Vol. 10, 2017, 57-65
ISSN: 2349-0632 (P), 2349-0640 (online) Published 11 December 2017
www.researchmathsci.org
DOI: http://dx.doi.org/10.22457/jmi.v10a8
57
Journal of
Observation on the Cubic Diophantine Equation with
Four Unknowns (x
3+y
3)+(x+y)(x+y+1)=zw
2T. Priyadharshini1 and S. Mallika2 1
Department of Mathematics, Shrimati Indira Gandhi College, Trichy-2 Tamilnadu INDIA, e.mail: jpdharshini16@gmail.com
2
Department of Mathematics, Shrimati Indira Gandhi College, Trichy-2 Tamilnadu INDIA, e.mail: msmallika65@gmail.com
Received 1 November 2017; accepted 06 December 2017
Abstract. The non-homogeneous cubic Diophantine equation represented by
(
3 3)
(
)(
)
21
zw
y
x
y
x
y
x
+
+
+
+
+
=
is analyzed for its non-zero distinct integer solutions. A few interesting relations between the solutions and Polygonal numbers, Pyramidal numbers are also presented.Keywords: Integer solutions, non-homogeneous cubic, Polygonal numbers, Pyramidal numbers.
AMS Mathematics Subject Classification (2010): 11D25
1. Introduction
Integral solution for the Homogeneous or Non- homogeneous Diophantine cubic equation is an interesting concept as it can be seen from [1,2,3]. In [4−11] a few special cases of cubic Diophantine equation with four unknowns are studied. In this communication we present the integral solutions of an interesting cubic equations with four unknowns
2 3
3
)
1
)(
(
)
(
x
+
y
+
x
+
y
x
+
y
+
=
zw
. A few remarkable relations between the solutions are also presented.2. Notations
= n m
t , nth term of a regular polygon with m sides
(
)(
)
− −
+ =
2 2 1
1 n m
n
Pr
n = Pronic number of rank n.=n(n+1)
n
S
= Star number of rank n.58
6 ,
n
CP = Centered Pyramidal Number of rank n with sides m
3 n
=
3. Method of analysis
The cubic Diophantine equation with four unknowns to be solved for its non-zero distinct integral solution is
(
3 3)
2)
1
)(
(
x
y
x
y
zw
y
x
+
+
+
+
+
=
(1)The substuition of the linear transformations
v
u
x
=
+
, y=u−v(
u≠v≠0)
(2) in (1) leads to2 2 2
3v w
U + = (3) where U =u+1 (4) Different patterns of solutions of (1) are presented below
3.1. PATTERN-1
Assume
2 2
3b
a
w= + (5)
where a and b are non-zero distinct integers. Substituting (5) in (3), we get
2 2 2 2 2
)
3
(
3
v
a
b
U
+
=
+
(6)Employing the method of factorization, we get
2 2
) 3 (
) 3 (
) 3 )(
3
(U +i v U −i v = a+i b a−i b
Equating the positive and negative factors, we get )
3
(U +i v = (a+i 3b)2 )
3
(U −i v = (a−i 3b)2 Equating the real and imaginary part, we get
2 2
3b
a
U = − (7)
ab
v=2 (8)
In view of (4)
1 3 2
2 − −
=a b
u (9)
Substituting (8) and (9) in (2), we get the distinct non-zero integer solutions of (1) are given below
( )
,
=
2−
3
2+
2
−
1
=
x
a
b
a
b
ab
x
( )
,
=
2−
3
2−
2
−
1
=
y
a
b
a
b
ab
y
2
6
2
)
,
(
=
2−
2−
=
z
a
b
a
b
z
(x3+y3)+(x+y)(x+y+1)=zw2
59 1. x
( ) ( )
a,1 + y a,1 +z(a,1)−4t4,a +16=0 2. y(
a,2a−1)
+t32,a +4=03. x
( ) ( ) ( )
a,a +ya,a +z a,a +w(a,a)+4t4,a +4=0 4. z( ) ( )
a,1 −wa,1 −t4,a +11=03.2. PATTERN-2
Write (5) as a2 +3b2 =w*1 (10) Now write 1 as
(
)(
)
4 3 1 3 1
1= +i −i (11) Using (10) and (11) in (3) and employing the method of factorization, the above equation (3) is written as
(
)(
) (
)(
)(
) (
2)
23 3
4 3 1 3 1 3
3v U i v i i a i b a i b i
U+ − = + − + −
Equating the positive and negative factors, we get
(
)(
)
23 2
3 1
3v i a i b i
U+ = + + (12)
(
)(
)
23 2
3 1
3v i a i b i
U− = − −
(13)
Equating the real and imaginary parts, we get
(
a b ab)
U 3 6
2
1 2 − 2 −
=
(14)
(
a b ab)
v 3 2
2
1 2 − 2 +
=
(15)
Replacing a,b by 2A,2B respectively, we get
AB B
A
U =2 2 −6 2 −12 (16)
AB B
A
v=2 2 −6 2 +4
(17) In view of (4),
1 12 6
2 2 − 2 − −
= A B AB
u (18)
Substituting the above values ofu and v in (2), we get
( )
,
=
4
2−
12
2−
8
−
1
=
x
A
B
A
B
AB
x
( )
, =−16 −160
2
24
12
4
)
,
(
=
−
−
−
=
z
A
B
A
B
AB
z
2 2
12
4
)
,
(
A
B
A
B
w
w
=
=
+
represents the non-zero distinct integral solutions of (1)
PROPERTIES
1.
