# IRJET-On the Homogeneous Ternary Quadratic Diophantine Equation 3(X+Y)2-2xy=12z2

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On the Homogeneous Ternary Quadratic Diophantine Equation
3x  y   2 xy  12 z 2
2
H. Ayesha Begum1, T.R. Usha Rani2
1PG
Scholar, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-2, Tamil Nadu, India.
Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-2, Tamil Nadu, India.
---------------------------------------------------------------------***----------------------------------------------------------------------
2Assistant
 12ab,2a
Abstract – The ternary quadratic equation given by
2
3x  y   2 xy  12 z 2 is considered and searched for its many
different integer solutions. Eight different choices of integer
solutions of the above equations are presented. A few
interesting relations between the solutions and special
polygonal numbers are presented.
2
 10b 2  8ab, a 2  5b 2  ,
  18 A 2  90 B 2  72 AB,


,
 24 A 2  120 B 2  12 AB,9 A 2  45B 2 


  112 A 2  560 B 2  308 AB,


,
126 A 2  630 B 2  168 AB,49 A 2  245B 2 


Key Words: ternary quadratic, integer solutions. MSC
subject classification: 11D09.
2a
 10b 2  8ab,12ab, a 2  5b 2
2
1. INTRODUCTION
The diophantine equations offer an unlimited field for
research due to their variety [1-3]. In particular, one may
refer [4-15] for quadratic equations with three unknowns.
This communication concerns with yet another interesting
equation
representing
3x  y   2 xy  12 z
homogeneous quadratic equation with three unknowns for
determining its infinitely many non-zero integral points.
Also, few interesting relations among the solutions are
presented.
2
Introduction of the linear transformations
u  v  0
x uv , y  u v
(2)
t m,n  n term of a regular polygon with m sides.
th
 n  1m  2 
 n1 

2



2
However, we have solutions for (1), which are
illustrated below:
2
2. Notations

,
  216 A  1080 B  108 AB ,


,
126 A 2  630 B 2  792 AB ,81A 2  405B 2 


2
2
  1134A  5670B  1512AB,



2
2
2
2 
 504A  2520B  4788AB,441A  2205B 
2
Triangular number of rank n, T3,n 
5u 2  v 2  6 z 2
Different patterns of solutions of (1) are presented below.
nn  1
2
3.1. PATTERN-1
Write ‘6’ as
3. Method of Analysis:

The ternary quadratic diophantine equation to be solved for
its non-zero distinct integral solution is
3x  y   2 xy  12z 2
2
(3)

6  1 i 5 1 i 5
Assume
(1)
Note that (1) is satisfied by the following non-zero integer
solutions.

(4)
z  a 2  5b 2
where a and
(5)
b are non- zero distinct integers.
Using (4) and (5) in (3), we get



5u 2  v 2  1  i 5 1  i 5 a 2  5b 2

2
Equating the positive and negative factors, the resulting
equations are
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 

5u  1  i 5 a  i 5b
v  i 5u  1  i 5 a  i 5b
2
v i
2
v
(7)
Replacing
Equating real and imaginary parts in (6), we get
v  a  5b  10ab
2
Substituting the values of
u and v in (2) we get,
x  xa, b  2a 2  10b 2  8ab
(8)
y  ya, b  12ab
(9)
a and b by 3 A and 3B
1
9 A2  45B 2  126 AB
3
v
1
63 A2  315B 2  90 AB
3
Substituting the values of

xa,1  ya,1  4t3,a  2 Pra  2t 4,a  0mod 2
xa,1  ya,1  za,1  4t3,a  23 Pra  23t 4,a  0mod 3

za, a  1  4t 4,a  10 Pra  0mod 5

(14)
PROPERTIES:



3.2. PATTERN-2
x1, a   z1, a   75 Pra  174t3,a  87t 4,a  0mod 3
xa  1,1  ya  1,1  12t3,a  90 Pra  90t 4,a  0mod 2
z1, a   y1, a   90t3,a  27 Pra  27t 4,a  27 is
a cubical integer.
Write ‘6’ as
3.3. PATTERN-3
6
7  i 5 7  i 5 
Write ‘6’ as
(10)
9
6
Using (5) and (10) in (3), we get
5u 2  v 2 
7  i 5 7  i 5  a
9
2
 5b 2

