International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064 Index Copernicus Value (2013): 6.14 | Impact Factor (2014): 5.611 Lattice Points on the Homogeneous Cone x2+y2= 26z2 Dr. P. Jayakumar1, R. Venkatraman2 1Professor of Mathematics, Periyar Maniammai University Vallam Thanjavur-613403, India 2Assistant Professor of Mathematics, SRM University Vadapalani Campus, Chennai – 600026, India Abstract: The ternary quadratic homogeneous equation representing cone x2+y2= 26z2 is considered and analyzed for finding its non zero distinct integral solutions. Five different patterns of integer points satisfying the cone under consideration are obtained. A few interesting relation between the solutions and special number patterns are presented. Keywords: Homogeneous cone, Ternary Quadratic, Integral solutions, special numbers. 2010 Mathematics Subject Classification: 11D09 Notations Used r Tm,n - Polygonal number of rank n with size m - Pn - Pyramidal number of rank n with size m - g -Gnomonic number of rank n- Prn - Pronic number of rank n - Ct16,n- Centered hexadecagonal pyramidal number of rank n- OHn - Octahedral number of rank n- SOn - Stella octangular number of rank n - kyn - kynea number - carln -carol number- wn - woodall number- Sn-Star number- Cn- Cullen number. 2 3. z [A (A + 1), A (A + 1)] – 2(PA) = 0 1. Introduction 2 2 4. x [1, A (2A –1)] –5 [1, A (2A –1)] + 52SOA = 0 5. y (1, 1) – PT3 0 (Mod 5) The theory of Diophantine equations offers a rich variety of 2 2 6. x [(2A +1), A] –5y [(2A + 1),A] +156OHA = 0 fascinating problems. This communication concerns with 2 2 2 7. y (A, 2A +1) + z( A, 2A +1) –2T4,A- 30OHA = 0 another interesting ternary quadratic equation representing x + 2 2 2 8. x (2A –1, A) –3y(2A –1, A) –2z(2A –1, A) +7T4,A+32SOA= 0 y2= 26z2 a cone for determining its infinitely many non-zero 9. x [(A + 1),(A + 2)] + CHA – 2T23,A + 20T4,A 0 (Mod 2) integral points. Also a few interesting properties among the 10. 3x (A, A) is a nasty number. solutions are presented. 2.2 Pattern: II 2. Method of Analysis Instead of (3), Write 26 as The ternary quadratic equation to be solved for its non-zero 26 = (1 + 5i) (1 – 5i) (6) integer solution is x2 + y2 = 26z2 (1) Following the procedure presented as in pattern : 1, the Assume z (a, b) = a2 + b2 , where a, b > 0 (2) corresponding values of x and y are x = x (a, b) = a2 – b2 – 10ab (7) We illustrate below five different patterns of non-zero distinct y = y (a, b) = 5a2 – 5b2 +2ab (8) integer solutions to (1) Properties:- 1. 5x (A, A) – y (A, A) + 52T4,A = 0 5 2.1 Pattern: I 2. y (1, 1) –푃2 ≡ 0 (mod 2) 4 3. y [A,(A + 1)(2A + 1)] – x [A, (A + 1)(2A + 1)] –312푃푎 = 0 Write 26 = (5 + i) (5 – i) (3) 4. y (1,1) – x(1,1) – OH3 = Carol number. 5 5. y (A,1) – 5x(A,1) – 5z(A,1) +5T4,A – G6A +푃2 ≡ 0 (mod 2) Substituting (2) & (3) in (1), and applying the method of 2 2 6. y (2A –1, A) –5x(2A –1, A) –52SOA= 0 factorization, define 2 (x + iy) (x – iy) = (5 + i) (5 – i) (a + ib)2 (a – ib)2 7. x (A ,A+1) –PTA+22T4,A+ 24OHA= carol number 8. x (A,A) + 8y (A, A) is a nasty number Equating real and imaginary parts, we get 9. x (1, 2A + 1) + Ct16,A–SA + 20T3,A – 8T4,A= 0 2 2 10. 6x (A, A) is a nasty number x = x (a, b) = 5a – 5b – 2ab (4) 2 2 y = y (a, b) = a – b + 10ab (5) 2.3 Pattern: III Properties Equation (1) is written in the form of ratio as 1. x (A,A) + 2T4,A = 0 2. y (A, A + 1) – Ct16,A – 2T4,A ≡ 0 (mod 2) Volume 5 Issue 3, March 2016 www.ijsr.net Paper ID: NOV162296 Licensed Under Creative Commons Attribution CC BY 1774 International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064 Index Copernicus Value (2013): 6.14 | Impact Factor (2014): 5.611 zx 5 yz A y(a, b) = 26a2 + b2 (21) , B 0 , 5 yz zx B z = z (a, b) = 2ab (22) which is equivalent to the system of equations Properties: B (x + z) – A (5z + y) = 0 (9) 2 1. z [A(A + 1), A(A + 1)] –2(PA) = 0 B (5z – y) – A(x – z) = 0 (10) 6 2. x [(A + 1), (A + 2)] –25T4,A – G24A +푃3 ≡ 0 (mod2) 3. y (A, A + 1) –27T4,A – GA ≡ 0 (mod2) 2 Applying the method of cross multiplication the integer 4 .z (A, 2A – 1) –2SOA = 0 4 solutions of (1) are given by 5. x [A,( A+ 1)(2A + 1)] + 52T4,A −푃퐴 = 0 2 2 x (a, b) = A –B +10AB (11) 6. z (A, 2A–1) –2T6,A= 0 y (a, b) = A2 – 5B2 – AB (12) 7. z (A, 8A – 7) –2T18,A= 0 2 2 8. 