Research Paper
Mathematics
E-ISSN No : 2454-9916 | Volume : 2 | Issue : 11 | Nov 2016
INTEGRAL SOLUTION OF THE TERNARY CUBIC EQUATION đ?&#x;“(đ?’™đ?&#x;? + đ?’™đ?’šđ?&#x;? ) − đ?&#x;—đ?’™đ?’š + đ?’™ + đ?’š + đ?&#x;? = đ?&#x;?đ?&#x;–đ?’›đ?&#x;‘
R. Anbuselvi 1 S. A. Shanmugavadivu 2 1 2
Associate Professor of Mathematics, ADM College for women (Autonomous), Nagapattinam, Tamilnadu, India. Assistant Professor of Mathematics, Thiru.Vi.Ka. Govt. Arts College, Thiruvarur, Tamilnadu, India.
ABSTRACT The Ternary cubic equation 5(đ?‘Ľ 2 + đ?‘Ś 2 ) − 9đ?‘Ľđ?‘Ś + đ?‘Ľ + đ?‘Ś + 1 = 28đ?‘§ 3 is considered for determining its non-zero distinct integral solutions. Employing the linear transformation đ?‘Ľ = đ?‘˘ + đ?‘Ł, đ?‘Ś = đ?‘˘ − đ?‘Ł, and employing the method of factorization in complex conjugates, different patterns of integral solution to the ternary cubic equation under consideration are obtained in each pattern, interesting relations among the solutions and some special polygonal numbers like pyramidal, central pyramidal numbers are exhibited. KEY WORDS: Ternary Cubic, Integral Solutions, Polygonal Number, Pyramidal number. 1. INTRODUCTION Diophantine equations are numerously rich because of its variety. The determination of integral solutions for Cubic (homogeneous or nonhomogeneous) diophantineequations with three variables has been an interest to mathematicians. Since antiquity as can be seen from [1-3]. In this context one may refer [4-24]. In this communication, the non-homogeneous ternary cubic diophantine equation represented by 5(đ?‘Ľ 2 + đ?‘Ś 2 ) − 9xy + x + y + 1 = 28đ?‘§ 3 is considered for its non-zero distinct integral solution. A few interesting relations between special polygonal numbers and Pyramidal numbers are exhibited. NOTATIONS USED 1. đ?‘ƒđ?‘&#x;đ?‘› 2. đ?‘†đ?‘› 3. đ?‘†đ?‘œđ?‘› 4. đ?‘‚đ??ťđ?‘› 5. đ??şđ?‘›đ?‘œđ?‘› 6. đ?‘‚đ?‘?đ?‘™đ?‘› 7. đ?‘ƒđ?‘ƒđ?‘› 8. đ??śđ?‘ƒđ?‘›6 9. đ??śđ?‘ƒđ?‘›3 10. đ?‘‚đ?‘› 11. 4đ??ˇđ??šđ?‘› 12. đ?‘‡đ?‘š,đ?‘› -
Pronic number of rank n Star number of rank n Stella octangularnumber of rank n Octangular number of rank n Gnomic number of rank n Oblong number of rank n Pentagonal Pyramidal number of rank n Centred hexagonal pyramidal number Centred hexagonal pyramidal number Octahedral number of rank n Four dimensional figurate number Polygonal number of rank n with size m
