484
A Search For Integral Solutions To The Ternary Bi-Quadratic Equation
(
)
2 2 4 3 2 2 3 4x
y
x
y
xy
y
x
y
1
z
x
+
+
+
+
=
+
+
+
S. Vidhyalakshmi1, T. Mahalakshmi2, M. A. Gopalan3
1Assistant Professor, Department of Mathematics, Shrimati Indira Gandhi College, Affiliated to Bharathidasan
University, Trichy, Tamil Nadu, India
2Assistant Professor, Department of Mathematics, Shrimati Indira Gandhi College, Affiliated to Bharathidasan
University, Trichy, Tamil Nadu, India
3Professor, Department of Mathematics, Shrimati Indira Gandhi College, Affiliated to Bharathidasan University,
Trichy, Tamil Nadu, India
Article History: Received: 10 January 2021; Revised: 12 February 2021; Accepted: 27 March 2021; Published
online: 16 April 2021
ABSTRACT :This paper deals with the problem of obtaining non-zero distinct integer solutions to the ternary bi-quadratic
equation
x
4+
x
3y
+
x
2y
2+
xy
3+
y
4=
(
x
+
y
)
2+
1 z
+
2. A few interesting relations among the solution are presented. Given on integer solution of the equation under consideration, integer solutions for various choices of hyperbola and parabolas are exhibited. The formulation of second order Ramanujan Numbers with base numbers as real integers and Gaussian integers is illustrated and also the sequence of Diophantine 3-tuples are exhibited.Keywords: Ternary bi-quadratic, integer solutions, parabolas, hyperbolas, Second order Ramanujan numbers, sequence of
Diophantine 3-tuples.
INTRODUCTION
In number theory, Diophantine equations play a significant role and have a marvellous effects on credulous people. They occupy a remarkable position due to unquestioned historical importance. The subject of Diophantine equation is quite difficult. Every century has seen the solution of more mathematical problem than the century before and yet many mathematical problem, both major and minor still remains unsolved. It is hard to tell whether a given equation has solution or not and when it does, there may be no method to find all of them. It is difficult to tell which are early solvable and which require advanced techniques. There is no well unified body of knowledge concerning general methods. A Diophantine problem is considered as solved if a method is available to decide whether the problem is solvable or not and in case of its solvability, to exhibit all integers satisfying the requirements set forth in the problem. Many researchers in the subjects of Diophantine equation exhibit great interest in homogeneous and non-homogeneous bi-quadratic Diophantine equations. In this context, are may refer [1-12]. This communication concerns yet another interesting ternary bi-quadratic equation given by
x
4+
x
3y
+
x
2y
2+
xy
3+
y
4=
(
x
+
y
)
2+
1
+
z
2and is studied for its non-zero distinct integer solution. A few interesting relations among the solution are presented. Given an integer solution of the equation under consideration, integer solutions for various choices of hyperbola and parabolas are exhibited. The formulation of second order Ramanujan Numbers with base numbers as real integers and Gaussian integers is illustrated and also the sequence of Diophantine 3-tuples are exhibited.METHOD OF ANALYSIS
The ternary bi-quadratic equation under consideration is
(
)
2 2 4 3 2 2 3 4x
y
x
y
xy
y
x
y
1
z
x
+
+
+
+
=
+
+
+
(1) Introduction of the transformations0
v
u
,
uv
4
z
,
v
u
y
,
v
u
x
=
+
=
−
=
(2) in (1) leads to0
1
u
4
u
5
v
u
6
v
4−
2 2+
4−
2−
=
(3) Treating (3) as a quadratic inv
2 and solving forv
2, we’ve1
u
5
v
2=
2+
(4) which is the well known pellian equation whose general solution given by,485 n n n n
g
5
2
1
u
f
2
1
v
=
=
(5) where...
3
,
2
,
1
,
0
n
,
)
5
4
9
(
)
5
4
9
(
g
)
5
4
9
(
)
5
4
9
(
f
1 n 1 n n 1 n 1 n n=
−
−
+
=
−
+
+
=
+ + + +In view of (2), the sequence of values of x , y and z satisfying (1) are represented by
n n n
u
v
x
=
+
n nf
2
1
g
5
2
1
+
=
n n ng
5
f
x
5
2
=
+
(6) n n nu
v
y
=
−
n nf
2
1
g
5
2
1
−
=
n n ng
5
f
y
5
2
=
−
(7)5
g
f
z
n=
n n
n
=
−
1
,
0
,
1
,
2
,...
