Integral Solutions of the Ternary Cubic Equation

Page 4

https://doi.org/10.22214/ijraset.2023.49685

11 III
2023
March

ISSN: 2321-9653; IC Value: 45.98; SJ Impact Factor: 7.538

Volume 11 Issue III Mar 2023- Available at www.ijraset.com

Integral Solutions of the Ternary Cubic Equation

1Associate Professor, PG and Research Department of Mathematics, Cauvery College for Women (Autonomous), Affiliated to Bharathidasan University, Trichy – 18.

2PG Student, PG and Research Department of Mathematics, Cauvery College for Women (Autonomous), Affiliated to Bharathidasan University, Trichy – 18.

Abstract: The non-homogenous cubic equation with three unknowns represented by the Diophantine equation ( + ) ( )+ ( + + )= is analyzed for its patterns of stringfy integral solutions. A few interesting properties among the solutions and some special polygonal numbers are presented.

Keywords: Cubic equation, integral solutions, polygonal number, square number, special number.

I. INTRODUCTION

Number theory is a vast and fascinating field of mathematics. Concerned with the properties of numbers in general and integers in particular as well as the wider classes of problems that arises from their study. The study of number theory is very important because of all other branches depends on this branch for their final results.Solving equations in integers is the central problem of Number theory. Number theory may be subdivided into several fields, according to the method use and the type of questions investigated. The term “arithmetic” is also used to refer to Number theory. The word Diophantine refers to the Hellenistic Mathematician of 3rd century, Diophantus of Alexandria who made of such equations and was one of the first Mathematician to introduce symbolism into algebra.In this communication, the non-homogeneous cubic equation with three unknowns represented by the equation 3( + ) 4( )+2( + +1)=522 is considered and in particular a few interesting relations among the solutions are presented.

 n m T , Polygonal number

 n O Octrahedral number

 n CS Centered square number

 n CC Centered cube number

 n Gno Gnomonic number

 n SO Stella Octangula number

 n CH Centered Hexagonal number

=Truncated tetrahedral number

II. METHOD OF ANALYSIS

The ternary cubic equation to be considered for its quasi integral solution is 3( + ) 4( )+2( + +1)=522 (1) After using the transformations, s r x   , s r y  (2) in (1) leads to ( +1) +5 =261 (3)

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)− ( )+
)=
( +
( + +

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ISSN: 2321-9653; IC Value: 45.98; SJ Impact Factor: 7.538

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We demonstrate five different patterns of quasi distinct integer solutions of (1) as below:

A.

(a,1) (a,1)+ +6 + , + + , +11 +2 , ≡0( 177)

4 (1,1) (1,1) 28 (1,1) is a Nasty number

522 (1,1) y(1,1) is a happy couple number

B. Pattern:2

Instead of (5), Write 261=(4+ 7√5)(4 7√5) (7)

Equating real and imaginary parts, we get = ( , )=4 60 105 +175 1

( , )=7 105 +12

substituting (7) and (4) in (3) and the factorization method, as shown in pattern1, the corresponding integral solutions of (1) are represented

Properties

1.6 (1,1) is a Ruth-Aaron number

2. (a,1) (a,1)+14a (a,1)+12 128 ≡0( 180)

3. (a,a)+ (a,a) is a cubical integer

4. (1,1) (1,1) is a palindrom number

517 (1,1) is a duck number

C. Pattern: 3

Instead of(5), Write 261= 9+ 6√5 9 6√5

Equating real and imaginary parts, we get

( , )=6 90 +27 45 substituting (8) and (4) are in (3) and the factorization method, as shown in pattern 1, the corresponding integral solutions of (1) are represented by

