Observation on the Positive Pell Equation
y
2
2
x
2
18
M.A.Gopalan
1and Sharadha Kumar
21Professor, Department of Mathematics, SIGC, Trichy, Tamilnadu, India. Email: mayilgopalan@gmail.com 2M.Phil Scholar, Department of Mathematics, SIGC, Trichy, Tamilnadu, India. Email: sharadhak12@gmail.com
Article Received: 04 October 2017 Article Accepted: 23 November 2017 Article Published: 12 December 2017
1. INTRODUCTION
The binary quadratic Diophantine equations of the form
ax
2
by
2
N
,
a
,
b
,
N
0
are rich in variety and havebeen analyzed by many Mathematicians for their respective integer solutions for particular values of a, andb N .
In this context, one may refer [1-13]. This communication concerns with the problem of obtaining non-zero distinct
integer solutions to the binary quadratic equation given by
y
2
2
x
2
18
representing hyperbola. A fewinteresting relations among its solutions are presented. Knowing an integral solution of the given hyperbola, integer
solutions for other choices of hyperbolas and parabolas are presented. Also, employing the solutions of the given
equation, special Pythagorean triangle is constructed.
2. METHOD OF ANALYSIS
The binary quadratic equation representing hyperbola is given by
18
2
2 2
x
y
(1)The smallest positive integer solution
(
x
0,
y
0)
of (1) is6
,
3
0 0
y
x
To obtain, the other solutions of (1), consider the pellian equation
1
2
2 2
x
y
(2)Whose smallest positive integer solution is
3 ~ , 2 ~
0 0 y x
The general solution
(
~
x
n,
~
y
n)
of (2) is given by
3
2
2
,
0
,
1
,
2
....
~
2
~
y
x
n1n
n
n (3)
A B S T R A C T
The hyperbola represented by the binary quadratic equation
y
2
2
x
2
18
is analyzed for finding its non-zero distinct integer solutions. A few interesting relations among its solutions are presented. Also, knowing an integral solution of the given hyperbola, integer solutions for other choices of hyperbolas and parabolas are presented. Also, employing the solutions of the given equation, special Pythagorean triangle is constructed.
3
2
2
,
0
,
1
,
2
....
~
2
~
y
x
n1n
n
n (4)
From (3) and (4), solving for
y
~
n,
~
x
n, we have
nn n
n f
y
2 1 2
2 3 2
2 3 2 1
~ 1 1
nn n
n
g
x
2
2
1
2
2
3
2
2
3
2
2
1
~
1 1
Applying Brahmagupta lemma between the solutions
(
x
0,
y
0)
and(
~
x
n,
~
y
n)
, the general solution(
x
n1,
y
n1)
is
n n
n x y y x
x 1 0~ 0~
n n
n y y x x
y 1 0~ 2 0~
n n
n
f
g
x
2
3
2
3
1
(5)n n
n
f
g
y
2
2
6
2
6
1
(6)
Thus, (5) and (6) represent the integer solutions of the hyperbola (1).
A few numerical examples are given in the following table 1
Table 1: Examples
n
1
n
x
y
n1 -1 3 60 21 30
1 123 174
2 717 1014
Note that the
x
values are odd and yvalues are even.Also, xn1 0
mod3
, yn10
mod6
Recurrence relations for
x
and y are:...
1
,
0
,
1
,
0
6
2 13
x
x
n
x
n n n...