2 n 2 n nx
y
z
=
−
(8) Replacingn
byn +
1
in (6) , we get(
) (
)
n n 1 n n n n n 1 n 1 n 1 nf
2
13
g
5
2
29
x
g
5
4
f
9
2
1
f
5
4
g
9
5
2
1
f
2
1
g
5
2
1
x
+
=
+
+
+
=
+
=
+ + + + n n 1 n29
g
13
5
f
x
5
2
+=
+
(9) Replacingn
byn +
1
in (9) , we get(
) (
)
n n 2 n n n n n 1 n 1 n 2 nf
2
233
g
5
2
521
x
g
5
4
f
9
2
13
f
5
4
g
9
5
2
29
f
2
13
g
5
2
29
x
+
=
+
+
+
=
+
=
+ + + + n n 2 n521
g
233
5
f
x
5
2
+=
+
(10) Eliminatingf
n andg
n between (6) , (9) and (10) , we have,...
3
,
2
,
1
n
,
0
x
x
18
x
n−
n+1+
n+2=
=
(11) In a similar manner, from (7) one obtainsn n 1 n
11
g
5
5
f
y
5
2
+=
−
−
(12) n n 2 n199
g
89
5
f
y
5
2
+=
−
−
(13)486
Eliminating
f
n andg
n between (7) , (11) and (12) , we have,...
3
,
2
,
1
n
,
0
y
y
18
y
n−
n+1+
n+2=
=
(14) Thus (11) and (14) represent recurrence relations satisfied by the values ofx
andy
respectively . A few numerical examples ofx
n ,y
n andz
n satisfying (1) are given in the Table 1.1 belowTable: 1.1 Numerical Examples
n
x
n
y
n
z
n
-1 1 -1 0 0 13 -5 144 1 233 -89 46368 2 4181 -1597 14930352 3 75025 -28657 4807526976From then above table, we observe some interesting relations among the solutions which are presented below: ➢ Both
x
n ,y
n values are odd andz
n values are even.➢ One can generate second order Ramanujan numbers with base integers as real integers by choosing
x
,y
and z values suitably.For illustrations, consider
( )
2 2 2 2 2 2 2 2 05
13
9
15
16
20
35
37
18
*
8
24
*
6
36
*
4
72
*
2
144
z
−
=
−
=
−
=
−
=
=
=
=
=
=
Now,250
9
13
5
15
5
13
9
15
425
16
13
5
20
5
13
16
20
481
16
15
9
20
9
15
16
20
1394
35
13
5
37
5
13
35
37
1450
35
15
9
37
9
15
35
37
1625
35
20
16
37
16
20
35
37
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2=
+
=
+
−
=
−
=
+
=
+
−
=
−
=
+
=
+
−
=
−
=
+
=
+
−
=
−
=
+
=
+
−
=
−
=
+
=
+
−
=
−
Note: 1487
(
) (
) (
) (
)
(
) (
)
(
) (
) (
) (
)
(
) (
)
(
) (
) (
) (
)
(
) (
)
(
) (
) (
) (
)
(
) (
)
(
) (
) (
) (
)
(
) (
)
(
) (
) (
) (
)
(
10
) (
18
)
26
1000
30
18
8
24
6
18
8
24
6
18
8
24
6
1700
26
32
10
40
18
8
36
4
18
8
36
4
18
8
36
4
1924
30
32
18
40
24
6
36
4
24
6
36
4
24
6
36
4
5576
26
70
10
74
18
8
72
2
18
8
72
2
18
8
72
2
5800
30
70
18
74
24
6
72
2
24
6
72
2
24
6
72
2
6500
40
70
32
74
36
4
72
2
36
4
72
2
36
4
72
2
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2=
+
−
=
−
+
→
+
+
−
=
−
+
+
→
=
=
+
−
=
−
+
→
+
+
−
=
−
+
+
→
=
=
+
−
=
−
+
→
+
+
−
=
−
+
+
→
=
=
+
−
=
−
+
→
+
+
−
=
−
+
+
→
=
=
+
−
=
−
+
→
+
+
−
=
−
+
+
→
=
=
+
−
=
−
+
→
+
+
−
=
−
+
+
→
=
Thus , 1625 , 1450 , 1394 , 481 , 425 , 250 , 6500, 5800, 5576, 1924, 1700, 1000 are second order Ramanujan numbers with base integers as real integers.
