Integral Points on the Ternary Quadratic
Diophantine Equation
y
2
33
x
2
4
t,
t
0
Dr. G. Sumathi1, W. Princy Dona 2
1
Assistant Professor,PG and Research Department of Mathematics, Shrimati Indira Gandhi College Trichy-2, Tamil Nadu, India.
2
M.Phil Scholar,PG and Research Department of Mathematics, Shrimati Indira Gandhi College Trichy-2, Tamil Nadu, India.
Abstract: The binary quadratic equation = + 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 solutions, generalized Fibonacci Sequences of numbers, generalized Lucas Sequences of numbers.
AMS Mathematics Subject Classification: 11D09 Notations
GF
n(
k
,
s
)
:
Generalized Fibonacci Sequences of rank n.
GL
n(
k
,
s
)
:
Generalized Lucas Sequences of rank n.
2
)
2
)(
1
(
1
,
m
n
n
t
mn
6
]
)
5
(
)(
2
(
)
1
(
[
n
n
m
n
m
P
nm
Pr
n
n
(
n
1
)
1
2
)
1
(
,
mn
n
Ct
mn
S
n
6
n
(
n
1
)
1
I. INTRODUCTION
The binary quadratic equation of the form = + 1, where D is non-square positive integer has been studied by various mathematicians 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 times the product of the generators added with unity are obtained by employing the non-integral solutions of binary quadratic equation = 3 + 1. In [5] a, special Pythagorean triangle is obtained by employing the integral solutions of = 182 + 14. In [6] different pattern of infinitely many Pythagorean triangles are obtained by employing the non-integral solutions of = 14 + 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 quadratic equation = 33 + 4 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. METHODS OF ANALYSIS
Consider the binary quadratic equation t
x
y
2
33
2
4
,
t
0
(1)with least positive integer solutions is
,
)
2
(
4
0 t
x
y
0
23
(
2
)
tconsider the Pell equation
1
33
22
x
y
(2)whose general solution
(
~
x
n,
~
y
n)
is represented byn n
n
n
g
y
f
x
2
1
~
,
33
2
1
~
in which,
1
133
4
23
33
4
23
n nn
f
g
n
23
4
33
n1
23
4
33
n1,
n
1
,
0
,
1
,
2
...
Applying Brahmagupta lemma between the solutions of
x
0,
y
0
and
~
x
n,
~
y
n
, the general solutions of equation (1) are found to ben
t n
t
n
f
g
y
2
(
2
)
33
2
)
2
(
23
1
n t
n t
n
f
g
x
33
2
)
2
(
23
)
2
(
2
1
The recurrence relations satisfied by the values of
x
n1 andy
n1 are respectively.0
46
2 13
n n
n
x
x
x
0
46
2 13
n n
n
y
y
y
[image:2.612.197.415.436.544.2]A few numerical examples are presented in the table 1 below:
Table 1: Numerical examples
n
1
n
x
y
n1-1 4(2)t 23(2)t
0 184(2)t 1057(2)t 1 8460(2)t 48599(2)t 2 388976(2)t 2234497(2)t 3 17884436(2)t 102738263(2)t 4 822295080(2)t 4723725601(2)t
