International Journal of Innovative Science and Modern Engineering (IJISME) ISSN: 2319-6386, Volume-3 Issue-6, May 2015 On the Ternary Cubic Equation 3 푥2 + 푦2 − 2푥푦 + 4 푥 + 푦 + 4 = 51푧3 Manju Somanath, V. Sangeetha, M. A. Gopalan, M. Bhuvaneshwari Abstract—The non-homogeneous ternary cubic Diophantine 퐶퐶(푛)- Centered Cube number equation given by ퟑ 풙ퟐ + 풚ퟐ − ퟐ풙풚 + ퟒ 풙 + 풚 + ퟒ = ퟓퟏ풛ퟑ is 푇푇퐻(푛)- Truncated Tetrahedral considered. Different patterns of non-zero distinct integer number solutions to the above equation are obtained. For each of these 푇푂퐻(푛)-Truncated Octahedral patterns, a few interesting relations between the solutions and the special figurate numbers are obtained. number 푔푚 (푛)- m-gram number of rank n Keywords: Non-homogeneous, ternary cubic Diophantine equation, integer solutions, polygonal numbers, pyramidal II. METHOD OF ANALYSIS numbers. 2010 Mathematics Subject Classification: 11 D 25 The ternary cubic equation to be solved is 2 2 3 I. INTRODUCTION 3 푥 + 푦 − 2푥푦 + 4 푥 + 푦 + 4 = 51푧 (1) Introducing the linear transformations Integral solutions for the homogeneous or non – 푥 = 푢 + 푣 ; 푦 = 푢 − 푣 (푢 ≠ 푣 ≠ 0) (2) homogeneous Diophantine equation is an interesting in (1), it is written as concept as it can be seen from [1-3].In particular, one may 2푢 + 2 2 + 8푣2 = 51푧3 (3) refer [4-23] for cubic Diophantine equation with three Now (3) is solved through different ways and using (2), unknowns. This communication concerns with yet another different patterns of integer solutions to (1) are obtained. interesting cubic equation with three unknowns PATTERN I 3 푥2 + 푦2 − 2푥푦 + 4 푥 + 푦 + 4 = 51푧3 for determining Assume 푧 = 푎2 + 8푏2 (4) its non-zero distinct integer solutions. A few interesting Write 51 as 51 = (7 + 푖 2)(7 − 푖 2) (5) relations among the solutions are presented. Using (4) and (5) in (3) and applying the method of NOTATIONS factorization, define 푡푚,푛 - Polygonal number of rank n 3 with sides m 2푢 + 2 푢 + 푖2 2푣 = 7 + 푖 2 푎 + 푖2 2푏 푐푝푚 ,푛 - Centered Polygonal number Equating the real and imaginary parts, the values of u and v of rank n with sides m are obtained as 푚 푃푛 - Pyramidal number of rank n 1 with sides m 푢 = [7푎3 − 12푎2푏 − 168푎푏2 + 32푏3 − 2] 퐶푃푚 - Centered Pyramidal number of 2 푛 1 rank n with sides m 푣 = [푎3 + 42푎2푏 − 24푎푏2 − 112푏3] 푃퐶푆푚 - Prism number of rank n with 2 푛 Substituting the above values of u and v in (2), the values of sides m x and y are given by 푃(푛)- Pronic number of rank n 푥 = 4푎3 + 15푎2푏 − 96푎푏2 − 40푏3 − 1 퐺(푛)- Gnomonic number 푦 = 3푎3 − 27푎2푏 − 72푎푏2 + 72푏3 − 1 (6) 퐶푂(푛)- Centered Octahedral number Thus (6) and (4) represent the non-zero distinct integer 퐼(푛)- Icosahedral number solutions to (1). 