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IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 6, August 2014.
www.ijiset.com
ISSN 2348 – 7968
On Ternary Quadratic Diophantine Equation
3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2
1
2
S.VIDHYALAKSHMI , M.A.GOPALAN , J.SHANTHI
1
3
Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy Tamilnadu, India- 620 002. Email:
[email protected],
2
Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy Tamilnadu, India- 620 002.
Email:[email protected]
3
Lecturer, Department of Mathematics, Shrimati Indira Gandhi College,Trichy
Tamilnadu, India -620002.Email:[email protected]
Abstract
The ternary quadratic equation representing non-homogeneous
cone given by 3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 is
analyzed for its non-zero distinct integer points on it. The
different patterns of integer points satisfying the cone under
consideration are obtained. A few interesting relations between
the solutions and special number patterns are presented.
Keywords: Ternary non-homogeneous quadratic, integral
solutions
2010 Mathematics Subject Classification: 11D09
Sn
- Star number of rank n.
2.METHOD OF ANALYSIS:
The ternary quadratic diophantine equation to be
solved for its non-zero distinct integral solutions is
3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2
The substitution of linear transformations
x = u + v, y = u − v
(1)
(2)
in (1) leads to
1. Introduction
The ternary quadratic Diophantine equations offer an
unlimited field for research due to their variety [1, 20]. For
an extensive review of various problems, one may refer
[2 – 19]. This communication concerns with yet another
interesting
ternary
quadratic
equation
3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 representing a cone
for determining its infinitely many non-zero integral
points. Also, a few interesting relations among the
solutions are presented.
NOTATIONS:
Tm,n - Polygonal number of rank n with size m.
Pnm - Pyramidal number of rank n with size m.
Ct m,n - Centered Polygonal number of rank n with size
m.
CPm,n - Centered Pyramidal number of rank n with size
m.
Prn
(3)
(u + 1) 2 + 11v 2 = 15 z 2
Different patterns of solutions of (1) are presented below.
2.1 PATTERN : 1
Write 15 as
15 = (2 + i 11)(2 − i 11)
(4)
Assume z= a + 11b
(5)
where a, b are non-zero distinct integers.
Using (4) & (5) in (3), and applying the method of
factorization,define
2
[(u + 1) + i
2
]
11v = (2 + i 11)(a + i 11b) 2
Equating the real imaginary parts, we have
(6)
u = u (a, b) = 2a 2 − 22b 2 − 22ab − 1
v = v(a, b) = a 2 − 11b 2 + 4ab
Substituting the above values of
the values of x and y are given by
u & v in equation (2),
x = x(a, b) = 3a 2 − 33b 2 − 18ab − 1
(7)
y = y (a, b) = a − 11b − 26ab − 1
(8)
2
2
- Pronic number of rank n.
212
IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 6, August 2014.
www.ijiset.com
ISSN 2348 – 7968
Thus (5), (7) & (8) represent a non-zero distinct integral
solutions of (1) in two parameters.
1.x(1, s ) − 3 y (1, s ) + 2t12,s ≡ −4(mod 68)
2.
PROPERTIES:
1.x(a, b) − 6t 3,a −1 ≡ −4(mod15)
3.x(2 n ,1) − y (2 n ,1) = 990 J 2n + 12 J n + 4 j n + 227
2.6[x(a,1) − y (a,1)] − 2 S a ≡ −14(mod 60)
Note:
In addition to (13), one may also consider the linear
transformations z = X − 11T , v = X − 15T .Following the
method presented above , different set of solutions are
obtained.
3.z (1,2 n ) − 45 J 2n + 6 J n + 2 j n = 6
2.2 PATTERN 2:
Instead of (4), we write 15 as
(7 + i 11)(7 − i 11)
(9)
4
Following the procedure presented above, the
corresponding non-zero distinct integer solutions to (1)
are obtained as
15 =
x = x(a, b) = 4a 2 − 44b 2 − 4ab − 1
(10)
y = y (a, b) = 3a 2 − 33b 2 − 18ab − 1
along with (5)
PROPERTIES:
(11)
1.x(2 n ,1) = 12 J 2n − 6 J n − 2 j n − 41.
2.x(2 n ,1) − y (2 n ,1) = j 2n + 7 j n +21J n − 12.
3.3 x(a,1) = 2 S a − 137.
2.3 PATTERN : 3
Equation (3) can be written as
(u + 1) 2 = 15 z 2 − 11v 2
Introducing the linear transformations,
z = X + 11T & v = X + 15T
in (12), we get
(12)
(13)
2
 u + 1
X 2 = 165T 2 + 
(14)