7  i 5  a  i 5b
3
2
7  i 5  a  i 5b
5u 
3
2
5u 2  v 2 

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(11)
17  i 5 17  i 5 a  5b 
49
2
2 2
v  i 5u 
17  i 5  a  i 5b
7
(16)
v  i 5u 
17  i 5  a  i 5b
7
(17)
(12)

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(15)
49
Equating the positive and negative factors, the resulting
equations are
Equating real and imaginary parts in (11), we get
1 2
a  5b 2  14ab
3
17  i 5 17  i 5 
Using (5) and (15) in (3), we get
2
Equating the positive and negative factors, the resulting
equations are
u
(13)
Thus (13) and (14) represent the distinct non-zero integral
solutions of (1) in two parameters.

v i
u and v in (2) we get,
2
2
and from (5) z  z A, B   9 A  5B
PROPERTIES:
v  i 5u 
respectively, we get
x  x A, B   24 A2  120 B 2  12 AB 

y  y A, B   18 A2  90 B 2  72 AB 
Thus (8), (9) and (5) represent the distinct non-zero integral
solutions of (1) in two parameters.


u
u  a 2  5b 2  2ab
2

1 2
7a  35b 2  10ab
3
(6)
2
2
Equating real and imaginary parts in (16), we get
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
u
1
17a 2  85b 2  10ab
7
v
1 2
a  5b 2  34ab
7

Replacing
a and b
by
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

v  i 5u 
17  i 5 2  i 5  a  i 5b
7
3
v  i 5u 
17  i 5 2  i 5 a  i 5b
7
3
2
2


u
1
833 A2  4165B 2  490 AB
7
u
1
19a 2  95b 2  58ab
21
v
1
49 A2  245B 2  1666 AB
7
v
1
29a 2  145b 2  190ab
21
u and v
in (2) we get,
x  x A, B   126 A2  630 B 2  168 AB 

y  y A, B   112 A2  560 B 2  308 AB 

and from z  z A, B  49 A 2  5B 2

(18)
(19)
b

by 21A and 21B respectively, we get
u
1
8379 A2  41895B 2  22578 AB
21
v
1
12789 A2  63945B 2  83790 AB
21
Substituting the values of
u and v in (2) we get,
x  x A, B   1008 A2  5040 B 2  2772 AB 

y  y A, B   210 A2  1050 B 2  520 AB 
PROPERTIES:


Replacing a and
Thus (18), and (19) represent the distinct non-zero integral
solutions of (1) in two parameters.

(23)
Equating real and imaginary parts in (22), we get
7 A and 7 B respectively, we get
Substituting the values of
(22)
ya, a  1  za, a  1  1918 Pra  917t 4,a  0mod 5

(24)

x1, a   z1, a   770t3,a  217 Pra  217t 4,a  0mod 3
and from z  z A, B  441 A 2  5B 2
is a cubical number.
Thus (24), and (25) represent the distinct non-zero integral
solutions of (1) in two parameters.
 10xa,1  ya,1  378t 4,a  140 Pra   0mod 119
3.4. PATTERN-4
PROPERTIES:
One may write (3) as
5u  v  6 z *1
2
2
2
(20)

x1, a   y1, a   z1, a   315t 4,a  15960t3,a  0mod 7

xa, a  1  za, a  1  8694t 4,a  17703 Pra  0mod 3

ya,1  za,1  462t3,a  4977 Pra  4977t 4,a  0mod 3
Write ‘1’ as
2  i 5 2  i 5 
1
9
(21)
3.5. PATTERN-5
Using (4), (5) and (21) in (20), we get
5u 2  v 2 
(25)
17  i 5 17  i 5 2  i 5 2  i 5  a
49
9
5u 2  v 2  6 z 2
2
 5b 2
z

2
2
 
 v2  5 u2  z 2

z  vz  v  5u  z u  z 
Equating the positive and negative factors, the resulting
equations are
(26)
Equation (26) is written in the form of ratio as
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zv
uz p

 ,q  0
5u  z  z  v q
vu
z u p

 ,q  0
6z  u  v  u q
(27)
From the First and third factors of (27), we have
From the First and third factors of (32), we have
zv
p

