[Vidhyalakshmi , 2(5): May, 2013] ISSN: 2277-9655
IJESRT INTERNATIONAL JOURNAL OF ENGINEERING SCIENCES & RESEARCH TECHNOLOGY On the Non-Homogeneous Equation of the Eighth Degree with Six Unknowns x5-y5+(x 3-y3)xy = p(z 2-w2)2T3 S.Vidhyalakshmi *1 , K.Lakshmi 2, M.A.Gopalan 3 *1,2,3 Department of Mathematics,Shrimati Indira Gandhi College,Trichy, India [email protected] Abstract We obtain infinitely many non-zero integer sextuples satisfying the non-homogeneous (,,,x y zw , pT ,) 55 33 2223 equation of degree eight with six unknowns given by −+− = − Various x y( x yxypz ) ( w ). T interesting relations between the solutions and special numbers, namely, polygonal numbers, Pyramidal numbers, Star numbers, Stella Octangular numbers, Octahedral numbers, Pronic number, Jacobsthal number, Jacobsthal-
Lucas number, keynea number, Centered pyramidal numbers are exhibited.
Keywords : Replica Non-homogeneous equation, integral solutions, polygonal numbers, Pyramidal numbers, Centered pyramidal numbers.
MSC 2000 Mathematics subject classification: 11D41.
NOTATIONS: Tm, n -Polygonal number of rank n with size m m Pn - Pyramidal number of rank n with size m SO n -Stella octangular number of rank n Sn -Star number of rank n PR n - Pronic number of rank n OH n - Octahedral number of rank n Jn -Jacobsthal number of rank of n jn - Jacobsthal-Lucas number of rank n
KY n -keynea number of rank n CP n,3 - Centered Triangular pyramidal number of rank n CP n,6 - Centered hexagonal pyramidal number of rank n CP n,7 - Centered heptagonal pyramidal number of rank n
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 [6-8] 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
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[Vidhyalakshmi , 2(5): May, 2013] ISSN: 2277-9655
55 33 2223 with six unknowns given by x−+− y( x yxypz ) = ( − w ) T . A few relations between the solutions and the special numbers are presented.
Method of Analysis The Diophantine equation representing the non- homogeneous equation of degree eight is given by
55 33 2223 x−+− y( x yxypz ) = ( − w ) T (1) Introduction of the transformations xuvyuvzu=+, =− , =+ 1, wu =− 1, pv = , v >1 (2) in (1) leads to u2+ v 2 = T 3 (3) The above equation (3) is solved through different approaches and thus, one obtains different sets of solutions to (1) Approach1: Let T= a2 + b 2 (4) Substituting (4) in (3) and using the method of factorisation, define (u+ iv )( = a + ib ) 3 (5) Equating real and imaginary parts in (5) we get u= a3 − 3 ab 2 (6) v=3 ab2 − b 3 In view of (2), (4) and (6), the corresponding values of x,,, y zw , pT , are represented by xa=+33 ab 2 − 3 ab 23 − b ya=−33 ab 2 − 3 ab 23 + b z= a3 −3 ab 2 + 1 (7) w= a3 −3 ab 2 − 1 p=3 ab2 − b 3 T= a2 + b 2 The above values of xyzwp, , , , and T satisfies the following properties: + + −++4 −−+ 5 = 1. xa(,1) ya (,1) Ta (,1)6 pa 6 T3, a 12 T 4, a 8 T 5, aaa 2 pCP ,6 1 + + −+−−4 + = 2. za(,1) wa (,1) Ta (,1)6 pSaa 6 T3, a 19 T 4, a 16 PR a 0 3. The following are nasty numbers: 22nn+ 22 nn + 22 nn + 22 nn + 22 nn + a) x(2 ,2 ) y (2 ,2 ) z (2 ,2 ) w (2 ,2 ) p (2 ,2 ) j 6n 22nn+ 22 nn −−+ b) 30((2,2)p T (2,2)3 J61n+ 2 KYj 2 n 4 2 n ) 2 4. 9(z−− w zw + xy + p ) is a cubic integer. 22 2422 5. 8 [2zw++ ( xyzw )( + )(1 − p ) + 2 p − 2 xy ] is a biquadratic integer 2 2 6. (x− y )( zw +− ) 8 pzw ( += 1) 0 2 7. zwz(+=+ w ) ( x yxy )( +− p 1)
