zero Point Theorem of mathematical analysis, we get that F(x) has s distinct positive real roots on the real interval [0,+ ). If 0 < u 1 < u 2 < < u s < + are the roots of F(x), then holds. holds. 0 m 1 < u 1 < m 2 < u 2 < < u s 1 < m s < u s < + (1.3) On the other hand, we note that (1.2) can be written as a 1 x 2 m 1 + a 2 x 2 m a s x 2 m s = 1. (1.4) From the above discussion and Lemma 1.2, we can deduce the following theorem Theorem 1.1 The tree S(r;m i )=S(a 1 + a a s ;m 1,m 2,,m s ) of diameter 4 is integral if and only if (i) a i = 1 must hold if m i is not a perfect square, (ii) all solutions of the following equation are integers: s a i x 2 m i = 1. (1.5)
4 4 WANG Ligong, ZHANG Zheng 15 ² Moreover, there exist positive integers u 1,u 2,,u s satisfying (1.3) such that the following linear equation system in a 1,a 2,,a s has positive integral solutions (a 1,a 2,,a s ), and such that a i = 1 must hold if m i is not a perfect square. a 1 a 2 u 2 1 m + 1 u 2 1 m u 2 1 m = 1, s. a 1 a 2 a s u 2 s m + 1 u 2 s m u 2 s m = 1. s a s (1.6) Theorem 1.2 [6,13] The tree S(r;m i )=S(a a s ;m 1,,m s ) of diameter 4 is integral if and only if there exist positive integers u i and nonnegative integers m i (i = 1,2,,s) such that 0 m 1 < u 1 < m 2 < u 2 < < u s 1 < m s < u s < +, and such that a k = (u 2 i m k),i k (m i m k ) (k = 1,2,,s) (1.7) are positive integers, and such that a i = 1 must hold if m i is not a perfect square. Theorem 1.3 [10] If the tree S(a 1 + a a s ;m 1,m 2,, m s ) of diameter 4 is integral, and m 1 ( 0),m 2 (> 0),,m s (> 0) are perfect squares, then for any positive integer n the tree S(a 1 n 2 + a 2 n 2 + +a s n 2 ;m 1 n 2,m 2 n 2,,m s n 2 ) is integral. Lemma 1.3 [10] For s = 3, integers m i ( 0),a i (> 0),u i (> 0) (i = 1,2,3) are given in Table 1. a i and u i (i = 1,2,3) are those of Theorem 1.2. Then for any positive integer n the tree S(a 1 n 2 + a 2 n 2 + +a 3 n 2 ;m 1 n 2,m 2 n 2,m 3 n 2 ) of diameter 4 is integral. Next we shall give a result on number theory. In the following symbol (a,b) = d denotes the greatest common divisor d of integers a, b, while a b (a b) means that a divides b (a does not divide b). Lemma 1.4 [14] Let a,b and c be integers with d = (a,b), we have (1) If d c, then the linear Diophantine equation in two variables ax + by = c (1.8) does not have integral solutions.
5 2 Some New Families of Integral Trees of Diameter Four 5 (2) If d c, then there are infinitely many integral solutions for (1.8). Moreover, if x = x 0,y = y 0 is a particular solution of (1.8), then all its solutions are given by x = x 0 + (b/d)t, y = y 0 (a/d)t, where t is an integer. Table 1 Integral trees S(a 1 n 2 + a 2 n 2 + a 3 n 2 ; m 1 n 2, m 2 n 2, m 3 n 2 ), where n 1. a 1 a 2 a 3 m 1 m 2 m 3 u 1 u 2 u Lemma 1.5 [15] Let m be a positive integer. If 2 m or 4 m, then there exist positive integral solutions for the Diophantine equation x 2 y 2 = m. (1.9) Remark 1.1 We can give a method for finding the solutions of (1.9). Suppose that m = m 1 m 2. Let x y = m 1, x + y = m 2 and 2 (m 1 + m 2 ). Then the solutions of (1.9) can be found easily (see [15]). 2 Integral trees of diameter four In this section, we shall construct infinitely many new classes of integral trees S(a 1 + a 2 + a 3 ;m 1,m 2, m 3 ) of diameter 4 from Lemma 1.3 and Theorem 1.2. They are different from those of [113].
