On Binary Diophantine Equation
3
x
2
5
y
2
12
S. Mallika
1, V. Surya
2 1Assistant Professor, 2M. Phil Scholar, Department of Mathematics, SIGC, Trichy-620002, Tamilnadu, India.
Abstract: Non-homogeneous binary quadratic equation representing hyperbola given by
3
x
2
5
y
2
12
is analyzed for 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 hyperbola and parabola are presented.Keywords: Binary quadratic, Hyperbola, Parabola, Integral solutions, Pell equation. AMS Mathematics subject Classification (2010):11D0
I. INTRODUCTION
The binary quadratic Diophantine equations of the form
ax
2
by
2
N
,
(
a
,
b
,
c
0
)
are rich in variety and have been analyzed by many mathematicians for their respective integer solutions for particular values ofa
,
b
andN
.
In this context, one may refer [1-18].This communication concerns with the problem of obtaining non-zero distinct integer solutions to the binary quadratic equation
given by,
3
x
2
5
y
2
12
representing hyperbola. 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.II. METHOD OF ANALYSIS
The Diophantine equation representing the binary quadratic equation to be solved for its non-zero distinct integral solution is
3
x
2
5
y
2
12
(1) Introduce the linear transformation
x
X
5
T
,
y
X
3
T
(2) From (1) & (2) we have,
X
2
15
T
2
6
(3) whose smallest positive integer solution is1
,
3
00
T
X
To obtain the other solutions of (3), consider the Pell equation
X
2
15
T
2
1
(4) whose smallest positive integer solution is1
~
,
4
~
0 0
T
X
whose general solution is given by
n n
n
n
g
X
f
T
2
1
~
,
15
2
1
~
where
1
115
4
15
4
n nn
f
1
115
4
15
4
n nn
g
Applying Brahmagupta lemma between
x
0,
y
0
and
x
ny
n
~
,
~
, the other integer solutions of (1) are given by
n n
n
f
g
x
4
15
15
15
1
n n
n
f
g
y
3
15
12
The recurrence relations satisfied by x and y are given by
0
8
2 13
n n
n
x
x
x
0
8
2 13
n n
n
y
y
y
Some numerical examples of x and y satisfying (1) are given in the Table: 1 below:
Table: 1 Numerical examples
n
1
n
x
y
n1-1 8 6
0 62 48
1 488 378
2 3842 2976
3 30248 23430
4 288142 184464
From the above table, we observe some interesting relations among the solutions which are presented below:
A.
x
n1 &y
n1values are always even .B. Relations among the solutions
1)
x
n3
8
x
n2
x
n1
0
2)
5
y
n1
x
n2
4
x
n1
0
3)
15
y
n3
93
x
n2
12
x
n1
0
4)
40
y
n3
x
n1
3
x
n3
0
5)
9
x
n2
36
x
n1
45
y
n1
0
6)
x
n3
31
x
n1
40
y
n1
0
7)
y
n3
24
x
n1
31
y
n1
0
8)
x
n3
x
n1
10
y
n2
0
9)
4
y
n1
3
x
n1
y
n2
0
10)
3
x
n1
24
x
n2
3
x
n3
0
11)
5
y
n1
31
x
n2
4
x
n3
0
12)
5
y
n2
4
x
n2
x
n3
0
13)
y
n3
6
x
n2
y
n1
0
14)
x
n1
4
x
n2
5
y
n2
0
15)
y
n1
3
x
n2
4
y
n2
0
16)
4
x
n1
31
x
n2
5
y
n3
0
17)
4
y
n2
3
x
n2
y
n3
0