6
[
w
( )
1
,
1
]
=
6
( )
4
2 is a Nasty Number2. y
(
n(n−1,1)
+Sn +t22,n ≡0(
modn))
3. w
( ) ( )
1,B −z1,B −t50,B −2≡1(
mod2)
4. w
(
A,A+1) (
−y A,A+1)
−16PRA −t34,A −13≡0(
mod13)
NOTE:
We note that, if we replace a,bby 2A+1,2B+1 respectively in (14) and (15) we get the different set of distinct non-zero integer solutions of (1) given by
( )
,
=
4
2−
12
2−
16
−
8
−
5
=
x
A
B
A
B
B
AB
x
( )
, =−8 −8 −16 −5= y A B A B AB
y
10
24
24
8
12
4
)
,
(
=
2−
2−
−
−
−
=
z
A
B
A
B
A
B
AB
z
4
12
4
12
4
)
,
(
=
2+
2+
+
+
=
w
A
B
A
B
A
B
w
PROPERTIES
1.
6
[
w
( )
1
,
1
]
=
6
( )
6
2 is a Nasty Number 2. w(A,1)− y( )
A,1 −t6,A −41≡1(
mod3))
3. x(A,1)−w( )
A,1 +61≡0(
mod3)
4. x(A,1)+ z(A,1)+w
( )
A,1 −t26,A +51≡0(
mod5)
3.3. PATTERN-3
We can also write 1 as,
(
)(
)
49
3 4 1 3 4 1
1= +i −i
(19)
Using (5) and (19) in (3) and employing the method of factorization, we get
(
)(
) (
)(
)(
) (
2)
23 3
49
3 4 1 3 4 1 3
3v U i v i i a i b a i b
i
U+ − = + − + −
Equating the positive and negative factors, we have
(
)(
)
23 7
3 4 1
3v i a i b
i
(x3+y3)+(x+y)(x+y+1)=zw2
61
(
)(
)
23 7
3 4 1
3v i a i b
i
U− = − −
Equating the real and imaginary parts ,
(
a b ab)
U 3 24
7
1 2 − 2 −
=
(20)
(
a b ab)
v 4 12 2
7
1 2 − 2 +
=
(21)
Replacing a,b by 7A ,7B respectively, we get
AB B
A
U =7 2 −21 2 −168 (22)
AB B
A
v=28 2−84 2+14
(23) In view of (4)
1
168
21
7
2−
2−
−
=
A
B
AB
u
(24)Substituting (23) and (24) in (2), we get
( )
,
=
35
2−
105
2−
154
−
1
=
x
A
B
A
B
AB
x
( )
,
=
−
21
2+
63
2−
182
−
1
=
y
A
B
A
B
AB
y
2
336
42
14
)
,
(
=
2−
2−
−
=
z
A
B
A
B
AB
z
2 2
147
49
)
,
(
A
B
A
B
w
w
=
=
+
represents the non-zero distinct integral solutions of (1).
PROPERTIES
1.
6
[
w
( )
1
,
1
]
=
6
( )
14
2 is a Nasty Number2. z(A,1)+ y
( )
A,1 +w(A,1)−t86,A −165≡0(
mod159)
3. z(AA)+364t4,A +2=0
4. x
( )
A,1 −t72,A +106≡0(
mod2)
3.4. PATTERN-4
(3) is written in the form of ratio as
0 ,
3 = − = ≠
+
n n m U w
v v
U w
(25)
which is equivalent to the system of equations, 0
3 =
−
+nU mv
nw (26) 0
= − −mU nv
mw (27)
62 3m n
U = −
mn v=2
2 2
3
m
n
w
=
+
In view of (4), 1 3 2 − 2 −
= m n
u
Substituting the values of u and v in (2), we get the non zero distinct integer solutions of (1) is given by
( )
,
=
3
2−
2+
2
−
1
=
x
m
n
m
n
mn
x
( )
,
=
3
2−
2−
2
−
1
=
y
m
n
m
n
mn
y
2
2
6
)
,
(
=
2−
2−
=
z
m
n
m
n
z
2 2
3
)
,
(
m
n
m
n
w
w
=
=
+
PROPERTIES
1.
