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

## Full text

(1)

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

2

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: kselvakeerthana273@gmail.com

2

e-mail: msmallika65@gmail.com

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.

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 +yxy+ 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

(2)

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 +yxy+ x+y + = z (1) The substitution of linear transformations

0 ,

, = − ≠ ≠

+

=u v y u v u v

x (2)

in

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= +ii (6)

Substituting

4, 5, 6 in

### ( )

3 and using the method of factorization we get,

U +i 11v

Ui 11v

= a+i 11b

2 ai 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

+ = =

− − − = =

− − − = =

(3)

2 2

### )

2

15 4 2

5

3 x +yxy+ x+y + = z

23

Properties:

1.x

a,a+1

−3y

a,a+1

−60PRa =0

2.y

### ( ) ( )

a,az a,a +48t4,a ≡0

mod2

3.x

### ( ) ( )

a,1 +z a,1 −t10,a ≡9

mod15

3.2. Pattern- 2

### ( )

6 , 15 can be written as

## )

4

11 7 11 7

15= +ii

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,aaSO2 ≡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

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

(4)

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

kGk −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

ky kt90,k ≡56

mod147

2.x

### ( ) ( )

k +yk +176t3,k ≡0

mod4

3.x

k +z kGk ≡5

mod50

System 3:

Consider 165

= +σ

X

2

T X−σ =

(5)

2 2

### )

2

15 4 2

5

3 x +yxy+ 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

ky kz kt6,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

kGk +1≡0

mod14

2.x

### ( ) ( )

ky 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 =

(6)

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

### ( ) ( )

ky 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

### ( ) ( ) ( )

ky kz k −110PRk ≡10

mod38

3.4. Pattern- 4

### ( )

3 can be rewritten as

2 2

2 2

15

4v z v

U − = − (21)

(7)

2 2

### )

2

15 4 2

5

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

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