On the Ternary Quadratic Diophantine Equation $3(x^2+y^2)-5xy+2(x+y)+4=15z^2$

Vol. 11, 2017, 21-28

ISSN: 2349-0632 (P), 2349-0640 (online) Published 11 December 2017

www.researchmathsci.org

DOI: http://dx.doi.org/10.22457/jmi.v11a4

21

Journal of

On the Ternary Quadratic Diophantine Equation

3(x

2

+y

2

)-5xy+2(x+y)+4=15z

2

K.Selva Keerthana1 and S.Mallika2

Department of Mathematics, Shrimati Indira Gandhi College Trichy-2, Tamilnadu, India. 1e-mail: [email protected]

2

e-mail: [email protected]

Received 1 November 2017; accepted 8 December 2017

Abstract. The ternary non-homogeneous quadratic equation is analysed for its non-zero distinct integer solutions. A few interesting relations between the solutions and special polygonal and pyramidal numbers are presented.

Keywords: Ternary quadratic, integer solutions, non-homogeneous quadratic, polygonal numbers, pyramidal numbers.

AMS Mathematics Subject Classification (2010):11D09

1. Introduction

The Diophantine equation offer an unlimited field for research due to their variety

[ ]

1−3 . In particular, one may refer

[

4−12

]

for quadratic equation with three unknowns. The communication concerns with yet another interesting equation

(

2 2

)

(

)

2

15 4 2

5

3x +y − xy+ x+y + = z representing non-homogeneous quadratic equation with three unknowns for determining its infinitely many non-zero integral points. Also, a few interesting relations among the solutions are presented.

2. Notations

1. Polygonal number of rank n with side m

(

)(

)

   



 − −

+ =

2 2 1 1

,

m n n tmn

2. Gnonomic number of rank n

(

2 −1

)

= n

Gn

3. Pronic number of rank n

(

+1

)

=nn PRn

4. Centered Hexagonal pyramidal number of rank n

3 6

, n

22

(

)

2 2 1

,

+ − = mn n Ctmn

6. Stella Octangular number of rank n

(

2 2−1

)

=n n SOn

3. Method of analysis

The equation representing the ternary quadratic equation to be solved for its non-zero distinct integer solution is

(

2 2

)

(

)

2

15 4 2

5

3x +y − xy+ x+y + = z (1) The substitution of linear transformations

0 ,

, = − ≠ ≠

+

=u v y u v u v

x (2)

in

( )

1 leads to

2 2 2

15

11v z

U + = (3)

where 2

− =u

U (4)

Assume that

2 2

b a

z= + (5)

3.1. Pattern-1

Write 15 as

(

2 11

)(

2 11

)

15= +i −i (6)

Substituting

( ) ( ) ( )

4, 5, 6 in

( )

3 and using the method of factorization we get,

(

U +i 11v

)(

U−i 11v

) (

= a+i 11b

) (

2 a−i 11b

) (

2 2+i 11

)(

2−i 11

)

(7) Equating the positive and negative factors, the resulting equations are

(

) (

)(

)

2

11 11

2

11v i a i b

i

U + = + + (8)

(

) (

)(

)

2

11 11

2

11v i a i b

i

U − = − − (9)

Equating the real and imaginary parts in

( )

8

ab b a v

ab b

a U

4 11

22 22 2

2 2

2 2

+ − =

− −

= ₍₁₀₎

In view of

( )

4

ab b a v

ab b

a u

4 11

2 22 22 2

2 2

2 2

+ − =

− − −

= ₍₁₁₎

Substituting

( )

11 in

( )

2 ,

( )

( )

( )

2 2

2 2

2 2

11 ,

2 26 11 ,

2 18 33 3 ,

b a b a z z

ab b

a b a y y

ab b

a b a x x

+ = =

− − − = =

− − − = =

(

2 2

)

(

)

2

15 4 2

5

3 x +y − xy+ x+y + = z

23

Properties:

1.x

(

a,a+1

)

−3y

(

a,a+1

)

−60PRa =0

2.y

( ) ( )

a,a −z a,a +48t4,a ≡0

(

mod2

)

3.x

( ) ( )

a,1 +z a,1 −t10,a ≡9

(

mod15

)

3.2. Pattern- 2

Instead of

( )

6 , 15 can be written as

(

)(

)

4

11 7 11 7

15= +i −i

Proceeding as in Pattern: 1, the non-zero distinct integral solutions to

( )

1 as

( )

( )

( )

2 2

2 2

2 2

11 ,

2 18 33 3 ,

2 4 44 4 ,

b a b a z z

ab b

a b a y y

ab b a b a x x

+ = =

− − − = =

− − − = =

Properties:

1.x

( )

a,1 +4z

( )

a,1 −t18,a ≡1

(

mod3

)

2.x

( ) ( )

a,1 −ya,1 −t4,a −aSO2 ≡0

(

mod11

)

3.z

( )

a,a −12t4,a =0

3.3. Pattern-3

The substitution of linear transformation

σ

2 , 15 ,

11 = + =

+

= X T V X T U

z (12)

in

( )

3 leads to

2 2 2

165 +σ

= T

X

2 2

2

165T

X −σ = (13)

Write

( )

13 as

(

)(

)

2

165T

X

X +σ −σ = (14)

The equation

( )

14 is written as the system of two equations as follows

System 1 2 3 4 5 6

σ

+

X 15 T 15 165 55 T 55 2

55T σ

−

X 11 T 2

11T 2

T 3 T 2

3T 3

System 1:

Consider T X+σ =15

T X−σ =11

24 

  

= = =

k T

k 2 4

σ (15)

Substituting

( )

15 in

( )

12 and

( )

2 , we get the corresponding non-zero distinct integer

solutions to

( )

1 as

( )

( )

( )

k k z

k k

y

k k x

48 2 48

2 64

− =

− − =

− =

Properties:

1.6

(

x

( ) ( )

1 +y1

)

is a nasty number 2.z

( )

k −Gk −1≡0

(

mod46

)

3.

( ) ( ) ( )

624, 0

(

mod2

)

2 2

2 + + − ≡

k

t k z k y k x

System 2:

Consider 15

= +σ

X

2

11T

X −σ =

And solving we get

    

+ =

+ − − =

+ + =

1 2

2 22 22

30 8 22

2 2

k T

k k

k k X

σ (16)

Substituting

( )

16 in

( )

12 and

( )

2 , we get the corresponding distinct non-zero integral solutions to

( )

1 as

( )

( )

( )

22 44 24 26 96 66

30 8 22

2 2 2

+ + =

− − − =

+ + − =

k k k z

k k k

y

k k k

x

Properties:

1.x

( ) ( )

k −y k −t90,k ≡56

(

mod147

)

2.x

( ) ( )

k +yk +176t3,k ≡0

(

mod4

)

3.x

( ) ( )

k +z k −Gk ≡5

(

mod50

)

System 3:

Consider 165

= +σ

X

2

T X−σ =

(

2 2

)

(

)

2

15 4 2

5

3 x +y − xy+ x+y + = z

25 

   

− =

+ + − =

+ − =

1 2

82 2 2

83 2 2

2 2

k T

k k

k k X

σ (17)

Substituting

( )

17 in

( )

12 and

( )

2 , we get the corresponding non-zero distinct integer solution to

( )

1 as

( )

=−2 2+32 +230

k k k x

( )

=−6 2 −24 +94

k k k y

( )

=2 2+20 +72

k k k z

Properties: 1.y

( )

1 −CP4,6 =0

2.x

( ) ( )

k +yk +16t3,k ≡4

(

mod16

)

3.x

( ) ( ) ( )

k −y k −z k −t6,k ≡27

(

mod37

)

System 4:

Consider T X+σ =55

T X−σ =3

And solving we get,

    

= = =

k T

k k X

2 52

58

σ (18)

Substituting

( )

18 in

( )

12 and

( )

2 , we get the corresponding distinct non-zero integral

solutions to

( )

1 as

( )

k =192k−2 x

( )

k =16k−2 y

( )

k k z =80

Properties:

1.y

( )

k −Gk +1≡0

(

mod14

)

2.x

( ) ( )

k −y k +2Ct176,k −176PRk ≡2

(

mod176

)

3.

