OBSERVATIONS ON THE HYPERBOLA
8
x
2−
5
y
2=
27
SHREEMATHI ADIGA
1*, N. ANUSHEELA
2AND M. A. GOPALAN
31Assistant Professor, Department of Mathematics,
Government First Grade College, Koteshwara, Kundapura Taluk, Udupi – 576 222, Karnataka, India.
2Assistant Professor, Department of Mathematics,
Government Arts College, Udhagamandalam, The Nilgiris-643 002, India.
3Professor, Department of Mathematics,
Shrimati Indira Gandhi College, Trichy-620 002, Tamil Nadu, India.
(Received On: 28-07-18; Revised & Accepted On: 02-09-18)
ABSTRACT
T
his paper concerns with the problem of finding non-zero distinct integer solutions to the Pell-like equation represented by8
x
2−
5
y
2=
27
. 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. Employing the solutions, a special Pythagorean triangle is constructed.Keywords: Binary quadratic, Hyperbola, Parabola, Pell equation, Integral solutions.
2010 Mathematics Subject Classification: 11D09.
INTRODUCTION
The diophantine equation offer an field for research due to their variety [1-3]. In particular, the binary quadratic diophantine equations of the form ax2 −by2 =N,
(
a b N, , ≠0)
are rich in variety and have been analyzed by many mathematicians for their respective integer solutions for particular values of a, b and N. In this context, one may refer [4-16].This communication concerns with the problem of obtaining non-zero distinct integer solutions to the binary quadratic equation representing hyperbola given by8x2−5y2 =27. A few interesting 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. Employing the solutions, a special Pythagorean triangle is constructed.
METHOD OF ANALYSIS
The binary quadratic equation representing hyperbola is given by
2 2
8x −5y =27 (1)
Taking
x
=
X
+
5
T
,y= X+8T (2) In (1), it simplifies the equation2 2
40 9
X = T + (3)
Corresponding Author: Shreemathi Adiga1*, 1Assistant Professor, Department of Mathematics,
© 2018, IJMA. All Rights Reserved 66
The smallest positive integer solution
( ,
T X
0 0)
of (3) is0
1 ,
07
T
=
X
=
To obtain the other solutions of (3), consider the pellian equation
2 2
40 1
X = T + (4) whose smallest positive integer solution is
0 3, 0 19
T = X =
The general solution ( ,T X n n)of (4) is given by
(
)
140 19 3 40 , 0,1, 2.... n
n n
X T n
+
+ = + =
(5)
Since irrational roots occur in pairs, we have
(
)
140 19 3 40 , 0,1, 2.... n
n n
X T n
+
− = − =
(6)
From (5) and (6), solving forX T n, n, we have
(
) (
1)
11 1
19 3 40 19 3 40
2 2
n n
n n
X =
+ + + − +
= f
(
) (
1)
11 1
19 3 40 19 3 40
2 40 2 40
n n
n n
T g
+ +
=
+ − −
=
Applying Brahmagupta lemma between the solutions
( ,
T X
0 0)
and ( ,T X n n), the general solution(
T
n+1,
X
n+1)
of (3) isfound to be
1
7 1
2 2 40
n n n
T+ g f
⇒ = + (7)
1
7 40
2 2
n n n
X + f g
⇒ = + (8)
Using (7) and (8) in (2) we have
1 1 1
75
5 6
2 40
n n n n n
x + = X + + T+ = f + g (9)
1 1 1
15 48
8
2 40
n n n n n
y + = X + + T+ = f + g (10)
Thus, (9) and (10) represent the integer solutions of the hyperbola (1).
