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Integral Points on the Ternary Quadratic

Diophantine Equation

y

2

33

x

2

4

t

,

t

0

Dr. G. Sumathi1, W. Princy Dona 2

1

Assistant Professor,PG and Research Department of Mathematics, Shrimati Indira Gandhi College Trichy-2, Tamil Nadu, India.

2

M.Phil Scholar,PG and Research Department of Mathematics, Shrimati Indira Gandhi College Trichy-2, Tamil Nadu, India.

Abstract: The binary quadratic equation = + representing hyperbola is considered for finding its integer solutions. A few interesting properties among the solutions are presented. Also, we present infinitely many positive integer solutions in terms of Generalized Fibonacci sequences of numbers, Generalized Lucas sequences of numbers.

Keywords: Binary quadratic integral solutions, generalized Fibonacci Sequences of numbers, generalized Lucas Sequences of numbers.

AMS Mathematics Subject Classification: 11D09 Notations

GF

n

(

k

,

s

)

:

Generalized Fibonacci Sequences of rank n.

GL

n

(

k

,

s

)

:

Generalized Lucas Sequences of rank n.

2

)

2

)(

1

(

1

,

m

n

n

t

mn

6

]

)

5

(

)(

2

(

)

1

(

[

n

n

m

n

m

P

nm

Pr

n

n

(

n

1

)

1

2

)

1

(

,

mn

n

Ct

mn

S

n

6

n

(

n

1

)

1

I. INTRODUCTION

The binary quadratic equation of the form = + 1, where D is non-square positive integer has been studied by various mathematicians for its non-trivial integral solutions when D takes different integral values [1,2,4]. In [3] infinitely many Pythagorean triangles in each of which hypotenuse is four times the product of the generators added with unity are obtained by employing the non-integral solutions of binary quadratic equation = 3 + 1. In [5] a, special Pythagorean triangle is obtained by employing the integral solutions of = 182 + 14. In [6] different pattern of infinitely many Pythagorean triangles are obtained by employing the non-integral solutions of = 14 + 4. In this context one may also refer [7,8]. These results have motivated us to search for the integral solutions of yet another binary quadratic equation = 33 + 4 representing a hyperbola. A few interesting properties among the solutions are presented. Employing the integral solutions of the equation under consideration a few patterns of Pythagorean triangles are obtained.

II. METHODS OF ANALYSIS

Consider the binary quadratic equation t

x

y

2

33

2

4

,

t

0

(1)

with least positive integer solutions is

,

)

2

(

4

0 t

x

y

0

23

(

2

)

t

(2)

consider the Pell equation

1

33

2

2

x

y

(2)

whose general solution

(

~

x

n

,

~

y

n

)

is represented by

n n

n

n

g

y

f

x

2

1

~

,

33

2

1

~

in which,

1

1

33

4

23

33

4

23

n n

n

f

g

n

23

4

33

n1

23

4

33

n1

,

n

1

,

0

,

1

,

2

...

Applying Brahmagupta lemma between the solutions of

x

0

,

y

0

and

~

x

n

,

~

y

n

, the general solutions of equation (1) are found to be

n

t n

t

n

f

g

y

2

(

2

)

33

2

)

2

(

23

1

n t

n t

n

f

g

x

33

2

)

2

(

23

)

2

(

2

1

The recurrence relations satisfied by the values of

x

n1 and

y

n1 are respectively.

0

46

2 1

3

n n

n

x

x

x

0

46

2 1

3

n n

n

y

y

y

[image:2.612.197.415.436.544.2]

A few numerical examples are presented in the table 1 below:

Table 1: Numerical examples

n

1

n

x

y

n1

-1 4(2)t 23(2)t

0 184(2)t 1057(2)t 1 8460(2)t 48599(2)t 2 388976(2)t 2234497(2)t 3 17884436(2)t 102738263(2)t 4 822295080(2)t 4723725601(2)t

A. A Few Interesting Relations Between The Solutions Are Given Below 1)

x

n3

1057

x

n1

184

y

n1

0

2)

23

x

n2

x

n1

4

y

n2

0

3)