6 6 WANG Ligong, ZHANG Zheng 15 ² The idea of constructing such integral tree is as follows: First, we properly choose integers m 1 ( 0), m 2 (> 0),,m s (> 0). Then, we try to find proper positive integers u i (i = 1,2,,s 1) satisfying (1.3) such that there are positive integral solutions (a 1,a 2,,a s ) for the linear equation system (1.6) (or such that all a k s of (1.7) are positive integers). Finally, we obtain positive integers a 1,a 2,,a s such that all the solutions a k s of (1.6) or (1.7) are integers. Thus, we have constructed many new classes of integral trees S(a a s ;m 1,,m s ) of diameter 4. From Lemma 1.3 and Theorem 1.2, we will construct infinitely many new classes of integral trees S(a 1 + a 2 + a 3 ;m 1,m 2,m 3 ) of diameter 4 and obtain the following theorem. Theorem 2.1 For s = 3, let integers m i ( 0), a i (> 0), u i (> 0) (i = 1,2,3) be those of Theorem 1.2, given in the following items, where t and k ( 0) are integers, then for any positive integer n the tree S(a 1 n 2 + a 2 n 2 + a 3 n 2 ;m 1 n 2,m 2 n 2,m 3 n 2 ) of diameter 4 is integral. (1) m 1 = 0, m 2 = 9, m 3 = 64, u 1 = 2, u 2 = 6, u 2 3 = 44t + 64 is a perfect square, a 1 = 11t + 16, a 2 = 12t + 15, a 3 = 21t. We have the following cases for (1), where k 1. (i) If t = k(11k 8), then u 2 3 = [2(11k 4)]2 is a perfect square, u 3 = 2(11k 4), a 1 = (11k 4) 2, a 2 = 3(2k 1)(22k 5), a 3 = 21k(11k 8). (ii) If t = k(11k + 8), then u 2 3 = [2(11k + 4)]2 is a perfect square, u 3 = 2(11k + 4), a 1 = (11k + 4) 2, a 2 = 3(2k + 1)(22k + 5), a 3 = 21k(11k + 8). (2) m 1 = 0, m 2 = 9, m 3 = 144, u 1 = 2, u 2 = 6, u 3 = 3k, a 1 = k 2, a 2 = k 2 1, a 3 = 7(k 4)(k + 4), where k 5. (3) m 1 = 0, m 2 = 25, m 3 = 81, u 1 = 2, u 2 = 6, u 2 3 = 225(8t + 1) is a perfect square, a 1 = 16(8t + 1), a 2 = 33(9t + 1), a 3 = 55(25t + 2). We have the following cases for (3), where k 1. (i) If t = k(2k 1), then u 2 3 = [15(4k 1)]2 is a perfect square, u 3 = 15(4k 1), a 1 = 16(4k 1) 2, a 2 = 33(6k 1)(3k 1), a 3 = 55(10k 1)(5k 2). (ii) If t = k(2k+1), then u 2 3 = [15(4k+1)]2 is a perfect square, u 3 = 15(4k+1), a 1 = 16(4k + 1) 2, a 2 = 33(6k + 1)(3k + 1), a 3 = 55(10k + 1)(5k + 2). (4) m 1 = 0, m 2 = 16, m 3 = 64, u 1 = 2, u 2 = 6, u 3 = 8k, a 1 = 9k 2, a 2 = 5(2k 1)(2k + 1), a 3 = 35(k 1)(k + 1), where k 2.