18)
4
y
n1
3
x
n3
31
y
n2
0
19)
x
n1
31
x
n3
40
y
n3
0
21)
y
n2
3
x
n3
4
y
n3
0
22)
24
x
n3
y
n1
31
y
n3
0
23)
8
y
n2
y
n1
y
n3
0
24)
3
x
n1
31
y
n2
4
y
n3
0
C. Each of the following expressions represents a nasty number
1)
48
x
2n2
6
x
2n3
12
2)
378
6
96
8
1
4 2 2
2n
x
n
x
3)
24
x
2n2
30
y
2n2
12
4)
558
90
144
12
1
3 2 2
2n
y
n
x
5)
1464
30
372
31
1
4 2 2
2n
y
n
x
6)
1134
144
36
3
1
4 2 3
2n
x
n
x
7)
6
x
2n3
60
y
2n2
12
8)
186
x
2n3
240
y
2n3
12
9)
366
x
2n3
60
y
2n4
12
10)
24
1890
372
31
1
2 2 4
2n
y
n
x
11)
186
1890
48
4
1
3 2 4
2n
y
n
x
12)
1464
x
2n4
1890
y
2n4
12
13)
24
186
36
3
1
2 2 3
2n
y
n
y
14)
6
366
72
6
1
2 2 4
2n
y
n
y
15)
186
1464
36
3
1
3 2 4
2n
y
n
y
D. Each of the following expressions represents a cubical integer
1)
8
x
3n3
x
3n4
24
x
n1
3
x
n2
2)
63
3 3 3 5189
13
3
8
1
n
n
nn
x
x
x
x
3)
4
x
3n3
5
y
3n3
12
x
n1
15
y
n1
4)
31
3 35
3 493
115
2
4
1
n
n
nn
y
x
y
5)
244
3 35
3 5732
115
3
31
1
n
n
nn
y
x
y
x
6)
63
x
3n4
8
x
3n5
189
x
n2
24
x
n3
7)
x
3n4
10
y
3n3
3
x
n2
30
y
n1
8)
31
x
3n4
40
y
3n4
93
x
n2
120
y
n2
9)
61
x
3n4
10
y
3n5
183
x
n2
30
y
n3
10)
4
3 5315
3 312
3945
1
31
1
n
n
nn
y
x
y
x
11)
31
3 5315
3 493
3945
2
4
1
n
n
nn
y
x
y
x
12)
244
x
3n5
315
y
3n5
732
x
n3
945
y
n3
13)
4
3 431
3 312
293
1
3
1
n
n
nn
y
y
y
y
14)
3 561
3 33
3183
1
6
1
n
n
nn
y
y
y
y
15)
31
3 5244
3 493
3732
2
3
1
n
n
nn
y
y
y
y
E. Each of the following expressions represents a bi-quadratic integer
1)
8
x
4n4
x
4n5
32
x
2n2
4
x
2n3
6
2)
63
252
4
48
8
1
4 2 2 2 6 4 44n
x
n
x
n
x
n
x
3)
4
x
4n4
5
y
4n4
16
x
2n2
20
y
2n2
6
4)
31
5
124
20
24
4
1
3 2 2 2 5 4 44n
y
n
x
n
y
n
x
5)
244
5
976
20
186
31
1
4 2 2 2 6 4 44n
y
n
x
n
y
n
x
6)
63
x
4n5
8
x
4n6
252
x
2n3
32
x
2n4
6
7)
x
4n5
10
y
4n4
4
x
2n3
40
y
2n2
6
8)
31
x
4n5
40
y
4n5
124
x
2n3
160
y
2n3
6
9)
61
x
4n5
10
y
4n6
244
x
2n3
40
y
2n4
16
10)
4
315
16
1260
186
31
1
2 2 4 2 4 4 64n
y
n
x
n
y
n
x
11)
31
315
124
1260
24
4
1
3 2 4 2 5 4 64n
y
n
x
n
y
n
x
12)
244
x
4n6
315
y
4n6
976
x
2n4
1260
y
2n4
6
13)
4
31
16
124
18
3
1
2 2 3 2 4 4 54n
y
n
y
n
y
n
14)
61
4
244
36
6
1
2 2 4
2 4 4 6
4n
y
n
y
n
y
n
y
15)
31
244
124
976
18
3
1
3 2 4
2 5
4 6
4n
y
n
y
n
y
n
y
F. Each of the following expressions represents a quintic integer
1)
8
x
5n5
x
5n6
40
x
3n3
5
x
3n4
80
x
n1
10
x
n2
2)
63
5 5 5 7315
3 35
3 5630
110
3
8
1
n
n
n
n
nn
x
x
x
x
x
x
3)
4
x
5n5
5
y
5n5
20
x
3n3
25
y
3n3
40
x
n1
50
y
n1
4)
31
5 55
5 6155
3 325
3 4310
150
2
4
1
n
n
n
n
nn
y
x
y
x
y
x
5)
244
5 55
5 71220
3 325
3 52440
150
3
31
1
n
n
n
n
nn
y
x
y
x
y
x
6)
63
x
5n6
8
x
5n7
315
x
3n4
40
x
3n5
630
x
n2
80
x
n3
7)
x
5n6
10
y
5n5
5
x
3n4
50
y
3n3
10
x
n2
100
y
n1
8)
31
x
5n6
40
y
5n6
155
x
3n4
200
y
3n4
310
x
n2
400
y
n2
9)
61
x
