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III. REMARKS
(1) Let ( x 0 , y0
, z0
) be the initial solution of (1)
Observation on the Ternary Cubic Equation
3
x 7 x
0 h
Let
3
(19)
y1
7 y0
h
2
z1
7 z o
be the first solution of (1).
Substituting (18) in (1), we get
3
h 7 ( x0
y0)
(20)
Using (20) in (19) we obtain the general solution as follows:
EVEN ORDERED SOLUTION:
6n
x 2 n 7 x ,
0
6n
y 2 n 7 y ,
0
4n
z 2 n 7 z ,
0
where n=1, 2, 3, …..
ODD ORDERED SOLUTION:
32n1
x
2 n1
7 y0
,
32n1
y 2
7 x ,
n1
0
22n1
z 2
7 z ,
n1
0
where n=1, 2, 3, …..
Properties:
1. z 1
0 N
2
48
x
n
1 0(mod 7)
x0
n0
z2n
2. 2 3 2 3
x0 z2n x2n
z0
3. Each of the following is a nasty number:
x
(i). 2n
y2n
z2n
6
, (ii). z
x0
y0
z
2n
6
, (iii).
z
0 z
0 y2n
z
6 , (iv).
2n1
y0
z
0 x2n
z
6 , (v).
2n
x0
z
2n
6
2n
z2n1
4. Each of the following is a cubic integer:
x
(i).
2n
y
0 x
, (ii).
x0
x
2n
y
, (iii).
2n1
y
2n
y
, (iv).
2n1
x
2n
x
0
2n1
y0
y2n1
II. Employing the solutions(x, y, z) of (1), the following relations among the special polygonal and pyramidal
numbers are obtained.
1.
2.
3.
4.
3P
t
2
2
3
3 5
5
2
3
x
P P
x2
y
Py
6
3, x2
6P
t
t
3, x2
t
3, y
t
3, y
7
t
3, y2
P
4
z1
3,2( z1)
2
2
4
4
3
3
1
6 3
1
2
3
x
P P
y
P
x
y2
7
3,2( x1)
2P
t
t
3,2( x1)
t
3, y2
2
2
8
8
4
4
6 6
1
2
x
P P
x1
y
Py
3
3,2 x3
6P
t
4
x
3,2 x1
2
t
3,2 x3
6P
t
4
x
3,2 x1
t
3,2 y1
6P
t
4
y1
3,2( y1)
t
t
3,2 y1
6P
t
4
y1
3,2( y1)
3
P
t
P
7
t
2
5
z
3, z
3
z
3, z1
3
3P
7
t
3
3
z2
3, z2
3
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