A. A Few Interesting Relations Between The Solutions Are Given Below 1)
x
n3
1057
x
n1
184
y
n1
0
2)
23
x
n2
x
n1
4
y
n2
0
3)
23
x
n2
23
x
n1
184
y
n2
0
4)
1057
x
n3
x
n1
184
y
n3
0
5)
1057
y
n1
6072
x
n1
y
n3
0
6)
1057
y
n2
132
x
n1
23
y
n3
0
7)
x
n1
46
x
n2
x
n3
0
9)
4
y
n3
x
n2
23
x
n3
0
10)
23
x
n1
x
n2
4
y
n1
0
11)
23
y
n2
132
x
n2
y
n1
0
12)
x
n3
23
x
n2
4
y
n2
0
13)
y
n3
132
x
n2
23
y
n2
0
14)
23
x
n1
1057
x
n2
4
y
n3
0
15)
y
n1
132
x
n2
23
y
n3
0
16)
23
y
n1
132
x
n3
1057
y
n2
0
17)
23
y
n3
132
x
n3
y
n2
0
18)
y
n1
6072
x
n3
1057
y
n3
0
19)
132
x
n1
23
y
n1
y
n2
0
20)
y
n3
y
n1
46
y
n2
0
B. Each Of The Following Expression Represents A Nasty Number
1)
n n t
t
276
x
12684
x
48
(
2
)
)
2
(
4
1
2 2 3
2
2)
n n t
t
276
x
583188
x
2208
(
2
)
)
2
(
184
1
2 2 4
2
3)
n n t
t
276
y
1584
x
12
(
2
)
2
1
2 2 2
2
4)
n n t
t
276
y
72864
x
276
(
2
)
)
2
(
23
1
2 2 3
2
5)
n n t
t
276
y
3350160
x
12684
(
2
)
)
2
(
1057
1
2 2 4
2
6)
n n t
t
12684
x
583188
x
48
(
2
)
)
2
(
4
1
3 2 4
2
7)
n n t
t
12684
y
1584
x
276
(
2
)
)
2
(
23
1
3 2 2
2
8)
n n t
t
12684
y
72864
x
12
(
2
)
2
1
3 2 3
2
9)
n n t
t
12684
y
3350160
x
276
(
2
)
)
2
(
23
1
3 2 4
2
10)
n n t
t
583188
y
1584
x
12684
(
2
)
)
2
(
1057
1
4 2 2
11)
n n t
t
583188
y
72864
x
276
(
2
)
)
2
(
23
1
4 2 3
2
12)
n n t
t
583188
y
3350160
x
12
(
2
)
2
1
4 2 4
2
13)
n n t
t
2208
y
48
y
48
(
2
)
)
2
(
4
1
3 2 2
2
14)
n n t
t
101520
y
48
y
2208
(
2
)
)
2
(
184
1
4 2 2
2
15)
n n t
t
101520
y
2208
y
48
(
2
)
)
2
(
4
1
4 2 3
2
C. Each Of The Following Expression Represents A Cubical Integer
1)
69
23171
123
3 41057
3 3
)
2
(
2
1
n
n
nn
t
x
x
x
x
2)
3
36339
1 3 52113
3 3
)
2
(
4
1
n
n
nn
t
x
x
x
x
3)
138
1792
146
3 3264
3 3
2
1
n
n
nn
t
y
x
y
x
4)
6
21584
12
3 4528
3 3
2
1
n
n
nn
t
y
x
y
x
5)
138
31675080
146
3 5558360
3 3
)
2
(
1057
1
n
n
nn
t
y
x
y
x
6)
3171
3145797
21057
3 548599
3 4
)
2
(
2
1
n
n
nn
t
x
x
x
x
7)
6342
1792
22114
3 3264
3 4
)
2
(
23
1
n
n
nn
t
y
x
y
x
8)
6342
236432
22114
3 412144
3 4
2
1
n
n
nn
t
y
x
y
x
9)
6342
31675080
22114
3 5558360
3 4
)
2
(
23
1
n
n
nn
t
y
x
y
x
10)
291594
1792
397198
3 3264
3 5
)
2
(
1057
1
n
n
nn
t
y
x
y
x
11)
12678
21584
34226
3 4528
3 5
2
1
n
n
nn
t
y
x
y
x
12)
291594
31675080
397198
3 51558360
3 5
2
1
n
n
nn
13)
276
16
292
3 32
3 4
2
1
n
n
nn
t
y
y
y
y
14)
6345
13
32115
3 3 3 5
)
2
(
23
1
n
n
nn
t
y
y
y
y
15)
12690
2276
34230
3 492
3 5
2
1
n
n
nn
t
y
y
y
y
D. Each Of The Following Expression Represents A Bi-Quadratic Integer
1)
n n n n t
t
92
x
4228
x
23
x
1057
