International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 07 | July -2017 www.irjet.net p-ISSN: 2395-0072 FORMATION OF TRIPLES CONSIST SOME SPECIAL NUMBERS WITH INTERESTING PROPERTY V. Pandichelvi1, P. Sivakamasundari2 1Assistant Professor, Department of Mathematics, UDC, Trichy. 2Guest Lecturer, Department of Mathematics, BDUCC,Lalgudi, Trichy. ---------------------------------------------------------------------***---------------------------------------------------------------- Abstract- In this communication, we discover the triple them is a perfect square is given in the following two a,b,c involving some figurate numbers such that the sum sections. of any two of them is a perfect square. Also, we find some interesting relations among the triples. Section- A Key words Diophantine m -tuples, polygonal numbers. : Let an 2CP2n1 bn Dec Gno INTRODUCTION: Let n be an integer. A set of positive 2n2 2n2 which are equivalent to the following two equations integers a1,a2,a3,......amis said to have the an 20n2 30n 12,bn 16n2 30n 13 the property D(n) if ai a j n is a perfect square for all Now, we assume that 1 i j m such a sets is called a Diophantine m-tuple. such an bn 2 sets were studied by Diophantus [1].The set of numbers 1,2,7is Diophantine triple with property D(2).For an Let cn be any non-zero integer such that extensive review of various articles one may refer [2-12]. In b(n) cn 2 (1) this communication, we find the triple a,b,c involving centered pentagonal numbers, decagonal numbers, a(n) cn 2 (2) Gnomonic numbers, Kynea numbers and Jacobsthal-lucas Subtracting (2) from (1), we get numbers such that the sum of any two of them is a perfect 2 2 b(n) an (3) square. Also, a few interesting relations among the triples are presented. Put A 1, A in (3), we get A 2n2 (4) Notations: Substituting (4) in (2), the values of c are represented by 4 2 5n2 5n 2 cn 4n 20n 30n 12 (5) Let CP be a centered pentagonal number of n 2 Hence, rank . 20n2 30n 12,16n2 30n 13,4n4 20n2 30n 12 is a 2 t10,n 4n 3n be a decagonal number of rank n . triple in which the sum of any two of them is a perfect square. Gnon 2n 1 be a Gnomonic number of rank n . 2n n1 Kn 2 2 1 be a Kynea number of rank n . Table-1: Some numerical examples are illustrated n n below: jn 2 1 be a Jacobshthal-lucas number of rank n . n an bn cn a(n) bn a(n) cn b(n) cn Method of analysis: 1 62 59 58 112 22 12 The procedure for finding the triple involving 152 137 88 2 2 2 some interesting numbers such that the sum of any two of 2 17 8 7 © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 2372 International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 07 | July -2017 www.irjet.net p-ISSN: 2395-0072 3 282 247 42 232 182 172 Table-2: Some numerical examples are illustrated below: 4 452 389 572 292 322 312 n an bn cn a(n) bn a(n) cn b(n) cn 5 662 563 1838 352 502 492 1 23 58 266 92 172 182 A few interesting relations among the numbers are given 287 154 4202 2 2 2 below: 2 21 67 66 4223 538 3392426 2 2 2 1. anbn 36 Pron 12Gnon1 13 69 1843 1842 66047 2074 572 2 2 2 2. (1/ 4)an cn is a bi-quadratic integer 261 31987 31986 1050623 8218 2 2 3. (1/ 4)[bncn36Pron 12Gnon 37] is a 2.7165151661011 1029 521203 5212022 bi-quadratic integer 4. (1/ 4)[ancn 40Pron 10Gnon 34]is a CONCLUSION: bi-quadratic integer In this communication, we discover the triple involving SECTION-B various special numbers in such a way that the sum of any two of them is a perfect square. In this manner, one may seek out other triples and quadruples satisfying some other Let an K 2n properties. bn 8j2n j1 j4 which are equivalent to the following two equations REFERENCES: 4n 2n 2n cn an 2 2.2 1,bn 8.2 26 [1]. Baker A.,Davenport H., The equations 3x2 2 y 2 Now, we assume that 2 2 an bn 2 and 8x 7 z , J.Math.Oxford Ser. 20 (1969),129-137. Let be any non-zero integer such that [2]. I.G Bashmakova(ed), Diophantus of Alexandria, Arithmetic and the Book of polygonal numbers, Nauka, b(n) cn 2 (6) Moscow, 1974. a(n) cn 2 (7) [3]. Dickson L.E; History of theory of numbers ,Vol.2, Chelsea , New York 1966; 513-520. Subtracting (7) from (6), we get 2 2 b(n) an (8) [4]. Thamotherampillai N; The set of numbers {1, 2, 7}.Bull of Calcutta math soc.,1980; 72:195-197. The choices A 1, A lead (8) to [5]. Brown E; sets in which xy k is always a square, Math A 3.22n 24n1 13 (9) comp., 1985; 45: 613-620. Substituting (9) in (7), the values of c are represented by 4n 8n2 6n1 2n 4n1 [6]. Gupta H, singh K; On K-triad sequences. Internet J Math cn 8.2 2 6.2 76.2 26.2 170 Sci., 1985; 5: 799 - 804. Hence, [7]. Dujella A., Complete solution of a family of simultaneous 24n 2.22n 1,8.22n 26,8.24n 28n2 6.26n1 76.22n 26.24n1 170 pellian equations, Acta Math. Inform. Univ. Ostraviensis is a triple in which the sum of any two of them is a perfect 6 (1998), 59-67. square. [8]. Beardon AF, Deshpande MN; Diophantine triples. The athematical Gazette, 2002, 86:258-260 [9]. Deshpande MN; Families of Diophantine triplets,Bulletin of the Marathwada Mathematical Society, 2003; 4: 19-21. © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 2373 International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 04 Issue: 07 | July -2017 www.irjet.net p-ISSN: 2395-0072 [10].Bugeaud Y, Dujella A, Mignotte M; On the family of Diophantine triples { k 1,k 1,16k 3 4k }. Glasgo Math J. 49(2007),333-334. [11].Gopalan M.A., Pandichelvi V; On the extendibility of the Diophantine triple involving Jacobsthal numbers J2n1, J2n1 3,2J2n J2n1 J2n1 3, International journal of Mathematics and Applications, 2009;2(1):1-3. [12]Pandichelvi V, Construction of the Diophantine triple involving polygonal numbers. Impact J SciTech., 2011; 5(1): 7-11. © 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 2374 .