 2 
which is satisfied by
T (r , s ) = 2rs


2
2
(15)
u (r , s ) = 330r − 2 s − 1

2
2

X (r , s ) = 165r + s
Substituting the values of (15) in (13) and using (2), the
corresponding integer solutions of (1) are given by
x = x(r , s ) = 495r 2 − s 2 + 30rs − 1
(16)
y = y (r , s ) = 165r − 3s − 30rs − 1
(17)
z = z (r , s ) = 165r 2 + s 2 + 22rs
(18)
2
PROPERTIES:
2
1
[x(1, n) − y(1, n)]2t3,n−1 ≡ 10(mod 31)
2
2.4 PATTERN : 4
Consider (3) as
(19)
(u + 1) 2 − 4v 2 = 15( z 2 − v 2 )
Write (19) in the form of ratio as
A
(u + 1) + 2v
15( z − v)
=
= ,B ≠ 0
z+v
(u + 1) − 2v B
which is equivalent to the following two equations
B(u + 1) + (2 B − A)v − Az = 0
A(u + 1) + (15B − 2 A) − 15Bz = 0
On employing the method of cross multiplication, we get
u = 2 A 2 − 30 AB + 30 B 2 − 1
(20)


v = A 2 − 15 B 2
(21)
z = A 2 − 4 AB + 15B 2
Substituting the values of u and v from (20) in (2), the nonzero distinct integral values of x, y are given by
x = x( A, B) = 3 A 2 + 15B 2 − 30 AB − 1
 (22)
y = y ( A, B) = A 2 + 45B 2 − 30 AB − 1 
Thus (21) and (22) represent non-zero distinct integral
solutions of (1) in two parameters.
PROPERTIES:
Each of the following expressions represents a Nasty
number
i )2[3 z (1, β ) − y (1, β )]
ii )2[ y (a,−a ) + 1]
2.18 x(1, β ) − 26 y (1, β ) + 60 z (1, β ) = 96
Note:
(19) can also be expressed in the form of ratio in two
different ways as follows.
A
(u + 1) + 2v
5( z − v)
=
= , B ≠ 0.
(i ).
3( z + v)
(u + 1) − 2v B
A
(u + 1) + 2v
3( z − v)
=
= , B ≠ 0.
(ii ).
5( z + v)
(u + 1) − 2v B
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IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 6, August 2014.
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ISSN 2348 – 7968
Repeating the analysis as above the corresponding integer
solutions along with properties are presented below.
Solutions of (i):
x = x( A, B) = −9 A 2 − 5B 2 + 30 AB − 1
Cone z 2 = 2 x 2 − 7 y 2 , Diophantus J. Math., 1(2), 109115, 2012.
5. M.A.Gopalan, J. Kalinga Rani, on ternary quadratic
y = y ( A, B) = −3 A 2 − 15B 2 + 30 AB − 1
z = z ( A, B) = 3 A 2 + 5B 2 − 4 AB
PROPERTIES :
1.x(1,2α − 5α 2 ) − y (1,2α − 5α 2 ) − 14t 3,( −5α 2 + 2α −1)
3. M.A. Gopalan, and V. Pandichelvi, Integral solutions of
ternary quadratic equation z ( x − y ) = 4 xy , Impact J.sci
TSech; Vol (5), No.1, 01-06, 2011.
4. M.A.Gopalan, S. Vidhyalakshmi and A.Kavitha,
Integral points on the homogeneous
is
a
perfect square.
2.5 x(1, n) + z (1, n) ≡ 27(mod 34)
3.x(1, n) + y (1, n) + z (1, n) + 22t 3,n−1 ≡ 19(mod 22)
Solutions of (ii):
x = x( A, B ) = 15 A 2 + 3B 2 − 30 AB − 1
y = y ( A, B ) = 5 A 2 + 9 B 2 − 30 AB − 1
z = z ( A, B ) = −5 A 2 − 3B 2 + 4 AB
PROPERTIES :
1.x(2 n ,1) = 45 J 2n−45 J n − 15 j n + 17
2.x(2 n ,1) − y (2 n ,1) = 30 J 2n + 4
3.6 x(a,1) − 15S a ≡ 87(mod 90)
3.CONCLUSION:
In this paper, we have presented six different
patterns of non-zero distinct integer solutions of the nonhomogeneous
cone
given
by
3( x 2 + y 2 ) − 5 xy + x + y + 1 = 15 z 2 .To conclude, one may
search for patterns of non-zero integer distinct solutions
and their corresponding properties for other choices of
ternary quadratic diophantine equations.
ACCKNOWNLEDGEMENT:
The financial support from the UGC-New Dehli
[ F.5122/(SERO/UGC) dated march 2014 ] for a part of
this work is greatfully acknowledged.
4. REFERENCES:
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of
two
sheets
x 2 − 6 xy + y 2 + 6 x − 2 y + 5 = z 2 + 4 , Impact J. sci tech;
Vol (4), No.1, 23-32, 2010.
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Integral points on the hyperboloid of two sheets
3 y 2 = 7 x 2 − z 2 + 21 , Diophantus J.math., 1(2),99107,2012.
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9.
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quadratic equation x 2 = (α 2 − 1)( y 2 − z 2 ), α  1 , Bessel
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Observations
on
the
ternary
Quadratic
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107-110, 2012.
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of
43 x 2 + y 2 = z 2
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IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 6, August 2014.
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ISSN 2348 – 7968
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