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[Vidhyalakshmi , 2(5): May, 2013] ISSN: 2277-9655
+++ + ++−+−3 5 = 8. xaa(,) yaa (,) zaa (,) waa (,) paa (,) Taa (,)36 pa 40 T3, a 16 p a 8 CP a ,6 0 +− + + −−+ −+5 = 9. yaa(,) zaa (,) waa (,) paa (,) Taa (,) SOa 4 T3, a 6( OH aa )2 p 6 CP a ,6 0(mod2) 2 2 2 10. x+− y2 zw − 2 p = 2 Approach2: Now, rewrite (3) as, u2+ v 2 = T 3 × 1 (8) Also 1 can be written as n n 1= ( − i )() i (9) Substituting (4) and (9) in (8) and using the method of factorisation, define, (u+ iv )( = in a + ib ) 3 (10) Equating real and imaginary parts in (10) we get π π n32 n 23 u=cos ( aab −− 3 )sin (3 abb − ) 2 2 (11) nπ n π v=cos (3 abb23 −+ )sin ( aab 32 − 3 ) 2 2 In view of (2), (4) and (11), the corresponding values of x,,, y zw , pT , are represented π π n3223 n 3223 x=cos( a −+−+ 3 ab 3 abb )sin( a −−+ 3 ab 3 abb ) 2 2 nπ n π y=cos( a3223 −−+− 3 ab 3 abb )sin( a 3223 −+− 3 ab 3 abb ) 2 2 nπ n π z=cos ( aab32 −− 3 )sin (3 abb 23 −+ )1 2 2 (12) nπ n π w=cos ( aab32 −− 3 )sin (3 abb 23 −− )1 2 2 π π n23 n 32 p=cos (3 abb −+ )sin ( aab − 3 ) 2 2 T= a2 + b 2 Approach3: 1 can also be written as ((m22−+ n ) imnm 2 )(( 22 −− n ) imn 2 ) 1 = (13) (m2+ n 2 ) 2 Following the same procedure as above we get the integral solution of (1) as x=+( m222223 n )[( m − na )(3 −+−+ ab 223 3 abb )2(3 mna 3 −−+ ab 223 3 abb )] y=+( m222223 n )[( m − na )(3 −−+− ab 223 3 abb )2(3 mna 3 −+− ab 223 3 abb )] z=+( m222223 n ) [( m −−− na )( 3 ab 2 ) 2 mnabb (3 23 −+ )] 1 (14) w=+( m222223 n ) [( m − na )( −− 3 ab 2 ) 2 mnabb (3 23 −− )] 1 =+2222223 − −+ 3 − 2 p( m n ) [( m n )(3 abb ) 2 mna ( 3 ab )] T=( m2 + n 222 )( ab + 2 )
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Approach4: Writing 1 as 2( mn + i(m2 − n2))( 2mn − i(m2 − n2)) 1 = (m2 + n )22 Following the same procedure as above we get the integral solution of (1) as x=+( m222 n )[2(3 mna 3 −+−+− ab 223 3 abb )( m 223 na )(3 −−+ ab 223 3 abb )] y=+( m222 n )[2(3 mna 3 −−+−− ab 223 3 abb )( m 223 na )(3 −+− ab 223 3 abb )] z=+( m222 n )[2( mna 3 −−− 3 ab 2 )( m 2223 n )(3 abb −+ )]1 (15) w=+( m222 n )[2( mna 3 −−− 3 ab 2 )( m 2223 n )(3 abb −− )]1 =+222 23 −+− 223 − 2 p( m n )[2(3 mnabb )( m na )( 3 ab )] T=( m2 + n 222 )( ab + 2 ) Approach5: The solution of (3) can also be obtained as 22 22 2 umm=( + n ), vnm = ( + n ), T =+ ( mn ) (16) In view of (16) and (2), the integral solutions of (1) is obtained as x=( m2 + n 2 )( mn + ) y=( m2 + n 2 )( mn − ) z= mm(2 + n 2 ) + 1 (17) w= mm(2 + n 2 ) − 1 p= nm(2 + n 2 ) T=( m2 + n 2 ) The above values of xyzwp, , , , and T satisfies the following properties: + + − −= 1. za(,1) wa (,1) pa (,1)4 CPa,3 T 4, a 1 22nn+ 22 nn + 22 nn − 22 nn −− 2. 6((2,2)p T (2,2) z (2,2) w (2,2) jj6n+ 1 4 n + 1 ) is a nasty number. + + −+−+5 = 3. xa(,1) ya (,1) pa (,1)4 Pa 2 T3, a 5 T 4, a 2 T 7, a 1 2121nn+++ 2121 nn ++ −−+ 4. 4((2p ,2 )(2 T ,2 )2 JKYj63n+ 2 21 n + 6 21 n + ) is a cubic integer. 5. The following are biquadratic integers: a) 2+ 2 − + + α 3 = 2 + 2 = 2 − 2 = > > 8((pa ,1) Ta ( ,1)48 F4,a ,3 24 CP a ,3 22 T 4, a ) r s , u r s v, 2rs, r s 0 + +− + −+−+4 b) 8 [(,)xaa yaa (,) zaa (,) waa (,) paa (,)18 pa 9 GN a 12 T3, a 6 T 4, a ] Approach6: 2 Assuming T = α (18) in (3), we have u2+ v 2 = (α 32 ) which is in the form of Pythagorean equation, whose solution is,
