OBSERVATIONS ON THE NON HOMOGENEOUS EQUATION OF THE EIGHTHDEGREE WITH FIVE UNKNOWNS

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1 Vol., Issue 5,May 013 ISSN: OBSERVATIONS ON THE NON HOMOGENEOUS EQUATION OF THE EIGHTHDEGREE WITH FIVE UNKNOWNS 4 4 x y ( k s )( z w ) p. Vidhyalakshmi.S 1, Lakshmi.K, Gopalan.M.A 3 Professor, Department of Mathematics, SIGC,Trichy,Tamil nadu, India 1 Lecturer, Department of Mathematics,SIGC,Trichy, Tamil nadu, India Professor, Department of Mathematics,SIGC,Trichy, Tamil nadu, India 3 ABSTRACT: We obtain infinitely many non-zero integer quintiples ( x, y, z, w, p ) satisfying the non- 4 4 homogeneous equation of degree eight with five unknowns given by x y ( k s )( z w ) p.various interesting relations between the solutions and special numbers, namely, polygonal numbers, Pyramidal numbers, Star numbers, Stella Octangular numbers, Octahedral numbers, Four Dimensional Figurative numbers, Five Dimensional Figurative numbers and six Dimensional Figurative numbers are exhibited. KEYWORDS : Non-homogeneous equation, integral solutions, -dimentional, 3-dimentional, 4- dimensional and 5- dimensional and - dimensional figurative numbers. MSC 000 Mathematics subject classification: 11D41. Tmn, ( n1)( m) n1 NOTATIONS -Polygonal number of rank n with size m m 1 Pn n ( n 1)(( m ) n 5 m ) - Pyramidal number of rank n with size m SO n(n 1) -Stella octangular number of rank n Sn n PRn n( n 1) 1 -Star number of rank n n( n 1) - Pronic number of rank n 1 OHn n(n 1) - Octahedral number of rank n 3 1 n ( 1) n Jn -Jacobsthal number of rank n 3 n n jn ( 1) - Jacobsthal-Lucas number of rank n KY ( n n 1) -keynea number. Copyright to IJIRSET

2 F, n,3 International Journal of Innovative Research in Science, Engineering and Technology Vol., Issue 5,May 013 n( n 1)( n )( n 3)( n 4)( n 5) - Six Dimensional Figurative number of rank n! whose generating polygon is a triangle. ISSN: n( n 1)( n )( n 3)( n 4) F5, n,3 - Five Dimensional Figurative number of rank n 5! whose generating polygon is a triangle. n( n 1)( n )( n 3) F4, n,3 - Four Dimensional Figurative number of rank n whose generating polygon is a 4! triangle n( n 1) ( n ) F4, n,4 - Four Dimensional Figurative number of rank n whose generating polygon is a square 1 3 n n CPn,3 - Centered Triangular pyramidal number of rank n n(7n 1) CPn,7 - Centered heptagonal pyramidal number of rank n I.INTRODUCTION The theory of diophantine equations offers a rich variety of fascinating problems. In particular, homogeneous and non-homogeneous equations of higher degree have aroused the interest of numerous mathematicians since antiquity[1-3]. Particularly in [4, 5] special equations of sixth degree with four and five unknowns are studied. In [-8] heptic equations with three and five unknowns are analysed. This paper concerns with the problem of determining non-trivial integral solution of the non- homogeneous equation of eighth degree with five unknowns given by 4 4 x y ( k s )( z w ) p. A few relations between the solutions and the special numbers are presented. II.METHOD OF ANALYSIS The Diophantine equation representing the non- homogeneous equation of degree eight under consideration is given by 4 4 x y ( k s )( z w ) p (1) Introduction of the transformations x u v, y u v, z uv 1, w uv 1 () in (1) leads to u v ( k s ) p (3) The above equation (3) is solved through different approaches and thus, one obtains different sets of solutions to (1) A. Case1: k s is not a perfect square 1)Approach1: Let p a b (4) Substituting (4) in (3) and using the method of factorisation, define ( u iv) ( k is)( a ib) (5) Copyright to IJIRSET