➢ Considering suitable values of
x
n andy
n, one generates second order Ramanujan numbers with base integers as Gaussian integers.For illustrations, consider again
z
0 represented by (*) •(
2
i
72
) (
4
i
36
) (
2
i
72
) (
4
i
36
)
6460
36
4
72
2
2 2 2 2+
−
=
−
+
+
=
−
+
→
=
and(
72
i
2
) (
36
i
4
) (
72
i
2
) (
36
i
4
)
6460
36
4
72
2
2 2 2 2+
−
=
−
+
+
=
+
→
=
•(
4
i
36
) (
6
i
24
) (
4
i
36
) (
6
i
24
)
1820
24
6
36
4
2 2 2 2+
−
=
−
+
+
=
−
+
→
=
and(
36
i
4
) (
24
i
6
) (
36
i
4
) (
24
i
6
)
1820
24
6
36
4
2 2 2 2+
−
=
−
+
+
=
+
→
=
•(
6
i
24
) (
8
i
18
) (
6
i
24
) (
8
i
18
)
800
18
8
24
6
2 2 2 2+
−
=
−
+
+
=
−
+
→
=
and(
24
i
6
) (
18
i
8
) (
24
i
6
) (
18
i
8
)
800
18
8
24
6
2 2 2 2+
−
=
−
+
+
=
+
→
=
488
Note that -6460, 6460, -1820, 1820, -800, 800 represent second order Ramanujan numbers with base integers as Gaussian integers.
In a similar manner, other second order Ramanujan numbers are obtained. ➢ Formulation of sequence of Diophantine 3-tuples:
Consider the solution to (1) given by
( )
say
c
13
x
,
5
y
0=
−
0=
=
0 It is observed that 2 2 0 0x
k
65
k
y
+
+
=
,a perfect squareThe pair
(
y
0,
x
0)
represents Diophantine 2-tuple with propertyD
(
k
2+
65
)
Ifc
1 is the3
rd tuple, then it is given by8
k
2
k
2
x
y
c
1=
0+
0+
=
+
Note that
(
−
5
,
13
,
2
k
+
8
)
represents diophantine 3-tuple with propertyD
(
k
2+
65
)
The process of obtaining sequence of diophantine 3-tuple with property
D
(
k
2+
65
)
is illustrated below: Let M be a3
3
square matrix given byM=
−
2
1
0
1
0
0
2
0
1
Now,(
−
5
,
13
,
2
k
+
8
)
M
=
(
−
5
,
2
k
+
8
,
4
k
−
7
)
Note that(
)
(
)
(
)
(
)
(
2
k
8
) (
4
k
7
)
k
65
(
3
k
3
)
perfect
square
square
perfect
5
k
65
k
8
k
2
5
square
perfect
10
k
65
k
7
k
4
5
2 2 2 2 2 2=
+
=
+
+
−
+
=
−
=
+
+
+
−
=
−
=
+
+
−
−
Therefore the triple
(
−
5
,
2
k
+
8
,
4
k
−
7
)
represents diophantine 3-tuple with propertyD
(
k
2+
65
)
. The repetition of the above process leads to sequences of diophantine 3-tuple whose general form(
−
5
,
c
s−1,
c
s)
is given by(
)
(
−
5
,
−
5
s
2+
2
k
+
10
s
−
2
k
+
8
,
−
5
s
2+
2
ks
+
13
)
,
s
=
1
,
2
,
3
...
A few numerical illustrations are given in Table below:Table: Numerical illustrations
k
(
)
1 0,
c
c
,
5
−
(
−
5
,
c
1,
c
2)
(
−
5
,
c
2,
c
3)
D
(
k
2+
65
)
0(
−
5
,
13
,
8
)
(
−
5
,
8
,
−
7
)
(
−
5
,
−
7
,
−
32
)
D
( )
65
1(
−
5
,
13
,
10
)
(
−
5
,
10
,
−
3
)
(
−
5
,
−
3
,
−
26
)
D
( )
66
2(
−
5
,
13
,
12
)
(
−
5
,
12
,
1
)
(
−
5
,
1
,
−
20
)
D
( )
69
It is note that the triple(
c
s−1,
c
s−
5
,
c
s+1)
,
s
=
1
,
2
,
3
...