23

x

n2

23

x

n1

184

y

n2

0

4)

1057

x

n3

x

n1

184

y

n3

0

5)

1057

y

n1

6072

x

n1

y

n3

0

6)

1057

y

n2

132

x

n1

23

y

n3

0

7)

x

n1

46

x

n2

x

n3

0

(3)

9)

4

y

n3

x

n2

23

x

n3

0

10)

23

x

n1

x

n2

4

y

n1

0

11)

23

y

n2

132

x

n2

y

n1

0

12)

x

n3

23

x

n2

4

y

n2

0

13)

y

n3

132

x

n2

23

y

n2

0

14)

23

x

n1

1057

x

n2

4

y

n3

0

15)

y

n1

132

x

n2

23

y

n3

0

16)

23

y

n1

132

x

n3

1057

y

n2

0

17)

23

y

n3

132

x

n3

y

n2

0

18)

y

n1

6072

x

n3

1057

y

n3

0

19)

132

x

n1

23

y

n1

y

n2

0

20)

y

n3

y

n1

46

y

n2

0

B. Each Of The Following Expression Represents A Nasty Number

1)

n n t

t

276

x

12684

x

48

(

2

)

)

2

(

4

1

2 2 3

2 

2)

n n t

t

276

x

583188

x

2208

(

2

)

)

2

(

184

1

2 2 4

2 

3)

n n t

t

276

y

1584

x

12

(

2

)

2

1

2 2 2

2 

4)

n n t

t

276

y

72864

x

276

(

2

)

)

2

(

23

1

2 2 3

2 

5)

n n t

t

276

y

3350160

x

12684

(

2

)

)

2

(

1057

1

2 2 4

2 

6)

n n t

t

12684

x

583188

x

48

(

2

)

)

2

(

4

1

3 2 4

2 

7)

n n t

t

12684

y

1584

x

276

(

2

)

)

2

(

23

1

3 2 2

2 

8)

n n t

t

12684

y

72864

x

12

(

2

)

2

1

3 2 3

2 

9)

n n t

t

12684

y

3350160

x

276

(

2

)

)

2

(

23

1

3 2 4

2 

10)

n n t

t

583188

y

1584

x

12684

(

2

)

)

2

(

1057

1

4 2 2

(4)

11)

n n t

t

583188

y

72864

x

276

(

2

)

)

2

(

23

1

4 2 3

2 

12)

n n t

t

583188

y

3350160

x

12

(

2

)

2

1

4 2 4

2 

13)

n n t

t

2208

y

48

y

48

(

2

)

)

2

(

4

1

3 2 2

2 

14)

n n t

t

101520

y

48

y

2208

(

2

)

)

2

(

184

1

4 2 2

2 

15)

n n t

t

101520

y

2208

y

48

(

2

)

)

2

(

4

1

4 2 3

2 

C. Each Of The Following Expression Represents A Cubical Integer

1)

69

2

3171

1

23

3 4

1057

3 3

)

2

(

2

1

 

n

n

n

n

t

x

x

x

x

2)

3

3

6339

1 3 5

2113

3 3

)

2

(

4

1

 

n

n

n

n

t

x

x

x

x

3)

138

1

792

1

46

3 3

264

3 3

2

1

 

n

n

n

n

t

y

x

y

x

4)

6

2

1584

1

2

3 4

528

3 3

2

1

 

n

n

n

n

t

y

x

y

x

5)

138

3

1675080

1

46

3 5

558360

3 3

)

2

(

1057

1

 

n

n

n

n

t

y

x

y

x

6)

3171

3

145797

2

1057

3 5

48599

3 4

)

2

(

2

1

 

n

n

n

n

t

x

x

x

x

7)

6342

1

792

2

2114

3 3

264

3 4

)

2

(

23

1

 

n

n

n

n

t

y

x

y

x

8)

6342

2

36432

2

2114

3 4

12144

3 4

2

1

 

n

n

n

n

t

y

x

y

x

9)

6342

3

1675080

2

2114

3 5

558360

3 4

)

2

(

23

1

 

n

n

n

n

t

y

x

y

x

10)

291594

1

792

3

97198

3 3

264

3 5

)