A few numerical examples are given in the Table: 1 below:
Table-1: Numerical Examples
n
1
+ n
x
y
n+1 -1 12 15 0 453 573 1 17202 21759 2 653223 826269 3 24805272 31376463 4 941947113 1191479325Recurrence relations for
x
and y are:3
38
2 10 ,
1, 0,1...
n n n
x
+−
x
++
x
+=
n
= −
3
38
2 10 ,
1, 0,1...
n n n
A few interesting relations among the solutions are given below: •
38
x
n+2−
x
n+1−
x
n+3=
0
•
x
n+2−
19
x
n+1−
15
y
n+1=
0
•
x
n+3−
721
x
n+1−
570
y
n+1=
0
•
y
n+2−
24
x
n+1−
19
y
n+1=
0
•
y
n+3−
912
x
n+1−
721
y
n+1=
0
•
19
x
n+2−
x
n+1−
15
y
n+2=
0
•
x
n+3−
x
n+1−
30
y
n+2=
0
•
19
y
n+3−
24
x
n+1−
721
y
n+2=
0
•
721
x
n+2−
19
x
n+1−
15
y
n+3=
0
•
721
x
n+3−
x
n+1−
570
y
n+3=
0
•
19
x
n+3−
721
x
n+2−
15
y
n+1=
0
•
19
y
n+2−
24
x
n+2−
y
n+1=
0
•
y
n+3−
48
x
n+2−
y
n+1=
0
•
x
n+3−
19
x
n+2−
15
y
n+2=
0
•
y
n+3−
24
x
n+2−
19
y
n+2=
0
•
19
y
n+2+
24
x
n+2−
y
n+3=
0
•
721
y
n+2−
24
x
n+3−
19
y
n+1=
0
•
721
y
n+3−
912
x
n+3−
y
n+1=
0
•
x
n+1−
721
x
n+3+
570
y
n+3=
0
•
38
y
n+2−
y
n+3−
y
n+1=
0
•
x
n+2−
19
x
n+3−
15
y
n+3=
0
•
y
n+2+
24
x
n+3−
19
y
n+3=
0
Each of the following expressions represent a cubical integer. • 1
[
5730 3 3 150 3 4 3 5730(
1 150 2)
]
405 xn+ − xn+ + xn+ − xn+
• 1
[
43518 3 3 30 3 5 3 43518(
1 30 3)
]
3078 xn+ − x n+ + xn+ − xn+• 1
[
192 3 3 150 3 3 3 192(
1 150 1)
]
27 xn+ − y n+ + xn+ − yn+• 1
[
7248 3 3 150 3 4 3 7248(
1 150 2)
]
513 xn+ − y n+ + xn+ − yn+• 1
[
91744 3 3 50 3 5 3 91744(
1 50 3)
]
6489 xn+ − y n+ + xn+ − yn+•
[
3 4 3 5(
2 3)
]
1
217590 5730 3 217590 5730
405 xn+ − xn+ + xn+ − xn+
• 1
[
192 3 4 5730 3 3 3 192(
2 5730 1)
]
513 xn+ − yn+ + xn+ − yn+© 2018, IJMA. All Rights Reserved 68
• 1
[
275232 3 4 5730 3 5 3 275232(
2 5730 3)
]
513 xn+ − yn+ + xn+ − yn+• 1
[
64 3 5 72530 3 3 3 64(
3 72530 1)
]
6489 xn+ − yn+ + xn+ − yn+• 1
[
275232 3 5 217590 3 5 3 275232(
3 217590 1)
]
27 xn+ − yn+ + xn+ − yn+
• 1
[
7248 3 5 217590 3 4 3 7248(
3 217590 2)
]
513 xn+ − yn+ + xn+ − yn+• 1
[
48 3 4 1812 3 3 3 48(
2 1812 1)
]
162 y n+ − y n+ + yn+ − yn+• 1
[
48 3 5 68808 3 3 3 48(
3 68808 1)
]
6156 yn+ − yn+ + yn+ − yn+• 1
[
7248 3 5 275232 3 4 3 7248(
3 275232 2)
]
648 y n+ − yn+ + yn+ − yn+ Each of the following expressions represent bi-quadratic integer:
•
(
)
24 4 4 5 1 2
2 1
2320650 60750 4 5730 150 328050
405 x n+ − x n+ + xn+ − xn+ −
•
(
)
24 4 4 6 1 3
2 1
133948404 92340 4 43518 30 18948168
3078 x n+ − xn+ + xn+ − xn+ −
•
(
)
24 4 4 4 1 2
2 1
5184 4050 4 192 150 1458
27 x n+ − y n+ + xn+ − yn+ −
•
(
)
24 4 4 5 1 2
2 1
3718224 76950 4 7248 150 526338
513 xn+ − y n+ + xn+ − yn+ −
•
(
)
24 4 4 6 1 3
2 1
595326816 324450 4 91744 50 84214242