7 2 Some New Families of Integral Trees of Diameter Four 7 (5) m 1 = 0, m 2 = 25, m 3 = 256, u 1 = 2, u 2 = 6, u 3 = 20k, a 1 = 9k 2, a 2 = (4k 1)(4k + 1), a 3 = 15(5k 4)(5k + 4), where k 1. (6) m 1 = 0, m 2 = 16, m 3 = 144, u 1 = 2, u 2 = 6, u 2 3 = 16(8t + 1) is a perfect square, a 1 = 8t + 1, a 2 = 15t, a 3 = 105(t 1). We have the following cases for (6), where k 1. (i) If t = k(2k 1), then u 2 3 = [4(4k 1)]2 is a perfect square, u 3 = 4(4k 1), a 1 = (4k 1) 2, a 2 = 15k(2k 1), a 3 = 105(k 1)(2k + 1). (ii) If t = k(2k+1), then u 2 3 = [4(4k+1)]2 is a perfect square, u 3 = 4(4k+1), a 1 = (4k + 1) 2, a 2 = 15k(2k + 1), a 3 = 105(k + 1)(2k 1). (7) m 1 = 0, m 2 = 16, m 3 = 196, u 1 = 2, u 2 = 7, u 2 3 = 16(5t + 1) is a perfect square, a 1 = 5t + 1, a 2 = 11t, a 3 = 16(4t 9). We have the following cases for (7), where k 1. (i) If t = k(5k 2), then u 2 3 = [4(5k 1)]2 is a perfect square, u 3 = 4(5k 1), a 1 = (5k 1) 2, a 2 = 11k(5k 2), a 3 = 16(10k 9)(2k + 1). (ii) If t = k(5k+2), then u 2 3 = [4(5k+1)]2 is a perfect square, u 3 = 4(5k+1), a 1 = (5k + 1) 2, a 2 = 11k(5k + 2), a 3 = 16(10k + 9)(2k 1). (8) m 1 = 0, m 2 = 16, m 3 = 100, u 1 = 3, u 2 = 6, u 2 3 = 400(3t + 1) is a perfect square, a 1 = 81(3t + 1), a 2 = 5(25t + 8), a 3 = 208(4t + 1). We have the following cases for (8), where k 1. (i) If t = k(3k 2), then u 2 3 = [20(3k 1)]2 is a perfect square, u 3 = 20(3k 1), a 1 = 81(3k 1) 2, a 2 = 5(5k 2)(15k 4), a 3 = 208(2k 1)(6k 1). (ii) If t = k(3k+2), then u 2 3 = [20(3k+1)]2 is a perfect square, u 3 = 20(3k+1), a 1 = 81(3k + 1) 2, a 2 = 5(5k + 2)(15k + 4), a 3 = 208(2k + 1)(6k + 1). (9) m 1 = 0, m 2 = 36, m 3 = 144, u 1 = 3, u 2 = 8, u 3 = 6k, a 1 = 4k 2, a 2 = 7(k 1)(k + 1), a 3 = 25(k 2)(k + 2), where k 3. (10) m 1 = 0, m 2 = 16, m 3 = 144, u 1 = 3, u 2 = 8, u 2 3 = 16(8t + 1) is a perfect square, a 1 = 4(8t + 1), a 2 = 21t, a 3 = 75(t 1). We have the following cases for (10), where k 1. (i) If t = k(2k 1), then u 2 3 = [4(4k 1)]2 is a perfect square, u 3 = 4(4k 1), a 1 = 4(4k 1) 2, a 2 = 21k(2k 1), a 3 = 75(k 1)(2k + 1), where k 2.
8 8 WANG Ligong, ZHANG Zheng 15 ² (ii) If t = k(2k+1), then u 2 3 = [4(4k+1)]2 is a perfect square, u 3 = 4(4k+1), a 1 = 4(4k + 1) 2, a 2 = 21k(2k + 1), a 3 = 75(k + 1)(2k 1). (11) m 1 = 0, m 2 = 16, m 3 = 100, u 1 = 3, u 2 = 8, u 3 = 10k, a 1 = 36k 2, a 2 = (5k 2)(5k + 2), a 3 = 39(k 1)(k + 1), where k 2. (12) m 1 = 0, m 2 = 36, m 3 = 225, u 1 = 3, u 2 = 9, u 2 3 = 100(7t + 4) is a perfect square, a 1 = 9(7t + 4), a 2 = 5(25t + 13), a 3 = 128(4t + 1). We have the following cases for (12), where k 0. (i) If t = k(7k 4), then u 2 3 = [10(7k 2)]2 is a perfect square, u 3 = 10(7k 2), a 1 = 9(7k 2) 2, a 2 = 5(5k 1)(35k 13), a 3 = 128(2k 1)(14k 1), where k 1. (ii) If t = k(7k+4), then u 2 3 = [10(7k+2)]2 is a perfect square, u 3 = 10(7k+2), a 1 = 9(7k + 2) 2, a 2 = 5(5k + 1)(35k + 13), a 3 = 128(2k + 1)(14k + 1). (13) m 1 = 0, m 2 = 25, m 3 = 