5u  z  q
v u
p

6z  u  q
z  vq  5u  v p  0
uq  6 p   vq  6 zp  0
(28)
From the second and third factors of (28), we have
z u
p

v  u  q
u  z q  z  v p  0
uq  p   vp  zq  0
(29)
Applying the method of cross multiplication for solving (28)
and (29),
Applying the method of cross multiplication for solving (33)
and (34),
u  6 p 2  q 2
v  5 p 2  q 2  10 pq
v  6 p 2  q 2  12 pq
z  5 p 2  q 2
z  6 p 2  q 2  2 pq
v in (2) we get
x  x p, q   10 p 2  2q 2  8 pq 

y  y p, q   12 pq

Thus (30) along with the value of
solutions to (1)
z
Substituting the values of
represent the integer


ya,1  za,1  24t3,a  7t 4,a  0mod1

z
(35)
represent the integer
xa,1  a   ya,1  a   28 Pra  10t 4,a  0mod 2
z1, a  1  8t3,a  3t 4,a  0mod 3
y1, a   z1, a   11t 4,a  14 Pra  0 is a nasty
number.
3.6. PATTERN-6
v
in (2) we get
v
PROPERTIES:
xa,1  a   ya,1  a   za,1  a   20t 4,a  0mod 1
 
u  6 z u
2
and
Thus (35) along with the value of
solutions to (1)
x1, a   z1, a   5 Pra  6t3,a  5t 4,a  0mod 5
2
u
x  x p, q   12 p 2  12 pq 


2
y  y p, q   2q  12 pq 

(30)
PROPERTIES:

(34)
u  5 p 2  q 2  2 pq
Substituting the values of u and

(33)
From the second and third factors of (32), we have
uz
p

z  v  q

(32)
2
2
3.7. PATTERN-7

v  u v  u   6z  u z  u 
Equation (3) is written in the form of ratio as
uz
3u  z  p

 ,q  0
2u  z  u  v
q
(31)
One may write equation (31) in the form of ratio as
(36)
From the First and third factors of (32), we have
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 6z  v 6z  v   6a  b  6a  b  6  1 6  1
uv
p

2u  v  q
2
 uq  2 p   vq  2 zp  0
Equating positive and negative factors, the resulting
equations are
(37)


From the second and third factors of (36), we have
3u  z  p

u  v  q
 
6z  v  
6z  v 
(38)
v
z  2 p 2  3q 2  2 pq
y  ya, b  2b 2  12ab
and
(39)
z
(46)
(47)
x1, a  1  y1, a  1  28 Pra  26t 4,a  0mod 2

represent the integer
y1, a   z1, a   9t 4,a  20t 3,a  0 is a nasty

number.
za, a  1  18t 3,a  5 Pra  5t 4,a  0mod 1

za,1  a   Pra  6t3,a  t 4,a  0mod 3
4. REMARKABLE OBSERVATIONS:
x1, a   y1, a   z1, a   26 Pra  35t 4,a  0 is a
Let
nasty number.
za,1  xa,1  20t3,a  12t 4,a  0mod 3
u , v , z  be any given integer solution of (3),
0
0
0
Then, each of the following triples of non-zero distinct
integers based on u0 , v0 , z0 also satisfies (1).
3.8. PATTERN-8
4.1.
Equation (3) can be written as
6 z 2  v 2  5u 2
u  6a 2  b 2

6 1
Triple 1:
u
0
 h, v0 , z0  h
Here,
(40)



 



 
1
n
n
12   1 10 u 0   12   1 12 z 0  v0
2
1
n
n
yn 
12   1 10 u 0   12   1 12 z 0  v0
2
1
n
n
zn 
10   1 10 u 0   10   1 12 z 0
2
xn 
(41)
Write ‘5’ as

in (2) we get
PROPERTIES:
PROPERTIES:
6 1
(44)
Thus (45), (46) and (47) represent the distinct non-zero
integral solutions of (1) in two parameters.
v in (2) we get
Thus (39) along with the value of
solutions to (1)

and
x  xa, b  12a 2  12ab
u
(43)
(45)
u
Substituting the values of
x  x p, q   4 p 2  12 pq 


2
y  y p, q   6q  12 pq 

5
2
v  2 p 2  3q 2  12 pq
Substituting the values of
Assume
6 1
z  6a 2  b 2  2ab
u  2 p 2  3q 2


6  1
2
v  6a 2  b 2  12ab
Applying the method of cross multiplication for solving (37)
and (38),


6a  b  
6a  b
Equating rational and irrational parts in (43), we get
 u3q  p   vp  3zq  0

2

(42)


 
Using (41), (42) in (40) and employing the method of
factorization the above equation (40) is written as
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4.2.
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h  6u
Triple 2:
0
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, h  6v0 ,6 z0 
forms of three dimensional surfaces, namely, nonhomogeneous cone, paraboloid, ellipsoid, hyperbolic
paraboloid and so on for finding integral points on them and
corresponding properties.
Here,
xn 
1
36
206


