IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 6, August 2014. www.ijiset.com ISSN 2348 – 7968 On Ternary Quadratic Diophantine Equation 3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 1 2 S.VIDHYALAKSHMI , M.A.GOPALAN , J.SHANTHI 1 3 Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy Tamilnadu, India- 620 002. Email: [email protected], 2 Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy Tamilnadu, India- 620 002. Email:[email protected] 3 Lecturer, Department of Mathematics, Shrimati Indira Gandhi College,Trichy Tamilnadu, India -620002.Email:[email protected] Abstract The ternary quadratic equation representing non-homogeneous cone given by 3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 is analyzed for its non-zero distinct integer points on it. The different patterns of integer points satisfying the cone under consideration are obtained. A few interesting relations between the solutions and special number patterns are presented. Keywords: Ternary non-homogeneous quadratic, integral solutions 2010 Mathematics Subject Classification: 11D09 Sn - Star number of rank n. 2.METHOD OF ANALYSIS: The ternary quadratic diophantine equation to be solved for its non-zero distinct integral solutions is 3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 The substitution of linear transformations x = u + v, y = u − v (1) (2) in (1) leads to 1. Introduction The ternary quadratic Diophantine equations offer an unlimited field for research due to their variety [1, 20]. For an extensive review of various problems, one may refer [2 – 19]. This communication concerns with yet another interesting ternary quadratic equation 3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 representing a cone for determining its infinitely many non-zero integral points. Also, a few interesting relations among the solutions are presented. NOTATIONS: Tm,n - Polygonal number of rank n with size m. Pnm - Pyramidal number of rank n with size m. Ct m,n - Centered Polygonal number of rank n with size m. CPm,n - Centered Pyramidal number of rank n with size m. Prn (3) (u + 1) 2 + 11v 2 = 15 z 2 Different patterns of solutions of (1) are presented below. 2.1 PATTERN : 1 Write 15 as 15 = (2 + i 11)(2 − i 11) (4) Assume z= a + 11b (5) where a, b are non-zero distinct integers. Using (4) & (5) in (3), and applying the method of factorization,define 2 [(u + 1) + i 2 ] 11v = (2 + i 11)(a + i 11b) 2 Equating the real imaginary parts, we have (6) u = u (a, b) = 2a 2 − 22b 2 − 22ab − 1 v = v(a, b) = a 2 − 11b 2 + 4ab Substituting the above values of the values of x and y are given by u & v in equation (2), x = x(a, b) = 3a 2 − 33b 2 − 18ab − 1 (7) y = y (a, b) = a − 11b − 26ab − 1 (8) 2 2 - Pronic number of rank n. 212 IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 6, August 2014. www.ijiset.com ISSN 2348 – 7968 Thus (5), (7) & (8) represent a non-zero distinct integral solutions of (1) in two parameters. 1.x(1, s ) − 3 y (1, s ) + 2t12,s ≡ −4(mod 68) 2. PROPERTIES: 1.x(a, b) − 6t 3,a −1 ≡ −4(mod15) 3.x(2 n ,1) − y (2 n ,1) = 990 J 2n + 12 J n + 4 j n + 227 2.6[x(a,1) − y (a,1)] − 2 S a ≡ −14(mod 60) Note: In addition to (13), one may also consider the linear transformations z = X − 11T , v = X − 15T .Following the method presented above , different set of solutions are obtained. 3.z (1,2 n ) − 45 J 2n + 6 J n + 2 j n = 6 2.2 PATTERN 2: Instead of (4), we write 15 as (7 + i 11)(7 − i 11) (9) 4 Following the procedure presented above, the corresponding non-zero distinct integer solutions to (1) are obtained as 15 = x = x(a, b) = 4a 2 − 44b 2 − 4ab − 1 (10) y = y (a, b) = 3a 2 − 33b 2 − 18ab − 1 along with (5) PROPERTIES: (11) 1.x(2 n ,1) = 12 J 2n − 6 J n − 2 j n − 41. 