https://doi.org/10.22214/ijraset.2023.49479
11 III
2023
March
ISSN: 2321-9653; IC Value: 45.98; SJ Impact Factor: 7.538
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On The Ternary Quadratic Diophantine Equation
1Assistant Professor, 2PG student, PG and Research Department of Mathematics Cauvery College for Women (Autonomous) Trichy-18, India (Affiliated Bharathidasan University)
Abstract: The non-zero unique integer solutions to the quadratic Diophantine equation with three unknowns
are examined. We derive integral solutions in four different patterns. A few intriguing relationships between the answers and a few unique polygonal integersare shown.
Keywords: Ternary quadraticequation, integral solutions.
I. INTRODUCTION
There is a wide range of ternary quadratic equations. One might refer to [1-8] for a thorough review of numerous issues. These findings inspired us to look for an endless number of non-zero integral solutions to another intriguing ternary quadratic problem provided by 2 2 2 14 z y xy x illustrating a cone for figuring out its many non-zero integral points. A few intriguing connections between the solutions are displayed.
II. CONNECTED WORK
a Pr Pronic number of the rank ‘n’
a Gno Gnomonic number of rank ‘n’
n m T , Polygonal number of rank ‘n’ with sides ‘m’
III. METHODOLOGY
The ternary quadratic Diophantine equation to be solved for its non-zero integral solutions is,
(1) Replacement of linear transformations
x and y
(1) results in
12 16 z
We present below different patterns of solving (3) and thus obtain different choices of integer solutions of (1)
A. Pattern: 1
Assume, ) , ( b a z z = 2 2 12 16 b a
Where a and b are non-zero integers. Substitute (4) in (3) we get,
Equating rational and irrational terms we get,
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2 2 2 14 z y xy x
1
T.
2
P. Sangeetha
,
Divyapriya
2 2 2 14 z y xy x
2 2 2 14 z y xy x
(2)
(3)
2 2 2
(4)
2 2 2 2 ) 12 (4 ) 12 (4 )
b a b a (5)
ab b a 8 ] 12 4[16 1 2 2
12 )(4 12 (4
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Replacing the preceding variables of and in equation (2), the equivalent answers to (1) are provided by
By using equation (9) and the value of
Replacing
preceding variables of and in equation (2), the equivalent integer answers to (1) are provided
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2 2 2 2 2 2 12 16 ) , ( 8 3 4 ) , ( 8 3 4 ) , ( b a b a z z ab b a b a y y ab b a b a x x Now put A a 4 and B b 4 we get, 2 2 2 2 2 2 192 256 128 48 64 128 48 64 B A z AB B A y AB B A x OBSERVATION: 1) 0(mod256) 256 ,2) ( ,2) ( A Gno A x A y 2) (1,1)] (1,1) (1,1) 2[ z y x is a Perfect square. 3) 0(mod592) 176 32 ,2) ( ,2) ( ,2) ( 26, A A Gno T A z A y A x 4) 0(mod308) 308 88 ) ,2 (3 ) , (2 18, a a Gno T a a x a a z 5) 0(mod116) 164 8 ,1) ( ,1) ( 2 18, A A Gno T A x A y 6) 0(mod352) 144 176 ) 1, ( 1) , ( 14, A A Gno T A A z A A x B. Pattern: 2 Equation (3) can be written as 1 12 16 2 2 2 z (8) Write 1 as 12) 2 12)(7 2 (7 1 (9)
z,
write 12) 2 12)(7 2 (7 ) 12 (4 ) 12 (4 ) 12 )(4 12 (4 2 2 b a b a Equating positive and negative terms we get, ab b a b a ab b a b a 56 24 32 ) , ( 48 21 28 ) , ( 2 2 2 2
we can
the
2 2 2 2 2 2 12 16 ) , ( 8 3 4 ) , ( 104 45 60 ) , ( b a b a z z ab b a b a y y ab b a b a x x
1) 0(mod128) 80 16 ,1) ( ,1) ( 10, a a Gno T a x a y 2) 0(mod102) 48 40 ,1) ( ,1) ( ,1) ( 4, a a Gno T a x a y a z
by
OBSERVATION:
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3) (1,1) 2 (1,1) 9 y z is a nasty number
4) ) , ( 4 ) , ( 10 a a y a a z is a Perfect square