 REFERENCES
 8 6  u 0  46   8 6  v0
n
n


n

n
 
1
n
n
12 1 u0  12 6 v0
2
yn 
[1]
L.E.Dickson, “History of Theory of Numbers and
Diophantine Analysis”, vol.2, Dover publications, New
York 2005.
[2]
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
Hyberboloid
of
two
sheets
Impact
J.
X 2  6 XY  Y 2  6 X  2Y  5  Z 2  4 ,
Sci.Tech., 4 (2010) 23-32.
[5]
M.A.Gopalan, S.Vidhyalakshmi and A.Kavitha, Integral
points on the Homogeneous cone Z 2  2 X 2  7Y 2 , The
Diophantus J Math., 1(2) (2012) 127-136.
[6]
M.A.Gopalan, S.Vidhyalakshmi and G.Sumathi, Lattice
points
on
the
Hyperboloid
one
sheet
4Z 2  2 X 2  3Y 2  4 , The Diophantus J Math., 1(2)
(2012) 109-115.
[7]
M.A.Gopalan, S.Vidhyalakshmi and K.Lakshmi, Integral
points on the Hyperboloid two
sheet
3Y 2  7 X 2  Z 2  21 , The Diophantus J Math., 1(2)
(2012) 99-107.
[8]
M.A.Gopalan,
Observations
[9]
M.A.Gopalan, S.Vidhyalakshmi, T.R.Usha Rani and
S.Mallika, Integral points on the Homogeneous cone
6Z 2  3Y 2  2 X 2  0 , 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
Z  9 X 2  4Y 2 ,
“Advanceds in Theoretical and Applied Mathematics”,
7(4) (2012) 349-385.
[11]
M.A.Gopalan, S.Vidhyalakshmi and T.R.Usha Rani,
Integral points on the Non-Homogeneous cone
2Z 2  4 XY  8 X  4Z  0 , “Global Journal of Mathematics
and Mathematical Science”, 2(1) (2012) 61-67.
zn  6n z0
4.3.
8u
Triple3:
,8v0  h,8z0  h 
0
Here,
1
16 6 An v0  96Bn z0 
32 6
1
16 6 An v0  96Bn z0 
yn  8n u0 
32 6
1
zn 
16 Bn v0  16 6 An z0
32 6
xn  8 n u 0 

where

 

 40  16 6   40  16 6 
n
An  40  16 6  40  16 6
n
Bn
4.4.

Triple 4:
 3u
Here,
1
xn 
12 30
yn 
zn 
12 30
12 30
Where
6
1
1
0
n
 h,3v0 ,3z 0  h
30 An u0  36 Bn z0  3n v0
6
30 An u0  36 Bn z0  3n v0
 30B u
n 0

 6 30 An z0

 

30    33  6 30 
n
An   33  6 30   33  6 30

Bn   33  6
5.
n
n
n
n
CONCLUSION
In this paper, we have presented infinitely many non-zero
distinct integer solutions to the ternary quadratic equation
3x  y   2 xy  12 z 2
2
S.Vidhyalakshmi
and
S.Mallika,
on Hyperboloid of
one sheet
X 2  2Y 2  Z 2  2 , Bessel J. Math., 2(3) (2012) 221-226.
representing a homogeneous cone. As diophantine equation
are rich in variety, to conclude, one may search for other
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International Research Journal of Engineering and Technology (IRJET)
e-ISSN: 2395-0056
Volume: 06 Issue: 03 | Mar 2019
p-ISSN: 2395-0072
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[12]
M.A.Gopalan, S.Vidhyalakshmi and K.Lakshmi, Lattice
points on the Elliptic paraboloid 16Y 2  9Z 2  4 X ,
Bessel J. Math., 3(2) (2013) 137-145.
[13]
K.Meena, S.Vidhyalaksmi E.Bhuvaneswari and
equation” 5 X 2  Y 2  6 XY  20Z 2 ,International Journal
of Advanced Scientific Research, 1(2) (2016) 59-61.

[14]
M.A.Gopalan, S.Vidhyalakshmi and U.K.Rajalakshmi, “On
ternary
Diophantine
equation”
5 X 2  Y 2  6 XY  196Z 2 , Journal of Mathematics, 3(5)
(2017) 1-10.

[15]


M.A.Gopalan, S.Vidhyalakshmi and S.Aarthy Thangam,