2.x(2 n ,1) − y (2 n ,1) = j 2n + 7 j n +21J n − 12. 3.3 x(a,1) = 2 S a − 137. 2.3 PATTERN : 3 Equation (3) can be written as (u + 1) 2 = 15 z 2 − 11v 2 Introducing the linear transformations, z = X + 11T & v = X + 15T in (12), we get (12) (13) 2 u + 1 X 2 = 165T 2 + (14) 2 which is satisfied by T (r , s ) = 2rs 2 2 (15) u (r , s ) = 330r − 2 s − 1 2 2 X (r , s ) = 165r + s Substituting the values of (15) in (13) and using (2), the corresponding integer solutions of (1) are given by x = x(r , s ) = 495r 2 − s 2 + 30rs − 1 (16) y = y (r , s ) = 165r − 3s − 30rs − 1 (17) z = z (r , s ) = 165r 2 + s 2 + 22rs (18) 2 PROPERTIES: 2 1 [x(1, n) − y(1, n)]2t3,n−1 ≡ 10(mod 31) 2 2.4 PATTERN : 4 Consider (3) as (19) (u + 1) 2 − 4v 2 = 15( z 2 − v 2 ) Write (19) in the form of ratio as A (u + 1) + 2v 15( z − v) = = ,B ≠ 0 z+v (u + 1) − 2v B which is equivalent to the following two equations B(u + 1) + (2 B − A)v − Az = 0 A(u + 1) + (15B − 2 A) − 15Bz = 0 On employing the method of cross multiplication, we get u = 2 A 2 − 30 AB + 30 B 2 − 1 (20) v = A 2 − 15 B 2 (21) z = A 2 − 4 AB + 15B 2 Substituting the values of u and v from (20) in (2), the nonzero distinct integral values of x, y are given by x = x( A, B) = 3 A 2 + 15B 2 − 30 AB − 1 (22) y = y ( A, B) = A 2 + 45B 2 − 30 AB − 1 Thus (21) and (22) represent non-zero distinct integral solutions of (1) in two parameters. PROPERTIES: Each of the following expressions represents a Nasty number i )2[3 z (1, β ) − y (1, β )] ii )2[ y (a,−a ) + 1] 2.18 x(1, β ) − 26 y (1, β ) + 60 z (1, β ) = 96 Note: (19) can also be expressed in the form of ratio in two different ways as follows. A (u + 1) + 2v 5( z − v) = = , B ≠ 0. (i ). 3( z + v) (u + 1) − 2v B A (u + 1) + 2v 3( z − v) = = , B ≠ 0. (ii ). 5( z + v) (u + 1) − 2v B 213 IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 6, August 2014. www.ijiset.com ISSN 2348 – 7968 Repeating the analysis as above the corresponding integer solutions along with properties are presented below. Solutions of (i): x = x( A, B) = −9 A 2 − 5B 2 + 30 AB − 1 Cone z 2 = 2 x 2 − 7 y 2 , Diophantus J. Math., 1(2), 109115, 2012. 5. M.A.Gopalan, J. Kalinga Rani, on ternary quadratic y = y ( A, B) = −3 A 2 − 15B 2 + 30 AB − 1 z = z ( A, B) = 3 A 2 + 5B 2 − 4 AB PROPERTIES : 1.x(1,2α − 5α 2 ) − y (1,2α − 5α 2 ) − 14t 3,( −5α 2 + 2α −1) 3. M.A. Gopalan, and V. Pandichelvi, Integral solutions of ternary quadratic equation z ( x − y ) = 4 xy , Impact J.sci TSech; Vol (5), No.1, 01-06, 2011. 4. M.A.Gopalan, S. Vidhyalakshmi and A.Kavitha, Integral points on the homogeneous is a perfect square. 2.5 x(1, n) + z (1, n) ≡ 27(mod 34) 3.x(1, n) + y (1, n) + z (1, n) + 22t 3,n−1 ≡ 19(mod 22) Solutions of (ii): x = x( A, B ) = 15 A 2 + 3B 2 − 30 AB − 1 y = y ( A, B ) = 5 A 2 + 9 B 2 − 30 AB − 1 z = z ( A, B ) = −5 A 2 − 3B 2 + 4 AB PROPERTIES : 1.x(2 n ,1) = 45 J 2n−45 J n − 15 j n + 17 2.x(2 n ,1) − y (2 n ,1) = 30 J 2n + 4 3.6 x(a,1) − 15S a ≡ 87(mod 90) 3.CONCLUSION: In this paper, we have presented six different patterns of non-zero distinct integer solutions of the nonhomogeneous cone given by 3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 .To conclude, one may search for patterns of non-zero integer distinct solutions and their corresponding properties for other choices of ternary quadratic diophantine equations. 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