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= ( , )= +5 (4) Write 261= 16+ √5 16 √5 (5) Applying the factorization
substituting (4) and (5) in
(( +1)+ √5 )(( +1) √5 )=(16+ √5)(16 6 √5)(( + √5 ) ( √5 ) ) (6) Equating real and imaginary parts,the values of r and s are = ( , )=16 240 15 +25 1 = ( , )= 15 +48 80 Substituting the values of u &v in equation (2), the non zero distinct integral solutions
are executed as = ( , )=17 255 33 55 1 = ( , )=15 225 63 +105 1 = ( , )= +5 Properties 16 (a,a)
2.42 (1,1)
3. - (a,1+
Pattern: 1 Assume
method and
(3),
of (1)
is a Perfect square
is a palindrom number
= ( , )=11 165 93 +155 1 = ( , )= 3 +45 117 +195 1 = ( , )= +5
=
20
by
(8)
=
1 =
= ( , )=15 255 63 105 1 = ( , )=3 45 117 +195 1
( , )=9 135 90 +150

Properties

1.24

3 z(a,a) (a,a) is a palindrom number

4.6 (1,1)is a duck number

5. 2y(1,1) (1,1) is a cubical integer

D.

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Since finding integral solutions is the topic we are interested in, we have choose a and b in such a way that r, s, and z are integers. Let us take =6 and =6 , we have

In terms of (2), the integral solutions of (1) are given

Properties

1. 6 (1,1) is a Perfect square

2. (1,1) is a cubical integer

3. 5 (1,1) is a duck number

4. 12 (1,1) is a palindrom number

5. x(a,1)+ (a,1)

Since finding integral solutions is the topic we are interested in, we have choose a and b in such a way that r, s, and z are integers. Let us take =3 and =3 , we have

In terms of (2), the integral solutions of (1) are given by

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1412 ©IJRASET: All Rights are Reserved | SJ Impact Factor 7.538 | ISRA Journal Impact Factor 7.894 | = ( , )= +5
(a,1)+54
87 ≡0( 447)
(1,1) is a perfect square 2 y(a,1)
(a,1)+6
(9)
= ( , )= 1 6(96 1440 90 +150 ) 1 = ( , )= 1 6(6 90 +288 480 ) = ( , )= +5
Pattern: 4 instead of (5), Write 261= ( √ )( √ )
Equating real and imaginary parts, we get
= ( , )=3456 3240 51840 +5400 1 = ( , )=216 3240 +10368 17280 = ( , )=36 +180
= ( , )=3672 6480 41472 +11880 1 = ( , )=3240 62208 +22680 1 = ( , )=36 +180
by
(a,1)+103680 , +72360 ≡0( 42118)
192a
261= ( √ )( √ ) (10)
= ( , )= 1 3(48 720 45 +75 ) 1 = ( , )= 1 3(3 45 +144 240 ) = ( , )= +5
E. Pattern: 5 instead of (5), Write
Equating real and imaginary parts, we get
= ( , )=432 6480 405 +675 1 = ( , )=27 405 +1296 2160 = ( , )=9 +45
= ( , )=459 6885 891 +1485 1 = ( , )=405 6075 1701 +2835 1 = ( , )=36 +180

for Research in Applied Science & Engineering Technology (IJRASET)

ISSN: 2321-9653; IC Value: 45.98; SJ Impact Factor: 7.538

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Properties

1. (1,1) 6

2.y(a,1) (a,1)+6a (a,1)+654 +441 ≡0( 4533)

3.y(a,1) (a,1)+a6 (a,1)+327stara+441 ≡0( 4206)

4. (1,1)+10 is a cubical integer

5 (1,1)+9 is a woodall number

III. CONCLUSION

In this paper, we have presented five different patterns of non-zero distinct integer solutions of the non-homogeneous cone given by 3( + ) 4( )+2( + +1)=522

To conclude, one may search for other patterns of non-zero integer distinct solutions and their corresponding properties for other choices of cubic Diophantine equations.

REFERENCES

[1] Dickson L.E., “History of the theory of numbers”, Chelsia Publishing Co., Sol II, New York, 1952

[2] Carmichael R.D., “The Theory of Numbers and Diophantine Analysis”, Dover Publications, New York, 1959

[3] Mordell L .J., “Diophantine Equations”, Academic Press, London 1969

[4] Telang S. G., “Number Theory”, Tata Mc Graw-Hill Publishing Company, New Delhi 1996

[5] Janaki G, Saranya C, “Observations on Ternary Quadratic Diophantine Equation

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Engineering Technology, vol 10, Issue III,Pg.No: 231-234, August 2022

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