On the special Diophantine Equation
y
2
7
x
2
9
t,
t
0
Dr. G. Sumathi 1, S. Maheswari 2
1
Assitant Professor, 2M.Phil Scholar, PG & Research Department of Mathematics, Shrimati Indira Gandhi college, Trichy-2, Tamilnadu, India.
Abstract: The binary quadratic equation
y
2
7
x
2
9
t,
t
0
representing hyperbola is considered for finding its integer solutions. A few interesting properties among the solutions are presented. Also, we present infinitely many positive integer solutions in terms of generalized Fibonacci sequences of numbers, generalized Lucas sequences of numbers.Keywords: Binary quadratic integral solution, generalized Fibonacci sequences of numbers, generalized Lucas sequences of numbers.
AMS Mathematics subject classification: 11D09 Notations
)
,
(
K
S
GF
n : Generalized Fibonacci sequences of rank n.)
,
(
K
S
GL
n : Generalized Lucas sequences of rank n.
2
)
2
)(
1
(
1
,
m
n
n
t
mn
6
))
5
(
)(
2
)((
1
(
n
m
n
m
n
p
nm
)
1
(
Pr
n
n
n
1
2
)
1
(
,
mn
n
Ct
mnI. INTRODUCTION
The binary quadratic equation of the form
y
2
Dx
2
1
where D non-square positive integer has been studied by various mathematician for its non-trivial integral solutions. When D takes different integral values [1,2,4]. In [3] infinitely many Pythagorean triangles in each of which hypotenuse is four time the product of the generators added with unity are obtained by employing the non-integral solutions of binary quadratic equationy
2
3
x
2
1
. In [5] a, special Pythagorean triangle is obtained by employing the integral solution ofy
2
182
x
2
14
. In [6] different pattern of infinitely many Pythagorean triangles are obtained by employing the non-integral solutions ofy
2
14
x
2
4
. In this context one may also refer [7,8]. These results have motivated us to search for the integral solutions of yet another binary quartic equation isy
2
7
x
2
9
t,
t
0
representing a hyperbola. A few interesting properties among the solutions are presented. Employing the integral solutions of the equation under consideration a few patterns of Pythagorean triangles are obtained.II. METHOD OF ANALYSIS
Consider the binary quadratic equation
y
2
7
x
2
9
t
,
t
0
(1) with least positive integer solution of (1),
0
3
(
3
),
0
8
(
3
),
D
7
t
y
t
x
y
2
7
x
2
1
(2) whose general solution(
~
x
n,
~
y
n)
is represented by,
x
n
g
n
y
n
f
n
2
1
~
,
7
2
1
~
In which,
f
n
(
8
3
7
)
n
1
(
8
3
7
)
n
1
g
n
(
8
3
7
)
n
1
(
8
3
7
)