6489 xn+ − y n+ + xn+ − yn+ −
•
(
)
24 5 4 6 2 3
2 1
88123950 2320650 4 217590 5730 328050
405
x n+ − x n+ + xn+ − xn+ −
•
(
)
24 5 4 4 2 1
2 1
98496 2939490 4 192 5730 526338
513 x n+ − y n+ + xn+ − yn+ −
•
(
)
24 5 4 5 2 2
2 1
195696 154710 4 7248 5730 1458
27 x n+ − y n+ + xn+ − yn+ −
•
(
)
24 5 4 6 2 3
2 1
141194016 2939490 4 275232 5730 526338
513 x n+ − y n+ + xn+ − yn+ −
•
(
)
24 6 4 4 3 1
2 1
415296 470647170 4 64 72530 84214242
6489 x n+ − y n+ + xn+ − yn+ −
•
(
)
24 6 4 6 3 3
2 1
7431264 5874930 4 275232 217590 1458
27 x n+ − y n+ + xn+ − yn+ −
•
(
)
24 5 4 4 2 1
2
1
7776
293544
4 48
1812
52488
162
y
n+−
y
n++
y
n+−
y
n+−
•
(
)
24 6 4 4 3 1
2
1
295488
423582048
4 48
68808
75792672
6156
y
n+−
y
n++
y
n+−
y
n+−
•
(
)
24 6 4 5 3 2
2 1
3718224 111623670 4 7248 217590 526338
513 xn+ − y n+ + xn+ − yn+ −
• 2 4 6 4 5
(
3 2)
21
4696704
178350336
4 7248
275232
839808
648
y
n+−
y
n++
y
n+−
y
n+−
•
(
)
24 5 4 6 2 3
2 1
141194016 2939490 4 275232 5730 526338
513 x n+ − y n+ + xn+ − yn+ −
Each of the following expressions represents Nasty number:
•
[
4860
34380
2 2150
2 3]
405
1
+ +
−
+
x
nx
n• 1
[
36936 261108 2 2 180 2 4]
3078 + x n+ − x n+
• 1
[
324 1152 2 2 900 2 2]
27 + xn+ − y n+• 1
[
6156 43488 2 2 900 2 3]
513 + xn+ − y n+• 1
[
77868 550464 2 2 300 2 4]
6489 + xn+ − y n+
• 1
[
4860 1305540 2 3 34380 2 4]
405 + x n+ − x n+
• 1
[
6156 1152 2 3 34380 2 2]
513 + xn+ − yn+• 1
[
324 43488 2 3 34380 2 3]
27 + xn+ − y n+
• 1
[
6156 1651392 2 3 34380 2 4]
513 + xn+ − y n+
• 1
[
77868 384 2 4 435180 2 2]
6489 + x n+ − y n+
• 1
[
324 1651392 2 4 1305540 2 4]
27 + xn+ − y n+
• 1
[
6156 43488 2 4 1305540 2 3]
513 + x n+ − y n+
• 1
[
1944 288 2 3 10872 2 2]
162 + y n+ − y n+• 1
[
73872 288 2 4 412848 2 2]
6156 + y n+ − y n+
• 1
[
7776 43488 2 4 1651392 2 3]
648 + y n+ − y n+
Each of the following expressions represents quintic integer
•
(
) (
)
(
)
−
−
−
+
−
+ +
+ +
+ +
2 1
3 2 1
6 5 5
5 3
24603750
939863250
5
150
5730
5
24603750
939863250
405
1
n n
n n
n n
x
x
x
x
x
x
•
(
) (
)
(
)
−
−
−
+
−
+ +
+ +
+ +
3 1
3 3 1
7 5 5
5 3
284222520
00
4122931875
5
30
43518
5
284222520
00
4122931875
3078
1
n n
n n
n n
x
x
x
x
x
x
•
(
) (
)
(
)
−
−
−
+
−
+ +
+ +
+ +
1 1
3 1 1
5 5 5
5 3
109350
139968
5
150
192
5
109350
139968
27
1
n n
n n
n n
y
x
y
x
y
x
•
(
) (
)
(
)
−
−
−
+
−
+ +
+ +
+ +
2 1
3 2 1
6 5 5
5 3
39475350
1907448912
5
150
7248
5
39475350
1907448912
513
1
n n
n n
n n
y
x
y
x
© 2018, IJMA. All Rights Reserved 70 •
(
) (
)
(
)
−
−
−
+
−
+ + + + + + 3 1 3 3 1 7 5 5 5 32105356050
000
3863075709
5
50
91744
5
2105356050
000
3863075709
6489
1
n n n n n ny
x
y
x
y
x