1
,
0
,
1
,
0
6
2 13
y
y
n
1. A few interesting relations among the solutions are given below
x
n2
3
x
n1
2
y
n1
0
3
x
n2
x
n1
2
y
n2
0
17
x
n2
3
x
n1
2
y
n3
0
12
y
n1
17
x
n1
x
n3
0
3
y
n1
4
x
n1
y
n2
0
17
y
n1
24
x
n1
y
n3
0
4
y
n2
x
n1
x
n3
0
y
n2
4
x
n1
3
y
n1
0
17
y
n2
4
x
n1
3
y
n3
0
2
y
n3
3
x
n1
17
x
n2
0
12
y
n3
x
n1
17
x
n3
0
3
x
n3
17
x
n2
2
y
n1
0
x
n3
3
x
n2
2
y
n2
0
3
x
n3
x
n2
2
y
n3
0
y
n1
4
x
n2
3
y
n2
0
y
n1
8
x
n2
y
n3
0
3
y
n1
4
x
n3
17
y
n2
0
3
y
n2
4
x
n2
y
n3
0
y
n2
4
x
n3
3
y
n3
0
y
n1
24
x
n3
17
y
n3
0
y
n3
3
y
n2
4
x
n2
0
3
y
n3
17
y
n2
4
x
n1
0
2. Each of the following expressions represents a cubical integer
12 3 4 60 3 3
312 2 60 1
36 1
n n n
n x x x
x
12 3 5 348 3 3
312 3 348 1
216 1
n n n
n x x x
2 3 3 2 3 3
32 1 2 1
3 1 n n n
n x y x
y
12 3 4 84 3 3
312 2 84 1
54 1
n n n
n x y x
y
12 3 5 492 3 3
312 3 492 1
306 1
n n n
n x y x
y
60 3 5 348 3 4
360 3 348 2
36 1
n n n
n x x x
x
60 3 3 12 3 4
360 1 12 2
54 1
n n n
n x y x
y
60 3 4 84 3 4
360 2 84 2
18 1
n n n
n x y x
y
60 3 5 492 3 4
360 3 492 2
54 1
n n n
n x y x
y
348 3 3 12 3 5
3348 1 12 3
306 1
n n n
n x y x
y
348 3 4 84 3 5
3348 2 84 3
54 1
n n n
n x y x
y
348 3 5 492 3 5
3348 3 492 3
18 1
n n n
n x y x
y
84 3 3 12 3 4
384 1 12 2
72 1
n n n
n y y y
y
492 3 3 12 3 5
3492 1 12 3
432 1
n n n
n y y y
y
492 3 4 84 3 5
3492 2 84 3
72 1
n n n
n y y y
y
3. Each of the following expressions represents bi-quadratic integer
432
2160
4
12
60
2592
36
1
2 1 2 4 4 5 42
x
n
x
n
x
n
x
n
2592
75168
4
12
348
93312
216
1
2 1 3 4 4 6 42
x
n
x
n
x
n
x
n
6
6
4
2
2
18
3
1
2 1 1 4 4 4 42
y
n
x
n
y
n
x
n
648
4536
4
12
84
5832
54
1
2 1 2 4 4 5 4
3672
150552
4
12
492
187272
306
1
2 1 3 4 4 6 42
y
n
x
n
y
n
x
n
2160
12528
4
60
348
2592
36
1
2 2 3 5 4 6 42
x
n
x
n
x
n
x
n
3240
648
4
60
12
5832
54
1
2 2 1 5 4 4 42
y
n
x
n
y
n
x
n
1080
1512
4
60
84
648
18
1
2 2 2 5 4 5 42
y
n
x
n
y
n
x
n
3240
26568
4
60
492
5832
54
1
2 2 3 5 4 6 42
y
n
x
n
y
n
x
n
106488
3672
4
348
12
187272
306
1
2 3 1 6 4 4 42
y
n
x
n
y
n
x
n
18792
4536
4
348
84
5832
54
1
2 3 2 6 4 5 42
y
n
x
n
y
n
x
n
6264
8856
4
348
492
648
18
1
2 3 3 6 4 6 42
y
n
x
n
y
n
x
n
6048
864
4
84
12
10368
72
1
2 2 1 5 4 4 42
y
n
y
n
y
n
y
n
212544
5184
4
492
12
373248
432
1
2 3 1 6 4 4 42
y
n
y
n
y
n
y
n
35424
6048
4
492
84
10368
72
1
2 3 2 6 4 5 42
y
n
y
n
y
n
y
n
4. Each of the following expressions represents Nasty number
432 72 2 3 360 2 2
36 1
x n x n
2592 72 2 4 2088 2 2
216 1
x n x n
36 12 2 2 12 2 2
3 1