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322 22 α =+r s, u = 2 rsvr , =− s , r > s > 0 (Or) (19) 322 22 α =+r sur, =− s , v = 2 rs r > s > 0 (20) Solving the first equation of (19) we have two choices of solutions, namely, rmm=(22 + nsnm ), = ( 22 + n ),α =+ m 22 nmn , >> 0 (21) rm=−33,3 mns 2 = mnn 23 − ,α =+ m 22 nmn , >> 0 (22) In view of (18), (19), and (21) and (2), we get the integral solution of (1) as x=+( m222 n )(2 mnm +− ( 22 n )) y=+( m222 n )(2 mnm −− ( 22 n )) z=2( mn m2 + n 2 )1 2 + (23) w=2( mn m2 + n 2 )1 2 − =2 + 22 2 − 2 p( m n )( m n ) T=( m2 + n 2 ) 2 In view of (18), (19), (22) and (2), we get a different integral solution of (1) as x=−2( m3 3)(3 mn 2 mnn 2 −+−+ 3 ) mn 66 15 mnn 2222 ( − m ) y=−2( m3 3)(3 mn 2 mnn 2 −−+− 3 ) m 66 n 15 mnnm 2222 ( − ) z=−2( m3 3 mn 2 )(3 mnn 2 −+ 3 ) 1 (24) w=−2( m3 3 mn 2 )(3 mnn 2 −− 3 ) 1 =−+6 6 222 − 2 pm n15 mnn ( m ) T=( m2 + n 2 ) 2 Similarly taking (20), instead of (19) and performing the same procedure we will get two more patterns. Approach7: Assuming u =UT , v =VT (25) 2 2 in (3),we get, U + V = T (26) n 2 2 Assume in (26), T= w , w= a + b (27) And Write (26) as 2 2 n U +V = w ×1 (28) Also Write 1 as m m 1= ( − i ) () i (29) Substituting (27) and (29) in (28) and using the method of factorisation define m n (U + iV ) = i (a + ib ) (30) π i( m+ n θ ) n 2 2 -1 b =re2 , wherer =+= a b , θ tan a Equating the real and imaginary parts, we have, n π Ur=cos( m + n θ ) 2 (31) π Vr=n sin( m + n θ ) 2 http: // www.ijesrt.com (C) International Journal of Engineering Sciences & Research Technology [1218-1223]
[Vidhyalakshmi , 2(5): May, 2013] ISSN: 2277-9655
In view of (31), (25) and (2), we get π π n m m 2 2 n xr=[cos( ++ nθ ) sin( + nab θ )]( + ) 2 2 mπ m π yr=n[cos( +− nθ ) sin( + nab θ )](2 + 2 ) n 2 2 mπ zr=n[cos( + nabθ )(2 ++ 2 ) n ] 1 2 (32) mπ wr=n[cos( + nabθ )(2 +− 2 ) n ] 1 2 π nm 2 2 n pr=sin( + nabθ )]( + ) 2 T=( a2 + b 2 ) n
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,S.Vidhyalakshmi and K.Lakshmi, On the non-homogeneous sextic equation 4 2 2244 x+2( x + wxy ) += y z ,IJAMA,4(2),171-173,Nov.2012 [5] M.A.Gopalan,S.Vidhyalakshmi and K.Lakshmi, Integral Solutions of the sextic equation with five 3333 5 unknowns x+ y =+ z w +3( xyT + ) , IJESRT,502-504, Dec.2012 [6] M.A.Gopalan and sangeetha.G, parametric integral solutions of the heptic equation with 5unknowns 44 33 225 xy−+2( xyxy + )( −= )2( XYz − ) ,Bessel Journal of Mathematics 1(1), 17-22, 2011. [7] M.A.Gopalan and sangeetha.G, On the heptic diophantine equations with 5 unknowns 44 2 2 5 x− y =( X − Yz ) ,Antarctica Journal of Mathematics, 9(5) 371-375, 2012 [8] Manjusomnath, G.sangeetha and M.A.Gopalan, On the non-homogeneous heptic equations with 3 3p 5 7 unknowns x+(2 − 1) y = z ,Diophantine journal of Mathematics, 1(2), 117-121, 2012
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