IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 05 Issue: 03 | Mar-2016, Available @ http://www.ijret.org 502 ON HOMOGENEOUS BIQUADRATIC DIOPHANTINEEQUATION: 2 2 2 4 4 ) ( 17 R w z y x    P.Jayakumar 1 , R.Venkatraman 2 1 Professor of Mathematics, Periyar Maniammai University, Vallam,Thanajvur- 613403,Tamil Nadu, India 2 Ph.D. Scholar, Assistant Professor of Mathematics, SRM University, Vadapalani Campus, Chennai–600026. Tamil Nadu, India. Abstract Five different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x 4 – y 4 = 17( z 2 – w 2 ) R 2 are obtained. Some interesting relations among the special numbers and the solutions are exposed. Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers --------------------------------------------------------------------***---------------------------------------------------------------------- 2010 Mathematics Subject Classification: 11D25 Symbols used: t m,, n = )] 4 ( ) 2 ( [ 2 1    m m n p n m = )] 5 ( ) 2 ( 3 [ 6 1 3 2     m n m n n G n = 2n-1 Ct 16 , n = 8n(n +1)+1 OH n = 1/3n(2n 2 +1) SO n = n (2n 2 -1) ky n = (2n +1) 2 -2carl n - carol number 1. INTRODUCTION The Mathematics is the Queen of all sciences. In particular, the Number theory is the King of Mathematics. The Number theory, in particular Diophantine equations have a blend of interesting problems. Many greatest Mathematicians was fascinated by problems in Diophantine equations. For a vide review, one may try to see [1-12]. In this work, we are observed a lot of infinitely the non –zero integer values of the cubic Diophantine equation. Some interesting relations among the special numbers and the solutions are found. 2. DESCRIPTION OF METHOD Let us consider the cubic Diophantine equation 2 2 2 4 4 ) ( 17 R w z y x    (1) Consider the transformations x = u +v, y = u – v, z = 2uv +1, w = 2uv – 1 (2) Using (2) in (1), it gives us the equation u 2 + v 2 = 17R 2 (3) 2.1 Method: I We can write 17 as17 = (4+i) (4 –i) (4) and R= a 2 + b 2 = (a +ib) (a – ib) (5) Using (4) and (5) in (3) and applying the process of factorization , it takes form as (u +iv) (u – iv) = (4 +i) (4 – i) (a +ib) 2 (a – ib) The above equations give us (u + i v) = (4 + i) (a + i b) 2 (u + i v) = (4 –i) (a – i b) Comparing both sides of above equations, we obtain u = u (a, b) = 4a 2 – 4b 2 – 2ab v = v(a, b) = a 2 – b 2 + 8ab Using the values of u and v in (2) we get the non-zero integer values of x, y, z and w and R of (1) are furnished by x = x (a, b) = 5a 2 – 5b 2 + 6ab y = y (a, b) = 3a 2 – 3b 2 – 10ab z = z (a, b) = 2(4a 4 + 4b 4 – 24a 2 b 2 + 30a 3 b – 30ab 3 ) +1 w = w (a, b) = 2(4a 4 + 4b 4 – 24a 2 b 2 + 30a 3 b – 30ab 3 ) – 1 R = R(a, b) = a 2 + b 2 Observations: 1.3x [a (a + 1),1] – 5y[a(a+1),1] – 68P a = 0 2.3x [a (a – 1),1] – 5y[a(a–1),1] – 68 t 4,a  0 (Mod 2) 3.z(b, 1) – w(b,1)  0 (Mod 2) 4.R(a+1, a+1) -2t 4,a – G 2a  0 (Mod 3)