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On The Negative Pell Equation

y

2

15

x

2

14

S. Vidhyalakhsmi1, T. Mahalakshmi2

1

Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620002, TamilNadu, India 2

Assistant Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620002, TamilNadu, India

Abstract: The binary quadratic equation represented by the negative pelliany2 15x214 is analyzed for its distinct integer solutions. A few interesting relations among the solutions are also given. Further, employing the solutions of the above hyperbola, we have obtained solutions of other choices of hyperbolas, parabolas.

Keywords:Binary quadratic, hyperbola, parabola, pell equation, integral solutions.

I. INTRODUCTION

Diophantine equation of the form y2Dx21, where D0and square free, is known as negative pell equation. In general, the general form of negative pell equation is represented by y2Dx2N ,N0,D0and square free. It is known that negative pell

equations y2 3x21,y2 7x24have no integer solutions whereas y2 65x21, y2202x21have integer solutions. It is observed that the negative pell equation do not always have integer solutions. For negative pell equations with integer solutions, one may refer [1-12].

In this communication, yet another negative pell equation given by 2 7 2 14

x

y is considered for its non-zero distinct integer

solutions. A few interesting relations among the solutions are also given. Further, employing the solutions of the above hyperbola, we have obtained solutions of other choices of hyperbolas, parabolas.

II. METHODOFANALYSIS The negative Pell equation representing hyperbola under consideration is

14 15 2 2

x

y (1)

whose smallest positive integer solution is 11 ,

3 0

0 y

x

To obtain the other solutions of (1), consider the Pell equation

1 15 2 2

x

y

whose general solution is given by

n n n

n g y f

x

2 1 ~ , 15 2

1

~

where

1

1

15 4 15

4    

n n

n

f

4 15

1

4 15

1 , 1,0,1,2...

   n

gn n n

Applying Brahmagupta lemma between

x0,y0

and

~xn,y~n

, the other integer solutions of (1) are given by

n n

n f g

x

15 2

11 2

3

1 

n n

n f g

y 15

2 3 2 11

1  

The recurrence relations satisfied by x and y are given by 0

8 2 1

3    

n n

n x x

x

0

8 2 1

3    

n n

n y y

y

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Table: 1 Numerical examples

n xn1 yn1

-1 3 11

0 23 89

1 181 701

2 1425 5519

3 11219 43451

4 88327 342089

From the above table, we observe some interesting relations among the solutions which are presented below: 1) xn1 and yn1 are always odd

2) Relations among the solutions a) 4xn1xn2 yn10

b) xn14xn2 yn20 c) 4xn131xn2yn30

d) 31xn1xn38yn10 e) xn131xn38yn30 f) 15xn14yn1yn20 g) 120xn131yn1yn30 h) 15xn131yn24yn30 i) 31xn24xn3yn10 j) 4xn2xn3yn20 k) xn24xn3yn30 l) 15xn2yn14yn2 0 m) 30xn2yn3yn10 n) 15xn24yn2yn30 o) 15xn14yn2yn30

p) 15

x

n3

4

y

n1

31

y

n2

0

q) 120xn3yn131yn30 r) 15xn3yn24yn30 s) xn1xn32yn20

3) Each of the following expressions represents a nasty number

a)

89 11 14

7 6

3 2 2

2n  xn  x

b)

701 11 112

28 3

4 2 2

2n  xn  x

c)

45 11 14

7 6

2 2 2

2n  yn  x

d)

345 11 56

28 6

3 2 2

2n  yn  x

e)

2715 11 434

217 6

4 2 2

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f)

701 89 14

7 6

4 2 3

2n  xn  x

g)

45 89 56

14 3

2 2 3

2n  yn  x

h)

345 89 14

7 6

3 2 3

2n  yn  x

i)

2715 89 56

14 3

4 2 3

2n  yn  x

j)

45 701 434

217 6

2 2 4

2n  y n  x

k)

345 701 56

14 3

3 2 4

2n  yn  x

l)