225, u 1 = 3, u 2 = 10, u 3 = 5k, a 1 = 4k 2, a 2 = 6(k 1)(k + 1), a 3 = 15(k 3)(k + 3), where k 4. (14) m 1 = 0, m 2 = 36, m 3 = 225, u 1 = 3, u 2 = 12, u 2 3 = 25(7t + 2) is a perfect square, a 1 = 4(7t + 2), a 2 = 3(25t + 2), a 3 = 72(t 1). We have the following cases for (14), where k 0. (i) If t = 7k 2 6k+1, then u 2 3 = [5(7k 3)]2 is a perfect square, u 3 = 5(7k 3), a 1 = 4(7k 3) 2, a 2 = 3(5k 3)(35k 9), a 3 = 72k(7k 6). (ii) If t = 7k 2 +6k+1, then u 2 3 = [5(7k+3)]2 is a perfect square, u 3 = 5(7k+3), a 1 = 4(7k + 3) 2, a 2 = 3(5k + 3)(35k + 9), a 3 = 72k(7k + 6). (iii) If t = 7k 2 8k+2, then u 2 3 = [5(7k 4)]2 is a perfect square, u 3 = 5(7k 4), a 1 = 4(7k 4) 2, a 2 = 3(5k 2)(35k 26), a 3 = 72(k 1)(7k 1). (iv) If t = 7k 2 +8k+2, then u 2 3 = [5(7k+4)]2 is a perfect square, u 3 = 5(7k+4), a 1 = 4(7k + 4) 2, a 2 = 3(5k + 2)(35k + 26), a 3 = 72(k + 1)(7k + 1). Proof We only prove (1), (2) (14) can be shown similarly to (1) by using Lemma 1.3 and Theorem 1.2. (1) For s = 3, by Lemma 1.3, we find m 1 = 0, m 2 = 9, m 3 = 64, u 1 = 2, u 2 = 6.
9 2 Some New Families of Integral Trees of Diameter Four 9 From Theorem 1.2, we get a 1 = (u2 1 m 1)(u 2 2 m 1)(u 2 3 m 1) (m 2 m 1 )(m 3 m 1 ) a 2 = (u2 1 m 2)(u 2 2 m 2)(u 2 3 m 2) (m 1 m 2 )(m 3 m 2 ) a 3 = (u2 1 m 3)(u 2 2 m 3)(u 2 3 m 3) (m 1 m 3 )(m 2 m 3 ) = u2 3 4, (2.1) = 3 11 (u2 3 9), (2.2) = (u2 3 64). (2.3) So, S(a 1 + a 2 + a 3 ;m 1,m 2,m 3 ) of diameter 4 is integral if and only if a 1, a 2, a 3 are positive integers. From (2.1) and (2.2), we get the Diophantine equation 12a 1 11a 2 = 27. (2.4) A result in elementary number theory (see also Lemma 1.4) yields that all positive integral solutions of (2.4) are given by a 1 = 11t + 16, a 2 = 12t + 15, where t( 1) is an integer. From (2.2) and (2.3), we have u 2 3 = 44t + 64, a 3 = 21t, where t is a nonnegative integer. Since u 2 3 (= 44t + 64) must be a perfect square, we deduce (u 3 + 8)(u 3 8) = 44t. (2.5) By Lemma 1.5 and Remark 1.1, we can obtain all positive integral solutions of (2.5) by discussing the following two cases for any positive integer k: Case 1 By (2.5), we assume that { u3 + 8 = 22k, u 3 8 = 2t/k. Then we get t = k(11k 8), u 3 = 2(11k 4). Hence, u 2 3 = [2(11k 4)]2 is a perfect square, a 1 = (11k 4) 2, a 2 = 3(2k 1)(22k 5), a 3 = 21k(11k 8). Case 2 By (2.5), we assume that { u3 8 = 22k, u = 2t/k. Then we get t = k(11k + 8), u 3 = 2(11k + 4). Hence, u 2 3 = [2(11k + 4)]2 is a perfect square, a 1 = (11k + 4) 2, a 2 = 3(2k + 1)(22k + 5), a 3 = 21k(11k + 8). Hence, when m 1 = 0, m 2 = 9, m 3 = 64, and a i (i = 1,2,3) are any case on the above two cases, then the tree S(a 1 + a 2 + a 3 ;m 1,m 2, m 3 ) of diameter 4 is integral. By Theorem 1.3, it is not difficult to prove that the tree S(a 1 n 2 + a 2 n 2 + a 3 n 2 ;m 1 n 2,m 2 n 2, m 3 n 2 ) of diameter 4 is also integral for any positive integer n.
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