...
forms an arithmetic progression. In a similar way one may generate sequences of diophantine 3-tuples with suitable property through the other solutions to (1).1. Relations among the solutions are given below.
❖
x
n+2−
18
x
n+1+
x
n=
0
❖8
y
n−
x
n+1+
21
x
n=
0
❖8
y
n+1+
3
x
n+1+
x
n=
0
❖8
y
n+2−
55
x
n+1−
3
x
n=
0
❖144
y
n−
x
n+2+
377
x
n=
0
❖48
y
n+1+
x
n+2+
7
x
n=
0
489 ❖
8
y
n−
21
x
n+2+
377
x
n+1=
0
❖8
y
n+1−
x
n+2+
21
x
n+1=
0
❖8
y
n+2+
3
x
n+2+
x
n+1=
0
❖21
y
n+1+
8
x
n+1−
y
n=
0
❖y
n+2+
8
x
n+1+
3
y
n+1=
0
❖377
y
n+1+
8
x
n+2−
21
y
n=
0
❖377
y
n+2+
144
x
n+2−
y
n=
0
❖21
y
n+2+
8
x
n+2−
y
n+1=
0
❖y
n−
18
y
n+1+
y
n+2=
0
❖144
y
n+2+
55
x
n+2+
x
n=
0
❖y
n+1+
8
x
n+
y
n=
0
❖y
n+2+
144
x
n+
5
x
n=
0
❖x
n+1−
7
y
n+1−
48
x
n=
0
2. Each of the following expressions is a nasty number:
Solving (6) and (9) , we get
n n 1
n29
x
x
8
1
f
=
−
+ (15)
n 1 n
nx
13
x
8
5
g
=
+−
(16) Replacingn
by2 +
n
1
in (15) we get
2n 1 2n 2
1 n 229
x
x
8
1
f
+=
+−
+ Now ,f
2n+1+
2
=
f
n2
2 n 2 n 2 1 n 2x
16
6
f
x
29
4
3
−
+
=
+ + , a nasty number (16a)For simplicity and clear understanding the other choices of nasty numbers are presented below: ❖
29
x
x
16
4
3
2 n 2 1 n 2 +−
++
❖
521
x
x
288
24
1
3 n 2 1 n 2 +−
++
❖6
x
2n+1−
y
2n+1+
2
❖6
22
x
2n+1+
2
y
2n+2+
12
❖
199
x
y
110
55
6
3 n 2 2 n 2 ++
++
❖
521
x
29
x
16
4
3
3 n 2 2 n 2 +−
++
❖
x
29
y
4
7
2
1 n 2 2 n 2 +−
++
❖6
−
11
x
2n+2−
29
y
2n+1+
2
❖2
199
x
2n+2+
29
y
2n+3+
6
❖
5
x
521
y
754
377
6
1 n 2 3 n 2 +−
++
490 ❖
11
x
521
y
42
7
2
2 n 2 3 n 2−
+
−
+ + ❖6
−
199
x
2n+3−
521
y
2n+3+
2
❖
11
y
y
16
4
3
2 n 2 1 n 2−
+
−
+ + ❖
199
y
y
288
24
1
3 n 2 1 n 2−
+
−
+ + ❖
11
y
199
y
16
4
3
2 n 2 3 n 2 +−
++
3. Each of the following expressions is a cubical integer:
Replacing
n
by3 +
n
2
in (15) we get
3n 2 3n 3
2 n 329
x
x
8
1
f
+=
+−
+ Now,f
3n+2=
f
n3−
3
f
n 3 n n 2 n 33
f
f
f
++
=
3 n 1 n n 3 n 3 2 n 3x
87
x
3
x
f
x
29
8
1
−
+
−
=
+ + + , a cubical integer.For simplicity and clear understanding the other choices of cubical integers are presented below: ❖
29
x
3n 2x
3n 387
x
n3
x
n 1
8
1
+ + +−
+
−
❖
521
x
3n 2x
3n 41563
x
n3
x
n 2
144
1
+ + +−
+
−
❖
x
3n+2−
y
3n+2+
3
x
n−
3
y
n
❖
11
x
3n 2y
3n 333
x
n3
y
n 2
3
1
+ + ++
+
+
❖
199
x
3n 2y
3n 4597
x
n3
y
n 1
