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-, 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, - 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 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 - 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 2 2 2 2 2 2 3 xyx()()() y  zwz  w  X  YT

The above values of x, y , z , w , X , Y and T satisfies the xp 42  2 (  2   2 )   following relations: 2 2 2 2  1. [(,)xab yab (,)][ Xab (,) Yab (,)] is a perfect yp 4  (    )   square. zp 42  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  824   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[4T x (,)    y (,)    T (,)]   is a 5. xaa(2,) yaa (2,)  Taa (2,)  12 CP 4,2 a,6 nasty number 63T3,aa SO 4. 22 [(,)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 x2 T p UVT (7) 2 y2 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 t2 , V     , U     ,     0 YT ()  2 2 2 2 (11) 2 2 2 U2 , 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 nn22  (u v )( u  v )  1  T Ucos ( a  b )  2 ab sin  22 (18) (24)  Writing (24) as a set of double equations in two different ways nn22 t2 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 u4 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  x4 k  6 k  3 k  1  p w p  a  b  nn y4 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 nn2 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 X8 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 ) Tk21 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

Published By: Retrieval Number: F0855053615/2015©BEIESP Blue Eyes Intelligence Engineering 14 & Sciences Publication On the Non-Homogeneous Quintic Equation with Seven Unknowns 2 2 2 2 2 2 3 xyx()()() y  zwz  w  X  YT

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 v3 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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