( ) ( )

644, 0

(

mod2

)

2

2 − − ≡

k

t k y k z

System 5:

Consider 55

= +σ

X

2

3T

X −σ =

26 

  

+ =

+ − − =

+ + =

1 2

26 6 6

29 6 6

2

k T

k k

k k X

σ (19)

Substituting

( )

19 in

( )

12 and

( )

2 , we get the corresponding non-zero distinct integer solutions to

( )

1 as

( )

=−6 2+24 +94

k k k x

( )

=−18 2−48 +6

k k k

y

( )

=6 2 +28 +40

k k k z

Properties:

1.x

( ) ( )

k −y k −24t3,k ≡ 28

(

mod60

)

2.y

( ) ( )

1 +z1 −SO2 =0

3.x

(

2k−1

) (

+z 2k−1

)

−52Gk ≡0

(

mod134

)

System 6:

Consider

2

55T

X +σ = 3

= −σ

X

And solving we get,

    

+ =

+ + =

+ + =

1 2

26 110 110

29 110 110

2 2

k T

k k

k k

X

σ (20)

Substituting

( )

20 in

( )

12 and

( )

2 , we get the corresponding distinct non-zero distinct integral solutions to

( )

1 as

( )

=330 2+330 +94

k k

k x

( )

=110 2+80 +6

k k k

y

( )

=110 2+132 +40

k k

k z

Properties:

1.z

(

k

(

2k−1

)

)

−y

(

k

(

2k−1

)

)

−52t6,k ≡0

(

mod2

)

2.y

( )

1 −t4,14 =0

3.x

( ) ( ) ( )

k −y k −z k −110PRk ≡10

(

mod38

)

3.4. Pattern- 4

( )

3 can be rewritten as

(

2 2

)

2 2

15

4v z v

U − = − (21)

(

2 2

)

(

)

2

15 4 2

5

3 x +y − xy+ x+y + = z

27

(

)

_{, 0}

2 15 2 ≠ = − − = + + _β β α v U v z v z v U

which is equivalent to the following two equations,

(

2 −

)

− =0

+ v z

Uβ β α α

(

15 −2

)

−15 =0

+ β α β

α v z

U

On employing the method of cross multiplication, we get

2 2

30 30

2α − αβ+ β

= U 2 2 15β α − = v 2 2 15

4αβ β

α − +

= z

In view of

( )

4 ,

2 2 2 2 2 2 15 4 15 2 30 30 2 β αβ α β α β αβ α + − = − = − + − = z v u (22)

Substituting

( )

22 in

( )

2 , we get

( )

( )

( )

     + − = = − + − = = − + − = = 2 2 2 2 2 2 15 4 , 2 45 30 , 2 15 30 3 , β αβ α β α β αβ α β α β αβ α β α z z y y x x (23)

Thus

( )

23 represent non-zero distinct integral solutions of

( )

1

Properties:

1.x

( ) ( )

α,1 −yα,1 −2t4,α ≡0

(

mod3

)

2.z

(

α,α

)

−12t4,α =0

3.3x

( ) ( )

α,1 −yα,1 −t18,α ≡49

(

mod53

)

4. Conclusion

In this paper, we have made an attempt to obtain all integer solutions to ternary quadratic

equation

(

2 2

)

(

)

2

15 4 2

5

3x +y − xy+ x+y + = z .One may search for other patterns of solutions and their corresponding properties.

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6 2 2

2 − + + − + = +

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7

2x y

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