ON TERNARY QUADRATIC DIOPHANTINE EQUATION

Research Paper

Mathematics

E-ISSN No : 2454-9916 | Volume : 3 | Issue : 2 | Feb 2017

ON TERNARY QUADRATIC DIOPHANTINE EQUATION đ?&#x;• đ?’™đ?&#x;? + đ?’šđ?&#x;? − đ?&#x;?đ?&#x;‘đ?’™đ?’š + đ?’™ + đ?’š + đ?&#x;? = đ?&#x;‘đ?&#x;?đ?’›đ?&#x;?

R. Anbuselvi 1 | K. Kannak 2 1

Associate Professor of Mathematics, ADM College for women (Autonomous), Nagapattinam, Tamilnadu, India.

2

Lecturer of Mathematics, Valivalam Desikar Polytechnic College, Nagapattinam, Tamilnadu, India.

ABSTRACT The ternary quadratic equation representing non-homogeneous cone given by 7 đ?‘Ľ 2 + đ?‘Ś 2 − 13đ?‘Ľđ?‘Ś + đ?‘Ľ + đ?‘Ś + 1 = 31đ?‘§ 2 is analyzed for its non-zero distinct integer points on it. The different patterns of integer points satisfying the cone under consideration are obtained. A few interesting relations between the solutions and special number patterns are presented.

KEYWORDS: Ternary non-homogeneous quadratic, integral solutions.

1. INTRODUCTION: The ternary quadratic Diophantine equations offer an unlimited field for research due to their variety [1]. For an extensive review of various problems, one may refer [2-18]. The communication concerns with yet another interesting ternary quadratic equation 7 đ?‘Ľ 2 + đ?‘Ś 2 − 13đ?‘Ľđ?‘Ś + đ?‘Ľ + đ?‘Ś + 1 = 31đ?‘§ 2 representing a cone for determining its infinitely many non-zero integral points. Also, a few interesting relations among the solutions are presented.

2. NOTATIONS: đ?‘‡đ?‘š ,đ?‘› – Polygonal number of rank n with size m. đ?‘ƒđ?‘›đ?‘š - Pyramidal number of rank n with size m. đ??śđ?‘Ąđ?‘š ,đ?‘› – Centered Polygonal number of rank n with size m. đ??śđ?‘ƒđ?‘š ,đ?‘› – Centered pyramidal number of rank n with size m. đ?‘ƒđ?‘&#x;đ?‘› – Pronic number of rank n. đ?‘†đ?‘› – Star number of rank n.

3. METHOD OF ANALYSIS: The ternary quadratic Diophantine equation to be solved for its non-zero distinct integral solution is 7 đ?‘Ľ 2 + đ?‘Ś 2 − 13đ?‘Ľđ?‘Ś + đ?‘Ľ + đ?‘Ś + 1 = 31đ?‘§ 2

(1)

The substitution of linear transformation đ?‘Ľ = đ?‘˘ + đ?‘Ł, đ?‘Ś = đ?‘˘ − đ?‘Ł

(2)

in (1) leads to (đ?‘˘ + 1)2 − 27đ?‘Ł 2 = 31đ?‘§ 2

(3)

Different patterns of solutions of (1) are presented below

CopyrightŠ 2016, IERJ. This open-access article is published under the terms of the Creative Commons Attribution-NonCommercial 4.0 International License which permits Share (copy and redistribute the material in any medium or format) and Adapt (remix, transform, and build upon the material) under the Attribution-NonCommercial terms.

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Pattern - I: Write 31 as 31 = 2 + đ?‘– 27 (2 − đ?‘– 27)

(4)

Assume that đ?‘§ = đ?‘Ž2 + 27đ?‘? 2

(5)

Where a, b are non-zero distinct integers. use (4) & (5) in (3) and applying the method of factorization, define đ?‘˘ + 1 + đ?‘– 27đ?‘Ł = 2 + đ?‘– 27 (đ?‘Ž + đ?‘– 27đ?‘?)2

(6)

Equating the real and imaginary parts, we have đ?‘˘ = đ?‘˘ đ?‘Ž, đ?‘? = 2đ?‘Ž2 − 54đ?‘Žđ?‘? − 54đ?‘? 2 − 1 đ?‘Ł = đ?‘Ł đ?‘Ž, đ?‘? = đ?‘Ž2 + 4đ?‘Žđ?‘? − 27đ?‘? 2 Substituting the above values of u and v in equation (2), the value of x and y are given by đ?‘Ľ = đ?‘Ľ đ?‘Ž, đ?‘? = 3đ?‘Ž2 − 50đ?‘Žđ?‘? − 81đ?‘? 2 − 1 đ?‘Ś = đ?‘Ś đ?‘Ž, đ?‘? = đ?‘Ž2 − 58đ?‘Žđ?‘? − 27đ?‘? 2 − 1

(7)

đ?‘§ = đ?‘§ đ?‘Ž, đ?‘? = đ?‘Ž2 + 27đ?‘? 2 Thus equation (7) represent a non-zero distinct integral solutions of (1) in two parameters.