55
1
+ + ++
+
+
❖
521
x
3n 329
x
3n 41563
x
n 187
x
n 2
8
1
+ + + +−
+
−
❖
x
3n 329
x
3n 23
x
n 187
y
n
21
1
−
+
−
+ + + ❖
−
11
x
3n+3−
29
y
3n+3−
33
x
n+1−
87
y
n+1
❖
199
x
3n 329
y
3n 4597
x
n 187
y
n 2
3
1
+ + + ++
+
+
❖
x
3n 4521
y
3n 23
x
n 21563
y
n
377
1
−
+
−
+ + + ❖
11
x
3n 4521
y
3n 333
x
n 21563
y
n 1
21
1
+ + + +−
−
−
−
❖
−
199
x
3n+4−
521
y
3n+4−
597
x
n+2−
1563
y
n+2
❖
11
y
3n 2y
3n 233
y
n3
y
n 1
8
1
+ + +−
−
−
−
❖
199
y
3n 2y
3n 4597
y
n3
y
n 2
144
1
+ + +−
−
−
−
491 ❖
11
y
3n 4199
y
3n 333
y
n 2597
y
n 1
8
1
+ + + +−
+
−
4. Each of the following expressions is a bi-quadratic integer.
Replacing
n
by4 +
n
3
in (15) we get
4n 3 4n 4
3 n 429
x
x
8
1
f
+=
+−
+ Now,f
4n+3+
4
f
n2−
2
=
f
n4
4 n 2 n 2 1 n 2 4 n 4 3 n 4x
116
x
4
x
48
f
x
29
8
1
−
+
−
+
=
+ + + + , a bi-quadratic integer.For simplicity and clear understanding the other choices of bi-quadratic integers are presented below:
❖
29
x
x
116
x
4
x
48
8
1
2 n 2 1 n 2 4 n 4 3 n 4 +−
++
+−
++
❖
521
x
x
2084
x
4
x
864
144
1
3 n 2 1 n 2 5 n 4 3 n 4 +−
++
+−
++
❖
x
4n+3−
y
4n+3+
4
x
2n+1−
4
y
2n+1+
6
❖
11
x
y
44
x
4
x
18
3
1
2 n 2 1 n 2 4 n 4 3 n 4 ++
++
++
++
❖
199
x
y
796
x
4
y
330
55
1
3 n 2 1 n 2 3 n 2 3 n 4 ++
++
++
++
❖
521
x
29
x
2084
x
116
x
48
8
1
3 n 2 2 n 2 5 n 4 4 n 4 +−
++
+−
++
❖
x
29
y
4
x
116
y
126
21
1
1 n 2 2 n 2 3 n 4 4 n 4 +−
++
+−
++
❖
−
11
x
4n+4−
29
y
4n+4−
44
x
2n+2−
116
y
2n+2+
6
❖
199
x
29
y
796
x
116
y
18
3
1
3 n 2 2 n 2 4 n 4 4 n 4 ++
++
++
++
❖
x
521
y
4
x
2084
y
2262
377
1
1 n 2 3 n 2 4 n 4 5 n 4 +−
++
+−
++
❖
11
x
521
y
44
x
2084
y
126
21
1
2 n 2 3 n 2 4 n 4 5 n 4−
−
−
+
−
+ + + + ❖
−
199
x
4n+5−
521
y
4n+5−
796
x
2n+3−
2084
y
2n+3+
6
❖
11
y
y
44
y
4
y
48
8
1
2 n 2 1 n 2 4 n 4 3 n 4−
−
−
+
−
+ + + + ❖
199
y
y
796
y
4
y
864
144
1
3 n 2 1 n 2 5 n 4 3 n 4−
−
−
+
−
+ + + + ❖
11
y
199
y
44
y
796
y
48
8
1
2 n 2 3 n 2 4 n 4 5 n 4 +−
++
+−
++
5. Each of the following expressions is a quintic integer:
Replacing
n
by5 +
n
4
in (15) we get
5n 4 5n 5
4 n 529
x
x
8
1
f
+=
+−
+ Now,f
5n+4=
f
n5−
5
f
n3+
5
f
n492 n 3 n 4 n 5 5 n
f
5
f
5
f
f
=
++
−
5 n 1 n n 3 n 3 2 n 3 5 n 5 4 n 5x
145
x
5
x
290
x
10
x
f
x
29
8
1
=
−
+
−
+
−
+ + + + + , a quintic integer.For simplicity and clear understanding the other choices of quintic integers are presented below: ❖