2

(

1057

1

 

n

n

n

n

t

y

x

y

x

11)

12678

2

1584

3

4226

3 4

528

3 5

2

1

 

n

n

n

n

t

y

x

y

x

12)

291594

3

1675080

3

97198

3 5

1558360

3 5

2

1

 

n

n

n

n

(5)

13)

276

1

6

2

92

3 3

2

3 4

2

1

  

n

n

n

n

t

y

y

y

y

14)

6345

1

3

3

2115

3 3 3 5

)

2

(

23

1

  

n

n

n

n

t

y

y

y

y

15)

12690

2

276

3

4230

3 4

92

3 5

2

1

 

n

n

n

n

t

y

y

y

y

D. Each Of The Following Expression Represents A Bi-Quadratic Integer

1)

n n n n t

t

92

x

4228

x

23

x

1057

x

12

(

2

)

)

2

(

2

1

4 4 5 4 2 2 3

2 

2)

n n n n t

t

23

x

8452

x

x

2113

x

24

(

2

)

)

2

(

4

1

4 4 6 4 2 2 4

2 

3)

n n n n t

t

184

y

1056

x

46

y

264

x

6

(

2

)

2

1

4 4 4 4 2 2 2

2 

4)

n n n n t

t

8

y

2112

x

2

y

528

x

6

(

2

)

2

1

4 4 5 4 2 2 3

2 

5)

n n n n t

t

1184

y

2233440

x

46

y

558360

x

6342

(

2

)

)

2

(

1057

1

4 4 6 4 2 2 4

2 

6)

n n n n t

t

4228

x

194396

x

1057

x

48599

x

12

(

2

)

)

2

(

2

1

5 4 6 4 3 2 4

2 

7)

n n n n t

t

8456

y

1056

x

2114

y

264

x

138

(

2

)

)

2

(

23

1

5 4 4 4 3 2 2

2 

8)

n n n n t

t

8456

y

48576

x

2114

y

12144

x

6

(

2

)

2

1

5 4 5 4 3 2 3

2 

9)

n n n n t

t

8456

y

2233440

x

2114

y

558360

x

138

(

2

)

)

2

(

23

1

5 4 6 4 3 2 4

2 

10)

n n n n t

t

388792

y

1056

x

97198

y

264

x

6342

(

2

)

)

2

(

1057

1

6 4 4 4 4 2 2

2 

11)

n n n n t

t

16904

y

2112

x

4226

y

528

x

6

(

2

)

2

1

6 4 5 4 4 2 3

2 

12)

n n n n t

t

388792

y

2233440

x

97198

y

558360

x

6

(

2

)

2

1

6 4 6 4 4 2 4

2 

13)

t

n n

n n

t 368 y 8y 92 y 2y 6(2)

2 1 5 4 4 4 3 2 2

2        

14)

t

n n

n n

t 8460 y 4y 2115 y y 138(2)

) 2 ( 23 1 6 4 4 4 4 2 2

2        

15)

t

n n

n n

t

16920

y

368

y

4230

y

92

y

6

(

2

)

2

1

6 4 5 4 4 2 3

(6)

E. Each Of The Following Expression Represents A Quintic Integer

1)

10

3

21130

1

5

3 5

10565

3 3 5 7

2113

5 5

)

2

(

4

1

    

n

n

n

n

n

n

t

x

x

x

x

x

x

2)

20

3

42260

1

10

3 5

21130

3 3

2

5 7

4226

5 5

)

2

(

8

1

    

n

n

n

n

n

n

t

x

x

x

x

x

x

3)

460

1

2640

1

230

3 3

1320

3 3

46

5 5

264

5 5

2

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

4)

20

2

5280

1

10

3 4

2640

3 3

2

5 6

528

5 5

2

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

5)

460

3

5583600

1

230

3 5

2791800

3 3

46

5 7

558360

5 5

)

2

(

1057

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

6)

10570

3

485990

2

5285

3 5

242995

3 4

1057

5 7

48599

5 6

)

2

(

2

1

    

n

n

n

n

n

n

t

x

x

x

x

x

x

7)