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 3 2 3 3 2 7 5 6 5 3939863250
0
3569019975
5
5730
217590
5
939863250
0
3569019975
405
1
n n n n n nx
x
x
x
x
x
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 1 2 3 1 2 5 5 6 5 31507958370
50528448
5
5730
192
5
1507958370
50528448
513
1
n n n n n ny
x
y
x
y
x
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 2 2 3 2 2 6 5 6 5 34177170
5283792
5
5730
7248
5
4177170
5283792
27
1
n n n n n ny
x
y
x
y
x
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 3 2 3 3 2 7 5 6 5 31507958370
0
7243253021
5
5730
275232
5
1507958370
0
7243253021
513
1
n n n n n ny
x
y
x
y
x
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 1 3 3 1 3 5 5 7 5 3000
3054029486
2694855744
5
72530
64
5
000
3054029486
2694855744
6489
1
n n n n n ny
x
y
x
y
x
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 3 3 3 3 3 7 5 7 5 3158623110
200644128
5
217590
275232
5
158623110
200644128
27
1
n n n n n ny
x
y
x
y
x
•(
) (
)
(
)
−
−
−
+
−
+ + = + + + 2 3 3 2 3 6 5 7 5 30
5726294271
1907448912
5
217590
7248
5
0
5726294271
1907448912
513
1
n n n n n ny
x
y
x
y
x
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 1 2 3 1 2 5 5 6 5 347554128
1259712
5
1812
48
5
47554128
1259712
162
1
n n n n n ny
y
y
y
y
y
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 1 3 3 1 3 5 5 7 5 3000
2607571087
1819024128
5
68808
48
5
000
2607571087
1819024128
6156
1
n n n n n ny
y
y
y
y
y
•(
) (
)
(
)
−
−
−
+
−
+ + + + + + 2 3 3 2 3 6 5 7 5 300
1155710177
3043464192
5
275232
7248
5
00
1155710177
3043464192
6483
1
n n n n n ny
y
y
y
y
y
REMARKABLE OBSERVATIONS1. 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: Hyperbolas
S.No Hyperbolas
(
,)
n n
X Y
1 2 2
40Xn −Yn =26244000
[
(
5730
x
n+1−
150
x
n+2) (
, 960
x
n+2−
36240
x
n+1)
]
2 2 2
40Xn −Yn =1515853440
[
(
43518
x
n+1−
30
x
n+3) (
, 192
x
n+3−
275232
x
n+1)
]
3 2 2
40
X
n−
Y
n=
116640
[
(
192
x
n+1−
150
y
n+1) (
, 960
y
n+1−
1200
x
n+1)
]
4 2 2
40
X
n−
Y
n=
42107040
[
(
7248
x
n+1−
150
y
n+2) (
, 960
y
n+2−
45840
x
n+1)
]
5 2 2
40
X
n−
Y
n=
6737139360
[
(
91744
x
n+1−
50
y
n+3) (
, 320
y
n+3−
580240
x
n+1)
]
6 2 2
40
X
n−
Y
n=
26244000
[
(
217590
x
n+2−
5730
x
n+3) (
, 36240
x
n+3−
1376160
x
n+2)
]
7 2 2
40
X
n−
Y
n=
42107040
[
(
192
x
n+2−
5730
y
n+1) (
, 36240
y
n+1−
1200
x
n+2)
]
8 2 2
9 2 2
40
X
n−
Y
n=
42107040
[
(
275232xn+2−5730yn+3) (
, 36240yn+3−1740720xn+2)
]
10 2 2
40
X
n−
Y
n=
6737139360
[
(
64
x
n+3−
72530
y
n+1) (
, 458720
y
n+1−
400
x
n+3)
]
11 2 2
40Xn −Yn =116640
[
(
275232xn+3−217590yn+3) (
, 1376160yn+3−1740720xn+3)
]
12 2 2
40Xn −Yn =42107040
[
(
7248
x
n+3−
217590
y
n+2) (