y n x n
648 72 2 3 504 2 2
54 1
y n x n
3672 72 2 4 2952 2 2
306 1
432 360 2 4 2088 2 3
361
x n x n
648 360 2 2 72 2 3
54 1
y n x n
216 360 2 3 504 2 3
18 1
y n x n
648 360 2 4 2952 2 3
54 1
y n x n
3672 2088 2 2 72 2 4
306 1
y n x n
648 2088 2 3 504 2 4
54 1
y n x n
216 2088 2 4 2952 2 4
18 1
y n x n
864 504 2 2 72 2 3
72 1
y n y n
5184 2952 2 2 72 2 4
432 1
y n y n
864 2952 2 3 504 2 4
72 1
y n y n
3. REMARKABLE OBSERVATIONS
3.1 Employing linear combinations among the solutions of (1), one may generate integer solutions for other choices of hyperbola which are presented in table 2 below
Table 2: Hyperbola
Sl.no Hyperbola
X
n,
Y
n
1
10368
2
X
n2
Y
n2
12
x
n2
60
x
n1
,
84
x
n1
12
x
n2
2
373248
2
X
n2
Y
n2
12
x
n3
348
x
n1
,
492
x
n1
12
x
n3
3
72
2
X
n2
Y
n2
2
y
n1
2
x
n1
,
4
x
n1
2
y
n1
4
23328
2
X
n2
Y
n2
12
y
n2
84
x
n1
,
120
x
n1
12
y
n2
5
749088
2
X
n2
Y
n2
12
y
n3
492
x
n1
,
696
x
n1
12
y
n3
6
10368
7
23328
2
X
n2
Y
n2
60
y
n1
12
x
n2
,
24
x
n2
84
y
n1
8
2592
2
X
n2
Y
n2
60
y
n2
84
x
n2
,
120
x
n2
84
y
n2
9
23328
2
X
n2
Y
n2
60
y
n3
492
x
n2
,
696
x
n2
84
y
n3
10
749088
2
X
n2
Y
n2
348
y
n1
12
x
n3
,
24
x
n3
492
y
n1
11
23328
2
X
n2
Y
n2
348
y
n2
84
x
n3
,
120
x
n3
492
y
n2
12
2592
2
X
n2
Y
n2
348
y
n3
492
x
n3
,
696
x
n3
492
y
n3
13
41472
2
X
n2
Y
n2
84
y
n1
12
y
n2
,
24
y
n2
120
y
n1
14
1492992
2
X
n2
Y
n2
492
y
n1
12
y
n3
,
24
y
n3
696
y
n1
15
41472
2
X
n2
Y
n2
492
y
n2
84
y
n3
,
120
y
n3
696
y
n2
3.2 Employing linear combination among the solutions for other choices of parabola which are presented in table 3 below
Table 3: Parabola
Sl.no
Parabola
X
n,
Y
n
1
10368
72
X
n
Y
n2
72
12
x
2n3
60
x
2n2
,
84
x
n1
12
x
n2
2
373248
432
X
n
Y
n2
432
12
x
2n4
348
x
2n2
,
492
x
n1
12
x
n3
3
72
6
X
n
Y
n2
6
2
y
2n2
2
x
2n2
,
4
x
n1
2
y
n1
4
23328
108
X
n
Y
n2
108
12
y
2n3
84
x
2n2
,
120
x
n1
12
y
n2
5
749088
612
X
n
Y
n2
612
12
y
2n4
492
x
2n2
,
696
x
n1
12
y
n3
6
10368
72
X
n
Y
n2
72
60
x
2n4
348
x
2n3
,
492
x
n2
84
x
n3
7
23328
108
X
n
Y
n2
108
60
y
2n2
12
x
2n3
,
24
x
n2
84
y
n1
8
2592
36
X
n
Y
n2
36
60
y
2n3
84
x
2n3
,
120
x
n2
84
y
n2
9
23328
108
X
n
Y
n2
108
60
y
2n4
492
x
2n3
,
696
x
n2
84
y
n3
10
749088
11
23328
108
X
n
Y
n2
108
348
y
2n3
84
x
2n4
,
120
x
n3
492
y
n2
12
2592
36
X
n
Y
n2
36
348
y
2n4
492
x
2n4
,
696
x
n3
492
y
n3
13
41472
144
X
n
Y
n2
144
84
y
2n2
12
y
2n3
,
24
y
n2
120
y
n1
14
1492992
864
X
n
Y
n2
864
492
y
2n2
12
y
2n4
,
24
y
n3
696
y
n1
15
41472
144
X
n
Y
n2
144
492
y
2n3
84
y
2n4
,
120
y