x
5n+4−
y
5n+4+
5
x
3n+2−
5
y
3n+2+
10
x
n−
10
y
n
❖
x
5n 529
y
5n 45
x
3n 3145
y
3n 210
x
n290
y
n
21
1
−
+
−
+
−
+ + + + ❖
x
5n 6521
y
5n 45
x
3n 42605
y
3n 210
x
n 15210
y
n
377
1
−
+
−
+
−
+ + + + + ❖
11
y
5n 4y
5n 555
y
3n 25
y
3n 3110
y
n10
y
n 1
8
1
+ + + + +−
−
−
−
−
−
❖
521
x
5n 4x
5n 62605
x
3n 25
x
3n 45210
x
n10
x
n 2
144
1
+ + + + +−
+
−
+
−
❖
199
x
5n+6−
521
y
5n+6−
995
x
3n+4−
2605
y
3n+4−
1990
x
n+2−
5210
y
n+2
❖
x
5n 6521
y
5n 45
x
3n 42605
y
3n 210
x
n5210
y
n 1
377
1
+ + + + +−
+
−
+
−
❖
11
x
5n 6521
y
5n 555
x
3n 42605
y
3n 3110
x
n 25210
y
n 1
21
1
+ + + + + +−
−
−
−
−
−
❖
199
y
5n 4y
5n 4995
y
3n 25
y
3n 41990
y
n10
y
n 2
144
1
+ + + + +−
−
−
−
−
−
❖
11
y
5n 6199
y
5n 555
y
3n 499
y
3n 3110
y
n 21990
y
n 1
8
1
+ + + + + +−
+
−
+
−
❖
11
x
5n 4y
5n 555
x
3n 25
y
3n 3110
x
n10
y
n 1
3
1
+ + + + ++
+
+
+
+
❖
199
x
5n 4y
5n 6995
x
3n 25
y
3n 41990
x
n10
y
n 2
55
1
+ + + + ++
+
+
+
+
❖
521
x
5n 529
x
5n 62605
x
3n 3145
x
3n 45210
x
n 1290
x
n 2
8
1
+ + + + + +−
+
−
+
−
❖
x
5n 529
y
5n 45
x
3n 3145
y
3n 210
x
n290
y
n
21
1
−
+
−
+
−
+ + + + ❖
−
11
x
5n+5−
29
y
5n+5−
55
x
3n+3−
145
y
3n+3−
110
x
n+1−
290
y
n+1
❖
199
x
5n 529
y
5n 6995
x
3n 3145
y
3n 41990
x
n 1290
y
n 2
3
1
+ + + + + ++
+
+
+
+
REMARKABLE OBSERVATIONSI. 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:
Illustration Let 1 n n n n 1 n n
x
x
29
Y
x
13
x
X
+ +−
=
−
=
8
Y
f
n=
n (17) n nX
8
5
g =
(18) W.K.T493
4
g
f
n2−
2n=
(19) Substituting (17) and (18) in (19) we have4
X
64
5
Y
64
1
2 n 2 n−
=
256
X
5
Y
n2−
2n=
which represents a hyperbola.For simplicity and clear understanding, the other choices of hyperbola are presented in the table 1.2 below:
Table:1. 2 Hyperbola S. NO Hyperbola (X,Y) 1
Y
2−
5
X
2=
256
(
x
n+1−
13
x
n,
29
x
n−
x
n+1)
2Y
2−
5
X
2=
82944
(
x
n+2−
233
x
n,
521
x
n−
x
n+2)
3Y
2−
5
X
2=
16
(
2
x
n+
2
y
n,
2
x
n−
2
y
n)
4Y
2−
5
X
2=
144
(
2
y
n+1+
10
x
n,
22
x
n+
2
y
n)
5Y
2−
5
X
2=
48400
(
2
y
n+2+
178
x
n,
398
x
n+2+
2
y
n+2)
6Y
2−
5
X
2=
256
(
1
x
n+2−
233
x
n+1,
521
x
n+1−
29
x
n+2)
7Y
2−
5
X
2=
1764
(
x
n+1+
13
y
n,
x
n+1−
29
y
n)
8Y
2−
5
X
2=
16
(
10
x
n+1+
26
y
n+1,
−
22
x
n+1−
58
y
n+1)
9Y
2−
5
X
2=
324
(
−
89
x
n+1−