3 Vol., Issue 5,May 013 Equating real and imaginary parts in (5) we get ISSN: u kf ( a, b) sg( a, b) v sf ( a, b) kg( a, b) () where 4 4 f ( a, b) ( a 15a b 15 a b b ) g( a, b) (a b 0a b ab ) (7) In view of (), (4), () and (7), the corresponding values of x, y, z, w and p are represented by x ( k s) f ( a, b) ( k s) g( a, b) y ( k s) f ( a, b) ( k s) g( a, b) z { kf ( a, b) sg( a, b)}{ sf ( a, b) kg( a, b)} 1 w { kf ( a, b) sg( a, b)}{ sf ( a, b) kg( a, b)} 1 (8) The above values of x, y, z, w and p satisfy the following properties: 1. ( 5 k s )( SOa. Pa T 3, 31 T, a 93 T a 4, a T5, a ) x ( a,1) 0(mod ). The following are nasty numbers ( a)3{ x( a, a) y( a, a) z( a, a) w( a, a) 1 s(70f, a, 3 5, a,3 4, a,3 3 a 3, a 5, a 4, a 1800F 150F 540P T T 3 T )} n n n n n n. (b) sk[5 sk{ p(, ).( j 1) KY } z(, ) w(, )] n 1n ( w1) x (w1) x y zw 1 n n n n ( k s) [8( k s)( KY J ) y(, )] is a cubic integer. 5. (k +s )[4T 4 T. P 1CP 8 T )] 0(mod8) 5 4, a 3, a1 a a,7 3, a. k{4( CP ) 04F 3T 1} s{4 T. P 1CP 8T 4 T } x( a,1) y( a,1) 0 5 a,3 4, a1,4 4, a 3, a1 a a,7 3, a 4, a 5 a,3 4, a1,4 4, a 3, a1 a a,7 3, a 4, a 7. kx( a,1) sy( a,1) ( k s )[4( CP ) 04F 3T 1 4 T. P 1CP 8T 4 T ] ) Remark1: Instead of (), taking the substitution in (1) as x u v, y u v, z uv, w uv We get the solution of (1) as Copyright to IJIRSET

4 Vol., Issue 5,May 013 ISSN: x ( k s) f ( a, b) ( k s) g( a, b) y ( k s) f ( a, b) ( k s) g( a, b) z { kf ( a, b) sg( a, b)}{ sf ( a, b) kg( a, b)} w { kf ( a, b) sg( a, b)}{ sf ( a, b) kg( a, b)} (9) 3)Approach: Now, rewrite (3) as, u v ( k s ) p 1 (10) Also 1 can be written as 1 ( i)( i) (11) Substituting (4) and (11) in (10) and using the method of factorisation, define, ( u iv) i( k is)( a ib) (1) Following the same procedure as in approach1 we get the integral solution of (1) as x ( k s) f ( a, b) ( k s) g( a, b) y ( k s) f ( a, b) ( k s) g( a, b) z { sf ( a, b) kg( a, b)}{ kf ( a, b) sg( a, b)} 1 w { sf ( a, b) kg( a, b)}{ kf ( a, b) sg( a, b)} 1 (13) 4)Approach3:1 can also be written as 1 n n (1 i) (1 i) (14) n Substituting (4) and (14) in (10) and using the method of factorisation, define (1 i) ( k is)( a ib) ( u iv) n n Equating real and imaginary parts in (5) we get n n u cos { kf ( a, b) sg( a, b)} sin { sf ( a, b) kg( a, b)} n n v sin { kf ( a, b) sg( a, b)} cos { sf ( a, b) kg( a, b)} Copyright to IJIRSET 179

