On the Integer Solutions of the Homogeneous Biquadratic Diophantine Equation

https://doi.org/10.22214/ijraset.2023.49586

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

On the Integer Solutions of the Homogeneous Biquadratic Diophantine Equation

1Associate Professor, 2PG Student, Department of Mathematics Cauvery College for Women (Autonomous), Trichy, Tamil Nadu, India

Abstract: The homogenous biquadratic Diophantine equation with five unknowns non-zero unique integer solutions

are found using several techniques. The unusual numbers and the solutions are found to have a few intriguing relationships. The relationships between the solutions’ recurrences are also shown.

Keywords: Five unknowns in a homogenous biquadratic equation, Numbers in polygons.

Notations



n T12, Dodecagonal number of rank n.

 n T18, Octadecagonal number of rank n.



n T22, Icosidigonal number of rank n.



n T26, Icosihexagonal number of rank n.



n T28, Icosioctagonal number of rank n.

 n Gno Gnomonic number of rank n.

I. INTRODUCTION

The biquadratic equation can be used to represent the quartic equation. These issues can be resolved using the quadratic formula since they can be reduced to quadratic equations [1-5], they are simple to solve. Several mathematicians have developed an interest in biquadratic Diophantine equations, both homogeneous and non-homogeneous. One can refer to [6-11] in the context for a variety of issues involving the two, three, and four variable Diophantine equations. This communication examines the non-zero unique integer solutions to the biquadratic equation with five unknowns provided by

.Also the recurrence linkages between the solutions are discovered.

II. METHOD OF ANALYSIS

In order to get the non-zero unique integral solution to the homogeneous biquadratic Diophantine problem with five unknowns,

(1)

An explanation of the linear transformation

Equation (1) is changed to

A. Pattern I

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  2 2 2 4 4 26 p w z y x 

G. Janaki1, M. Krishnaveni2

  2 2 2 4 4 26 p w z y x 

v u w v u z v u y v u x       2 , 2 , , (2)

2 2 2

p v u  

26

(3)

Assume

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Using (4) and (5) in (3) and employing the method of factorization, we get,

Equating

Equation (2) non-zero unique integer solutions are as follows when u and v are substituted:

B. Pattern II

26 can also be expressed as,

(6)

Applying (5) and (6) in (3) and the factorization approach, we obtain,

Similar to pattern 1, the non- zero distinct integer answers of (1) are

1120 ©IJRASET: All Rights are Reserved | SJ Impact Factor 7.538 | ISRA Journal Impact Factor 7.894 |   i i   5 5 26 (4) and   ib a ib a b a p     2 2 (5)

        2 2 5 5 ib a ib a i i iv u iv u    

the like factors, we

    2 5 ib a i iv u         2 5 ib a i iv u 

get,

and

ab b a u 2 5 5 2 2  ab b a v 10 2 2  

Equating real

imaginary parts, we get,

  ab b a b a x x 8 6 6 , 2 2      ab b a b a y y 12 4 4 , 2 2     ab b a b a z z 6 11 11 , 2 2      ab b a b a w w 14 9 9 , 2 2     2 2 , b a b a p p    Properties 1)        mod15 0 5 ,1 ,1 2 22,   a a Gno T a y a x . 2)         40 1,1 1,1 1,1 1,1   w z y x is Duck number. 3)        mod15 0 5 ,1 ,1 2 18,   a a Gno T a p a w . 4)   2 6 , A A A z  is Nasty number. 5)        mod15 0

2 126,    a a Gno T a p a z .

5 ,1 ,1

  

 

5 1 5 1 26 i i

        2 2 5 1 5 1 ib a ib a i i iv u iv u    

  ab b a b a x x 8 6 6 , 2 2     ab b a b a y y 12 4 4 , 2 2      ab b a b a z z 18 7 7 , 2 2     ab b a b a w w 22 3 3 , 2 2      2 2 , b a b a p p   

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Using (8) and (9) in (7) and employing the method of factorization, we get,

Equating rational and irrational parts, we get,

Equation (2) non-zero unique integer solutions are as follows when u and v are substituted:

25 1,1 1,1

is Square number. 5) For all values of a and b, y x  is divisible by 2

1121 ©IJRASET: All Rights are Reserved | SJ Impact Factor 7.538 | ISRA Journal Impact Factor 7.894 | Properties: 1)          mod15 0 5 ,1 ,1 ,1 2 22,    a a Gno T a w a z a x 2)     0 20 7 ,1 ,1 28,     a a Gno T a z a x 3)        mod15 0 5 ,1 ,1 2 22,  a a Gno T a w a z 4)     2 6 , , A A A z A A y  is Nasty number. 5)     242 1,1 11   w is Palindrom number C. Pattern III Rewriting equation (3) as 2 2 2 26 1 v p u   (7) Assume   b a b a b a u    26 26 26 2 2 (8) Write 1 as,   5 26 5 26 1   (9)

        v p v p b a b a     26 26 26 26 5 26 5 26 2 2 Equating the like factors, we get,     v p b a     26 26 5 26 2     v p b a  26 26 5 26 2

ab b a p 10 26 2 2    ab b a v 52 5 130 2 2   

  ab b a b a x x 52 4 156 , 2 2       ab b a b a y y 52 6 10 , 2 2     ab b a b a z z 52 3 182 , 2 2       ab b a b a w w 52 7 78 , 2 2   ab b a p 10 26 2 2    Properties: 1)          291 mod 0 31 2 1, 1, 1, 1, 4,     a b Gno T b y b p b z b x 2)  

 x

3)     50

  y x

   



w

212 1,1

is palindrom number.

1,1 1,1

is Duck number. 4)

y

D.

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Applying (11) and (12) in (10) and the factorization approach, we obtain,

By equating the rational and irrational components, we obtain,

In order to discover only integer solutions, we must substitute

B b 5 in equations (12) and (13).

Equation (2) non-zero unique integer solutions are as follows when u and v are

is Duck number.

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Pattern

IV

2 2 2 26 1 u p v   (10) Write1 as,    25 1 26 1 26 1   (11) Assume   b a b a b a v     26 26 26 2 2 2 (12)

   25 1 26 1 26        v p v p b a b a    26 26 26 26 2 2

the

factorization, we get,   5 1 26     v p b a    26 26 2   5 1 26    v p b a  26 26 2

Rewriting equation (3) as,

Equating

like

 ab b a p 2 26 5 1 2 2    (13)  ab b a u 52 26 5 1 2 2    (13)

A a 5 and

AB B A u 260 5 130 2 2    2 2 25 650 B A v 

  AB B A B A x x 260 20 780 , 2 2      AB B A B A y y 260 30 520 , 2 2       AB B A B A z z 520 15 910 , 2 2      AB B A B A w w 520 35 390 , 2 2     AB B A p 10 5 130 2 2   

   

  y x

2)   144 1 1,1  p

3)        mod649 0 389 2 1, 1, 1, 12,    a a Gno T a x a w a p 4)       625

1,1   y w x

substituted:

Properties: 1)

790 1,1 1,1

is Square number.

1,1 1,1

is Square number.

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E. Pattern V

Equation (3) can be written as,

Which is represented in the form of ratio as,

This is equivalent to the following two equations

Solving the above equation by cross ratio method, we get,

Substituting u and v in equation (2), the non – zero distinct integer solutions are

III. CONCLUSION

Other non-zero unique integer solutions to the multivariable biquadratic equations under consideration may be sought after.

REFERENCES

[1] Carmicheal R.D, The Theory of Numbers and Diophantine Analysis, Dover Publications, New York, 1959.

[2] Dickson.L.E., “History of the Theory of Numbers”, Chelsea Publishing Company, New York. Vol II. 1952.

[3] Mordell. L.J, “Diophantine equations”, Academic Press, New York, 1969.

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

[5] Nigel D. Smart, The Algorithmic Resolutions of Diophantine Equations, Cambridge University Press, London, 1999.

[6] Gopalan M.A and Janaki G, Integral solutions of the ternary quartic equation

[7] Gopalan M.A and Kalingarani J, Quartic equation in five unknowns

issue 2, 305-311, 2009.

2 2 2 2 25 v p p u       v p v p p u p u    5 5 (14)

0 , 5 5      B B A p u v p v p p u

  0 5   Av p A B Bu   0 5    Bv p B A Au

AB B A u 10 2 2   AB B A v 2 5 5 2 2    2 2 B A p  

  AB B A B A x x 12 4 4 , 2 2       AB B A B A y y 8 6 6 , 2 2      AB B A B A z z 22 3 3 , 2 2       AB B A B A w w 18 7 7 , 2 2    2 2 B A p   Properties 1)        mod15 0 11 ,1 ,1 2 28,   a a Gno T a w a y 2)     20 1,1 1,1   w p is Duck number 3) For

values of A and B, y x  is divisible by 2. 4)        mod15 0 5 ,1 ,1 2 22,   a a Gno T a x a y 5)           64 1,1 2 1,1 1,1 1,1 1,1      p w z y x is Cube number.

all

4 2 2 z xy y x   , Impact J.Sci. Tech, Vol 2, issue 2, 71-79, 2008.

  2 2 2 4 4 2 p w z y x  , Bulletin of pure and applied sciences, Volume 28,