5) 0 567 ) , ( 4 ) , ( 3 4, a T a a y a a x
6) ) , ( 2 ) , ( ) , ( a a z a a y a a x is a cubical integer
C. Pattern: 3 Equation (3) can be written as
Replacing the preceding variables of z and in equation (2), the equivalent answers to (1) are provided by
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2 2 2 16 12 z (10) Assume, 2 2 12b a (11) 4 12) 2 12)(4 2 (4 16 i i (12) In equations (11)
2 2 2 2 2 ) 7 ( 4 12) 2 12)(4 2 (4 12 b a i i z (13) Equation (13) as, 2 2 ) 12 ( ) 12 )( 2 12 2 4 )( 2 12 2 4 ( ) 12 )( 12 ( b i a b i a i i i z i z Equating positive and negative terms we get, ab b a b a ab b a b a z z 4 12 ) , ( 24 24 2 ) , ( 2 2 2 2
and (12) in (10) we get,
ab b a b a z z ab b b a y y ab a b a x x 24 24 2 ) , ( 4 24 ) , ( 4 2 ) , ( 2 2 2 2 OBSERVATION: 1) 0(mod60) 8 4Pr ,1) ( ,1) ( ,1) ( a a Gno a z a y a x 2) 0(mod40) 10 48Pr ) (1, ) (1, 2 a a Gno a z a y 3) 0 88 ) , (2 ) , (2 4, a T a a y a a x 4) (1,1) 2 (1,1) z x is a Perfect square 5) 0 14 ) , ( ) , ( 4, a T a a y a a x 6) 0(mod32) 10 2 ,1) ( ) (1, 4, a a Gno T a z a x D. Pattern: 4 Equation (3) as, 2 2 2 12 16 z (14) Write equation (14) as
International Journal for Research in Applied Science & Engineering Technology (IJRASET)
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Equation (15) is written in the form of a ratio as
This is equivalent to the following two equations
By using cross-multiplication, we get
Replacing the preceding variables of
and in equation (2), the equivalent integer answers to (1) are provided by
IV. CONCLUSION
In this paper, We have found an endless number of non-zero distinct integer solutions to the ternary quadratic Diophantine equation
16 3 )
To sum up, one can look for other solution patterns and their accompanying attributes.
REFERENCES
[1] Batta. B and Singh. A.N, History of Hindu Mathematics, Asia Publishing House 1938.
[2] Carmichael, R.D., “The Theory of Numbers and Diophantine Analysis”, Dover Publications,New York, 1959.
[3] Dickson. L.E., “History of the theory of numbers”, Chelsia Publishing Co., Vol.II, New York, 1952.
[4] G. Janaki and C. Saranya, Observations on the ternary quadratic Diophantine equation , International Journal of Innovative Research in Science, Engineering, and Technology 5(2) (2016), 2060-2065.
[5] G. Janaki and S. Vidhya, on the integer solutions of the homogeneous biquadratic Diophantine equation , International Journal of Engineering Science and Computing 6(6) (2016), 7275-7278.
[6] M. A. Gopalan and G. Janaki, Integral solutions of
[7] G. Janaki and P. Saranya, on the ternary cubic Diophantine equation
, Impact J. Sci. Tech. 4(1) (2010), 97-102.
, International Journal of Science and Research-online 5(3) (2016), 227-229.
[8] G. Janaki and C. Saranya, Integral solutions of the ternary cubic equation , International Research Journal of Engineering and Technology 4(3) (2017), 665-669.
729 ©IJRASET: All Rights are Reserved | SJ Impact Factor 7.538 | ISRA Journal Impact Factor 7.894 | 12 ) )(4 (4 z z (15)
q p z z 4 12 4
0 4 p zq q (16) 0 12 4 q zp p (17)
2 2 2 2 48 4 ) , ( 8 ) , ( 12 ) , ( q p b a z z pq b a q p b a
2 2 2 2 2 2 48 4 ) , ( 8 12 ) , ( 8 12 ) , ( q p q p z z pq q p q p y y pq q p q p x x OBSERVATION: 1) ) , ( ) , ( ) , ( p p z p p y p p x is a Perfect square 2) (1,1) 16 (1,1) 2 y z is a cubical integer 3) 0(mod30) 6Pr ) , ( ) , ( ) , ( p p p z p p y p p x 4) 0(mod39) 9 2 ,1) ( ,1) ( ,1) ( 6, p p Gno T p z p y p x 5) 0 23 ) , ( ) , ( 2 p p p z p p x 6) 0(mod4) 8 60 ) (2, ) (2, 4, p p Gno T p z p x
2 2 2
2(
z xy y x