x
12
(
2
)
)
2
(
2
1
4 4 5 4 2 2 32
2)
n n n n t
t
23
x
8452
x
x
2113
x
24
(
2
)
)
2
(
4
1
4 4 6 4 2 2 42
3)
n n n n t
t
184
y
1056
x
46
y
264
x
6
(
2
)
2
1
4 4 4 4 2 2 22
4)
n n n n t
t
8
y
2112
x
2
y
528
x
6
(
2
)
2
1
4 4 5 4 2 2 32
5)
n n n n t
t
1184
y
2233440
x
46
y
558360
x
6342
(
2
)
)
2
(
1057
1
4 4 6 4 2 2 42
6)
n n n n t
t
4228
x
194396
x
1057
x
48599
x
12
(
2
)
)
2
(
2
1
5 4 6 4 3 2 42
7)
n n n n t
t
8456
y
1056
x
2114
y
264
x
138
(
2
)
)
2
(
23
1
5 4 4 4 3 2 22
8)
n n n n t
t
8456
y
48576
x
2114
y
12144
x
6
(
2
)
2
1
5 4 5 4 3 2 32
9)
n n n n t
t
8456
y
2233440
x
2114
y
558360
x
138
(
2
)
)
2
(
23
1
5 4 6 4 3 2 42
10)
n n n n t
t
388792
y
1056
x
97198
y
264
x
6342
(
2
)
)
2
(
1057
1
6 4 4 4 4 2 22
11)
n n n n t
t
16904
y
2112
x
4226
y
528
x
6
(
2
)
2
1
6 4 5 4 4 2 32
12)
n n n n t
t
388792
y
2233440
x
97198
y
558360
x
6
(
2
)
2
1
6 4 6 4 4 2 42
13)
t
n n
n n
t 368 y 8y 92 y 2y 6(2)
2 1 5 4 4 4 3 2 2
2
14)
t
n n
n n
t 8460 y 4y 2115 y y 138(2)
) 2 ( 23 1 6 4 4 4 4 2 2
2
15)
t
n n
n n
t
16920
y
368
y
4230
y
92
y
6
(
2
)
2
1
6 4 5 4 4 2 3E. Each Of The Following Expression Represents A Quintic Integer
1)
10
321130
15
3 510565
3 3 5 72113
5 5
)
2
(
4
1
n
n
n
n
nn
t
x
x
x
x
x
x
2)
20
342260
110
3 521130
3 32
5 74226
5 5
)
2
(
8
1
n
n
n
n
nn
t
x
x
x
x
x
x
3)
460
12640
1230
3 31320
3 346
5 5264
5 5
2
1
n
n
n
n
nn
t
y
x
y
x
y
x
4)
20
25280
110
3 42640
3 32
5 6528
5 5
2
1
n
n
n
n
nn
t
y
x
y
x
y
x
5)
460
35583600
1230
3 52791800
3 346
5 7558360
5 5
)
2
(
1057
1
n
n
n
n
nn
t
y
x
y
x
y
x
6)
10570
3485990
25285
3 5242995
3 41057
5 748599
5 6
)
2
(
2
1
n
n
n
n
nn
t
x
x
x
x
x
x
7)
21140
12640
210570
3 31320
3 42114
5 5264
5 6
)
2
(
23
1
n
n
n
n
nn
t
y
x
y
x
y
x
8)
21140
2121440
210570
3 460720
3 42114
5 612144
5 6
2
1
n
n
n
n
nn
t
y
x
y
x
y
x
9)
21140
35583600
210570
3 52791800
3 42114
5 7558360
5 6
)
2
(
23
1
n
n
n
n
nn
t
y
x
y
x
y
x
10)
971980
12640
3485990
3 31320
3 597198
5 5264
5 7
)
2
(
1057
1
n
n
n
n
nn
t
y
x
y
x
y
x
11)
84520
25280
321130
3 42640
3 54226
5 6528
5 7
2
1
n
n
n
n
nn
t
y
x
y
x
y
x
12)
971980
35583600
3485990
3 52791800
3 597198
5 7558360
5 7
2
1
n
n
n
n
nn
t
y
x
y
x
y
x
13)
920
120
2460
3 310
3 492
5 52
5 6
2
1
n
n
n
n
nn
t
y
y
y
y
y
y
14)
7 5 5 5 5 3 3 3 3
1 10 10575 5 2115
21150 ) 2 ( 8 1
n n n n n
n
t y y y y y y
15)
7 5 6 5 5 3 4 3 3
2 920 21150 460 4230 92
42300 2 1
n n n n n
n
t y y y y y y
III. REMARKABLE OBSERVATIONS
1) The solutions of (1) in terms of special integers namely, generalized Fibonacci
GF
n andLucas
GL
n are exhibited below.)