2. Methods of Analysis
Equating real and imaginary parts of (6) on both sides, we get đ?‘˘ + 1 = 3đ?‘Ž3 − 57đ?‘Ž2 đ?‘? − 171đ?‘Žđ?‘?2 + 361đ?‘?3 đ?‘Ł = đ?‘Ž3 + 9đ?‘Ž2 đ?‘? − 57đ?‘Žđ?‘?2 − 57đ?‘?3 Substituting the values of đ?‘˘, đ?‘Ł in (2), the non-zero distinct integralsolutions to (1) are given by đ?‘Ľ = 4đ?‘Ž3 − 48đ?‘Ž2 đ?‘? − 228đ?‘Žđ?‘?2 + 304đ?‘?3 − 1 đ?‘Ś = 2đ?‘Ž3 − 66đ?‘Ž2 đ?‘? − 114đ?‘Žđ?‘?2 + 418đ?‘?3 − 1 đ?‘§ = đ?‘Ž2 − 19đ?‘?2 PROPERTIES 1. đ?‘Ľ(đ?‘Ž, 1) − đ?‘Ś(đ?‘Ž, 1) − đ?‘†đ?‘œđ?‘› − 2đ?‘Ą4,3đ?‘Ž ≥ 114 (đ?‘šđ?‘œđ?‘‘ 113) 2. đ?‘Ľ(đ?‘Ž, 1) − 4đ??śđ?‘ƒ6,đ?‘Ž + 48đ?‘Ą4,đ?‘Ž − 303 ≥ 0 (đ?‘šđ?‘œđ?‘‘ 228) 3. đ?‘Ś (đ?‘Ž, 1) − đ?‘Ľ (đ?‘Ž, 1) + 4đ??śđ?‘ƒđ?‘Ž3 + 36đ?‘Ą2,đ?‘Ž ≥ 114 (đ?‘šđ?‘œđ?‘‘ 134) 4. đ?‘Ľ (đ?‘Ž, 1) + đ?‘§ (đ?‘Ž, 1) − 4đ??śđ?‘˘đ?‘?đ?‘› − đ?‘†đ?‘› + 53đ?‘Ą4,đ?‘Ž ≥ 93 (đ?‘šđ?‘œđ?‘‘ 16) 5. đ?‘§ (đ?‘Ž, đ?‘Ž(đ?‘Ž + 1)) − 228đ??šđ?‘ đ?‘›4 + 38đ??śđ?‘ƒđ?‘›6 – 39 đ?‘Ą4,đ?‘Ž = 0 6. đ?‘Ľ (đ?‘Ž, đ?‘Ž + 1) − 16 đ?‘†đ?‘œđ?‘› − 408đ?‘ƒđ?‘&#x;đ?‘Ž ≥ 303 (đ?‘šđ?‘œđ?‘‘ 292) 7. đ?‘§ (đ?‘Ž, đ?‘Ž + 1) − 5đ?‘Ą4,2đ?‘Ž ≥ 19 (đ?‘šđ?‘œđ?‘‘ 38) Note: Re-writing (5) as 28 = (−3 + đ?‘–√19)(−3 − đ?‘–√19)
The Ternary Cubic Diophantine equation under consideration is 5(đ?‘Ľ 2 + đ?‘Ś 2 ) − 9đ?‘Ľđ?‘Ś + đ?‘Ľ + đ?‘Ś + 1 = 28đ?‘§ 3
(1)
Introduction of the transformations x = u + v, y = u − v In (1) leads to
(2)
(� + 1)2 + 19� 2 = 28� 3
(3)
Equation (3) is solved through different methods and thus,we obtain different patterns of Solutions to (1).
đ?‘Ľ = −2đ?‘Ž3 − 66đ?‘Ž2 đ?‘? + 114đ?‘Žđ?‘?2 + 418đ?‘?3 − 1 đ?‘Ś = −4đ?‘Ž3 − 48đ?‘Ž2 đ?‘? + 228đ?‘Žđ?‘?2 + 304đ?‘?3 − 1 đ?‘§ = đ?‘Ž2 + 19đ?‘?2 2.2 Method 2 Equation (3) can be written as (đ?‘˘ + 1)2 + 19 = 28đ?‘§ 3 ∗ 1
2.1 Method 1 Take đ?‘§ = đ?‘Ž2 + 19đ?‘?2
(4)
Write28 = (3 + đ?‘–√19)(3 − đ?‘–√19)
(5)
Using (4) and (5) in (3) and employing the method of factorization, define (đ?‘˘ + 1) + đ?‘–√19đ?‘Ł = (3 + đ?‘–√19)(đ?‘Ž + đ?‘–√19đ?‘?)3
And proceeding as in method 1, the non-zero distinct integralsolutions to (1) are given by
Write ‘1’ as 1=
(9 + đ?‘– √19)(9 − đ?‘–√19) 102
Define (6) (đ?‘˘ + 1 + đ?‘–√19 đ?‘Ł) =
(3 + đ?‘– √19)(9 + đ?‘–√19 đ?‘?)3 (9 + đ?‘– √19) (7) 10
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Research Paper
E-ISSN No : 2454-9916 | Volume : 2 | Issue : 11 | Nov 2016
