INTEGRAL SOLUTION OF THE TERNARY CUBIC EQUATION (+)−+++=

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]

[17] [18]

Dickson, L.E. “History of Theory of Numbers and Diophantine Analysis�, Vol.2, Dover Publications, New York 2005.

Mordell L.J., “Diophantine Equations� Academic Press, new York, 1970 RD. Carmicheal, “The Theory of Numbers and Diophantine Analysis�, Dover Publications, New York 1959 M.A. Gopalan, ManjuSomnath and N.Vanitha, “On ternary cubic Diophantine equation x2+y2 = 2z3 Advances in Theoretical and Applied Mathematics�, Vol. 1, no.3, 227-231, 2006.

M.A. Gopalan, ManjuSomnath and N. Vanitha, “Ternary cubic Diophantine equation x2 – y2 = z3�, ActaCienciaIndica�, Vol.XXXIIIM, No. 3, 705-707, 2007.

M.A. Gopalan and R.Anbuselvi, “Integral Solutions of ternary cubic Diophantine equation x2 + y2 + 4N = zxy�, Pure and Applied Mathematics Sciences Vol.LXVII, No. 1-2, 107-111, March 2008.

M.A. Gopalan, ManjuSomnath and N. Vanitha, “Ternary cubic Diophantine equation 22a-1 (x2+y2) = z3, actaCienciaIndia�, Vol.XXXIVM, No. 3, 1135-1137, 2008.

M.A. Gopalan, J.Kaligarani Integral solutions x3 + y3 + 8k(x+y) = 2k+1)z3 Bulletin of Pure and Applied Sciences, vol 29E, No.1, 95-99,2010 M.A.Gopalan, J.Kaligarani Integral solutions x3 + y3 + 8k(x+y) = (2k+1)z3 Bulletin of Pure and Applied Sciences, Vol. 29E, No.1, 95-99, 2010 M.A.Gopalan, S.Premalatha On the ternary cubic equationx3 + x2 + y3 – y2 = 4(z3 + z2 ) Cauvery Research Journal Vol. 4, iss 1&2 87-89, July 2010-2011.

M.A.Gopalan, V.Pandichelvi, observations on the ternary cubic diophantine equation x3 + y3 + x2 - y3= 4(z3 + z2 ) Archimedes J.Math 1(1), 31-37, 2011 M.A.Gopalan, G.Srividhya Integral solutions of ternary cubic diophantine equation x3 + y3 = z2ActaCienciaIndica, Vol XXXVII No.4, 805-808, 2011 M.A.Gopalan, A.Vijayashankar, S.Vidhyalakshmi Integral solutions of ternary cubic equation x2+y2– xy+2(x+y+z) = (k2+3)z3 Archimedes J.Math 1(1), 59-65, 2011 M.A.Gopalan, G.Sangeetha on the ternary cubic diophantine equation y2=Dx2+z3 Archimedes J.Math 1(1), 7-14, 2011 M.A.Gopalan and B.Sivakami, “Integral Solutions of the ternary cubic Diophantine equation 4x2 – 4xy + 6y2 = [(k+1)2+5] w3� Impact J.Sci. Tech., vol.

6, No.1, 15-22, 2012 M.A.Gopalan, S.Vidhyalakshmi and G.Sumathi, on the non-homogeneous equation with three unknowns “x3+y3 = 14z3+3(x+y)� Discovery Science, Vol.2, No:4,37-39. Oct.2012 M.A.Gopalan, B.Sivakami Integral solutions of the ternary cubic equation 4x2 – 4xy + 6y2 = ((k+1)2+5)w3 Impact J. Sci. Tech., Vol 6 No. 1, 15-22, 2012 M.A.Gopalan, B.Sivakami on the ternary cubic diophantine equation 2xz = y2(x+z) Bessel. J.Math 2(3), 171-177, 2012.

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Research Paper [19]

[20] [21] [22]

[23]

[24]

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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