Properties: đ?‘– đ?‘Ľ đ?‘Ž, 1 − đ?‘Ś đ?‘Ž, 1 − 2đ?‘‚đ?‘?đ?‘™đ?‘› ≥ 0 đ?‘šđ?‘œđ?‘‘ 6 đ?‘–đ?‘– đ?‘§ đ?‘Ž, đ?‘Ž + 1 − 22đ?‘Ą4,đ?‘Ž − đ?‘ đ?‘› ≥ 26 (đ?‘šđ?‘œđ?‘‘ 60) đ?‘–đ?‘–đ?‘– đ?‘Ľ đ?‘Ž, đ?‘Ž + 1 + 50đ?‘?đ?‘&#x;đ?‘› + 39đ?‘ đ?‘œđ?‘› ≥ −82 (đ?‘šđ?‘œđ?‘‘ 20) đ?‘–đ?‘Ł đ?‘§ đ?‘Ž, đ?‘Ž đ?‘Ž + 1

− 324đ??šđ?‘ đ?‘›4 − 108đ?‘?đ?‘Ž5 − đ?‘Ą4,đ?‘Ž = 0

đ?‘Ł đ?‘Ś đ?‘Ž, đ?‘Ž + 1 − đ?‘Ą4,đ?‘Ž + 58đ?‘?đ?‘&#x;đ?‘› − đ?‘ đ?‘› ≥ −29 (đ?‘šđ?‘œđ?‘‘ 48) đ?‘Łđ?‘– đ?‘Ľ đ?‘Ž, 1 − 3đ?‘§ đ?‘Ž, 1 ≥ −13(đ?‘šđ?‘œđ?‘‘ 50) đ?‘Łđ?‘–đ?‘– đ?‘Ľ đ?‘Ž, đ?‘Ž + 1 − 3đ?‘§ đ?‘Ž, đ?‘Ž + 1 + 44đ?‘ƒđ?‘&#x;đ?‘› ≥ −1(đ?‘šđ?‘œđ?‘‘ 6) Pattern – II: Instead of (4) we write as 31 =

4+2đ?‘– 27 (4−2đ?‘– 27)

(8)

4

Write 1 đ?‘Žđ?‘ 1 =

3+đ?‘– 27 (3−đ?‘– 27) 36

Assume that đ?‘§ = đ?‘Ž2 + 27đ?‘? 2

(9) (10)

equation (3) can be written as

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Research Paper �+1

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+ 27đ?‘Ł 2 = 31đ?‘§ 2 ∗ 1

(11)

use (8) & (9) in (11) and applying the method of factorization define �+1

2

+ đ?‘– 27đ?‘Ł =

1 12

(4 + 2đ?‘– 27)(3 + đ?‘– 27)(đ?‘Ž + đ?‘– 27đ?‘?)2

(12)

Equating the real and imaginary part, we have đ?‘˘ = đ?‘˘ đ?‘Ž, đ?‘? = đ?‘Ł = đ?‘Ł đ?‘Ž, đ?‘? =

1 12 1 12

{−42đ?‘Ž2 + 1134đ?‘? 2 − 540đ?‘Žđ?‘? − 12

(13)

{10đ?‘Ž2 − 270đ?‘? 2 − 84đ?‘Žđ?‘?

As our interest is on finding integer solutions, we choose a and suitably so that the values of u and v are in integers.

Replace a by 12A and b by 12B in (10) and (13) we get đ?‘˘ = đ?‘˘ đ?‘Ž, đ?‘? = −504đ??´2 + 13608đ??ľ2 − 6480đ??´đ??ľ − 1 đ?‘Ł = đ?‘Ł đ?‘Ž, đ?‘? = 120đ??´2 − 3240đ??ľ2 − 1008đ??´đ??ľ

(14)

đ?‘§ = 144đ??´2 − 3888đ??ľ2 Substituting the values of u and v in equation (2), the value of x and y are given by đ?‘Ľ = đ?‘Ľ đ??´, đ??ľ = −384đ??´2 + 10368đ??ľ2 − 7488đ??´đ??ľ − 1 đ?‘Ś = đ?‘Ś đ??´, đ??ľ = −624đ??´2 + 16848đ??ľ2 − 5472đ??´đ??ľ − 1

(15)

