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 p 2 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 p 2 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.
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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 wp 4 ( ) (13) 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 )] 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 T4,a ] 2n 2 n 2 n 2 n w(2 ,2 ) +(2 T ,2 ) 48 J64nn 4 J 12 0 4. w(,1) a z (,1) a T (,1) a T4,a 1 3. 3[4T x (,) y (,) T (,)] is a 5. xaa(2,) yaa (2,) Taa (2,) 12 CP 4,2 a,6 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
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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 Tk 2 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. z p ()() m2 n 2 2 A 2 B 2 T w p ()() m2 n 2 2 A 2 B 2 T D. .Approach4 X( m2 n 2 ) T [( m 2 n 2 )( A 2 B 2 ) 4 mnAB 22 Substituting T a b 2 2 2 2 (m n )( A B )] in (3) and writing it as a system of double equations as Y( m2 n 2 ) T [( m 2 n 2 )( A 2 B 2 ) 4 mnAB u v () a b 3 (m2 n 2 )( A 2 B 2 )] (23) u v () a b 3 2 2 2 2 2 2 2 2 T ( m n ) [2 AB ( m n ) 2 mn ( A B )] and solving, we get
C. Approach3 Equation (3) can be written as
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32 Journal of Engineering, Science and Mathematics (IJESM), Vol 2(2), u a3 ab 40-45, June 2013. (26) 9. M.A.Gopalan & A.Vijayashankar, Integral solutions of v3 a23 b b non-homogeneous quintic equation with five 5 unknowns xy zw R , Bessel J.Math.,1(1),23- 30,2011. In view of (26) and (2), the corresponding solutions to (1) 10. M.A.Gopalan & A.Vijayashankar, solutions of uintic equation with 4 4 2 2 3 are represented as shown below: five unknowns x y 2( z w ) P , Accepted for 32 Publication in International Review of Pure and Applied x a 3 ab p Mathematics. 32 11. M.A.Gopalan, G. Sumathi & S.Vidhyalakshmi, On the y a 3 ab p non-homogenous quintic equation with five knowns 3 3 3 3 5 23 12. x y z w 6 T , IJMIE, Vol.3 (4), 501-506, April- z p 3 a b b 2013. 23 w p 3 a b b
X a3 33 ab 2 a 2 b b 3
3 2 2 3 Y a 33 ab a b b 22 T a b Properties:
1. 4[(,)xaa yaa (,) zaa (,) waa (,)] is a cubic
integer 2. X( a ,1) Y ( a ,1) 4 CPa,3 14 T 4, a 4 T 9, a 0
3. xaa(,) yaa (,) zaa (,) waa (,) Xaa (,)
Y ( a , a ) 12 SOa 36 T4, a 24 T 5, a 0 4. 16j T (22n ,2 2 n ) X (2 2 n ,2 2 n ) Y (2 2 n ,2 2 n ) 6n xy(22n ,2 2 n ) (2 2 n ,2 2 n ) is a biquadratic integer
5. XaaYaa (2,) (2,) 2(2,) xaa 2(2,) yaa 2T (2 a , a ) 0(mod 20)
III. CONCLUSION In conclusion, one may search for different patterns of solutions to (1) and their corresponding properties.
REFERENCES
1. L.E.Dickson, History of Theory of Numbers, Vol.11, Chelsea Publishing company, New York (1952). 2. L.J.Mordell, Diophantine equations, Academic Press, London(1969). 3. Carmichael,R.D.The theory of numbers and Diophantine Analysis,Dover Publications, New York (1959) 4. M.A.Gopalan & A.Vijayashankar, An Interesting Diophantine 3 3 5 problem x y2 z , Advances in Mathematics, Scientific Developments and Engineering Application, Narosa Publishing House, Pp 1-6, 2010. 5. M.A.Gopalan & A.Vijayashankar, Integral solutions of ternary quintic 2 2 5 Diophantine equation x(2 k 1) y z ,International Journal of Mathematical Sciences 19(1-2), 165-169, Jan-June 2010 6. M.A.Gopalan,G.Sumathi & S.Vidhyalakshmi, Integral solutions of non-homogeneous ternary quintic equation in terms of pells sequence x3 y 3 xy( x y ) 2 Z 5 7. , JAMS (Research India Publication), Vol.6 (1), 59-62, 2013 8. S.Vidhyalakshmi, K.Lakshmi and M.A.Gopalan, Observations on the homogeneous quintic equation with four unknowns 5 5 5 2 2 2 x y 2 z 5( x y )( x y ) w ,International
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