Research Paper
Mathematics
E-ISSN No : 2454-9916 | Volume : 3 | Issue : 1 | Jan 2017
ON THE TERNARY QUADRATIC EQUATION x 2 y 2 xy 3z 2
Dr. R. Anbuselvi 1 | K. S. Araththi 2 1
Associate Professor, Department of Mathematics, A.D.M. College for women, Nagapattinam, India.
2
Research Scholar, Department of Mathematics, A.D.M. College for women, Nagapattinam, India.
ABSTRACT The ternary Quadratic Diophantine equation x y xy 3z is analyzed for its non – zero distinct integer solutions. Five different patterns of non-zero distinct integer solutions to the equation under consideration are obtained. A few interesting relations between the solutions and special numbers are exhibited 2
2
2
KEYWORDS: Integral solutions, Ternary quadratic.
I. INTRODUCTION The Ternary Quadratic Diophantine Equation offers an unlimited field for research because of their variety [1-2]. For an extensive review of various problems, one may refer [3-10]. This communication concerns with yet another interesting Also a few interesting relations among the solutions have been presented.
u 6ab; v a 2 3b 2 ; …… (5) z 3b 2 a 2 Using (5) in (2), we obtain the integer solutions to (1) as given below:
x x(a, b) a 2 3b 2 6ab
II. NOTATIONS obl n- Oblong number of rank ‘n’
y y (a, b) a 2 3b 2 6ab
tm,n- Polygonal number of rank ‘n’ with sides’m’
z z (a, b) a 2 3b 2
III. METHOD OF ANALYSIS The ternary quadratic equation to be solved in integers is
PROPERTIES (i) x(1, n) y (1, n) 0(mod 2) (ii) y (1, n) z (1, n) 6obln 0(mod 6)
x 2 y 2 xy 3z 2 ……………………
(1)
(iv)
Now, introducing the linear transformations
x u v ; y u v …………………
(iii)
(2)
x(1, n) z (1, n) 0(mod 6)
3x(1, n) 6 y (1, n) 9n(n 2) 0(mod 9) 4 y (1, n) 2 z (1, n) 0(mod 6)
(v) It is observed that (3) may also be written in the following three ways
In (1), it leads to
u 3( z v) A ………….……… zv u B
a.
x x(a, b) 3b 2 a 2 6ab
(3)
y y (a, b) a 2 3b 2 6ab z z (a, b) a 2 3b 2
We solve equation (3) in five different methods and obtain five sets of solutions A. PATTERN:1 (3) is equivalent to the system of equations
WAY: 1
PROPERTIES x(1, n) z (1, n) 6obln 0 (i) (ii)
bu az av 0 au 3bz 3bv 0
(iii)
from which we get
(v)
(iv)
x(1, n) y (1, n) 0(mod12) 3 y (1, n) z (1, n) 6obln 0 6 x(1, n) 3 y (1, n) 0(mod 3) x(1, n) z (1, n) `12n 6obln
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International Educational Scientific Research Journal [IESRJ]
122
Research Paper b.
E-ISSN No : 2454-9916 | Volume : 3 | Issue : 1 | Jan 2017
WAY: 2
REFERENCES 1. R. Anbuselvi K S Araththi on ternary quadratic Equations 2 y 2 xy 4 z 2 , Indian journal for research analysis, 2016, 9(6) 379-381.
x(a, b) b 2 3a 2 6ab
y (a, b) b 2 3a 2 6ab z (a, b) 3a 2 b 2
2.
R. Anbuselvi K S Araththi on ternary quadratic Equations x 4 y 40z , Global journal for research analysis, 2016, 8(5)256-258. 2
PROPERTIES x(1, n) y (1, n) 0(mod 6) (i) (ii) (iii) (iv) (v) c.
x(n,1) 3 z (n,1) 6n 0(mod10) 6 x(n,1) 2 y (n,1) 0(mod 6) 3x(n,1) 3nz(n,1) 18obln y (n,1) z (n,1) 6obln 0(mod 6)
Meena K, Vidhyalakshmi S,Gopalan M.A,Priya K,Integral points on the cone 3( x 2 y 2 ) 5 xy 47 z 2 , Bulletin of Mathematics and Statistics and Research,2014,2(1),65-70.
4.
Gopalan M.A,Vidhyalakshmi S,Nivetha S,on Ternary Quadratic Equation 4( x 2 y 2 ) 7 xy 31z 2 Diophantus J.Math,2014,3(1),1-7.
5.
Gopalan M. A, Vidhyalakshmi S, Kavitha A, Observation on the Ternary Cubic Equation x 2 y 2 xy 12 z 3 Antarctica J. Math, 2013; 10(5):453-460.
6.
Gopalan M. A,Vidhyalakshmi S, Lakshmi K, Lattice points on the Elliptic Paraboloid, 16 y 2 9 z 2 4 x 2 Bessel J. Math, 2013, 3(2), 137-145.
7.
Gopalan M. A, Vidhyalakshmi S, Umarani J, Integral points on the Homogenous Cone x 2 4 y 2 37 z 2 , Cayley J. Math, 2013, 2(2), 101-107.
8.
Gopalan M.A,Vidhyalakshmi S,Sumathi G,Lattice points on the Hyperboloid of one sheet
y y (a, b) 3a 2 b 2 6ab z z (a, b) 3a 2 b 2 PROPERTIES (i) (ii) (iii) (iv)
x(1, n) y (1, n) 2n 2 0(mod 6) y(1, n) z(1, n) 2t 4,n 0(mod 6)
x(1, n) z(1, n) 2t 4,n 0(mod5) y (1, n) z (1, n) 2n(n 3) 0
B. PATTERN:2 Assume u 3v 3 z 2
2
2
2
3.
WAY: 3
x x(a, b) 3a 2 b 2 6ab
2
4 z 2 2 x 2 3 y 2 4 , The Diophantus J. Math, 2012,
(7)
1(2), 109-115. Also, write 4 as
9.
3 (i 3) * (i 3)
(8)
z a 2 3b 2
(9)
Substitute (8) and (9) in (7) and employing the method of factorization, define
u i 3v (i 3)[a i 3b]2
Gopalan M. A, Vidhyalakshmi S, Lakshmi K, Integral points on the Hyperboloid of two sheets 3 y 2 7 x 2 z 2 21 , Diophantus J. Math, 2012, 1(2), 99-107.
10. Gopalan M. A, Vidhyalakshmi S, Mallika S, Observation on Hyperboloid of one sheet x 2 2 y 2 z 2 2 Bessel J. Math, 2012, 2(3),
(9)
Equating the real and imaginary parts, we get
u 6ab v a 2 3b 2 z a 2 3b 2
and hence x = x(a,b) = a 3b 6ab 2
y = y(a,b) = a
2
3b 2 6ab 2 2 z = z(a,b) = a 3b 2
IV. CONCLUSION To conclude, one may search for other patterns of solutions to the equation under consideration.
International Educational Scientific Research Journal [IESRJ]
123