퐷(푛)- Dodecahedral number Properties 푆푂(푛)- Stella octangula number 푥 푎, 1 + 15푧 푎, 1 − 2푆푂 푎 − 푡 ≡ 푅퐷(푛)- Rhombic Dodecahedral 62,푎 number 14 (푚표푑 65) 퐶퐷(푛)- Centered Dodecahedral 푥 1, 푎 + 1 + 푦 1, 푎 + 1 − 16푆푂 푎 + 144푡3,푎 ≡ number −143(푚표푑 164) 푥 푎, 1 + 푧 푎, 1 − 2푆푂 푎 − 8푡6,푎 ≡ −33 (푚표푑 86) 30 24 푥 1, 푏 − 푦 1, 푏 + 푧 1, 푏 + 6퐶푃푏 + 6퐶푃푏 + 6 58퐶푃푏 is a Kynea prime. Manuscript Received on May 2015. 푦 1, 푏 + 9푧 1, 푏 − 27푆푂 푏 − 18퐶푃6 is a Dr. Manju Somanath, Asst. Prof., Department of Mathematics, 푎 National College,Trichy-620 001,Tamilnadu, India. perfect square. Prof. V. Sangeetha, Asst. Prof., Department of Mathematics, National College,Trichy-620 001,Tamilnadu, India. Dr. M. A. Gopalan, Professor, Department of Mathematics, SIGC, Trichy-620 002, Tamilnadu, India. M. Bhuvaneshwari, M.Phil., Scholar, Department of Mathematics, SIGC , Trichy-620 002,Tamilnadu, India. Published By: Retrieval Number: F0859053615/2015©BEIESP Blue Eyes Intelligence Engineering 29 & Sciences Publication On the Ternary Cubic Equation ퟑ 풙ퟐ + 풚ퟐ − ퟐ풙풚 + ퟒ 풙 + 풚 + ퟒ = ퟓퟏ풛ퟑ 5 PATTERN II 푥 퐴, 1 + 푧 퐴, 1 − 32푃퐴 + 276푃 퐴 ≡ Write (3) as 2푢 + 2 2 + 8푣2 = 51푧3 ∗ 1 (7) 87 (푚표푑 108). 5 Write 1 as 푦 1, 퐵 − 1152푃퐵 ≡ −25 (푚표푑 216). (1+푖2 2)(1−푖2 2) 1 = (8) 9 PATTERN V Using (5) and (8) in (7) and repeating the procedure as in It is noted that 51 is also represented by pattern I, the corresponding integer solutions (1) are given by 51 = (1 + 푖5 2)(1 − 푖5 2) 푥 = 3푎3 − 27푎2푏 − 72푎푏2 + 72푏3 − 1 (11) Using (8) and (11) in (7) and following the procedure as in 푦 = −2푎3 − 33푎2푏 + 48푎푏2 + 88푏3 − 1 2 2 pattern II, we obtain the integer solutions to (1) as 푧 = 푎 + 8푏 3 2 2 3 Properties 푥 = −54퐴 − 891퐴 퐵 + 1296퐴퐵 + 2376퐵 − 1 푦 = −117퐴3 + 135퐴2퐵 + 2808퐴퐵2 − 360퐵3 − 1 푦 1, 푏 − 푥 1, 푏 − 푧 1, 푏 − 60푡 − 3,푏−1 푧 = 9퐴2 + 72퐵2 115푡 + 1 is a Truncated Octahedral number of 4,푏 Properties rank a. 푥 퐴, 1 − 푦 퐴, 1 − 31푆푂 퐴 − 퐶푃6 + 푥 1, 푏 + 푦 1, 푏 − 2푃퐶푆30 − 6푃30 − 6푡 ≡ 퐴 푏 푏 3,푏−1 2052푡 ≡ 6 (푚표푑 455). −1(푚표푑 19). 3,퐴 6 푦 퐴, 1 − 푧 퐴 + 1, 퐴 + 180푆푂 퐴 − 126푡3,퐴−1 + 푦 푎, 푎 + 1 − 10퐶퐷 푎 − 퐶푃푎 − 177푃 푎 ≡ 2664푡4,퐴 = −127. 12 푚표푑 77 . 5 2 푥 퐴, 1 − 2푧 퐴 + 1, 퐴 + 108푃퐴 + 1998푡3,퐴−1 ≡ 푧 푎 푎 + 1 , 푎 + 푎 + 1 − 8퐶푃 − 28푃(푎) 9,푎 푎+1 8(푚표푑 261). is a Nasty number. 푥 1, 퐵 퐵 + 1 + 푦 1, 퐵 퐵 + 1 + 푧 1, 퐵 퐵 + 푧 푎 푎 + 1 , 푎2 + 푎 + 1 − 푥 푎 푎 + 1 , 푎2 + 푎 + 1−6푅퐷푎푎+1+152푃푎=−57 1−30퐶푃퐵퐵+130−2퐶푃퐵퐵+13 PATTERN III +푔7 퐵 퐵 + 1 − 77푃 퐵 + 1 is a Dodecahedral Write 1 as number of rank 퐵(퐵 + 1). (7+푖6 2)(7−푖6 2) 1 = (9) 푧 퐴, 퐴 + 1 − 72퐺(퐴 + 1) is a perfect square. 121 PATTERN VI Using (5) and (9) in (7) and following the procedure similar Consider (9) and (11) in (7).The repetition of process as in to pattern II, the corresponding integer solutions to (1) are pattern III, the integer solutions to (1) are given by obtained as 3 2 2 3 2 2 3 푥 = −5808퐴 − 392040퐴 퐵 + 139392퐴퐵 푥 = 5203퐴 − 22143퐴 퐵 − 124872퐴퐵 + 59048퐵 − 1 3 3 2 2 3 + 1045440퐵 − 1 푦 = −726퐴 − 49005퐴 퐵 + 17424퐴퐵 + 130680퐵 − 1 3 2 2 2 3 푦 = −45496퐴 − 84216퐴 퐵 + 1091904퐴퐵 푧 = 121퐴 + 968퐵 + 224576퐵3 − 1 Properties 2 2 13 푧 = 484퐴 + 3872퐵 푧 퐴, 1 − 푦 퐴, 1 − 396푃퐴 − 푔17416 퐴 − Properties 31707푡4,퐴 = −129712. 