2715 701 14

7 6

4 2 4

2n  yn  x

m)

3 23 14

7 6

2 2 3

2n  yn  y

n)

3 181 112

28 3

2 2 4

2n  yn  y

o)

23 181 14

7 6

3 2 4

2n  yn  y

4) Each of the following expressions represents a cubical integer

a)

267 1 33 2 89 3 3 11 3 4

7 1

 

  nnn

n x x x

x

b)

2103 1 33 3 701 3 3 11 3 5

56 1

 

  nnn

n x x x

x

c)

135 1 33 1 45 3 3 11 3 3

7 1

 

  nnn

n y x y

x

d)

1035 1 33 2 345 3 3 11 3 4

28 1

 

  nnn

n y x y

x

e)

8145 1 33 3 2715 3 3 11 3 5

217 1

 

  nnn

n y x y

x

f)

2103 2 267 3 701 3 4 89 3 5

7 1

 

  nnn

n x x x

x

g)

135 2 267 1 45 3 4 89 3 3

28 1

 

  nnn

n y x y

x

h)

1035 2 267 2 345 3 4 89 3 4

7 1

 

  nnn

n y x y

x

i)

8145 2 267 3 2715 3 4 89 3 5

28 1

 

  nnn

n y x y

x

j)

135 3 2103 1 45 3 5 701 3 3

217 1

 

  nnn

n y x y

x

k)

1035 3 2103 2 345 3 5 701 3 4

28 1

 

  nnn

n y x y

x

l)

8145 3 2103 3 2715 3 5 701 3 5

7 1

 

  nnn

n y x y

x

m)

9 2 69 1 3 3 4 23 3 3

7 1

 

  nnn

n y y y

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n)

9 3 543 1 3 3 5 181 3 3

56 1

 

  nnn

n y y y

y

o)

69 3 543 2 23 3 5 181 3 4

7 1

 

  nnn

n y y y

y

5) Each of the following expressions represents a bi-quadratic integer

a)

89 11 356 44 42

7 1 3 2 2 2 5 4 4

4n  xn  xn  xn  x

b)

701 11 2804 44 336

56 1 4 2 2 2 6 4 4

4n  xn  xn  xn  x

c)

45 11 180 44 42

7 1 2 2 2 2 4 4 4

4n  yn  xn  yn  x

d)

345 11 1380 44 168

28 1 3 2 2 2 5 4 4

4n  yn  xn  yn  x

e)

2715 11 10860 44 1302

217 1 4 2 2 2 6 4 4

4n  yn  xn  y n  x

f)

701 89 2804 356 42

7 1 4 2 3 2 6 4 5

4n  xn  xn  xn  x

g)

45 89 180 356 168

28 1 2 2 3 2 4 4 5

4n  yn  xn  yn  x

h)

345 89 1380 356 42

7 1 3 2 3 2 5 4 5

4n  yn  xn  yn  x

i)

2715 89 10860 356 168

28 1 4 2 3 2 6 4 5

4n  y n  xn  yn  x

j)

45 701 180 2804 1302

217 1 2 2 4 2 4 4 6

4n  y n  xn  y n  x

k)

345 701 1380 2804 168

28 1 3 2 4 2 5 4 6

4n  yn  xn  yn 

x

l)

2715 701 10860 2804 42

7 1 3 3 6 4 6

4n  y n  xn  yn 

x

m)

3 23 12 92 42

7 1 2 2 3 2 4 4 5

4n  yn  yn  yn  y

n)

3 181 12 724 336

56 1 2 2 4 2 4 4 6

4n  yn  yn  yn  y

o)

23 181 92 724 42

7 1 3 2 4 2 5 4 6

4n  yn  yn  yn  y

6) Each of the following expressions represents a quintic integer

a)

890 1 110 2 445 3 3 55 3 4 89 5 5 11 5 6

7 1     

  nnnnn

n x x x x x

x

b)