n
1
Applying Brahmagupta lemma between
(
x
0,
y
o)
and(
x
~
n,
~
y
n)
the general solution of equation (1) are given to ben
g
t
n
f
t
n
x
7
)
3
(
4
2
)
3
(
3
1
,
g
n
t
n
f
t
n
y
2
)
3
(
7
3
)
3
(
4
1
,The recurrence relation satisfied by the values xn+1 and yn+1 are respectively,
[image:2.612.41.281.81.219.2]
x
n
3
16
x
n
2
x
n
1
0
,y
n
3
16
y
n
2
y
n
1
0
, A few numerical examples are in the table 1 below:Table 1: Numerical Examples
n
1
n
x
y
n
1
0
)
3
(
48
t
127
(
3
t
)
1
)
3
(
765
t
2024
(
3
t
)
2
)
3
(
12192
t
32257
(
3
t
)
3
)
3
(
194307
t
514088
(
3
t
)
4
)
3
(
3096720
t
8193151
(
3
t
)
5
49353213
(
3
t
)
130576328
(
3
t
)
6
786554688
(
3
t
)
A. A Few Interesting Relations Between The Solutions Are Given Below 1)
x
n2
8
x
n1
3
y
n1
0
2)
x
n1
8
x
n2
3
y
n2
0
3)
8
x
n1
127
x
n2
3
y
n3
0
4)
48
y
n1
x
n3
127
x
n1
0
5)
x
n1
6
y
n2
x
n3
0
6)
48
y
n3127
x
n3
x
n1
0
7)
3
y
n1
8
x
n3
127
x
n2
0
8)
3
y
n2
8
x
n2
x
n3
0
9)
x
n2
8
x
n3
3
y
n3
0
10)
127
y
n1
y
n3
336
x
n1
0
11)
42
x
n2
y
n3
y
n1
0
12)
336
x
n3
127
y
n3
y
n1
0
13)
y
n2
21
x
n1
8
y
n1
0
14)
21
x
n2
8
y
n2
y
n1
0
15)
21
x
n3
127
y
n2
8
y
n1
0
16)
127
y
n2
8
y
n3
21
x
n1
0
17)
8
y
n2
y
n3
21
x
n2
0
18)
y
n2
8
y
n3
21
x
n3
0
19)
x
n1
16
x
n2
x
n3
0
20)
y
n1
16
y
n2
y
n3
0
B. Each Of The Following Expressions Represents A Nasty Number
1)
96
1524
36
(
3
)
)
3
(
3
1
2 2 3
2n n t
t
x
x
2)
6
1518
36
(
3
)
)
3
(
3
1
2 2 4
2n n t
t
x
x
3)
96
252
12
(
3
)
3
1
2 2 2
2n n t
t
y
x
4)
12
504
12
(
3
)
3
1
2 2 3
2n n t
t
y
x
5)
96
64260
1524
(
3
)
)
3
(
127
1
2 2 4
2n n t
t
y
x
6)
1524
24288
36
(
3
)
)
3
(
3
1
3 2 4
2n n t
7)
1524
252
96
(
3
)
)
3
(
8
1
3 2 2
2n n t
t
y
x
8)
1524
4032
12
(
3
)
3
1
3 2 3
2n n t
t
y
x
9)
1524
64260
96
(
3
)
)
3
(
8
1
3 2 4
2n n t
t
y
x
10)
24288
252
1524
(
3
)
)
3
(
127
1
4 2 2
2n n t
t
y
x
11)
3036
504
12
(
3
)
3
1
4 2 3
2n n t
t
y
x
12)
24288
2 464260
2 412
(
3
)
3
1
tn n
t
y
x
13)
192
12
12
(
3
)
3
1
3 2 2
2n n t
t
y
y
14)
3060
12
192
(
3
)
)
3
(
16
1
4 2 2
2n n t
t
y
y
15)
9180
576
36
(
3
)
)
3
(
3
1
4 2 3
2n n t
t
y
y
C. Each Of The Following Expressions Is A Cubical Integer
1)
[
16
254
48
762
]
)
3
(
3
1
1 2
3 3 4
3n
n
n
nt
x
x
x
x
2)
[
16
42
48
126
]
3
1
1 1
3 3 3
3n
n
n
nt
y
x
y
x
3)
[
2
84
6
252
]
3
1
1 2
3 3 4
3n
n
n
nt
y
x
y
x
4)
[
16
10710
48
32130
]
)
3
(
127
1
1 3
3 3 5
3n
n
n
nt
y
x
y
x
5)
[
254
4048
762
12144
]
)
3
(
3
1
2 3
4 3 5
3n
n
n
nt
x
x
x
x
6)
[
254
42
762
126
]
)
3
(