, 1376160
y
n+3−
45840
x
n+3)
]
13 2 2
40Xn −Yn =4199040
[
(
48
y
n+2−
1812
y
n+1) (
, 11460
y
n+1−
300
y
n+2)
]
14 2 2
40Xn −Yn =6063413760
[
(
48
y
n+3−
68808
y
n+1) (
, 435180
y
n+1−
300
y
n+3)
]
15 2 2
40Xn −Yn =67184640
[
(
7248
y
n+3−
275232
y
n+2) (
, 1740720
y
n+2−
45840
y
n+3)
]
2. Employing linear combination among the solutions for other choices of parabola which are presented in Table 3 below:
Table-3: Parabolas
S.No
Parabolas
(
X
n,
Y
n)
1 2
16200
X
n−
Y
n=
26244000
(
)
(
)
2 2 2 3
2 1
810 5730
150
,
960
36240
n n
n n
x
x
x
x
+ +
+ +
+
−
−
2 2
123120
X
n−
Y
n=
1515853440
(
)
(
)
2 2 2 4
3 1
6156
43518
30
,
192
275232
n n
n n
x
x
x
x
+ +
+ +
+
−
−
3 2
1080
X
n−
Y
n=
116640
(
)
(
1 2 2 1)
2 254 192
150
,
960
1200
n n
n n
x
y
y
x
+ +
+ +
+
−
−
4 2
20520
X
n−
Y
n=
42107040
(
)
(
)
2 2 2 3
2 1
1026 7248
150
,
960
45840
n n
n n
x
y
y
x
+ +
+ +
+
−
−
5 2
259560
X
n−
Y
n=
6737139360
(
)
(
)
2 2 2 4
3 1
12978 91744
50
,
320
580240
n n
n n
x
y
y
x
+ +
+ +
+
−
−
6 2
16200
X
n−
Y
n=
26244000
(
)
(
)
2 3 2 4
3 2
810 217590 5730 ,
36240 1376160
n n
n n
x x
x x
+ +
+ +
+ −
−
7 2
20520
X
n−
Y
n=
42107040
(
)
(
1 2 3 2)
2 21026 192
5730
,
36240
1200
n n
n n
x
y
y
x
+ +
+ +
+
−
−
8 2
1080
X
n−
Y
n=
116640
(
)
(
)
2 3 2 3
2 2
54 7248
5730
,
36240
45840
n n
n n
x
y
y
x
+ +
+ +
+
−
−
9 2
20520
X
n−
Y
n=
42107040
(
)
(
)
2 3 2 4
3 2
1026 275232 5730 ,
36240 1740720
n n
n n
x y
y x
+ +
+ +
+ −
−
10 2
259560
X
n−
Y
n=
6737139360
(
)
(
)
2 4 2 2
1 3
12978 64
72530
,
458720
400
n n
n n
x
y
y
x
+ +
+ +
+
−
−
11 2
1080
X
n−
Y
n=
116640
(
)
(
)
2 4 2 4
3 3
54 275232 217590 ,
1376160 1740720
n n
n n
x y
y x
+ +
+ +
+ −
−
© 2018, IJMA. All Rights Reserved 72
12
20520
−
2=
42107040
n
n
Y
X
(
)
(
3 2 4 3)
2 31026 7248
217590
,
1376160
45840
n n
n n
x
y
y
x
+ +
+ +
+
−
−
13 2
6480
X
n−
Y
n=
4199040
(
)
(
)
2 3 2 2
1 2
324 48
1812
,
11460
300
n n
n n
y
y
y
y
+ +
+ +
+
−
−
14 2
246240
X
n−
Y
n=
6063413760
(
)
(
)
2 4 2 2
1 3
12312 48
68808
,
435180
300
n n
n n
y
y
y
y
+ +
+ +
+
−
−
15 2
25920
X
n−
Y
n=
67184640
(
)
(
)
2 4 2 3
2 3
1296 7248 275232 ,
1740720 45840
n n
n n
y y
y y
+ +
+ +
+ −
−
3. GENERATORS OF THE PYTHAGOREAN TRIANGLE
Let p, q be the non-zero distinct integers such thatp=xn+1+yn+1 , q=xn+1
Note that p> >q 0. Treat p, q as the generators of the Pythagorean triangle T X Y Z
(
, ,)
2 2 2 2
2 , , , 0
X = pq Y = p −q Z = p +q p> >q
Let A,P represent the area and perimeter of T
Then the following interesting relations are observed. • 5X −4Y− =Z 27
• 4Z 9Y 20A 27
P
− + =
• 2A xn 1yn 1 P = + +
Suppose p, q be the non-zero distinct integers such that
1 1
n n
p=x+ +y + , q= yn+1