21140

1

2640

2

10570

3 3

1320

3 4

2114

5 5

264

5 6

)

2

(

23

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

8)

21140

2

121440

2

10570

3 4

60720

3 4

2114

5 6

12144

5 6

2

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

9)

21140

3

5583600

2

10570

3 5

2791800

3 4

2114

5 7

558360

5 6

)

2

(

23

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

10)

971980

1

2640

3

485990

3 3

1320

3 5

97198

5 5

264

5 7

)

2

(

1057

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

11)

84520

2

5280

3

21130

3 4

2640

3 5

4226

5 6

528

5 7

2

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

12)

971980

3

5583600

3

485990

3 5

2791800

3 5

97198

5 7

558360

5 7

2

1

    

n

n

n

n

n

n

t

y

x

y

x

y

x

13)

920

1

20

2

460

3 3

10

3 4

92

5 5

2

5 6

2

1

    

n

n

n

n

n

n

t

y

y

y

y

y

y

14)

7 5 5 5 5 3 3 3 3

1 10 10575 5 2115

21150 ) 2 ( 8 1     

  nnnnn

n

t y y y y y y

15)

7 5 6 5 5 3 4 3 3

2 920 21150 460 4230 92

42300 2 1     

  nnnnn

n

t y y y y y y

III. REMARKABLE OBSERVATIONS

1) The solutions of (1) in terms of special integers namely, generalized Fibonacci

GF

n and

Lucas

GL

n are exhibited below.

)

1

,

46

(

)

2

(

92

)

1

,

46

(

)

2

(

2

1 1

1

t n t n

n

GL

GF

x

)

1

,

46

(

)

2

(

528

)

1

,

46

(

2

)

2

(

23

1 1

1

n t n

t

n

GL

GF

(7)

2) Employing The 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 2 2

)

2

(

64

33

Y

t

X

46

x

n2

2114

x

n1

,

368

x

n1

8

x

n2

2 2 2 2

)

2

(

135424

33

Y

t

X

46

x

n3

97198

x

n1

,

16920

x

n1

8

x

n3

3 2 2 2

)

2

(

4

33

Y

t

X

46

y

n1

264

x

n1

,

46

x

n1

8

y

n1

4 2 2 2

)

2

(

2116

33

Y

t

X

46

y

n2

12144

x

n1

,

2114

x

n1

8

y

n2

5 2 2 2

)

2

(

4468996

33

Y

t

X

(

46

y

n3

558360

x

n1

,

97198

x

n1

8

y

n3)

6 2 2 2

)

2

(

64

33

t

Y

X

2114

x

n3

97198

x

n2

,

16920

x

n2

368

x

n3

7 2 2 2

)

2

(

2116

33

Y

t

X

2114

y

n1

264

x

n2

,

46

x

n2

368

y

n1

8 2 2 2

)

2

(

4

33

Y

t

X

2114

y

n2

12144

x

n2

,

2114

x

n2

368

y

n2

9 2 2 2

)

2

(

2116

33

t

Y

X

2114

y

n3

558360

x

n2

,

97198

x

n2

368

y

n3

10 2 2 2

)

2

(

4468996

33

Y

t

X

97198

y

n1

264

x

n3

,

46

x

n3

16920

y

n1

11 2 2 2

)

2

(

2116

33

Y

t

X

97198

y

n2

12144

x

n3

,

2114

x

n3

16920

y

n2

12 2 2 2

)

2

(

4

33

Y

t

X

97198

y

n3

558360

x

n3

,

97198

x

n3

16920

y

n3

13 2 2 2

)

2

(

2112

33

X

Y

t

368

y

n1

8

y

n2

,

46

y

n2

2114

y

n1

14 2 2 2

)

2

(

4468992

33

X

Y

t

16920

y

n1

8

y

n3

,

46

y

n3

97198

y

n1

15 2 2 2

)

2

(

2112

(8)

3) 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

2

t

X

132

Y

2

8

(

4

)

t

2 1

2 2 3

2

2114

,

368

8

46

x

n

x

n

x

n

x

n

2

184

(

2

)

t

X

33

Y

2

67712

(

4

)

t

3 1

2 2 4

2

97198

,

16920

8

46

x

n

x

n

x

n

x

n

3 t t

Y

X

33

2

(

4

)

