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Observation on the Ternary Cubic Equation
n
(5 * 2
7
n n n
i 2 3)(5*2 i2
3)
(5)
2n2
2
Using (4) and (5) in (3) and applying the method of factorization, define
n n
(5 * 2 i2
3)
3
u i 3v
( a i 3b)
n1
2
Equating real and imaginary parts, we get
n
2 3 2 2 3
u [5( a 9ab
) 9( a b b )]
n1
2
n
2 3 2 2 3
v [( a 9ab
) 15(
a b b )]
n1
2
Using (6) and (7) in (2), we have
n
2
3 2 2 3
x(
a,
b)
[6( a 9ab
) 6( a b b )]
n1
2
n
2 3 2 2 3
y(
a,
b)
[4( a 9ab
) 24( a b b )]
n1
2
2 2
z(
a,
b)
( a 3b
)
Where n= 0,1, 2……..
(6)
(7)
(8)
Properties:
(1) n
3n
n 2n
x (2 ,1) 3[ j3n
9 jn
3J2n
( 1)
9( 1)
( 1)
1]
(2) n
3n1
n1
y (2 ,1) 3J
9 j 3 j ( 1)
9( 1)
15
(3) n
z ( 1,2 ) 3 j2 2
3n1
n1
2n2
n
For illustration and clear understanding, substituting n=1 in (8), the corresponding non-zero distinct
integral solutions to (1) are given by
3 2 2 3
x(
a,
b)
3( a 9ab
) 3( a b b )]
3 2 2 3
y(
a,
b)
2( a 9ab
) 12(
a b b )]
2 2
z(
a,
b)
( a 3b
)
Properties:
(1) 9
x( 1, n)
3 29t
4,
n 4t3,
n 2CPn
(2) 4
y ( 1, n)
3CPn
62t3,
n 31t
4,
n Sn
14
(3) 12
x ( 1, n)
y(1,
n)
z(1,
n)
CPn
16t3,
n 37t4,
n 6
(4) 7
x ( 1, n)
y(1,
n)
6P n 6t6,
n 9(mod49)
(5) 8
y ( n,1)
2P n t28,
n 12(mod29)
2.2Pattern:2
Introducing the linear transformations
u 3T , v T
(9)
Substituting in (3) we get
2 2 3
4 12T
7z
2
z a 12b
(10)
Let
2
(11)
Write 7 as
(4 i 12 )(4 i 12 )
7 (12)
4
Using (11) and (12) in (10), we get
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