1
,
46
(
)
2
(
92
)
1
,
46
(
)
2
(
2
1 11
t n t n
n
GL
GF
x
)
1
,
46
(
)
2
(
528
)
1
,
46
(
2
)
2
(
23
1 11
n t n
t
n
GL
GF
2) Employing The 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 2 2 2
)
2
(
64
33
Y
tX
46
x
n2
2114
x
n1,
368
x
n1
8
x
n2
2 2 2 2
)
2
(
135424
33
Y
tX
46
x
n3
97198
x
n1,
16920
x
n1
8
x
n3
3 2 2 2
)
2
(
4
33
Y
tX
46
y
n1
264
x
n1,
46
x
n1
8
y
n1
4 2 2 2
)
2
(
2116
33
Y
tX
46
y
n2
12144
x
n1,
2114
x
n1
8
y
n2
5 2 2 2
)
2
(
4468996
33
Y
tX
(46
y
n3
558360
x
n1,
97198
x
n1
8
y
n3)6 2 2 2
)
2
(
64
33
tY
X
2114
x
n3
97198
x
n2,
16920
x
n2
368
x
n3
7 2 2 2
)
2
(
2116
33
Y
tX
2114
y
n1
264
x
n2,
46
x
n2
368
y
n1
8 2 2 2
)
2
(
4
33
Y
tX
2114
y
n2
12144
x
n2,
2114
x
n2
368
y
n2
9 2 2 2
)
2
(
2116
33
tY
X
2114
y
n3
558360
x
n2,
97198
x
n2
368
y
n3
10 2 2 2
)
2
(
4468996
33
Y
tX
97198
y
n1
264
x
n3,
46
x
n3
16920
y
n1
11 2 2 2
)
2
(
2116
33
Y
tX
97198
y
n2
12144
x
n3,
2114
x
n3
16920
y
n2
12 2 2 2
)
2
(
4
33
Y
tX
97198
y
n3
558360
x
n3,
97198
x
n3
16920
y
n3
13 2 2 2
)
2
(
2112
33
X
Y
t
368
y
n1
8
y
n2,
46
y
n2
2114
y
n1
14 2 2 2
)
2
(
4468992
33
X
Y
t
16920
y
n1
8
y
n3,
46
y
n3
97198
y
n1
15 2 2 2
)
2
(
2112
3) 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
2
tX
132
Y
2
8
(
4
)
t
2 1
2 2 3
2
2114
,
368
8
46
x
n
x
nx
n
x
n2
184
(
2
)
tX
33
Y
2
67712
(
4
)
t
3 1
2 2 4
2
97198
,
16920
8
46
x
n
x
nx
n
x
n3 t t
Y
X
33
2
(
4
)
2
2
46
y
2n2
264
x
2n2,
46
x
n1
8
y
n1
4 t t
Y
X
33
1058
(
4
)
)
2
(
23
2
46
y
2n3
12144
x
2n2,
2114
x
n1
8
y
n2
5 t t
Y
X
33
2234498
(
4
)
)
2
(
1057
2
46
y
2n4
558360
x
2n2,
97198
x
n1
8
y
n3
6
4
(
2
)
tX
33
Y
2
32
(
4
)
t
2114
2 497198
2 3,
16920
2368
3
n n
nn
x
x
x
x
7 t t
Y
X
33
1058
(
2
)
)
2
(
23
2
2114
y
2n2
264
x
2n3,
46
x
n2
368
y
n1
8 t t
Y
X
33
2
(
4
)
2
2
2114
y
2n3
12144
x
2n3,
2114
x
n2
368
y
n2
9
23
(
2
)
tX
33
Y
2
1058
(
4
)
t
3 2
3 2 4
2
558360
,
97198
368
2114
y
n
x
nx
n
y
n10 t t
Y
X
33
2234498
(
4
)
)
2
(
1057
2
97198
y
2n2
264
x
2n4,
46
x
n3
16920
y
n1
11 t t
Y
X
33
1058
(
4
)
)
2
(
23
2
97198
y
2n3
12144
x
2n4,
2114
x
n3
16920
y
n2
12 t t
Y
X
33
2
(
4
)
2
2
97198
y
2n4
558360
x
2n4,
97198
x
n3
16920
y
n313
132
(
2
)
tX
Y
2
1056
(
4
)
t
1 2
3 2 2
2
8
,
46
2114
368
y
n
y
ny
n
y
n14 t t
Y
X
2234496
(
4
)
)
2
(
6072
2
16920
y
2n2
8
y
2n4,
46
y
n3
97198
y
n1
15 t t
Y
X
1056
(
4
)
)
2
(
4) Employing the following solution
(
x
,
y
)
, each of the following expressions among the special polygonal, pyramidal, star, pronic and centered polygonal number is congruent to under modulo 4.2
1 3 2
2 , 3
3 2
Pr
198
3
x x y
y
P
t
P
2
2 3
2 2
1 5
1
1188
1
12
x
x y
y
S
P
S
P
2
1 5 2
, 3
5
1
396
x
x y
y
S
P
t
P
2
, 6
5 2
, 4
5
198
1
2
xx y
y
Ct
P
Ct
P
2
, 4
5 2
3 3
3
1
132
4
x x
n n
Ct
P
P
Pt
IV. CONCLUSION
To conclude, one may search for other choices of positive Pell equations for finding their integer solutions 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, (1970).
[3] M.A.Gopalan and G.Janaki, Observation On
y
2
3
x
2
1
, 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] Dr.G.Sumathi “Observations on the hyperbola”
y
2
182
x
2
14
,
Journal of mathematics and Informatics,Vol 11, 73-81, 2017.[6] 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.[7] Dr.G.Sumathi “Observations on the Equation