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Observations on the Non- Homogeneous Sextic ISSN: 2319-8753 International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 5, May 2013 OBSERVATIONS ON THE NON- HOMOGENEOUS SEXTIC EQUATION WITH FOUR UNKNOWNS 3 3 2 5 x y 2(k 3)z w S.Vidhyalakshmi1 ,M.A.Gopalan 2 , A.Kavitha 3 Professor, PG and Research Department of Mathematics,Shrimati Indira Gandhi college, Trichy-620 002, Tamilnadu, 1 India Professor, PG and Research Department of Mathematics,Shrimati Indira Gandhi college, Trichy-620 002, Tamilnadu, 2 India Lecturer, PG and Research Department of Mathematics,Shrimati Indira Gandhi college, Trichy-620 002, Tamilnadu, 3 India Abstract: The sextic non-homogeneous equation with four unknowns represented by the Diophantine equation is analyzed for its patterns of non-zero distinct integral solutions are illustrated. Various interesting relations between the solutions and special numbers, namely polygonal numbers, Pyramidal numbers, Jacobsthal numbers, Jacobsthal-Lucas number, Pronic numbers, Star numbers are exhibited. Keywords: Integral solutions, sextic non-homogeneous equation. M.sc 2000 mathematics subject classification: 11D41 NOTATIONS tm,n : Polygonal number of rank n with size m CPn,m : Centred polygonal number of rank n Kyn : Kynea number of rank n m CPn : Centred pyramidal number of rank n F4,n,6 : Four dimensional figurative number of rank n whose generating polygon is hexagon F4,n,7 : Four dimensional figurative number of rank n whose generating polygon is heptagon S n : Star number of rank n Prn : Pronic number of rank n jn : Jacobsthal lucas number of rank n J n : Jacobsthal number of rank n I. INTRODUCTION The theory of Diophantine equations offers a rich variety of fascinating problems [1-4]. Particularly, in [5-6], sextic equations with 3 unknowns are studied for their integral solutions. [7-9] analyze sextic equations with 4 unknowns for their non-zero integer solutions. This communication analyses another sextic equation with 4 unknowns given by x3 y3 2(k 2 3)z 5w Copyright to IJIRSET www.ijirset.com 1301 ISSN: 2319-8753 International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 5, May 2013 . Various interesting properties among theProperties among the values of x,y,z,w are presented. II. METHOD OF ANALYSIS The equation under consideration is (1) where k is a given non-zero integers 3 3 2 5 x y 2(k 3)z w A.Case:1 Introduction of the transformations 2s2 x u v, y u v, w 2 u (2) In (1) leads to 2 2 2 5 2s2 u 3v (k 3)z 2 (3) 2 2 Let z a 3b (4) Using (4) in (3) and