*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**

**3(x**

**2**

**+y**

**+y**

**2**

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

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

**2****K.Selva Keerthana****1 ****and S.Mallika****2**

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### )

### (

### )

215 4 2

5

3*x* +*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### )

### (

### )

215 4 2

5

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

0 ,

, = − ≠ ≠

+

=*u* *v* *y* *u* *v* *u* *v*

*x* (2)

in

### ( )

1 leads to2 2 2

15

11*v* *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*11

*v*

## )(

*U*−

*i*11

*v*

## ) (

=*a*+

*i*11

*b*

## ) (

2*a*−

*i*11

*b*

## ) (

2 2+*i*11

## )(

2−*i*11

## )

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

## ) (

## )(

## )

211 11

2

11*v* *i* *a* *i* *b*

*i*

*U* + = + + (8)

## (

## ) (

## )(

## )

211 11

2

11*v* *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 22 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### )

### (

### )

215 4 2

5

3 *x* +*y* − *xy*+ *x*+*y* + = *z*

23

**Properties: **

1.*x*

### (

*a*,

*a*+1

### )

−3*y*

### (

*a*,

*a*+1

### )

−60*PRa*=0

2.*y*

### ( ) ( )

*a*,

*a*−

*z*

*a*,

*a*+48

*t*4,

*a*≡0

### (

mod2### )

3.*x*

### ( ) ( )

*a*,1 +

*z*

*a*,1 −

*t*10,

*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 22 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 +4

*z*

### ( )

*a*,1 −

*t*18,

*a*≡1

### (

mod3### )

2.*x*

### ( ) ( )

*a*,1 −

*ya*,1 −

*t*4,

*a*−

*aSO*2 ≡0

### (

mod11### )

3.*z*

### ( )

*a*,

*a*−12

*t*4,

*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 to2 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 followsSystem 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 integersolutions to

### ( )

1 as### ( )

### ( )

### ( )

*k*

*k*

*z*

*k*
*k*

*y*

*k*
*k*
*x*

48 2 48

2 64

− =

− − =

− =

**Properties: **

1.6

### (

*x*

### ( ) ( )

1 +*y*1

### )

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 6630 8 22

2 2 2

+ + =

− − − =

+ + − =

*k*
*k*
*k*
*z*

*k*
*k*
*k*

*y*

*k*
*k*
*k*

*x*

**Properties: **

1.*x*

### ( ) ( )

*k*−

*y*

*k*−

*t*90,

*k*≡56

### (

mod147### )

2.*x*

### ( ) ( )

*k*+

*yk*+176

*t*3,

*k*≡0

### (

mod4### )

3.*x*

### ( ) ( )

*k*+

*z*

*k*−

*Gk*≡5

### (

mod50### )

**System 3: **

Consider 165

= +σ

*X*

2

*T*
*X*−σ =

### (

2 2### )

### (

### )

215 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 −*CP*4,6 =0

2.*x*

### ( ) ( )

*k*+

*yk*+16

*t*3,

*k*≡4

### (

mod16### )

3.*x*

### ( ) ( ) ( )

*k*−

*y*

*k*−

*z*

*k*−

*t*6,

*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 integralsolutions to

### ( )

1 as### ( )

*k*=192

*k*−2

*x*

### ( )

*k*=16

*k*−2

*y*

### ( )

*k*

*k*

*z*=80

**Properties: **

1.*y*

### ( )

*k*−

*Gk*+1≡0

### (

mod14### )

2.*x*

### ( ) ( )

*k*−

*y*

*k*+2

*Ct*176,

*k*−176

*PRk*≡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*−24

*t*3,

*k*≡ 28

### (

mod60### )

2.*y*

### ( ) ( )

1 +*z*1 −

*SO*2 =0

3.*x*

### (

2*k*−1

### ) (

+*z*2

*k*−1

### )

−52*Gk*≡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*

### (

2*k*−1

### )

### )

−*y*

### (

*k*

### (

2*k*−1

### )

### )

−52*t*6,

*k*≡0

### (

mod2### )

2.*y*

### ( )

1 −*t*4,14 =0

3.*x*

### ( ) ( ) ( )

*k*−

*y*

*k*−

*z*

*k*−110

*PRk*≡10

### (

mod38### )

**3.4. Pattern- 4 **

### ( )

3 can be rewritten as### (

2 2### )

2 2

15

4*v* *z* *v*

*U* − = − (21)

### (

2 2### )

### (

### )

215 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*

*U*β 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 −*y*α,1 −2

*t*4,α ≡0

### (

mod3### )

2.*z*

### (

α,α### )

−12*t*4,α =0

3.3*x*

### ( ) ( )

α,1 −*y*α,1 −

*t*18,α ≡49

### (

mod53### )

**4. Conclusion **

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

equation

### (

2 2### )

### (

### )

215 4 2

5

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

**REFERENCES **

1. *L.E.Dickson, History of Theory of Numbers and Diophantine Analysis, Dover *
Publications, New York 2005.

2. *L.J.Mordell, Diophantine Equations, Academic Press, New York 1970. *

3. *R.D Carmicheal, The Theory of Numbers and Diophantine Analysis, Dover *
Publications, New York 1959.

4. M.A.Gopalan and D.Geetha , Lattice Points on the Hyperboloid of Two Sheets

4 5

2 6

6 2 2

2 − + + − + = +

*z*
*y*
*x*
*y*
*xy*

*x* *, Impact J Sci Tech, 4 (2010) 23-32. *

5. 5.M.A.Gopalan , S.Vidhyalakshmi, A. Kavitha, Integral points on the homogeneous

cone 2 2 2

7

2*x* *y*

28 of one sheet4 2 =2 2+3 2−4

*y*
*x*

*z* *. The Diophantus J Math 1(2) 2012 109-115. *

7. M.A.Gopalan , S.Vidhyalakshmi.,K. Lakshmi, Integral points on the hyperboloid of two sheets 3 2=7 2− 2+21

*z*
*x*

*y* *. The Diophantus J Math 1(2) (2012) 99-107. *

8. M.A.Gopalan, S.Vidhyalakshmi and S.Mallika, Observations on hyperboloid of one sheet 2+2 2− 2 =2

*z*
*y*

*x* *. Bessel J Math., 2(3) (2012) 221-226. *

9. M.A.Gopalan , S.Vidhyalakshmi,T.R. Usha Rani ,S. Mallika, Integral points on the homogeneous cone6 2 +3 2−2 2 =0

*x*
*y*

*z* *, The Impact J Sci Tech., 6(1) (2012) 7-13. *
10. M.A.Gopalan, S.Vidhyalakshmi and G.Sumathi, Lattice points on the elliptic

paraboloid 2 2

4

9*x* *y*

*z*= + *, Advances in theoretical and Applied Mathematics, 7(4) *
(2012) 379-385.

11. M.A.Gopalan, S.Vidhyalakshm and T.R.Usha Rani, Integral points on the non-homogeneous2 2 +4 +8 −4 =0

*z*
*x*
*xy*

*z* *. Global Journal of Mathematics and *
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12. M.A.Gopalan, S.Vidhyalakshmi and K. Lakshmi, Lattice points on the elliptic
paraboloid 16*y*2+9*z*2 =4*x*