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On the Homogeneous Biquadratic Equation with 5 Unknowns ( International Journal of Innovative Science and Modern Engineering (IJISME) ISSN: 2319-6386, Volume-3 Issue-8, July 2015 On the Homogeneous Biquadratic Equation with 5 2 2 2 2 2 2 2 Unknowns xy4 kxy 1 4 kxy 2 2 4 kpqz 1 M. A. Gopalan, K. Geetha, Manju Somanath Abstract: The homogeneous biquadratic equation with five In this context one may refer [4-12] for various problems on unknowns given by the biquadratic Diophantine equations. However, often we come across non-homogeneous biquadratic equations and as such one may require its integral solution in its most general is considered and analyzed for finding its non zero distinct form. This paper concern with the homogeneous biquadratic integral solutions. Introducing the linear transformations equation with five unknowns x u v, y u v, p2, u v q2 u v and employing the method of factorization different patterns of non zero distinct integer solutions of the equation under the above for determining its infinitely many non-zero integral equation are obtained. A few interesting relations between the solutions. Also a few interesting properties among the integral solutions and the special numbers namely Polygonal numbers, Pyramidal numbers, Centered Polygonal numbers, solutions are presented. Centered Pyramidal numbers, Thabit-ibn-Kurrah number, Carol number, Mersenne number are exhibited. II. METHOD OF ANALYSIS Keywords: Homogeneous equation, Integral solutions, Polygonal The biquadratic equation with five unknowns to be solved numbers, Pyramidal numbers and Special numbers. 2010 Mathematics Subject Classification Code: 11D25 for its non-zero distinct integral solution is Notations: tmn, = Polygonal number of rank n with sides m n (1) pm = Pyramidal number of rank n with sides m Consider the transformations ctmn, = Centered Polygonal number of rank n with sides m x u v n y u v cpm = Centered Pyramidal number of rank n with sides m (2) p2 u v gn = Gnomonic number q2 u v Tha = Thabit-ibn-Kurrah number n On substituting (2) in (1), we get car ln = Carol number cu l = Cullen number 2 2 2 n ku v3 k 1 4 k 1 z (3) mern = Mersenne number Pattern 1: wo = Woodhall number 22 n Assume z ka 31 k b (4) p = Pronic number n Write (4k-1) as I. INTRODUCTION 4k 1 k i 3 k 1 k i 3 k 1 The theory of Diophantine equations offers a rich variety of (5) fascinating problems. In particular biquadratic Diophantine Substituting (4) and (5) in (3) and employing the method of equations, homogeneous and non-homogeneous have factorization, aroused the interest of numerous mathematicians since antiquity [1-3]. Revised Version Manuscript Received on July 07, 2015. M. A. Gopalan, Professor, Department of Mathematics, Shrimathi Indira Gandhi College, Trichy, Tamil Nadu, India. K. Geetha, Asst. Prof., Department of Mathematics, Cauvery College for Women, Trichy, Tamil Nadu, India. Manju Somanath, Asst. Prof., Department of Mathematics, National College, Trichy. Tamil Nadu, India. Published By: Retrieval Number: H0914073815/2015©BEIESP Blue Eyes Intelligence Engineering 6 & Sciences Publication On the Homogeneous Biquadratic Equation with 5 Unknowns 222 kuikvkuikv3 1 3 1 kik 3 1 kik 3 1 ka 3 k 1 b Equating the real and imaginary parts, we get 22 22 u ka 3 k 1 b 2 3 k 1 ab v ka 2 ab 3 k 1 b Substituting u and v in (2) the values of x, y, p and q are given by 22 xxabk(,,)2 ka 2 b 31231 k k abab 2 y y( a , b , k ) 6 kab 22 ppabk(,,)3 ka 3 b 3112 k kab 6 ab (6) 22 qqabk( , , ) 3 ka 3 b 3 k 1 12 kab Thus (4) and 6 represent the non-zero distinct integer solutions of 1 Properties: n 1) x1,1,2 1 10 mern Nasty number 2) p a1,1, k 9 t6,aa 2 t 3, 3) q(1,2 b ,1) t18,b 3 mod31 2 4 22 4) znn 1,,1 ynn 21,,23 cpn 2 cp n t32, n 1mod31 qn2 1,2,1 pnn 2 1,2 1,1 ynn 2,2 