applying the method of factorization, define 5 s s u i 3v (k i 3)(a i 3b) (2 i2 3) Equating real and imaginary parts, we get s s 5 3 2 4 s s 4 2 3 5 u (2 k 2 3)(a 30a b 45ab ) 3(2 k 2 )(5a b 30a b 9b ) (5) s s 5 3 2 4 s s 4 2 3 5 v (2 k 2 )(a 30a b 45ab ) (2 k 2 3)(5a b 30a b 9b ) (6) Using (5) and (6) in (2), we have x(a,b) 2s1(k 1)(a5 30a3b2 45ab4 ) 2s1(k 3)(5a4b 30a2b3 9b5 ) s2 5 3 2 4 s2 4 2 3 5 y(a,b) 2 (a 30a b 45ab ) k2 (5a b 30a b 9b ) (7) w(a,b) 23s2{(k 3)(a5 30a3b2 45ab4 ) 3(k 1)(5a4b 30a2b3 9b5 )} Thus (7) and (4) represent the non-trivial integral solutions of (1). B.Properties: s (1)x(2 ,1) (k 1)[j6s1 90J4s1 45 j2s1 76] (k 3)[5( j (1)5s1) 30(3J (1)3s1) 9( j (1)s1)] 5s1 3s1 s1 3s2 26 (2)w(1,n) 2 {(k 3)[5(24F4,n,7 24F4,n,6 6CPn 18Prn 2t4,n ] 1 3(k 1)[t (6CP8 3CP10) 6CP27 16Pr 16t ] 4,n n n n n 4,n Copyright to IJIRSET www.ijirset.com 1302 ISSN: 2319-8753 International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 5, May 2013 n n (3)y(2 ,2 ) 0(mod2) For illustration and clear understanding, substituting s=1 in (7), the corresponding non-zero distinct integral solutions to (1) are given by 5 3 2 4 4 2 3 5 x(a,b) (2k 2)(a 30a b 45ab ) (2k 6)(5a b 30a b 9b ) 5 3 2 4 4 2 3 5 y(a,b) 4(a 30a b 45ab ) 4k(5a b 30a b 9b ) 2 2 z(a,b) a 3b 5 3 2 4 4 2 3 5 w(a,b) 4[(k 3)(a 30a b 45ab ) (3k 3)(5a b 30a b 9b ) C.Properties: n n n1 (1)x(1,2 ) y(1,2 ) (2k 6)[1 30(ky2n jn1 (1) 1) 45(3J4,n 1)] (6k 6)[2n (5 30( j (1)2n ) 9(ky 3J )] 2n 2n n n n (2)x(1,2 ) y(1,2 ) (2k 6)[30(ky2n jn1) 135J4n 16] (6k 6)[2n (30 j 35) 9(ky 3J )] 2n 2n n 6 28 4 (3)w(n,1) 4(k 3)[(t4,n )(CPn ) CPn 3CPn 48t3,n 24t4,n ] 4(3k 3)[24F 6CP11 2CP6 8t 33t 9] 4,n,7 n n 3,n 4,n 16 (4)x(1,n) y(1,n) w(1,n) 6[(k 3)[5t4,n (t20,n ) 15CPn 5t14,n 1] (3k 3)[t (3CP16 2CP9 ) 6CP24 26t 13t ] 4,n n n n 3,n 4,n D.Case:2 Consider the transformations x u v , y u v , w u (8) Using (8) in (1), we have 2 2 2 5 u 3v (k 3)z (9) Employing the factorization method, define 5 u i 3v (k i 3)(a i 3b) Equating real and imaginary parts, one has u k(a5 30a3b2 45ab4) 3(5a4b 30a2b3 9b5) 5 3 2 4 4 2 3 5 v (a 30a b 45ab ) k(5a b 30a b 9b ) Thus, the non-zero distinct integral solutions to (1) are given by x(a,b) (k 1)(a5 30a3b2 45ab4) (k 3)(5a4b 30a2b3 9b5) y(a,b) (k 1)(a5 30a3b2 45ab4) (k 3)(5a4b 30a2b3 9b5) 2 2 z(a,b) a 3b 5 3 2 4 4 2 3 5 w(a,b) k(a 30a b 45ab ) 3(5a b 30a b 9b ) E.Properties: Copyright to IJIRSET www.ijirset.com 1303 ISSN: 2319-8753 International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 5, May 2013 3 28 9 (1)x(n,1) (k 1)[t4,n (2CPn ) 6CPn 2CPn 22t3,n 22t4,n ] (k 3)[t (2t ) 6CP8 