8
1
2 1
4 3 3
3n
n
n
nt
y
x
y
x
7)
[
254
672
762
2016
]
3
1
2 2
4 3 4
3n
n
n
nt
y
x
y
x
8)
[
254
10710
762
10710
]
)
3
(
8
1
2 3
4 3 5
3n
n
n
n9)
[
4048
42
12144
126
]
)
3
(
127
1
3 1
5 3 3
3n
n
n
nt
y
x
y
x
10)
[
506
84
1518
252
]
3
1
3 2
5 3 4
3n
n
n
nt
y
x
y
x
11)
[
4048
10710
12144
32130
]
3
1
3 3
5 3 5
3n
n
n
nt
y
x
y
x
12)
[
32
2
96
6
]
3
1
2 1
4 3 3
3n
n
n
nt
y
y
y
y
13)
[
510
2
1530
6
]
)
3
(
16
1
3 1
5 3 3
3n
n
n
nt
y
y
y
y
14)
[
1530
96
4590
288
]
)
3
(
3
1
3 2
5 3 4
3n
n
n
nt
y
y
y
y
D. Each Of The Following Expressions Represent A Bi-Quadratic Integer
1)
[
16
254
64
1016
18
(
3
)]
)
3
(
3
1
2 2 3
2 4
4 5
4n n n n t
t
x
x
x
x
2)
[
253
4
1012
18
(
3
)]
)
3
(
3
1
2 2 4
2 4 4 6
4n n n n t
t
x
x
x
x
3)
[
16
42
64
168
3
]
3
1
2 2 2
2 4
4 4
4n n n n t
t
y
x
y
x
4)
[
2
84
8
336
3
]
3
1
2 2 3
2 4 4 5
4n n n n t
t
y
x
y
x
5)
[
16
10710
64
42480
762
(
3
)]
)
3
(
127
1
2 2 4
2 4
4 6
4n n n n t
t
y
x
y
x
6)
[
254
4048
1016
16192
18
(
3
)]
)
3
(
3
1
3 2 4
2 5
4 6
4n n n n t
t
x
x
x
x
7)
[
254
42
1016
168
48
(
3
)]
)
3
(
8
1
2 1
5 4 4
4n n n n t
t
y
x
y
x
8)
[
254
672
1016
672
3
]
3
1
3 2 3
2 5
4 5
4n n n n t
t
y
x
y
x
9)
[
254
10710
1016
4284
48
(
3
)]
)
3
(
8
1
3 2 4
2 5
4 6
4n n n n t
t
y
x
y
x
10)
[
4048
42
16192
168
762
(
3
)]
)
3
(
127
1
4 2 2
2 6
4 4
4n n n n t
t
y
x
y
x
11)
[
506
84
2024
336
3
]
3
1
4 2 3
2 6
4 5
4n n n n t
12)
[
4048
10710
16192
42840
3
]
3
1
4 2 4 2 6 4 64n n n n t
t
y
x
y
x
13)
[
32
2
128
8
3
]
3
1
3 2 2 2 5 4 44n n n n t
t
y
y
y
y
14)
[
510
2
2040
8
96
(
3
)]
)
3
(
16
1
4 2 2 2 6 4 44n n n n t
t
y
y
y
y
15)
[
1530
96
6120
384
18
(
3
)]
)
3
(
3
1
3 2 6 4 54n n n n t
t
y
y
y
y
E. Each Of The Following Expression Represents A Quantic Integer
1)
[
16
254
80
1270
160
2540
]
)
3
(
3
1
1 2 3 3 4 3 5 5 65n
n
n
n
n
nt
x
x
x
x
x
x
2)
[
253
5
1265
10
2530
]
)
3
(
3
1
1 3 3 3 5 3 5 5 75n
n
n
n
n
nt
x
x
x
x
x
x
3)
[
16
21
80
210
160
420
]
3
1
1 1 3 3 3 3 5 5 55n
n
n
n
n
nt
y
x
y
x
y
x
4)
[
2
84
10
420
20
840
]
3
1
1 2 3 3 4 3 5 5 65n
n
n
n
n
nt
y
x
y
x
y
x
5)
[
16
10710
80
53550
160
107100
]
)
3
(
127
1
1 3 3 3 5 3 5 5 75n
n
n
n
n
nt
y
x
y
x
y
x
6)
[
254
4048
1270
20240
2540
40480
]
)
3
(
3
1
2 3 4 3 5 3 6 5 75n
n
n
n
n
nt
x
x
x
x
x
x
7)
[
254
42
1270
210
2540
420
]
)
3
(
8
1
2 1 4 3 3 3 6 5 55n
n
n
n
n
nt
y
x
y
x
y
x
8)
[
254
672
1270
3360
2540
10020
]
3
1
2 2 4 3 4 3 7 5 65n
n
n
n
n
nt
y
x
y
x
y
x
9)
[
254
10710
1270
53550
460
107100
]
)
3
(
8
1