n3
696
y
n2
3.3 Let
p
,
q
be two non-zero distinct integers such that pq0.Treatp
,
q
as the generators of the Pythagorean triangle T
,
,
Where
2
pq
,
p
2
q
2,
p
2
q
2,
p
q
0
Taking
p
x
n1
y
n1,
q
x
n1, it is observed that T
,
,
is satisfied by the following relations:18
P A
4
1 1
2 xn yn P
A
Where A,P represent the area and perimeter of T
,
,
.4. CONCLUSION
In this paper, we have presented infinitely many integer solutions for the Diophantine equation, represented by
hyperbola is given by
y
2
2
x
2
18
. As the binary quadratic Diophantine equations are rich in variety, one maysearch for the other choices of equations and determine their integer solutions along with suitable properties.
REFERENCES
[1] L.E.Disckson, History of theory of Numbers, Vol.II, Chelsea Publishing co., New York (1952).
[2] L.J.Mordell, Diophantine Equations, Academic Press, London (1969).
[3] Gopalan et al., Integral points on the hyperbola
a
2
x
2
ay
2
4
a
k
1
2
k
2,
a
,
k
0
, Indianjournal of science,1(2), 2012, 25-126.
[5] S.Vidhyalakshmi, et al., Observations on the hyperbola
ax
2
a
1
y
2
3
a
1
Discovery, 4(10), 2013,22-24.
[6] K.Meena, S.Vidhyalakshmi, A.Nivetha, On the Binary Quadratic Diophantine Equation
0
14
4
22
xy
y
x
x
, Sch.J.Phys.Math.Stat., Volume-3, Issue-1, 2016, Page no:15-19.[7] K.Meena, S.vidhyalakshmi, R.Janani, On the Binary quadratic Diophantine Equation
0
21
9
22
xy
y
x
x
, IJETER, Volume-4, Issue-7, 2016, Page no:1-4.[8] K.Meena, MA.Gopalan, S.Nandhini, On the binary quadratic Diophantine equation
y
2
68
x
2
13
International Journal of Advanced Education and Research, Volume 2, Issue 1,2017, Page no.59-63.
[9] K.Meena, S.Vidhyalakshmi, R.Sobana Devi, On the binary quadratic equation
y
2
7
x
2
32
InternationalJournal of Advanced Science and Research, Volume 2, Issue 1, 2017, Page no.18-22 .
[10]K.Meena, MA.Gopalan, S.Hemalatha, On the hyperbola
y
2
8
x
2
16
National Journal ofMultidisciplinary Research and Developement, Volume 2, Issue 1, 2017, Page no.01-05.
[11]MA.Gopalan, KK.Viswanathan, G Ramya, On the Positive Pell equation
y
2
12
x
2
13
InternationalJournal of Advanced Education and Research, Volume 2, Issue 1, 2017, Page no.04-08.
[12]K.Meena, MA.Gopalan, V Sivaranjani, On the Positive Pell equation
y
2
102
x
2
33
International Journalof Advanced Education and Research, Volume 2, Issue 1, 2017, Page no.91-96 .
[13]K.Meena, S.Vidhyalakshmi, N.Bhuvaneswari, On the binary quadratic Diophantine equation
24
10
22