Equating real and imaginary parts, we have �+1 =
đ?‘Ł=
1 10
1 10
�+1 =
[8đ?‘Ž3 + 4332đ?‘?3 − 456đ?‘Žđ?‘?2 − 684đ?‘Ž2 b] (8)
đ?‘Ł=
[12đ?‘Ž3 − 152đ?‘?3 − 684đ?‘Žđ?‘?2 + 24đ?‘Ž2 b]
(9)
Since our aim is to find integer solution, assumingđ?‘Ž = 10đ??´and đ?‘? = 10đ??ľ in (4), (8) and (9) and substituting the values of đ?‘˘, đ?‘Ł in (2), we obtain the distinct nonzero integral solutions to (1) as đ?‘Ľ = 102 [20đ??´3 + 4180đ??ľ 3 − 1140đ??´đ??ľ 2 − 660đ??´2 B] − 1 đ?‘Ś = 102 [−4đ??´3 + 4484đ??ľ 3 − 228đ??´đ??ľ 2 − 708đ??´2 B] − 1 đ?‘§ = 102 [đ??´2 + 19đ??ľ 2 ] PROPERTIES 1. đ?‘§(đ??´, đ??´(đ??´ + 1)) − 22800đ??šđ?‘ đ??´4 + 3800đ??śđ?‘ƒđ??´6 + 39đ?‘‡4,10đ??´ = 0 (đ??´, 1) + 600 đ?‘‚đ??´ + 70800 (0đ?‘?1)đ??´ ≥ 2. đ?‘Ś 448399 (đ?‘šđ?‘œđ?‘‘ 48200) 3. đ?‘§(đ??´, đ??´2 ) − 91200đ??ˇđ??šđ??´ − 20đ?‘Ą4,10đ??´ =0 4. đ?‘§ (đ??´, đ??´ + 1) + 2500đ?‘ƒđ?‘&#x;đ??´ + đ?‘‡1002,đ??´ ≥ 1900 (mod 801) 5. đ?‘Ľ (đ??´, 1)) + 2000đ??śđ?‘ƒđ?‘›6 + 13200đ?‘‡12,đ??´ ≥ 417999 (đ?‘šđ?‘œđ?‘‘ 166800) 6. đ?‘Ľ (đ??´, 1) − 5đ?‘Ś(đ??´, 1) − 2880 đ?‘‡4,10đ??´ = 1823996 2.3 Method 3 Instead of (5), 28 can be written as 28 =
[24đ?‘Ž3 − 304đ?‘?3 + 48đ?‘Ž2 b − 1368đ?‘Žđ?‘?2 ]
(12) (13)
Since our aim is to find integer solutions, assuming a=20A, b=20B in (4), (12) and (13) and substituting the values of đ?‘˘, đ?‘Ł in (2), we obtain the distinct non-zero inegral solutions to (1) as đ?‘Ľ = 202 [40đ??´3 + 8360đ??ľ 3 − 1320đ??´2 đ??ľ − 2280đ??´đ??ľ 2 ] − 1 đ?‘Ś = 202 [−8đ??´3 + 8968đ??ľ 3 − 1416đ??´2 đ??ľ + 456đ??´đ??ľ 2 ] − 1 đ?‘§ = 202 [đ??´2 + 19đ??ľ 2 ] PROPERTIES 1. đ?‘Ľ (đ??´, 1) − 40đ??´đ?‘§ (đ??´, 1) + 105600 đ?‘‡12,đ??´ ≥ 3343999 (đ?‘šđ?‘œđ?‘‘ 1321600) 2. đ?‘§ (đ??´, đ??´ + 1) + 4000 đ?‘‡6,đ??´ ≥ 7600 (đ?‘šđ?‘œđ?‘‘ 19200) 3. đ?‘Ś(đ??´, 1) − 1600 đ?‘†0 đ??´ + 1416đ?‘‡4,20đ??´ ≥ 3587199 (đ?‘šđ?‘œđ?‘‘ 180800) 4. đ?‘Ľ(đ??´, 1) − đ?‘Ś(đ??´, 1) − 28800 đ?‘‚đ??ťđ??´ − 9600 đ?‘‡10,đ??´ ≥ 243200 (đ?‘šđ?‘œđ?‘‘ 1075200) 5. đ?‘Ľ (đ??´, đ??´ + 1) − 3840000 đ?‘ƒđ?‘ƒđ??´ − 3600 đ?‘‡4,40đ??´ − 4560000 đ??şđ?‘›đ?‘œđ??´ = 37999999 Conclusion In this paper, we have made an attempt to obtain a complete set of non-trivial distinct integral solutions for the non-homogeneous ternary cubic equation. To conclude, one may search for other choices of solutions to the considered cubic equation and further cubic equations with multivariables.
