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Observation on the Ternary Cubic Equation
4 i 12
3
2
i 12T
( a i 12b)
2
Equating real and imaginary parts, we obtained
3 2 2 3
2
2( a 36ab
) 6(3a
b 12b
)
3 2 2 3
2T
( a 36ab
) 4(3a
b 12b
)
Hence the values of x and y satisfies (1) are given by
3 2 2 3
x(
a,
b)
3( a 36ab
) 2(3a
b 12b
)
3 2 2 3
y(
a,
b)
2( a 36ab
) 8(3a
b 12b
)
2 2
z(
a,
b)
( a 12b
)
Properties:
(1) 25 24
x ( 1, n)
3[6CPn
6CPn
70t3,
n 33t
4,
n 8]
(2) z 1, n)
t 20t
9t
1
( 24,
n 3, n 4,
n
(3) 4
y n,1)
3CP
2t
2t
106t
53t
96
( n 23, n 5, n 3, n 4,
n
5
x ( n,1)
y(
n,1)
10P n t28,
n 72(mod168
(4) )
(5) 5
y n,
n)
z(
n,
n)
284P n 43t
0(mod86)
( 8,
n
2.3Pattern:3
Instead of (12) we write 7 as
(10 i 12 )(10 i 12 )
7
16
Following the procedure as presented in pattern 2 the corresponding non-zero distinct integral solutions to (1)
are obtained as
3 2 2 3
x(
a,
b)
3( a 36ab
) 2(3a
b 12b
)
3 2 2 3
y(
a,
b)
3( a 36ab
) 10(3a
b 12b
)
2 2
z(
a,
b)
( a 12b
)
Properties:
(1) 3
y n,1)
2CP
t 12t
20t
10t
120
( n 24, n 21, n 3, n 4,
n
(2) 16
x 1, n)
3[ 3CP
CP 33t
2]
( n 6, n 4,
n
(3) 5
x n,1)
6P n t 24(mod106)
( 8,
n
(4) 5
x n,1)
z(
n,1)
6P n t 36(mod107)
( 6,
n
5
z ( 1, n)
x(1,
n)
48P n 48t
6,
n 2(mod42
(5) )
2.4Pattern:4
Introducing the linear transformations
u 3T , v T
(13)
Let
2 2
z a 3b
(14)
Write 7 as
7 (2 i 3)(2 i 3)
(15)
Substituting (13), (14) and (15) in (3) and repeating the process as in pattern2, the non-zero distinct integral
solutions to (1) are obtained as
3 2 2 3
x(
a,
b)
( a 9ab
) 15(
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) 3
x n,1)
y(
n,1)
3[2CP
CP t 8t
2]
( n 4, n 20, n 4,
n
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