đ?‘§ = đ?‘§ đ??´, đ??ľ = 144đ??´2 − 3888đ??ľ2 Thus the equation (15) represents a non-zero distinct integral solutions of (1) in two parameters Properties: đ?‘– đ?‘Ľ đ??´, 1 − đ?‘‡10,đ??´ − đ??şđ?‘›đ?‘œđ?‘› + 388đ?‘?đ?‘Ą2,đ??´ ≥ 3268 đ?‘šđ?‘œđ?‘‘ 7099 đ?‘–đ?‘– đ?‘§ đ??´, đ??´ + 1 − 2016đ?‘‡6,đ??´ ≥ 3888 đ?‘šđ?‘œđ?‘‘ 9792 đ?‘–đ?‘–đ?‘– đ?‘Ś đ??´, 1 − đ?‘‡12,đ??´ − đ?‘‡14,đ??´ + đ?‘‡6,đ??´ − đ?‘‡4,đ??´ ≥ 16847 đ?‘šđ?‘œđ?‘‘ 5164 đ?‘–đ?‘Ł đ?‘Ľ đ??´, đ??´ + 1 − 9990đ?‘‡4,đ??´ − 10371đ??şđ?‘›đ?‘œđ?‘› − đ?‘†đ?‘› + 7488đ?‘ƒđ?‘&#x;đ?‘› = 20737 đ?‘Ł đ?‘§ đ??´, 1 − 144đ?‘‡4,đ??´ = 3888 Pattern – III: Equation (3) can be written as đ?‘˘+1

2

= 31đ?‘§ 2 − 27đ?‘Ł 2

(16)

Introducing the linear transformation � = � + 27� ��� � = � + 31�

(17)

in (16) we get

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Research Paper � 2 = 837� 2 +

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�+1 2

(18)

2

which is satisfied by đ?‘‡ đ?‘?, đ?‘ž = 2đ?‘?đ?‘ž đ?‘˘ đ?‘?, đ?‘ž = 1674đ?‘ƒ2 − 2đ?‘ž2 − 1

(19)

đ?‘‹ đ?‘?, đ?‘ž = 837đ?‘ƒ2 + đ?‘ž2 Substituting the values of (19) in (17) and using (2) the corresponding integer solutions of (1) are given by đ?‘Ľ = đ?‘Ľ đ?‘?, đ?‘ž = 2511đ?‘ƒ2 − đ?‘ž2 + 62đ?‘?đ?‘ž − 1 đ?‘Ś = đ?‘Ś đ?‘?, đ?‘ž = 837đ?‘ƒ2 − 3đ?‘ž2 − 62đ?‘?đ?‘ž − 1

(20)

đ?‘§ = đ?‘§ đ?‘?, đ?‘ž = 837đ?‘ƒ2 + đ?‘ž2 + 54đ?‘?đ?‘ž Thus the equation (20) represent a non-zero distinct integral solutions of (1) in two parameters.

Properties: đ?‘– đ?‘Ľ đ?‘ƒ, 1 + đ?‘Ś đ?‘ƒ, 1 − 3348đ?‘‡4,đ?‘ƒ + 6 = 0 đ?‘–đ?‘– đ?‘Ś đ?‘ƒ, 1 − đ?‘§ đ?‘ƒ, 1 ≥ −3 đ?‘šđ?‘œđ?‘‘ 8 đ?‘–đ?‘–đ?‘– đ?‘Ľ đ?‘ƒ, đ?‘ƒ + 1 + 2514đ?‘‡4,đ?‘ƒ − đ?‘‡6,đ?‘ƒ − 62đ?‘ƒđ?‘&#x;đ?‘› ≥ 0 đ?‘šđ?‘œđ?‘‘ 1 đ?‘–đ?‘Ł đ?‘Ś đ?‘ƒ, đ?‘ƒ + 1 − 823đ?‘ƒ2 − đ?‘‡24,đ?‘ƒ + 62đ?‘ƒđ?‘&#x;đ?‘› ≥ 0 đ?‘šđ?‘œđ?‘‘ 4 đ?‘Ł đ?‘Ľ đ?‘ƒ, 1 + đ?‘Ś đ?‘ƒ, 1 − 3đ?‘? đ?‘ƒ, 1 − 83đ?‘‡4,đ?‘ƒ ≥ −7 đ?‘šđ?‘œđ?‘‘ 54 Note: In addition to (17) one may also consider the linear transformations đ?‘§ = đ?‘‹ − 27đ?‘‡ and đ?‘Ł = đ?‘‹ − 31đ?‘‡ following the method presented above different set of solutions are obtained.

