International Journal of Innovative Science and Modern Engineering (IJISME) ISSN: 2319-6386, Volume-3 Issue-6, May 2015 On the Non-Homogeneous Quintic Equation with Seven Unknowns S. Vidhyalakshmi, M. A. Gopalan, K. Lakshmi by . A few Abstract— We obtain infinitely many non-zero integer solutions (,,,,,,)x y z w X Y T satisfying the non- homogeneous relations between the solutions and the special numbers are presented quintic equation with seven unknowns given by Initially, the following two sets of solutions in 2 2 2 2 2 2 3 . Various xyx()()() y zwz w X YT (,,,,,,)x y z w X Y T satisfy the given equation: interesting relations between the solutions and special numbers, 2 2 2 namely, polygonal numbers, Pyramidal numbers, Stella (2k k p ,2 k k p , p 2 k k , , Octangular numbers, Octahedral numbers,, Jacobsthal number, 22 Jacobsthal-Lucas number, keynea number, Centered pyramidal p2 k k ,2 k ,4 k ,2 k ) numbers are presented (2k2 3 k p 1,2 k 2 3 k p 1,2 p k 2 k , Index Terms— Centered pyramidal numbers, Integral p2 k22 k ,2 k 1,4 k 4 k 1,2 k 1) solutions, Non-homogeneous Quintic equation, Polygonal numbers, Pyramidal numbers However we have other patterns of solutions, which are MSC 2010 Mathematics subject classification: 11D41. illustrated below: Notations: II. METHOD OF ANALYSIS Tmn, - Polygonal number of rank n with size m The Diophantine equation representing the non- SO - Stella Octangular number of rank n homogeneous quintic equation is given by n (1) Jn - Jacobsthal number of rank of Introduction of the transformations KY - Keynea number of rank n xupyupz ,,,, pvwpv CP - Centered hexagonal pyramidal number of rank n,6 X u v, Y u v (2) m in (1) leads to Pn - Pyramidal number of rank with size u2 v 2 T 3 (3) OH - Octahedral number of rank n The above equation (3) is solved through different approaches j - Jacobsthal-Lucas number of rank and thus, one obtains different sets of solutions to (1) n A. Approach1 I. INTRODUCTION The solution to (3) is obtained as The theory of Diophantine equations offers a rich variety of u a() a22 b fascinating problems. In particular, quintic equations, 22 homogeneous and non-homogeneous have aroused the v b() a b (4) interest of numerous mathematicians since antiquity [1-3]. T() a22 b For illustration, one may refer [4-10] for homogeneous and non-homogeneous quintic equations with three, four and five In view of (2) and (4), the corresponding values of unknowns. This paper concerns with the problem of x,,,,,, y z w X Y T are represented by determining non-trivial integral solution of the non- x a() a22 b p (5) homogeneous quintic equation with seven unknowns given y a() a22 b p z p b() a22 b 22 w p b() a b 22 X( a b )( a b ) 22 Y( a b )( a b ) Manuscript Received on April 2015. T a22 b Prof. Dr. S. Vidhyalakshmi, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-2, Tamil Nadu, India. Prof. Dr. M. A. Gopalan,, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-2, Tamil Nadu, India. K. Lakshmi, Lecturer,Department of Mathematics, Shrimati Indira Gandhi College, Trichy-2,Tamil Nadu, India. Published By: Retrieval Number: F0855053615/2015©BEIESP Blue Eyes Intelligence Engineering 12 & Sciences Publication On the Non-Homogeneous Quintic Equation with Seven Unknowns The above values of x, y , z , w , X , Y and T satisfies the xp 42 2 ( 2 2 ) following relations: 2 2 2 2 1. [(,)xab yab (,)][ Xab (,) Yab (,)] is a perfect yp 4 ( ) square. zp 42 2 ( 2 2 ) 2. The following expressions are nasty numbers: 2 2 2 2 (13) wp 4 ( ) 42 ()a 3[(( Xaa 1),) aYaa (( 1),)4 a T2 6 CP ] X 8 3,a a,6 ()b 3[(( xaa 1),) a yaa (( 1),) azaa (( 1),) a Y 824 w( a ( a 1), a )] 22 T 4 (c) 2[(2,)x a a y (2,)6 a a SO 12 T ]. aa3, 3. The following expressions are cubic integers Properties: ()a 9[(2,) xaa yaa (2,) zaa (2,) waa (2,) 1. x(2,) a a y (2,)96(2 a a P5 2 T T )0 aa3,a2 4, X(2 a , a) Y (2 a , a )] xyx ()()()2 y 2 zwz 2 w 2 X 2 YT 2 3 2. X(22n ,2 2 n ) Y (2 2 n ,2 2 n ) z (2 2 n ,2 2 n ) (b) 4[(,1)Y a X (,1) a 2 T ] 4,a w(22n ,2 2 n ) +(2 T 2 n ,2 2 n ) 48 J 4 J 12 0 4. w(,1) a z (,1) a T (,1) a T 1 64nn 4,a 3. 