13
y
n+2,
199
x
n+1+
29
y
n+2)
10Y
2−
5
X
2=
568516
(
x
n+2+
233
y
n,
521
y
n−
x
n+2)
11Y
2−
5
X
2=
1764
(
5
x
n+2−
233
y
n+1,
−
11
x
n−
521
y
n+1)
12Y
2−
5
X
2=
16
(
178
x
n+2+
466
y
n+2,
−
398
x
n+2−
1042
y
n+2)
13Y
2−
5
X
2=
256
(
5
y
n−
y
n+1,
−
11
y
n−
y
n+1)
14Y
2−
5
X
2=
82944
(
89
y
n−
y
n+2,
−
199
y
n−
n+2)
15Y
2−
5
X
2=
256
(
89
y
n+1−
5
y
n+2,
−
11
y
n+2−
199
y
n+2)
II. Employing linear combinations among the solutions of (1), one may generate integer solutions forother choices of parabola which are presented in Table1. 3 below:
Illustration Let
16
x
x
29
Y
n=
2n+1−
2n+2+
From (16a), n 2 nY
8
1
f
=
In view of (19), one has
256
X
5
Y
8
4
X
64
5
Y
8
1
2 n n 2 n n=
−
=
−
which represents a parabola.
For simplicity and clear understanding the other choices of parabola are presented below in Table 1.3
Table:1. 3 parabola
494 1
8
Y
−
5
X
2=
256
+
−
−
+ + +16
x
x
29
,
x
13
x
2 n 2 1 n 2 n 1 n 2144
Y
−
5
X
2=
82944
+
−
−
+ + +288
x
x
521
,
x
233
x
3 n 2 1 n 2 n 2 n 32
Y
−
5
X
2=
16
+
−
+
+ +2
y
4
x
2
,
y
2
x
2
1 n 2 1 n 2 n n 46
Y
−
5
X
2=
144
+
+
+
+ + +12
y
2
x
22
,
x
10
y
2
2 n 2 1 n 2 n 1 n 5110
Y
−
5
X
2=
48400
+
+
+
+ + +220
y
2
x
398
,
x
178
y
2
3 n 2 1 n 2 n 2 n 68
Y
−
5
X
2=
256
+
−
−
+ + + +16
x
29
x
521
,
x
233
x
13
3 n 2 2 n 2 1 n 2 n 721
Y
−
5
X
2=
1764
+
−
+
+ + +42
y
29
x
,
y
13
x
1 n 2 2 n 2 n 1 n 82
Y
−
5
X
2=
16
+
−
−
+
+ + + +4
y
58
x
22
,
y
26
x
10
2 n 2 2 n 2 1 n 1 n 93
Y
−
5
X
2=
36
+
+
−
−
+ + + +6
y
29
x
199
,
y
13
x
89
3 n 2 2 n 2 2 n 1 n 10377
Y
−
5
X
2=
568516
+
−
+
+ = +754
y
521
x
,
y
233
x
1 n 2 3 n 2 n 2 n 1121
Y
−
5
X
2=
1764
+
−
−
−
+ + + +42
y
521
x
11
,
y
233
x
5
2 n 2 3 n 2 1 n 2 n 122
Y
−
5
X
2=
16
+
−
−
+
+ + + +4
y
1042
x
398
,
y
466
x
178
3 n 2 3 n 2 2 n 2 n 138
Y
−
5
X
2=
256
+
−
−
−
+ + +16
y
y
11
,
y
y
5
2 n 2 1 n 2 1 n n 14144
Y
−
5
X
2=
82944
+
−
−
−
+ + +288
y
y
199
,
y
y
89
3 n 2 1 n 2 2 n n 158
Y
−
5
X
2=
256
+
−
−
−
+ + + +16
y
199
y
11
,
y
5
y
89
2 n 2 3 n 2 2 n 1 n CONCLUSIONIn this paper an attempt has been made to obtained integer solutions to the ternary bi-quadratic equations. Since these equations are rich in verity, one may search for integer solutions to other choices of bi-quadratic equations with multiple variables.