Pattern – IV: Consider (3) as (đ?‘˘ + 1)2 − 4đ?‘Ł 2 = 31(đ?‘§ 2 − đ?‘Ł 2 )

(21)

Write (21) in the form of ratio as đ?‘˘ + 1 + 2đ?‘Ł 31(đ?‘§ − đ?‘Ł) đ??´ = = ,đ??ľ ≠0 đ?‘§+đ?‘Ł đ?‘˘ + 1 − 2đ?‘Ł đ??ľ Which is equivalent to the following two equations đ??ľ đ?‘˘ + 1 + 2đ??ľ − đ??´ đ?‘Ł − đ??´đ?‘? = 0 đ??´ đ?‘˘ + 1 + 31đ??ľ − 2đ??´ đ?‘Ł − 31đ??ľđ?‘? = 0 on employing the method of cross multiplication, we get đ?‘˘ = −2đ??´2 − 62đ??ľ2 + 62đ??´đ??ľ − 1

(22)

đ?‘Ł = đ??´2 − 31đ??ľ2 đ?‘§ = đ??´2 + 31đ??ľ2 − 4đ??´đ??ľ

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Substituting the values of u and v from 22 in (2), the non-zero distinct integral values of x and y are given by đ?‘Ľ = −đ??´2 − 93đ??ľ2 + 62đ??´đ??ľ − 1 đ?‘Ś = −3đ??´2 − 31đ??ľ2 + 62đ??´đ??ľ − 1

(24)

đ?‘§ = đ??´2 + 31đ??ľ2 − 4đ??´đ??ľ Thus the equation (24) represent non-zero distinct integral solutions of (1) in two parameters.

Properties: đ?‘– đ?‘Ľ đ??´, 1 + đ?‘§ đ??´, 1 ≥ −3 đ?‘šđ?‘œđ?‘‘ 58 đ?‘–đ?‘– đ?‘Ľ đ??´, 1 + đ?‘Ś đ??´, 1 + đ?‘§ đ??´, 1 + 98đ?‘‡4,đ??´ − đ?‘‡12,đ??´ + 5 = 0 đ?‘–đ?‘–đ?‘– đ?‘Ś đ??´, 1 + 3đ?‘§(đ??´, 1) ≥ 1 đ?‘šđ?‘œđ?‘‘ 50 đ?‘–đ?‘Ł 3đ?‘Ľ đ??´, 1 − đ?‘Ś đ??´, 1 − 2 = 0 đ?‘Ł đ?‘Ś đ??´, 1 + đ?‘§ đ??´, 1 − 2đ?‘Ľ(đ??´, 1) ≥ 4 đ?‘šđ?‘œđ?‘‘ 66 Note: Equation (21) can also be expressed in the form of ratios in three different ways that are presented below.

Way 1: đ?‘˘ + 1 + 2đ?‘Ł đ?‘§âˆ’đ?‘Ł đ??´ = = ,đ??ľ ≠0 31(đ?‘§ + đ?‘Ł) đ?‘˘ + 1 − 2đ?‘Ł đ??ľ

Solution for Way 1: đ?‘Ľ = −31đ??´2 − 3đ??ľ2 + 62đ??´đ??ľ − 1 đ?‘Ś = −93đ??´2 − đ??ľ2 + 62đ??´đ??ľ − 1 đ?‘§ = 31đ??´2 + đ??ľ2 − 4đ??´đ??ľ Way 2: đ?‘˘ + 1 + 2đ?‘Ł đ?‘§+đ?‘Ł đ??´ = = ,đ??ľ ≠0 31(đ?‘§ − đ?‘Ł) đ?‘˘ + 1 − 2đ?‘Ł đ??ľ Solution for Way 2: đ?‘Ľ = 31đ??´2 + 3đ??ľ2 + 62đ??´đ??ľ − 1 đ?‘Ś = 93đ??´2 + đ??ľ2 + 62đ??´đ??ľ − 1 đ?‘§ = 31đ??´2 + đ??ľ2 + 4đ??´đ??ľ Way 3: đ?‘˘ + 1 + 2đ?‘Ł 31(đ?‘§ + đ?‘Ł) đ??´ = = ,đ??ľ ≠0 đ?‘§âˆ’đ?‘Ł đ?‘˘ + 1 − 2đ?‘Ł đ??ľ Solution for Way 3:

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đ?‘Ľ = đ??´2 + 93đ??ľ2 + 62đ??´đ??ľ − 1 đ?‘Ś = 3đ??´2 + 31đ??ľ2 + 62đ??´đ??ľ − 1 đ?‘§ = đ??´2 + 31đ??ľ2 + 4đ??´đ??ľ

4. CONCLUSION: In this paper, we have presented four different patterns of non-zero distinct integer solutions of the non-homogeneous cone given by 7 đ?‘Ľ 2 + đ?‘Ś 2 − 13đ?‘Ľđ?‘Ś + đ?‘Ľ + đ?‘Ś + 1 = 31đ?‘§ 2 . To conclude, one may search for patterns of non-zero distinct integer solutions and their corresponding properties for other choices of ternary quadratic diophantine equation.

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