2 3 4 3 5 3 6 5 75n
n
n
n
n
nt
y
x
y
x
y
x
10)
[
4048
42
20240
210
40480
420
]
)
3
(
127
1
3 1 5 3 3 3 7 5 55n
n
n
n
n
nt
y
x
y
x
y
x
11)
[
506
84
2530
420
5060
840
]
3
1
3 2 5 3 4 3 7 5 65
n
n
n
n
nt
y
x
y
x
y
x
12)
[
4048
10710
20240
53550
40480
107100
]
3
1
3 3 5 3 5 3 7 5 75n
n
n
n
n
nt
y
x
y
x
y
x
13)
[
32
2
160
10
320
25
]
3
1
2 1 4 3 3 3 6 5 55n
n
n
n
n
n14)
[
510
2
2550
10
5100
25
]
)
3
(
16
1
3 1
5 3 3
7 5 5
5n
n
n
n
n
nt
y
y
y
y
y
y
15)
[
1530
96
7650
480
15300
960
]
)
3
(
3
1
3 2
5 3 4
3 7
5 6
5n
n
n
n
n
nt
y
y
y
y
y
y
F. The solution of (1) in terms of special integers namely , generalized fibonacci sequence and generalized Lucas sequence are exhibited below
)
1
,
16
(
)
3
(
24
)
1
,
16
(
2
)
3
(
3
1 1
1
n t n
t
n
GL
GF
x
)
1
,
16
(
)
3
(
63
)
1
,
16
(
)
3
(
4
1 11
t n t n
n
GL
GF
y
III. REMARKABLE OBSERVATION
[image:7.612.54.561.318.699.2]Employing linear combinations among the solutions of (1), one may generate integer solutions for other choices of hyperbolas which are presented in table 2 below
Table 2: Hyperbola S.NO Hyperbola (X,Y)
1
)
3
(
36
63
4
X
2
Y
2
2t(
8
x
n2
127
x
n1,
32
x
n1
2
x
n2)
2
)
3
(
2304
63
64
X
2
Y
2
2t(
x
n3
253
x
n1,
255
x
n1
x
n3)
3
)
3
(
4
14
4
X
2
Y
2
2t(
8
y
n1
21
x
n1,
8
x
n1
3
y
n1)
4
16
27
264
(
3
2t)
Y
X
(
2
y
n2
84
x
n1,
127
x
n1
3
y
n2)
5
)
3
(
64516
7
2 22
Y
tX
(
16
y
n3
10710
x
n1,
4048
x
n1
6
y
n3)
6 2
63
236
(
3
2t)
Y
X
(
254
x
n3
4048
x
n2,
510
x
n2
32
x
n3)
7
)
3
(
256
448
2 22
Y
tX
(
254
y
n1
42
x
n2,
2
x
n2
12
y
n1)
8
)
3
(
4
28
4
X
2
Y
2
2t(
127
y
n2
336
x
n2,
127
x
n2
48
y
n2)
9 2
7
2256
(
3
2t)
Y
X
(
254
y
n3
10710
x
n2,
4048
x
n2
96
y
n3)
10
)
3
(
64516
7
2 22
Y
tX
(
4048
y
n1
42
x
n3,
16
x
n3
1530
y
n1)
11
)
3
(
256
7
64
X
2
Y
2
2t(
506
y
n2
84
x
n3,
254
x
n3
1530
y
n2)
12 2
28
24
(
3
2t)
Y
X
(
4048
y
n3
1530
y
n2,
2024
x
n3
765
y
n3)
13
)
3
(
252
63
X
2
Y
2
2t(
32
y
n1
2
y
n2,
16
y
n2
254
y
n1)
14
252
2256
2258048
(
3
2t)
Y
X
(
510
y
n1
2
y
n3,
2
y
n3
506
y
n1)
15
)
3
(
2268
9
1) Employing linear combination among the solutions of (1), one may generate integer solutions for other choices of parabolas which are presented in table 3 below
Table 3: Parabola S.NO Parabola (X,Y) 1
)
3
(
6
7
)
3
(
2
tX
Y
2
2t(
8
x
n2
127
x
n1,
32
x
n1
2
x
n2)
2
)
3
(
384
7
)
3
(
64
tX
Y
2
2t(
x
n3
253
x
n1,
255
x
n1
x
n3)
3
)
3
(
2
28
)
3
(
2
tX
Y
2
2t(
8
y
n1
21
x
n1,
8
x
n1
3
y
n1)
4
)
3
(
32
7
)
3
(
16
tX
Y
2
2t(
2
y
n2
84
x