7010 1 110 3 3505 3 3 55 3 5 701 5 5 11 5 7

56 1     

  nnnnn

n x x x x x

x

c)

450 1 110 1 225 3 3 55 3 3 45 5 5 11 5 5

7 1     

  nnnnn

n y x y x y

x

d)

3450 1 110 2 1725 3 3 55 3 4 345 5 5 11 5 6

28 1     

  nnnnn

n y x y x y

x

e)

27150 1 110 3 13575 3 3 55 3 5 2715 5 5 11 5 7

217 1     

  nnnnn

n y x y x y

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f)

7010 2 890 3 3505 3 4 445 3 5 701 5 6 89 5 7

7 1

 

 

  nnnnn

n x x x x x

x

g)

450 2 890 1 225 3 4 445 3 3 45 5 6 89 5 5

28 1

 

 

  nnnnn

n y x y x y

x

h)

3450 2 890 2 1725 3 4 445 3 4 345 5 6 89 5 6

7 1

 

 

  nnnnn

n y x y x y

x

i)

27150 2 890 3 13575 3 4 445 3 5 2715 5 6 89 5 7

28 1

 

 

  nnnnn

n y x y x y

x

j)

450 3 7010 1 225 3 5 3505 3 3 45 5 7 701 5 5

217 1

 

 

  nnnnn

n y x y x y

x

k)

3450 3 7010 2 1725 3 5 3505 3 4 345 5 7 701 5 6

28 1

 

 

  nnnnn

n y x y x y

x

l)

27150 3 7010 3 13575 3 5 3505 3 5 2715 5 7 701 5 7

7 1

 

 

  nnnnn

n y x y x y

x

m)

30 2 230 1 15 3 4 115 3 3 3 5 6 23 5 5

7 1

 

 

  nnnnn

n y y y y y

y

n)

30 3 1810 1 15 3 5 905 3 3 3 5 7 181 5 5

56 1

 

 

  nnnnn

n y y y y y

y

o)

230 3 1810 2 115 3 5 905 3 4 23 5 7 181 5 6

7 1

 

 

  nnnnn

n y y y y y

y

III. REMARKABLEOBSERVATIONS

1) Employing linear combinations among the solutions of (1), one may generate integer solutions for other choices of hyperbola which are presented in the Table: 2 below:

Table: 2 Hyperbolas

S. No Hyperbola

X,Y

1 2 15 2 196

Y

X

89xn111xn2,3xn223xn1

2 2 15 2 12544

Y

X

701xn111xn3,3xn3181xn1

3 2 15 2 196

Y

X

45xn111yn1,3yn111xn1

4 2 15 2 3136

Y

X

345xn111yn2,3yn289xn1

5 2 15 2 188356

Y

X (2715xn111yn3,3yn3701xn1)

6 2 15 2 196

Y

X

701xn289xn3,23xn3181xn2

7 2 15 2 3136

Y

X

45xn289yn1,23yn111xn2

8 2 15 2 196

Y

X

345xn289yn2,23yn289xn2

9 2 15 2 3136

Y

X

2715xn289yn3,23yn3701xn2

10 2 15 2 188356

Y

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11 2 15 2 3136

Y

X

345xn3701yn2,181yn289xn3

12 2 15 2 196

Y

X

2715xn3701yn3,181yn3701xn3

13 735 2 49 2 144060

Y

X

3yn223yn1,89yn111yn2

14 47040 2 3136 2 5990069760

Y

X

3yn3181yn1,701yn111yn3

15 735 2 49 2 144060

Y

X

23yn3181yn2,701yn289yn3

2) Employing linear combinations among the solutions of (1), one may generate integer solutions for other choices of parabola which are presented in the Table: 3 below:

Table: 3 Parabolas

S. No Parabola

X,Y

1 7 15 2 98

Y

X

89x2n211x2n3,3xn223xn1

2 56 15 2 6272

Y

X

701x2n211x2n4,3xn3181xn1

3 7 15 2 98

Y

X

45x2n211y2n2,3yn111xn1

4 28 15 2 1568

Y

X

345x2n211y2n3,3yn289xn1

5 217 15 2 94178

Y

X

2715x2n211y2n4,3yn3701xn1

6 7 15 2 98

Y

X

701x2n389x2n4,23xn3181xn2

7 28 15 2 1568

Y

X

45x2n389y2n2,23yn111xn2

8 7 15 2 98

Y

X

345x2n389y2n3,23yn289xn2

9 28 15 2 1568

Y

X

2715x2n389y2n4,23yn3701xn2

10 217 15 2 94178

Y

X

45x2n4701y2n2,181yn111xn3

11 28 15 2 1568

Y

X

345x2n4701y2n3,181yn289xn3

12 7 15 2 98

Y

X

2715x2n4701y2n4,181yn3701xn3

13 735 7 2 10290

Y

X

3y2n323y2n2,89yn111yn2

14 47040 56 2 5268480

Y

X

3y2n4181y2n2,701yn111yn3

15 735 7 10290

2 

Y

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897

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IV. CONCLUSION

In this paper, we have presented infinitely many integer solutions for the hyperbola represented by the negative pell equation

14 15 2 2

x

y . As the binary quadratic Diophantine equation are rich in variety, one may search for the other choices of negative

pell equations and determine their integer solutions along with suitable properties. Employing the solutions of the above hyperbola, we have obtained solutions of other choices of hyperbolas, parabolas.

REFERENCES

[1] M.A. Gopalan, V. Geetha, S. Sumithra, “Observations on the Hyperbola y255x26”, International Research Journal of Engineering and Technology (IRJET) volume: 02 , Issue: 04, 1962-1964, July-2015.

[2] M.A. Gopalan, S. Vidhyalakshmi, T.R. Usharani, M. Arulmozhi, “On the negative pell equationy235x219”, International Journal of Research and Review, vol.2, Issue:4, pages 183-188, April 2015

[3] M.A. Gopalan, S. Vidhyalakshmi, J. Shanthi, D. Kanaka, “On the Negative Pell Equation y215x26”, Scholars Journal of Physics, Mathematics and Statistics, Sch.J.Phys.Math.Stat.; 2(2A), 123-128, 2015.

[4] K. Meena, S. Vidhyalakshmi, A. Rukmani, “ On the Negative pell Equation y231x26”, Universe of Emerging Technologies and Science, 2(12), 1-4, 2015.

[5] M.A. Gopalan, S. Vidhyalakshmi, V. Pandichelvi, P. Sivakamasundari, C.Priyadharsini., “ On The Negative pell Equationy245x211”, International Journal of Pure Mathematical Science, volume-16, 30-36, 2016.

[6] S. Vidhyalakshmi, M.A. Gopalan, E. Premalatha, S. Sofiachristinal, “ On The Negative pell Equationy272x28” , International Journal of Emerging Technologies in Engineering Research (IJETER),volume 4, Issue 2, 25-28, February 2016.

[7] M. Devi, T.R. Usharani, “On the Binary Quadratic Diophantine Equationy280x216”, Journal of Mathematics and Informatics ,vol.10V, 75-81, 2017. [8] R. Suganya, D. Maheswari, “On the Negative pellian Equationy2110x229”, Journal of Mathematics and Informatics, vol.1, 163-71, 2017.

[9] P. Abinaya, S. Mallika, “On the Negative pellian Equation y240x215”, Journal of Mathematics and Informatics, vol.1, 195-102, 2017.

[10] K. Meena, M.A. Gopalan, T. Swetha, “On the negative pell equationy240x24”, International Journal of Emerging Technologies in Engineering Research (IJETER), volume5, Issue 1, 6-11, January 2017.

[11] K. Meena, S. Vidhyalakshmi, G. Dhanalakshmi, “On the negative pell equation y25x24 ”, Asian Journal of Applied Science and Technology(AJAST)volume 1, Issue 7, pages 98-103, August 2017.

References

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