In this case, the corresponding T X Y Z
(
, ,)
Pythagorean satisfies the relation 5Y−16X +11Z =54REFERENCES
1. R.D. Carmichael, The Theory of Numbers and Diophantine Analysis, Dover Publications, New York 1950. 2. L.E. Dickson, History of theory of Numbers, Vol.II, Chelsea Publishing co., New York 1952.
3. L.J. Mordell, Diophantine Equations, Academic Press, London 1969.
4. K. Meena, M.A. Gopalan, S. Nandhini, On the binary quadratic Diophantine equation y2 =68x2 +13
,International Journal of Advanced Education and Research, 2(1), 2017, 59-63.
5. K. Meena, S. Vidhyalakshmi, R. Sobana Devi, On the binary quadratic equation y2 =7x2+32, International Journal of Advanced Science and Research, 2(1), 2017, 18-22.
6. K. Meena, M.A. Gopalan, S. Hemalatha, On the hyperbolay2 =8x2+16,National Journal of Multidisciplinary Research and Development, 2(1), 2017, 01-05.
7. M.A. Gopalan, K.K. Viswanathan, G. Ramya, On the Positive Pell equation y2 =12x2 +13, International Journal of Advanced Education and Research, 2(1), 2017, 04-08.
8. K. Meena, M.A. Gopalan, V. Sivaranjani, On the Positive Pell equation y2 =102x2+33, International Journal of Advanced Education and Research, 2(1), 2017, 91-96.
9. K. Meena, S. Vidhyalakshmi, N. Bhuvaneswari, On the binary quadratic Diophantine equation
2 2
10 24
y = x + , International Journal of Multidisciplinary Education and Research, 2(1), 2017, 34-39. 10. M.A. Gopalan, A. Vijayasankar, V. Krithika, Sharadha Kumar, On the Hyperbola 2x2−3y2 =15, Indo –
11. A. Vijayasankar, M.A. Gopalan, V. Krithika, On the Negative Pell Equation 2=112 2−7
x
y , International Journal of Engineering and Applied Sciences, 4(7), 2017, 11-14.
12. T.R. Usha Rani, K. Dhivya, On the positive pell equation y2 =14x2 +18, International Journal Of Academic Research and Development (IJARD), 3(3), May 2018, 43-50.
13. D. Maheswari, R. Suganya, Observations on the pell equationy2 =11x2 +5, International Journal of Emerging Technologies in Engineering Research (IJETER), 6(5), May 2018, 45-50.
14. S. Mallika, D. Hema, On negative pell equationy2 =20x2−11, International Journal of Academic Research and Development (IJARD), 3(3), May 2018, 33-40.
15. S. Mallika, D. Hema, Observations on the Hyperbola8x2 −3y2 =20, International Journal on Research Innovations in Engineering Science and Technology (IJRIEST), 3(4), April 2018, 575-582.
16. D. Maheswari, S. Dharuna, Observations on the Hyperbola 2x2−5y2 =27, International Journal of Emerging Technologies in Engineering Research (IJETER), 6(5), May 2018, 7-12.
Source of support: Nil, Conflict of interest: None Declared.