2

2

46

y

2n2

264

x

2n2

,

46

x

n1

8

y

n1

4 t t

Y

X

33

1058

(

4

)

)

2

(

23

2

46

y

2n3

12144

x

2n2

,

2114

x

n1

8

y

n2

5 t t

Y

X

33

2234498

(

4

)

)

2

(

1057

2

46

y

2n4

558360

x

2n2

,

97198

x

n1

8

y

n3

6

4

(

2

)

t

X

33

Y

2

32

(

4

)

t

2114

2 4

97198

2 3

,

16920

2

368

3

 

n n

n

n

x

x

x

x

7 t t

Y

X

33

1058

(

2

)

)

2

(

23

2

2114

y

2n2

264

x

2n3

,

46

x

n2

368

y

n1

8 t t

Y

X

33

2

(

4

)

2

2

2114

y

2n3

12144

x

2n3

,

2114

x

n2

368

y

n2

9

23

(

2

)

t

X

33

Y

2

1058

(

4

)

t

3 2

3 2 4

2

558360

,

97198

368

2114

y

n

x

n

x

n

y

n

10 t t

Y

X

33

2234498

(

4

)

)

2

(

1057

2

97198

y

2n2

264

x

2n4

,

46

x

n3

16920

y

n1

11 t t

Y

X

33

1058

(

4

)

)

2

(

23

2

97198

y

2n3

12144

x

2n4

,

2114

x

n3

16920

y

n2

12 t t

Y

X

33

2

(

4

)

2

2

97198

y

2n4

558360

x

2n4

,

97198

x

n3

16920

y

n3

13

132

(

2

)

t

X

Y

2

1056

(

4

)

t

1 2

3 2 2

2

8

,

46

2114

368

y

n

y

n

y

n

y

n

14 t t

Y

X

2234496

(

4

)

)

2

(

6072

2

16920

y

2n2

8

y

2n4

,

46

y

n3

97198

y

n1

15 t t

Y

X

1056

(

4

)

)

2

(

(9)

4) Employing the following solution

(

x

,

y

)

, each of the following expressions among the special polygonal, pyramidal, star, pronic and centered polygonal number is congruent to under modulo 4.

2

1 3 2

2 , 3

3 2

Pr

198

3

 

x x y

y

P

t

P

2

2 3

2 2

1 5

1

1188

1

12

x

x y

y

S

P

S

P

2

1 5 2

, 3

5

1

396

x

x y

y

S

P

t

P

2

, 6

5 2

, 4

5

198

1

2

x

x y

y

Ct

P

Ct

P

2

, 4

5 2

3 3

3

1

132

4

 

x x

n n

Ct

P

P

Pt

IV. CONCLUSION

To conclude, one may search for other choices of positive Pell equations for finding their integer solutions with suitable properties.

REFERENCES

[1] L.E.Dickson, History of Theory of numbers, Vol.2, Chelsea Publishing Company, New York, (1952). [2] L.J.Mordell, Diophantine Equations, Academic Press, New York, (1970).

[3] M.A.Gopalan and G.Janaki, Observation On

y

2

3

x

2

1

, Acta Ciancia Indica, XXXIVM(2) (2008) 639-696. [4] L.E.Dickson, History of theory of numbers and Diophantine analysis, Vol 2, Dover publications, New York,(2005).

[5] Dr.G.Sumathi “Observations on the hyperbola”

y

2

182

x

2

14

,

Journal of mathematics and Informatics,Vol 11, 73-81, 2017.

[6] Dr.G.Sumathi “Observations on the Pell Equation

y

2

14

x

2

4

” International Journal of Creative Research Thoughts, Vol 6, Issue 1, Pp1074-1084,March 2018.

[7] Dr.G.Sumathi “Observations on the Equation

y

2

312

x

2

1

” International Journal of Mathematics Trends and Technology, Vol 50, Issue4, 31-34, Oct 2017.

Figure

Table 1: Numerical examples

References

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