3[4T2 x (,) y (,) T (,)] is a 5. xaa(2,) yaa (2,) Taa (2,) 12 CPa,6 4, nasty number 63T3,aa SO 4. 22 [(,)x y (,) z (,) w (,)] is a 6. T(2,) a a z (2,) a a w (2,) a a 3 T4,a Cubic number 0(mod 2) 5. 2(,)[(,)T y x (,) X (,) Y (,) 2nn 1 2 7. T(2 ,2 ) 3 J2nn 1 3 KY 2 zw( , ) ( , )] is a quintic integer B. Approach2 The assumptions Remark1: (6) Similarly by considering (6), (8), (11) and (2), we get the u UT, v VT in (3) yields to corresponding integral solution to (1) as 22 x2 T p UVT (7) 2 y2 T p (i) Taking Tt (8) 22 in (7), we get z p () T 2 2 2 22 U t V (9) w p () T Then the solution to (9) is given by XT()2 2 2 2 2 (10) 2 t2 , V , U , 0 YT () 2 2 2 2 (11) 2 2 2 U2 , V , t , 0 T () From (6), (8) and (10) we get, ii) Now, rewrite (9) as, U2 t 21* V 2 (14) u(6), (8) 4 and2 (10) 2 ( we 2 get 2 ) Also 1 can be written as 2 2 2 2 nn v 4 ( ) (12) 1 (ii ) ( ) (15) 22 22 T 4 Let V a b (16) Substituting (15) and (16) in (14) and using the method of In view of (12) and (2), we get the corresponding integral factorization, define, solution of (1).as ()()U it in a ib 2 (17) Equating real and imaginary parts in (17) we get Published By: Retrieval Number: F0855053615/2015©BEIESP Blue Eyes Intelligence Engineering 13 & Sciences Publication International Journal of Innovative Science and Modern Engineering (IJISME) ISSN: 2319-6386, Volume-3 Issue-6, May 2015 3 nn22 (u v )( u v ) 1 T Ucos ( a b ) 2 ab sin 22 (18) (24) Writing (24) as a set of double equations in two different ways nn22 t2 ab cos sin ( a b ) as shown below, we get 22 3 Set1: u v T, u v 1 In view of (2), (6), (8) and (18), the corresponding values of 3 x,,,,,, y z w X Y T are represented as Set2: u v T, u v 1 Solving set1, the corresponding values of u, v and T are nn given by x T[cos ( a22 b ) 2 ab sin ] p 22 u4 k3 6 k 2 3 k 1, v 4 k 3 6 k 2 3 k , (25) nn y T[cos ( a22 b ) 2 ab sin ] p Tk2 1 22 In view of (25) and (2), the corresponding solutions to (1) z p a22 b obtained from set1 are represented as shown below: (19) 32 22 x4 k 6 k 3 k 1 p w p a b nn y4 k32 6 k 3 k 1 p X T[cos ( a2 b 2 ) 2 ab sin a 2 b 2 ] 22 z p 4 k32 6 k 3 k nn2 2 2 2 Y T[cos ( a b ) 2 ab sin a b ] 32 22 w p 4 k 6 k 3 k nn 32 T (sin ( a2 b 2 ) 2 ab cos ) 2 X8 k 12 k 6 k 1 22 (iii) 1 can also be written as Y 1 ((m2 n 2 ) i 2 mn )(( m 2 n 2 ) i 2 mn ) Tk21 1 (20) 2 2 2 ()mn Properties: 1. x( a ) y ( a ) 24 P4 0 (mod 2) Substituting (16) and (20) in (14) and using the method of a factorization, define, 2.2[()x a y () a z () a w () a T () a (m22 n ) i 2 mn ) 24PTT4 6 2 2 ] ()()U it a ib 2 (21) a4, a 8, a 2 2 2 is a nasty number. ()mn 8 3. 4[()X a Y ()8 a Pa 16 T3, a 6( OH ) a 4 CP a ,6 ] is Equating real and imaginary parts in (21) we get a cubic integer. 1 U{( m2 n 2 )( a 2 b 2 ) 4 mnab } 4. Xa() Ya () xa () ya () za () wa () 22 mn (22) 4 1 24P 4 CP 2 SO 0(mod 4) t{2 ab ( m2 n 2 ) 2 mn ( a 2 b 2 )} a a,6 a 22 mn 5. 4[()x a y () a Y () a T () a 48 P3 . a In view of (2), (6), (8) and (22), the corresponding values of 24T3,a 12( OH a ) 8 CP a ,6 ] are represented as is a biquadratic integer x( m2 n 2 ) T [( A 2 B 2 )( m 2 n 2 ) 4 mnAB ] p Remark2: 2 2 2 2 2 2 y( m n ) T [( A B )( m n ) 4 mnAB ] p Similarly, the solutions corresponding to set2 can also be obtained.