12z (1, 1) is a Nasty number z (a, b) = A + B (13) 9. x (A, A) –y (A, A) + 2T4,A = 0 10. z( A, A + 1) – 4T3,A = 0 Properties: 2 2 5 1. x [1,A(2A –1)] + z[1, A(2A –1)] –10SOA– 푃1 = 0 3. Conclusion 2. y (2A, 2A) – 5x (2A, 2A) + 220T4,A = 0 3. x (A,1) + 10(A, 1) –11T4,A +O4 + Ky1 = 0 4. z (1,1) – W2 ≡ 0 (mod5) In this work, the ternary quadratic Diophantine equations 5. x (2A, 1) – y(2A, 1) – 20T4,A– G11A = 0 referring a conies are analysed for its non-zero distinct integral 6. y (A, A + 1) + z (A, A + 1) – 6T4,A+ 2T3,A= 0 points. A few interesting properties between the solutions and ≡ 7. y [A(A + 1), 1] –z [A(A + 1), 1] –22T3,A 0 (mod 2) special numbers are presented. To conclude, one may search 8. y (2A - 1, A) – x (2A-1, A) + 4T4,A+ 11T6,A= 0 9. 3x (A, A) + 2y (A,A) + 2z(A, A) is a Nasty number. for other patterns of solutions and their corresponding 10. 3z (A, A) is a Nasty number. properties for the cone under consideration. 2.4 Pattern: IV References Equation (1) is written as [1] Dickson, L.E., History of theory of numbers, 2 2 2 26z – y = x * 1 (14) Vol.11Chelsea publishing company, New-York (1952). 2 2 Assume x (a, b) = 26a – b ) (15) [2] Mordell, L.J., Diophantine equation, Academic press, Write (1) as 1 = ( 26 + 5) ( 26 − 5) (16) London (1969) Journal of Science and Research, Vol (3), Substituting (15) & (16) in (14) and applying the method of Issue 12, 20-22 (December -14) factorization, define [3] Jayakumar. P, Sangeetha, K “Lattice points on the cone 2 2 2 ( 26 푧 + y) = ( 26 a + b) 2 ( 26 –5) x + 9y =50z ” International Journal of Science and Equating the rational and irrational factors we get Research, Vol (3), Issue 12, 20-22 (December -2014) x = x (A, B) = 26a2 – b2 [4] Jayakumar P, Kanaga Dhurga, C,” On Quadratic 2 2 2 y = y (A, B) = 130a2 + 5b2 –52ab (17) Diopphantine equation x +16y = 20z ” Galois J. Maths, z = z (A, B) = 26a2 + b2 –10ab (18) 1(1) (2014), 17-23. [5] Jayakumar. P, Kanaga Dhurga. C, “Lattice points on the 2 2 2 Properties cone x +9y =50z ” Diophantus J.Math,3(2) (2014), 61- 1. y (A, A) – 5z(A, A) +102 T4,A = 0 71 2 2. z [1, (A (2A –1)] +10SOA=0 [6] Jayakumar. P, Prabha. S “On Ternary Quadratic 2 2 2 3. z (A, A + 1) –27T4,A – GA + 10PA ≡ 0 (mod2) Diophantine equation x +15y =14z ” Archimedes J. 2 2 2 5 4. y [(2A –1),1 ] – 5 [(2A –1),1] +52 (GA) −푃3 ≡ 0 (mod2) Math., 4(3) (2014), 159-164. 5 5. y (1, A) –5T4,A +G26A −푃6 ≡ 0 (mod5) 2 2 [7] Jayakumar, P, Meena, J “Integral solutions of the Ternary 6. 5x (2A + 1, A) –y (2A + 1) + 156OHA ≡ 0 (mod2) Quadratic Diophantine equation: x2 +7y2 = 16z2” 7. x (2A, 2A) is a Perfect square International Journal of Science and Technology, Vol.4, 8. 1 [y (A, A) – 5x(A, A)] is a Cubic integer Issue 4, 1-4, Dec 2014. 42 [8] Jayakumar. P, Shankarakalidoss, G “Lattice points on 9. y (1, 1) –5x(1,1) + z(1, 1)+C3= 0 Homogenous cone x2 +9y2 =50z2” International journal of ≡ 10. 6x (1, 1) + z(1, 1) – Carl4 0 (mod2) Science and Research, Vol (4), Issue 1, 2053-2055, January -2015. 2.5 Pattern: V [9] Jayakumar. P, Shankarakalidoss. G “Integral points on 2 2 2 The Homogenous cone x + y =10z International Journal Assume x (a, b) = 26a2 – b2) (19) for Scienctific Research and Development, Vol (2), Issue Substituting (19) in (1) and applying the method of 11, 234-235, January -2015 factorization, define 2 2 [10] Jayakumar.P, Prapha.S “Integral points on the cone x ( 26 a + b) = ( 26 푧 + y) +25y2 =17z2” International Journal of Science and Research Vol(4), Issue 1, 2050-2052, January-2015. Equating rational and irrational, we get x(a, b) = (26a2 – b2) (20) Volume 5 Issue 3, March 2016 www.ijsr.net Paper ID: NOV162296 Licensed Under Creative Commons Attribution CC BY 1775 International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064 Index Copernicus Value (2013): 6.14 | Impact Factor (2014): 5.611 [11] Jayakumar.P, Prabha.