This document analyzes the ternary quadratic Diophantine equation 222 237 zyx  , which represents a homogeneous cone. Several methods are presented to find non-zero distinct integral solutions to the equation. Method I involves writing the equation as a ratio and solving a system of double equations. Method II uses factorization and representing constants as sums of squares. Method III is similar to Method II but uses a different representation of 1. Method IV introduces linear transformations to transform the equation into an equivalent form. Each method yields different sets of integer solutions along with interesting properties between the solutions and special polygonal numbers.

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A Study on the Homogeneous Cone 222
237 zyx 
S. Vidhyalakshmi1, T. Mahalakshmi2
1Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620002, TamilNadu, India
2Assistant Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620002,
TamilNadu, India
---------------------------------------------------------------------***----------------------------------------------------------------------
Abstract - The cone represented by the ternary quadratic Diophantine equation 222
237 zyx  is analyzed for itspatternsof
non-zero distinct integral solutions. A few interesting properties between the solutions and special polygonal numbers are
exhibited.
Key Words: Ternary quadratic, cone, integral solutions. 2010 Mathematics Subject Classification: 11D09
1. INTRODUCTION
The Diophantine equationoffers an unlimited field for research due to theirvariety[1-3].Inparticular,one
may refer [4-14] for quadratic equations with three unknowns. This communication concerns with yet another
interesting equation 222
237 zyx  representing non-homogeneous quadratic with three unknowns for
determining its infinitely many non-zero integral points. Also, a few interesting relations among the solutions are
presented.
2. METHOD OF ANALYSIS
The Ternary Quadratic Diophantine equation representing homogeneous cone under consideration is
222
237 zyx  (1)
We present below different methods of solving (1).
Method I:
Equation (1) is written in the form of ratio as
0,
4
)(74









zx
yz
yz
zx
(2)
which is equivalent to the system of double equations
0)4(  zyx 
0)47(7  zyx 
Applying the method of cross multiplication, the corresponding values of zyx ,, satisfying (1) are given by
 14284),( 22
x
 87),( 22
y
22
7),(  z

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Properties:
 )17(mod028)1,( ,10
 
 tx
  
 ,3
16),1(),1(21 tyz  is a nasty number.
 
 ,3
92)1,()1,(4 tzy 
Note:
Apart from (2), (1) is also written in the form of ratio as presented below:
(i)








zx
yz
yz
zx
4)(7
4
(ii)








zx
yz
yz
zx
4)(7
4
Following the above procedure, the solutions of (1) for choices (i) and (ii) are presented below:
Solutions for choice (i)
 14428),( 22
x
 87),( 22
y
22
7),(  z
Solutions for choice (ii)
 14428),( 22
x
 87),( 22
y
22
7),(  z
Method II:
Assume 22
7),( babaz  (3)
Write 23 as
16
)719)(719(
23
ii 
 (4)
Using (3) and (4) in (1) and employing the method of factorization, consider
2
)7(
4
719
7 bia
i
yix 



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Equating real and imaginary parts and replacing a by 2A, b by 2B, we have





ABBABAy
ABBABAx
387),(
1413319),(
22
22
(5)
and from (3), we have
22
284),( BABAz  (6)
Thus (5) and (6) represent the integer solutions to (1).
Properties:
 )4(mod0133)1,( ,40
 A
tAx
 ]133)1,([6 2,40
2
 
 tx is a nasty number.
 )11(mod0133)1,( ,10,32
 AA
ttAx
 )7(mod0),1(4),1( ,80
 B
tByBz
 ]),1(4),1([102 ,80 B
tByBz  is a nasty number.
 A
tAAxAAy ,3
1472)1,()1,(19 
Note:
It is seen that 23 is also represented as follows:
(iii)
64
)71317)(71317(
23
ii 
 (7)
(iv) )74)(74(23 ii  (8)
Following the above procedure, the solutions of (1) for choices (iii) and (iv) are presented below:
Solutions for choice (iii)
ABBABAx 36423834),( 22

ABBABAy 6818226),( 22

22
11216),( BABAz 
Solutions for choice (iv)
abbabax 14284),( 22

abbabay 87),( 22


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22
7),( babaz 
Method III:
Equation (1) is written as
1*237 222
zyx  (9)
Write 1 as
16
)73)(73(
1
ii 
 (10)
Substituting (3), ( 8) and (10) in (1) and following the procedure as above, the corresponding solutions to (1)
are given by
ABBABAx 98355),( 22

ABBABAy 10497),( 22

22
284),( BABAz 
Properties:
 )94(mod035)1,( ,12
 A
tAx
 ]35)1,([564 2,12
2
 
 tx is a nasty number.
 )97(mod035)1,( ,6,8
 AA
ttAx
 A
tAAxAAy ,3
1472)1,(7)1,(5 
 )351(mod0),1(4),1(7 ,84
 B
tByBz
Note:
It is seen that 1 is also represented as follows:
(v)
64
)731)(731(
1
ii 
 (11)
(vi)
121
)743)(743(
1
ii 
 (12)
Following the above procedure, the solutions of (1) for choices (v) and (vi) are presented below:

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Solutions for choice (v)
ABBABAx 36423834),( 22

ABBABAy 6818226),( 22

22
11216),( BABAz 
Solutions for choice (vi)
ABBABAx 29261232176),( 22

ABBABAy 3521463209),( 22

22
847121),( BABAz 
Method IV:
Introduction of the linear transformations
TXzTXyPx 7,23,4  (13)
in (1) leads to
222
161 PTX  (14)
which is satisfied by
2222
161,161,2 srXsrPrsT 
In view of (13), the corresponding integer solutions to (1) are given by
22
4644 srx 
rssry 46161 22

rssrz 14161 22

Also, (14) is written as the system of double equations as presented below in Table 1:
Table 1: System of double equations
System 1 2 3 4 5
PX  2
T 2
23T 2
7T 23T 161T
PX  161 7 23 7T T

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Solving each of the above systems, the values of X, P and T are obtained. Substituting these in (13), the
corresponding solutions to (1) are found. For simplicity, we present the solutions below:
Solutions for system 1:
32088 2
 KKx
104482 2
 KKy
88162 2
 KKz
Solutions for system 2:
32184184 2
 KKx
389246 2
 KKy
226046 2
 KKz
Solutions for system 3:
325656 2
 KKx
386014 2
 KKy
222814 2
 KKz
Solutions for system 4:
Tx 32
Ty 38
Tz 22
Solutions for system 5:
Tx 320
Ty 104
Tz 88
Note:
In addition to (13), one may also consider the linear transformations as
TxzTxypx 7,23,4 
The repetition of the above process leads to different sets of solutions to (1) that are exhibited below:
Set 1:

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32088 2
 KKx
58442 2
 KKy
74122 2
 KKz
Set 2:
32184184 2
 KKx
846 2
 Ky
83246 2
 KKz
Set 3:
325656 2
 KKx
83214 2
 KKy
814 2
 Kz
Set 4:
Tx 32
Ty 8
Tz 8
Set 5:
Tx 320
Ty 58
Tz 74
3. CONCLUSION
In this paper, we have made an attempt to obtain all integer solutionsto(1).As(1)issymmetricin zyx ,, ,itistobe
noted that, if ),,( zyx is any positive integer solution to (1),then the triples ),,( zyx , ),,( zyx  , ),,( zyx  ,
),,( zyx  , ),,( zyx  , ),,( zyx  , ),,( zyx  also satisfy (1). To conclude, onemaysearchforintegersolutions
to other choices of homogeneous cones along with suitable properties.
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