5 Vol., Issue 5,May 013 In view of () and (4), the corresponding values of x, y, z and w are represented by n n n n x (cos sin )( kf ( a, b) sg( a, b)) (cos sin )( sf ( a, b) kg( a, b)) n n n n y (cos sin )( kf ( a, b) sg( a, b)) (cos sin )( sf ( a, b) kg( a, b)) n n z [cos ( kf ( a, b) sg( a, b)) sin ( sf ( a, b) kg( a, b))] n n [sin ( kf ( a, b) sg( a, b)) cos ( sf ( a, b) kg( a, b))] 1 n n w [cos ( kf ( a, b) sg( a, b)) sin ( sf ( a, b) kg( a, b))] n n [sin ( kf ( a, b) sg( a, b)) cos ( sf ( a, b) kg( a, b))] 1 ISSN: (15) 5) Approach4: 1 can also be written as (( m n ) i mn)(( m n ) i mn) 1 ( m n ) (1) Following the same procedure as above we get the integral solution of (1) as x m n m n mn kf a b sg a b m n mn sf a b kg a b 5 ( ) [( ){ (, ) (, )} ( ){ (, ) (, )}] y m n m n mn kf a b sg a b m n mn sf a b kg a b 5 ( ) [( ){ (, ) (, )} ( ){ (, ) (, )}] 10 z m n m n kf a b sg a b mn sf a b kg a b ) [( ){ (, ) (, )} { (, ) (, )}] [{ mn{ kf ( a, b) sg( a, b)} ( m n ){ sf ( a, b) kg( a, b)}] 1 10 w m n m n kf a b sg a b mn sf a b kg a b ) [( ){ (, ) (, )} { (, ) (, )}] [{ mn{ kf ( a, b) sg( a, b)} ( m n ){ sf ( a, b) kg( a, b)}] 1 (17) )Approach5: Writing 1 as ( mn i( m n )( mn i( m n ) 1 ( m n ) (18) Following the same procedure as above we get the integral solution of (1) as Copyright to IJIRSET

6 Vol., Issue 5,May 013 ISSN: x m n mn m n kf a b sg a b mn m n sf a b kg a b 5 ( ) [( )( (, ) (, )) ( )( (, ) (, ))] y m n mn m n kf a b sg a b mn m n sf a b kg a b 5 ( ) [( )( (, ) (, )) ( )( (, ) (, ))] z m n mn kf a b sg a b m n sf a b kg a b 10 ( ) [ ( (, ) (, )) ( )( (, ) (, ))] [( m n ){ kf ( a, b) sg( a, b)) mn( sf ( a, b) kg( a, b))] 1 w m n mn kf a b sg a b m n sf a b kg a b 10 ( ) [ ( (, ) (, )) ( )( (, ) (, ))] [( ){ ( m n kf a, b) sg( a, b)) mn( sf ( a, b) kg( a, b))] 1 7)Approach:Rewriting (3) as u k p s p v (0) Let p. Using the method of factorisation, writing (0) as a system of double equations, solving it and using (), we get the solution of (1) as (19) 3 x ( s k) y s k 3 ( ) z sk 1 w sk 1 p (1) B. Case: k s is a perfect square 1)Approach1: Choose k and s such that k s d. () Substituting () in (3) we get u v d p (3) Assuming u p du, v p dv (4) in (3), we get u v p (5) which is in the form of Pythagorean equation, whose solution is, p u v,,, 0 () Using (4), () and (), we get the integral solution of (1) as Copyright to IJIRSET

7 Vol., Issue 5,May 013 ISSN: x d y ( ) ( ( )) d( ) ( ( )) 4 z d ( ) ( ) 1 w p 4 d ( ) ( ) 1 (7) It is to be noted that, the solutions of (5) may also be written as p, v, u, 0 Hence we get a different solution of (1) as x d y ( ) (( ) ) d( ) (( ) ) 4 z d ( ) ( ) 1 w p 4 d ( ) ( ) 1 (8) )Approach:Assuming 3 3 u p u, v p v (9) In (3) we get u v d (30) which is in the form of Pythagorean equation, whose solution is, d m n u mn v m n n,,, m 0 Performing the same procedure as above we get the integral solution of (1) as 3 x (( m n ) mn) 3 y (( m n ) mn) z mn m n 4 ( ) 1 w 4 nm( m n ) 1 p (3) It is to be noted that, the solutions of (30) may also be written as d m n u mn v m n n,,, m 0. Then we get a different solution to (1) Copyright to IJIRSET