[1]
Proceeding as in method 1 and performing some algebra, we obtain the distinct non-zero integral solutions to (1) as
[2] [3]
đ?‘Ľ = 22 [8đ??´3 + 608đ??ľ 3 − 96đ??´2 đ??ľ − 456đ??´đ??ľ 2 ] − 1 đ?‘Ś = 22 [4đ??´3 + 836đ??ľ 3 − 132đ??´2 đ??ľ − 228đ??´đ??ľ 2 ] − 1 đ?‘§ = 22 [đ??´2 + 19đ??ľ 2 ]
[4]
[5]
PROPERTIES 1. đ?‘Ľ (đ??´, 1) + 32 đ??śđ?‘ƒđ??´6 + 192 đ?‘‡6,đ??´ ≥ 2431 (đ?‘šđ?‘œđ?‘‘ 1632) 2. đ?‘Ľ (đ??´, đ??´ + 1) + 896 đ?‘†đ?‘‚đ??´ + 1216 đ?‘ƒđ?‘&#x;đ??´ + 768đ?‘ƒđ??´5 ≥ 2431 (đ?‘šđ?‘œđ?‘‘ 3360) 3. đ?‘Ś (đ??´, 1) − 4đ??´đ?‘§(đ??´, 1) + 33đ?‘‡4,4đ??´ ≥ 3343 (đ?‘šđ?‘œđ?‘‘ 1216) 4. đ?‘Ľ (đ??´, 1) − 2đ?‘Ś(đ??´, 1) − 672đ?‘‡4,đ??´ = 9119 5. đ?‘§ (đ??´, đ??´(đ??´ + 1)) − 912đ??šđ?‘ đ??´4 + 152đ??śđ?‘ƒđ??´6 + 39đ?‘‡4,2đ??´ =0 6. đ?‘§ (đ??´, đ??´ + 1) − 40 đ?‘‡6,đ??´ ≥ 76 (đ?‘šđ?‘œđ?‘‘ 192)
[6]
[7]
[8]
Note: Re-write (7) as
[9]
9 + đ?‘–√19 (đ?‘˘ + 1) + đ?‘–√19đ?‘Ł = (−3 + đ?‘–√19)(đ?‘Ž + đ?‘–√19đ?‘?)3 ( ) 10
[10]
and proceeding as in method 2, the non-zero distinct integralsolution to (1) are given by
[11] [12] [13]
đ?‘Ľ = 22 [−4đ??´3 + 836đ??ľ 3 − 132đ??´2 đ??ľ + 228đ??´đ??ľ 2 ] − 1 đ?‘Ś = 22 [−8đ??´3 + 608đ??ľ 3 − 96đ??´2 đ??ľ + 456đ??´đ??ľ 2 ] − 1 đ?‘§ = 22 [đ??´2 + 19đ??ľ 2 ]
[14]
2.4 Method 4 Equation (7) can be written as (6 + 2đ?‘– √19) (đ?‘Ž 2 (9 + đ?‘– √19) + đ?‘– √19đ?‘?)3 10
[16đ?‘Ž3 + 8664đ?‘?3 − 1368đ?‘Ž2 b − 912đ?‘Žđ?‘?2 ]
REFERENCES
(6 + 2đ?‘– √19)(6 − 2đ?‘–√19) (10) 22
(đ?‘˘ + 1) + đ?‘– √19đ?‘Ł =
1 20
1 20
[15]
[16]
(11)
Equating real and imaginary Parts on both sides
International Educational Scientific Research Journal [IESRJ]
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E-ISSN No : 2454-9916 | Volume : 2 | Issue : 11 | Nov 2016
S.Vidhyalakshmi, T.R.Usha Rani and M.A.Gopalan Integral solutions of nonhomogenous ternary cubit equation ax2+by2 = (a+b)z3DiophantusJ.Math, 2(1), 3138, 2013 M.A.Gopalan, K.Geetha On the ternary cubic diophantine equation x2 + y2 – xy = z3 Bessel J.Math., 3(2), 119-123, 2013 M.A.Gopalan, S.Vidhyalakshmi and A.Kavitha Observations on the ternary cubic equation x2 + y2 + xy = 12z3Antartica J.Math.10(5), 453-460, 2013 M.A.Gopalan, S.Vidhyalakshmi and K.Lakshmi Lattice points on the non homogeneous ternary cubic equation x3 + y3 + z3 (x+y+z) = 0 Impact J. Sci. Tech., Vol 7 No.1, 21-25,2013 M.A.Gopalan, S.Vidhyalakshmi and K.Lakshmi Lattice points on the non homogeneous ternary cubic equation x3 + y3 + z3 (x+y+z) = 0 Impact J. Sci. Tech., Vol 7 No.1, 51-55,2013 M.A.Gopalan, S.Vidhyalakshmi and S.Mallika on the ternary non homogeneous cubic equation x3 + y3 - 3 (x+y) = 2(3k2-2)z3 Impact J. Sci. Tech., Vol 7 No.1, 4155,2013
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