5n6
10
y
5n7
305
x
3n4
50
y
3n5
610
x
n2
100
y
n3
10)
4
5 7315
5 520
3 51575
3 340
33150
1
31
1
n
n
n
n
nn
y
x
y
x
y
x
11)
31
5 7315
5 6155
3 51575
3 4310
33150
2
4
1
n
n
n
n
nn
y
x
y
x
y
x
12)
244
x
5n7
315
y
5n7
1220
x
3n5
1575
y
3n5
2440
x
n3
3150
y
n3
13)
4
5 631
5 520
3 4155
3 340
2310
1
3
1
n
n
n
n
nn
y
y
y
y
y
y
14)
5 761
5 55
3 5305
3 310
3610
1
6
1
n
n
n
n
nn
y
y
y
y
y
y
15)
31
5 7244
5 6155
3 51220
3 4310
32440
2
31
1
n
n
n
n
nn
y
y
y
y
y
III. REMARKABLE OBSERVATIONS
A. Employing linear combinations among the solutions of (1), one may generate integer solutions for other choices of hyperbola which are presented in the Table: 2 below:
Table: 2 Hyperbolas
S. No Hyperbola
X
,
Y
1
15
2 260
Y
X
8
x
n1
x
n2,
31
x
n1
4
x
n2
2
15
216
23840
Y
X
63
x
n1
x
n3,
61
x
n1
x
n3
3
3
25
212
Y
X
4
x
n1
5
y
n1,
3
x
n1
4
y
n1
4 2
240
2576
Y
X
93
x
n1
15
y
n2,
6
x
n1
y
n2
5
3
25
211532
Y
X
244
x
n1
5
y
n3,
189
x
n1
4
y
n3
6
5
23
2180
Y
X
189
x
n2
24
x
n3,
244
x
n2
31
x
n3
7
48
25
2192
Y
X
x
n2
10
y
n1,
3
x
n2
31
y
n1
8
3
25
212
Y
X
31
x
n2
40
y
n2,
24
x
n2
31
y
n2
9
48
25
2192
Y
X
61
x
n2
10
y
n3,
189
x
n2
31
y
n3
10
3
25
211532
Y
X
4
x
n3
315
y
n1,
3
x
n3
244
y
n1
11
3
25
2192
Y
X
31
x
n3
315
y
n2,
24
x
n3
244
y
n2
12
3
25
212
Y
X
244
x
n3
315
y
n3,
189
x
n3
244
y
n3
13 2
15
236
Y
X
4
y
n2
31
y
n1,
8
y
n1
y
n2
14
16
215
22304
Y
X
y
n3
61
y
n1,
63
y
n1
y
n3
15 2
15
236
Y
B. Employing linear combinations among the solutions of (1), one may generate integer solutions for other choices of parabola which are presented in the Table: 3 below:
Table: 3 Parabolas
S. No Parabola
X
,
Y
1
15
230
Y
X
8
x
2n2
x
2n3,
31
x
n1
4
x
n2
2
15
2
2240
Y
X
63
x
2n2
x
2n4,
61
x
n1
x
n3
3
3
5
26
Y
X
4
x
2n2
5
y
2n2,
3
x
n1
4
y
n1
4
20
224
Y
X
93
x
2n2
15
y
2n3,
6
x
n1
y
n2
5
93
5
25766
Y
X
244
x
2n2
5
y
2n4,
189
x
n1
4
y
n3
6
5
230
Y
X
189
x
2n3
24
x
2n4,
244
x
n2
31
x
n3
7
48
5
296
Y
X
x
2n3
10
y
2n2,
3
x
n2
31
y
n1
8
3
5
26
Y
X
31
x
2n3
40
y
2n3,
24
x
n2
31
y
n2
9
48
5
296
Y
X
61
x
2n3
10
y
2n4,
189
x
n2
31
y
n3
10
93
5
25766
Y
X
4
x
2n4
315
y
2n2,
3
x
n3
244
y
n1
11
12
5
296
Y
X
31
x
2n4
315
y
2n3,
24
x
n3
244
y
n2
12
3
5
26
Y
X
244
x
2n4
315
y
2n4,
189
x
n3
244
y
n3
13
5
26
Y
X
4
y
2n3
31
y
2n2,
8
y
n1
y
n2
14
32
5
2384
Y
X
y
2n4
61
y
2n2,
63
y
n1
y
n3
15
5
26
Y
X
31
y
2n4
244
y
2n3,
63
y
n2
8
y
n3
IV. CONCLUSION
In this paper, we have presented infinitely many integer solutions for the Diophantine equation, represented by hyperbola is given
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(
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a
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,
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