8 Vol., Issue 5,May 013 ISSN: x ( mn( m n )) y mn m n 3 ( ( )) z 4 mn( m n ) 1 w nm m n 4 ( ) 1 p (33) 3) Approach3: Assuming u pdu, v pdv (34) in (3),we get, 4 u v p ( p ) (35) which is in the form of Pythagorean equation, whose solution is, p u v r s,,, 0 (3) Solving the first result of (3) we have mn, m n, p m n, m n 0 (or) (37) mn, m n, p m n, m n 0 (38) Using (34), (3) and (37), we get u mnd m n ( ) 4 4 v d( m n )( m n m n ) (39) Using (39) and (), we get the integral solution of (1) as x d m n mn m n m n m n 4 4 ( )(4 ( ) ( ) y d m n mn m n m n m n 4 4 ( )(4 ( ) ( ) 4 4 z 8 mnd ( m n ) ( m n )( m n m n ) 1 w mnd m n m n m n m n ( ) ( )( ) 1 ( ) p m n (40) Similarly using (34), (3), (38) and () we get a distinct solution to (1) It is to be noted that, the solutions of (35) may also be written as p, v, u, 0 Again performing the same procedure as above, we will get two more different integral solutions to (1) 4) Approach4: Also, taking in (35) and applying the method of factorisation define, ( u iv) ( m in) p m n (41) 4 (4) Copyright to IJIRSET 179

9 Vol., Issue 5,May 013 Equating real and imaginary parts in (4) we get ISSN: u m m n n v 4m n 4mn (43) Using (34), (41), (43) and (), we get the integral solution of (1) as x d( m n )( m 4m n m n 4 mn n ) y d( m n )( m 4m n m n 4 mn n ) z d ( m n ) ( m m n )(4m n 4 mn ) w d ( m n ) ( m m n )(4m n 4 mn ) 1 p m n (44) By using the same procedure as in approaches -5 we get 4 more patterns of solutions to (1) 5) Approach5: Assuming u du, v dv (45) in (3),we get, 3 u v p ( p ) (4) which is in the form of Pythagorean equation, whose solution is, 3 p, u, v, 0 (Or) (47) 3 p, v, u, 0 (48) Solving the first result of (47), we have m( m n ), n( m n ), p m n, m n 0 (Or) (49) 3 3 m 3 mn, 3 m n n, p m n, m n 0 (50) In view of (49), (47), (45) and (), we get the integral solution of (1) as x d m n mn m n ( ) ( ( )) y d m n mn m n ( ) ( ( )) 4 z 4 mnd ( m n ) ( m n ) 1 w mnd m n m n 4 4 ( ) ( ) 1 ( ) p m n (51) In view of (50), (47), (45) and (), we get a different integral solution of (1) as Copyright to IJIRSET

10 Vol., Issue 5,May 013 ISSN: x d m mn m n n m n m n n m 3 3 {( 3 )(3 ) 15 ( )} y d m mn m n n m n m n n m 3 3 {( 3 )(3 ) 15 ( )} 3 3 z d {( m 3 mn )(3 m n n )( m n 15 m n ( n m )} 1 w d m mn m n n m n m n n m 3 3 {( 3 )(3 )( 15 ( )} 1 (5) )Remark: Similarly taking (48) and performing the same procedure we will get two more patterns. 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 (195). [] L.J.Mordell, Diophantine equations, Academic Press, London(199) [3] Carmichael,R.D.,The theory of numbers and Diophantine Analysis,Dover Publications, New York (1959) [4] M.A.Gopalan,S.Vidhyalakshmi and K.Lakshmi, On the non-homogeneous sextic equation x ( x w) x y y z,ijama,4(), ,dec.01 [5] M.A.Gopalan,S.Vidhyalakshmi and K.Lakshmi, Integral Solutions of the sextic equation with five unknowns x y z w 3( x y) T, IJESRT, 1(10),50-504,Nov..01 [] M.A.Gopalan and sangeetha.g, parametric integral solutions of the heptic equation with 5unknowns x y ( x y )( x y) ( X Y ) z,bessel Journal of Mathematics 1(1), 17-, 011. [7] M.A.Gopalan and sangeetha.g, On the heptic diophantine equations with 5 unknowns x y ( X Y ) z,antarctica Journal of Mathematics, 9(5), , 01 [8] Manjusomnath, G.sangeetha and M.A.Gopalan, On the non-homogeneous heptic equations with 3 unknowns 3 p 5 7 x ( 1) y z,diophantine journal of Mathematics, 1(), , 01 Copyright to IJIRSET

Integral Solutions of the Sextic Equation with Five Unknowns

International Journal of Scientific and Research Publications, Volume 4, Issue 7, July 014 1 Integral Solutions of the Sextic Equation with Five Unknowns 6 6 4 x 6 w ( xy z) y ( y w) M.A.Gopalan *, S.Vidhyalakshmi