푧 퐴, 1 − 3872is a perfect square. 푥 퐴 퐴 + 1 , 1 − 푧 퐴 퐴 + 1 , 1 − 푥 퐴, 1 + 11616푃5 + 139392푃 퐴 − 1 + 17 퐴 1836퐶푃 + 37312푡 + 106216푃 퐴 − 퐴 퐴+1 3,퐴 퐴+1 246840푡4,퐴 = 1045439. 58079is a Centered hexagonal pyramidal number 10 푦 퐴 퐴 + 1 , 1 + 34122푃퐴 퐴+1 + of rank퐴(퐴 + 1). 134310푡3,퐴 퐴+1 −1 − 996314푃 퐴 = 224575. 푥 퐴 퐴 + 1 , 1 + 푦 퐴 퐴 + 1 , 1 − 11 PATTERN VII 2442퐶푃퐴 퐴+1 + 142296푡3,퐴 퐴+1 + Considering (10) and (11) in (7) and repeating the 34265푃 퐴 = 18972. process in pattern IV, the corresponding non-zero distinct 16 푦 1, 퐵 − 푥 1, 퐵 − 26862퐶푃퐵 − 17908푃 퐵 − integer solution to (1) are obtained as 1−124388푡4,퐵=−5929. 푥 = −24퐴3 − 216퐴2퐵 + 576퐴퐵2 + 576퐵3 − 1 푥 1, 퐵 − 15396푇푇퐻 퐵 − 15푆푂 퐵 + 푦 = −32퐴3 + 120퐴2퐵 + 768퐴퐵2 − 320퐵3 − 1 2 2 47803푃 퐵 + 7772푡4,퐵 is three times a nasty 푧 = 4퐴 + 32퐵 number. Properties 8 PATTERN IV 푥 퐴, 1 − 푦 퐴, 1 − 6퐶푃퐴 + 380푡3,퐴−1 + 1 can also be written as 146푡4,퐴 = 496. (1+푖12 2)(1−푖12 2) 1 = (10) 푧 퐴, 1 − 32 is a perfect square. 289 18 Using (5) and (10) in (7) and repeating the process as above, 푥 1, 퐵 + 푦 1, 퐵 − 96푃퐵 − 112푃 퐵 − we obtain the distinct non-zero solutions to (1) as 1184푡4,퐵 = −58. 푥 = 16퐴3 − 264퐴2퐵 − 384퐴퐵2 + 704퐵3 − 1 PATTERN VIII 푦 = −24퐴3 − 216퐴2퐵 + 576퐴퐵2 + 576퐵3 − 1 Replacing 푧 by 2푤 in (3), it is written as 2 2 3 푧 = 4퐴2 + 32퐵2 푢 + 1 + 2푣 = 102푤 Properties 6 푥 퐴, 1 − 푦 퐴, 1 − 60푃퐴 + 78푃 퐴 ≡ 128 (푚표푑 872). Published By: Retrieval Number: F0859053615/2015©BEIESP Blue Eyes Intelligence Engineering 30 & Sciences Publication International Journal of Innovative Science and Modern Engineering (IJISME) ISSN: 2319-6386, Volume-3 Issue-6, May 2015 13. GopalanM.A.,.Vidhyalakshmi S, LakshmiK.,2013, ―Lattice points on Writing 102 as 102 = (10 + 푖 2)(10 − 푖 2) and the non- homogeneous cubic equation 푥3 + 푦3 + 푧3 − 푥 + 푦 + performing the analysis as given above, the corresponding 푧=0‖, Impact J.Sci.Tech.,Vol 7, No.1, 51-55. integer solutions to (1) are found to be 14. Gopalan M.A., Vidhyalakshmi S.,Lakshmi K ,2013, ―Lattice points 푥 = 11푎3 + 24푎2푏 − 66푎푏2 − 16푏3 − 1 on the non-homogeneous cubic equation 푥3 + 푦3 + 푧3 + 푥 + 푦 + 3 2 2 3 푧 = 0‖, Impact J.Sci.Tech.,Vol 7, No.1, 21-25. 푦 = 9푎 − 36푎 푏 − 54푎푏 + 24푏 − 1 15. Gopalan M.A.,Vidhyalakshmi S.,Kavitha A,2013, ―Observation On 2 2 푧 = 2푎 + 4푏 The Ternary CubicEquation 푥2 + 푦2 + 푥푦 = 12푧3‖,Antarctica Properties J.Math.,10(5),453-460. 26 16. Gopalan M.A.and Geetha K., 2013, ―On the ternary cubic diophantine 푥 푎 푎 + 1 , 1 + 푦 푎 푎 + 1 , 1 − 6퐶푃푎 푎+1 + equation푥2 + 푦2 − 푥푦 = 푧3‖ , Bessels J.Math., 3(2),119-123.