n1,
127
x
n1
3
y
n2)
5
)
3
(
32258
7
)
3
(
127
tX
Y
2
2t(
16
y
n3
10710
x
n1,
4048
x
n1
6
y
n3)
6
)
3
(
18
7
)
3
(
3
tX
Y
2
2t(
254
x
n3
4048
x
n2,
510
x
n2
32
x
n3)
7
)
3
(
16
7
)
3
(
tX
Y
2
2t(
254
y
n1
42
x
n2,
2
x
n2
12
y
n1)
8)
3
(
2
28
)
3
(
2
tX
Y
2
2t(
127
y
n2
336
x
n2,
127
x
n2
48
y
n2)
9
)
3
(
128
7
)
3
(
8
tX
Y
2
2t(
254
y
n3
10710
x
n2,
4048
x
n2
96
y
n3)
10
127
(
3
t)
7
232258
(
3
2t)
Y
X
(
4048
y
n1
42
x
n3,
16
x
n3
1530
y
n1)
11
64
(
3
t)
7
2128
(
3
2t)
Y
X
(
506
y
n2
84
x
n3,
254
x
n3
1530
y
n2)
12
)
3
(
2
28
)
3
(
tX
Y
2
2t(
4048
y
n3
10710
x
n3,
2024
x
n3
765
y
n3)
13
63
(
3
t)
2126
(
3
2t)
Y
X
(
32
y
n1
2
y
n2,
16
y
n2
254
y
n1)
14
)
3
(
8064
16
)
3
(
252
tX
Y
2
2t(
510
y
n1
2
y
n3,
2
y
n3
506
y
n1)15
)
3
(
378
3
)
3
(
63
tX
Y
2
2t(
1530
y
n2
96
y
n3,
254
y
n3
4048
y
n2)
2) Employing the following solutions(x,y), each of the following expressions among the special polygonal, star, pyramidal, pronic and centered polygonal number id congruent under modulo 9
2
, 6
5 2
, 4
5
42
1
2
xx
y y
ct
p
ct
p
2
, 4
5 2
3 3
3
1
28
4
x x n
n
ct
p
p
pt
IV. CONCLUSION
In this paper, we have presented infinitely many integer solutions for the negative Pell Equations y = 7x + 9 , t≥0 . As the binary quadratic Diophantine equations are rich in variety, one may search for the other choices of Pell Equations and determine their integer solutions along with suitable properties.
REFERENCES
[1] L.E.Dickson, History of Theory of numbers, Vol.2, Chelsea Publishing Company, New York, (1952). [2] L.J.Mordell, Diophantine Equations, Academic Press, New York, 1969.
[3] M.A.Gopalan and G.Janaki, Observation On
y
2
33
x
2
4
t, Acta Ciancia Indica, XXXIVM(2) (2008) 639-696. [4] L.E.Dickson, History of theory of numbers and Diophantine analysis, Vol 2, Dover publications, New York,(2005).[5] G.Sumathi “Integral Points on the cone”
7
x
2
3
y
2
16
z
2, Journal of mathematics and Informatics, Vol 11, 47-54, 2017. [6] G.Sumathi “Observations on the hyperbola”y
2
182
x
2
14
,
Journal of mathematics and Informatics,Vol 11, 73-81, 2017.[7] Dr.G.Sumathi “Observations on the Pell Equation
y
2
14
x
2
4
” International Journal of Creative Research Thoughts, Vol 6, Issue 1, Pp1074-1084,March 2018.[8] Dr.G.Sumathi “Observations on the Equation
y
2
312
x
2
1
” International Journal of Mathematics Trends and Technology, Vol 50, Issue4, 31-34, Oct 2017.[9] Dr.G.Sumathi “Observations on the hyberbola