2t 31t 9] 4,n 12,n n 3,n 4,n 3 28 9 (2)x(n,1) y(n,1) 2[(2t4,n )(CPn ) 6CPn 2CPn 44t3,n 22t4,n ] 2k[(10t )(t ) 5CP6 30t 9] 4,n 3,n n 4,n (3)y(1,n) w(1,n) [t4,n (2t22,n 38t3,n 5t4,n ) 30t4,n 1] k[t (6CP11 CP12) 26CP6 10t 5t ] 4,n n n n 3,n 4,n 12 3 28 (4)x(n,1) y(n,1) w(n,1) 3k[t4,n (CPn 2CPn ) 6CPn 46t3,n 23t4,n ] 9[(t )(t ) 6CP8 2t 31t 9] 4,n 12,n n 3,n 4,n 28 17 (5)x(1,n) (k 1)[t4,n (Sn ) 6CPn 6CPn 2t23,n 2t12,n 12t3,n 6t4,n 1] (k 3)[(6t )(CP9 ) 27CP6 10t 5t ] 4,n n n 3,n 4,n F.Case:3 Write (9) as 2 2 2 5 u 3v (k 3)z *1 (10) Write 1 as (1 i 3)(1 i 3) 1 4 (11) Substituting (4) and (11) in (10) and employing the factorization method, define (1 i 3) 5 u i 3v (k i 3)(a i 3b) 2 Equating real and imaginary parts, we get 1 u [(k 3)(a5 30a3b2 45ab4 ) 3(k 1)(5a4b 30a2b3 9b5)] 2 1 v [(k 1)(a5 30a3b2 45ab4 ) (k 3)(5a4b 30a2b3 9b5)] 2 Thus, taking a=2A, b=2B the non zero distinct integral solutions to (1) are given by x(A, B) 25[(k 1)(A5 30A3B2 45AB4) (K 3)(5A4B 30A2B3 9B5)] 6 5 3 2 4 4 2 3 5 y(A, B) 2 [(A 30A B 45AB ) K(5A B 30A B 9B )] z(A, B) 22(A2 3B2) 4 5 3 2 4 4 2 3 5 w(A, B) 2 [(k 3)(A 30A B 45AB ) 3(K 1)(5A B 30A B 9B )] Copyright to IJIRSET www.ijirset.com 1304 ISSN: 2319-8753 International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 5, May 2013 It is to be noted that one may get integral solutions to (1) when k takes odd values. G.Properties: 5 9 (1)x(1,n) 2 {(k 1)[45(6F4,n,6 2CPn 3t4,n ) 30t4,n 1] (k 3)[t (3CP16 2CP3 52t 26t ) 10t 5t ]} 4,n n n 3,n 4,n 3,n 4,n 6 3 28 9 (2)y(n,1) 2 {[t4,n (2CPn ) 6CPn 2CPn 44t3,n 22t4,n ] k[30F 10CP9 10t 35t 9]} 4,n,6 n 3,n 4,n 4 6 (3)w(1,n) 2 {(k 3)(15t4,n (t8,n ) 30CPn 30t4,n 1) 3(k 1)[6t (CP9 ) 6CP27 32t 16t ]} 4,n n n 3,n 4,n 4 13 3 (4)x(1,n) y(1,n) w(1,n) 2 {(3k 9)[9(F4,n,7 6Cpn 2CPn 8t3,n ) 57t4,n 1] 6 23 (9k 9)[(CPn )(t20,n ) (t4,n )(t18,n ) 6CPn 24t3,n 12t4,n ]} H.Case:4 Instead of (11) we write 1 as (1 i4 3)(1 i4 3) 1 49 Following the procedure as presented in pattern 4 the corresponding non- zero distinct integral solutions to (1) are obtained as x(A, B) 74[(5k 11)(A5 30A2B2 45AB4) (11k 15)(5A4B 30A2B3 9B5)] y(A, B) 74[(3k 13)(A5 30A2B2 45AB4) (13k 9)(5A4B 30A2B3 9B5)] z(A, B) 72(A2 3B2) w(A, B) 74[(k 12)(A5 30A2B2 45AB4) 3(4k 1)(5A4B 30A2B3 9B5)] I.Properties: 4 3 28 8 (1)x(n,1) 7 [(5k 11)(t4,n (6CPn ) 6CPn 6CPn 44t3,n 22t4,n ) (11k 15)[5(12F 6CP8 CP 18t 9t ) 14] 4,n,4 n 24,n 3,n 4,n 4 10 (2)y(1,n) 7 {(3k 13)[15t4,n (CP6,n ) 15(3CPn 4t3,n t4,n ) 1] (13k 9)[6t (CP9 ) 6CP2 32t 16t ]} 4,n n n 3,n 4,n n n (3)z(2 ,2 ) is a perfect square.
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