1,1 2 tt 36 g 5) 58,n 42, n n 6) Pattern 2: 2 2 2 2 v a 32 k b kb kab The assumption u X 31 k T and v X kT (7) Substituting u and v in (2) the values of x, y, p and q are 2 2 2 given by in (3) gives X k31 k T z 2 2 2 2 (8) xxabk( , , ) 2 a 6 kb 2 kb 4 abk 2 ab Case 1: y y( a , b , k ) 8 abk 2 ab kk31 square (9) ppabk( , , ) 3 a2 9 kb 2 2 3 kb 2 10 abk 4 ab 22 Assume z a k31 k b (10) (11) 2 2 2 2 qqabk( , , ) a 3 kb kb 14 abk 4 ab From (8) and (10), we have, on factorization 2 Thus (10) and 11 represent the non-zero distinct integer X i k3 k 1 T a i k 3 k 1 b solutions of 2 X i k3 k 1 T a i k 3 k 1 b On equating real and imaginary parts, in either of the above two equations we obtain 22 X a k31 k b T 2 ab On substituting X and T in (4) we get the values of u and v to be 2 2 2 2 u a 3 k b kb 6 abk 2 ab Published By: Retrieval Number: H0914073815/2015©BEIESP Blue Eyes Intelligence Engineering 7 & Sciences Publication International Journal of Innovative Science and Modern Engineering (IJISME) ISSN: 2319-6386, Volume-3 Issue-8, July 2015 Properties: 1) y a, a ,1 2 t8,aa g 1 2) q1,1, k p 1,1, k 2 gk 1 xb1, ,1 3) 1 t6,b 2 4) p a,1,1 q a ,1,1 y a ,1,1 4 t18,aa t 14, 5) ynn 1, 2,2 xnn 1, 2,2 ct28,n ct 8, n 31 g n 87 Case 2: 2 Choosing k31 k w in (5), it is satisfied by 2 22 22 X 2 w RS , T w R S and z w R1 S Substituting X and T in (4) we get the values of u and v to be 2 2 2 2 2 u2 w RS 3 kwR 3 kwS wR wS 2 2 2 v2 w RS kwR kwS Substituting u and v in (2) we get the values of x, y, p and q are given by 2 2 2 2 2 x x R, S , w , k 4 w RS 2 kwR 2 kwS wR ws 2 2 2 2 y y R, S , w , k 4 kwR 4 kwS wR wS 2 2 2 2 2 p p R, S , w , k 6 w RS 5 kwR 5 kwS 2 wR 2 wS (12) 2 2 2 2 q p R, S , w , k 2 w RS 7 kwR 6 wS 2 wR Thus (10) and (12) represent the non-zero distinct integer solutions of Properties: nn 1) z2 ,2 ,1,1 2 mer2n 1 nn 2) p1,1,2,1 x 1,1,2,1 2 jal2n 1 n 3) q2 ,1,1,1 Tha22n ky n mer n 3 4) x1,1, wky , 1,1, wk , 2 q 1,1, wk , 1 t10,w g w 7 wg k 2 5) p2,1,1, n 1 q 2,1,1,2 n 2 t17,n 1 mod9 Pattern 3: Rewrite (5) as 2 2 2 z X k31 k T (13) z X z X k31 k T2 (14) Case 1: (14) can be written as the system of double equations as z X 31 k T (11) Published By: Retrieval Number: H0914073815/2015©BEIESP Blue Eyes Intelligence Engineering 8 & Sciences Publication On the Homogeneous Biquadratic Equation with 5 Unknowns z X kT (12) Which is satisfied by z41 k t X21 k t Tt 2 , , On substituting X and T in (7) we get the values of u and v which are given by u83 k t , v21 k t On substituting u and v in (2) we get the values of x, y, z, p and q. The non-zero distinct integrals values of x, y, z, p and q satisfying (1) are given by x x k, t 10 k 2 t y y k, t 6 k 4 t z z( k , t ) 4 k 1 t p p k, t 18 k 5 t q q k, t 14 k 7 t Properties: 26 1) x( n , n ) y ( n , n ) n 10 cpnn t14, 2 18 2) pnn( , ) qnn ( , ) 5 6 cpn 2 ct11, n 3 g n 3) x n,1 y n ,1 2 t62,nn 3 g 11 2 2 23 4) pnn , 2 nnzn 1, ,26 cpnn 2 ct25, 1mod12 2 20 16 5) q n,2 n 13 y n 1, n 6 cpnn 3 cp 7mod37 Case 2: Also, (10) can be written as the system of double equations as z X 3 k2 k T (13) z X T (14) Applying the same procedure as in case 1 we get the non-zero distinct integrals values of x, y, z, p and q satisfying (1) are given by x x( k , t ) 6 k2 3 t y y( k , t ) 4 k2 1 t z z( k , t ) 3 k2 k 1 t p p( k , t ) 9 k2 2 k 5 t q q( k , t ) 3 k2 6 k 3 t Properties: x n,1 y n ,1 t2 t 12,n 3 mod 4 1) 50,n Published By: Retrieval Number: H0914073815/2015©BEIESP Blue Eyes Intelligence Engineering 9 & Sciences Publication International Journal of Innovative Science and Modern Engineering (IJISME) ISSN: 2319-6386, Volume-3 Issue-8, July 2015 n 2) z(2 ,1) Tha2nn mer 1 3) p n1,2 t38,n 17 mod23 42 7p k , t 2 q k , t 2 z k , t 27 4) = Nasty Number 22,ynn n z 2,4 n wo cul Tha mer 1mod2 5) 2n 2 n 2 n n Case 3: 3.
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