495
REFERENCES
1. Gopalan.M.A., Vidhyalakshmi.S., Thiruniraiselvi.N., On the Biquadratic equation with three unknowns
(
2 2)
4z
220
xy
10
y
x
7
+
−
=
, Proceeding of the National Conference (UGC Sponsored). On Recent Developments on Emerging Fields in Pure and Applied Mathematics ReDeEM, (2015, March) 125-131.2. Gopalan.M.A., Sumathi.G., and Vidhyalakshmi.S., Integral solutions of ternary Biquadratic
non-homogeneous equation
(
k
+
1
)
(
x
2+
y
2)
−
(
2
k
+
1
)
xy
=
z
4, Archimedes J.Math, 3(1), (2013) 67-71.3. Gopalan.M.A., Vidhyalakshmi.S., and Sumathi.G., Integral solutions of ternary Biquadratic
non-homogeneous equation
(
+
1
)
(
x
2+
y
2)
+
(
2
+
1
)
xy
=
z
4, JARCE, 6(2), (2012, July-Dec) 97-98.4. Gopalan.M.A., Vidhyalakshmi.S., and Sumathi.G., Integral solutions of ternary Biquadratic
non-homogeneous equation
(
2
k
+
1
)
(
x
2+
y
2+
xy
)
=
z
4, Indian Journal of Engineering, 11(1), (2012) 37-40.5. Gopalan.M.A., Vidhyalakshmi.S., Shanthi.J., and Bhuvaneshwari.M, On biquadratic equation with three
unknowns
10
(
x
2+
y
2)
−
16
xy
=
65
z
4, International Journal of Research and Current Development, 1(2), (2015, June), 48-52.6. Gopalan.M.A., Vidhyalakshmi.S, and Kavitha.A., Observation on
3
(
x
2+
y
2)
−
5
xy
=
15
z
4, Cayley J.Math., 3(1), (2014), 1-5.7. Vidhyalakshmi.S., Kavitha.A., and Presenna.R., On Ternary bi-quadratic equation
(
x
2y
2)
3
xy
23
z
42
+
−
=
, Paper Presented at the International conference on Mathematical Methods and computations, (Jamal Mohammed college, Trichy) Jamal Academic Research Journal, (2015,January), 283-286.8. Dr. Jayakumar. P., Venkatraman. R., on Non-Homogeneous Biquadratic Diophantine Equation
(
x
2y
2)
15
xy
40
z
48
+
−
=
, IJSRD,4(9), (2016), 181-183.9. Venkatraman. R., and Dr. Jayakumar. P., On Non-Homogeneous Bi-Quadratic Diophantine Equation
(
x
2y
2)
5
xy
20
z
43
+
−
=
, IJPAM, 114(6), (2017), 1185-1192.10. Gopalan, M.A., Vidhyalakshmi, S., Thiruniraiselvi, N. On the bi-quadratic equation with three unknowns 4 2 2
220
10
)
(
7
x
+
y
−
xy
=
z
“Proceeding of the National Conference (UGC Sponsored) on Recent Developments on Emerging Fields in pure and Applied Mathematics ReDeEM March 2015, 125-131.11. Gopalan, M.A., Vidhyalakshmi, S., Thiruniraiselvi, N. On the bi-quadratic equation with three unknowns
))
(
3
(
)
(
x
+
y
+
z
3=
z
2xy
−
x
2+
y
2 Paper presented on the International Conference on Mathematical models and computation, (Jamal Mohammed College,Trichy) Jamal Academic Research journal,Pp:232-235 ,Feb-201412. Shreemathi Adiga, Anusheela N. and Gopalan M.A., Non-Homogeneous Bi-quadratic equation with three