x
(
m
,
m
+
1
) (
−
y
m
,
m
+
1
)
−
4
PR
m=
0
2. x
(
m,m+1) (
+ ym,m+1)
−t12,m + PRm +4≡0(
modm)
3. x
( ) ( )
m,1 + y m,1 +z(m,1)+w(m,1)−15t4,m +7=04. x(m,m+1)+ y
(
m,m+1)
+z(m,m+1)−t20,m +PRm +8≡0(
modm)
3.5. PATTERN-5
One may write (3) in the form of ratio as
0 ,
3 = ≠
− = +
n n m U w
v v
U w
(28)
which is equivalent to the system of equations 0
= − +nU mv
nw (29) 0
3 =
−
−mU nv
mw (30)
Applying the method of cross multiplication for solving (29) and (30), we have
2 2
3n
m
U
=
+
mn v=2
2 2
3n
m w= +
In view of (4),
1
3
22
−
−
=
m
n
u
Substituting the values of
u
andv
in (2), we have( )
,
=
2−
3
2+
2
−
1
=
x
m
n
m
n
mn
x
( )
,
=
2−
3
2−
2
−
1
=
y
m
n
m
n
mn
y
2
6
2
)
,
(
=
2−
2−
=
z
m
n
m
n
z
2 2
3
)
,
(
m
n
n
m
w
(x3+y3)+(x+y)(x+y+1)=zw2
63
represents the distinct non zero integral solutions of (1)
PROPERTIES
1.
x
(
m
,
1
)
−
w
(
m
,
1
)
+
7
≡
0
(mod
2
)
2. z(m,m+1)− y
(
m,m+1)
−2PRm +t6,m +4≡0(
mod7)
3. x(
1,n+1) (
+ y1,n+1) (
+z1,n+1)
+w(1,n+1)−t12,m +36=0 4. x( ) ( )
m,1 + y m,1 +w(m,1)−3t4,m +5=03.6. PATTERN-6
Introduction of the linear transformations T
X
U = +3 v= X −T (31) in (3) gives,
2 2 2
12
4
X
+
T
=
w
(32) which is satisfied by,2 2
6 2a b X = −
ab T =4
b
a
w
=
4
2+
12
Substituting the values of X,T in (31), we get
ab
b
a
U
=
2
2−
6
2+
12
(33)
ab
b
a
v
=
2
2−
6
2−
4
(34) In view of (4),
1 12 6
2 2 − 2 + −
= a b ab
u
(35) Substituting (34) and (35) in (2), we get the distinct non- zero integer solutions of (1) is given by
( )
,
=
4
2−
12
2+
8
−
1
=
x
a
b
a
b
ab
x
( )
, =16 −1= y a b ab y
( )
2 22 2
12 4
,
2 24 12
4 ) , (
b a b a w w
ab b
a b a z z
+ = =
− +
− = =
PROPERTIES
1. x
( )
a,a +z(a,a)−16t4,a +3=064 4.
6
[
w
( )
1
,
1
]
=
6
( )
4
is a Nasty Number3.7. PATTERN-7
Introduction of the linear transformations T
X
U = +3 v= X −T in (3) gives,
2 2 2
12
4
X
+
T
=
w
which can be written as1 * 12
4X2 + T2 =w2
(36) Write 1 as
16
) 12 2 )( 12 2 (
1= +i −i
(37) Substituting (37) in (36) and employing the method of factorization,
we get
ab b
a
X = 2 −3 2 −6
ab
b
a
T
=
2−
3
2+
2
Substituting the values of X,T in (31),we get
2 2
12 4a b
U = − (38)
ab
v=−8 (39) In view of (4),
1 12 4 2 − 2 −
= a b
u (40) Substituting (39) and (40) in (2), we get the distinct non- zero integer solutions of (1) is given by,
( )
,
=
4
2−
12
2−
8
−
1
=
x
a
b
a
b
ab
x
( )
,
=
4
2−
12
2+
8
−
1
=
y
a
b
a
b
ab
y
( )
,
=
8
2−
24
2−
2
=
z
a
b
a
b
z
( )
2 212
4
,
b
a
b
a
w
w
=
=
+
PROPERTIES
1. x
(
a,a(a+1))
−y(a,a(a+1))+16CPn,6 +16t4,a =0 2. w(
2a+1,b)
−28t4,b −4≡0(
mod2)
3.
[
( ) ( ) ( )
]
( )
28 6 4 1 , 1 1 , 1 1 , 1
6x +y +z − = is a Nasty Number
4. x
(
a,a+1) (
+y a,a+1) (
+z a,a+1) (
+wa,a+1)
−t42,a +t74,a +40≡0(
mod4)
4. Conclusion
(x3+y3)+(x+y)(x+y+1)=zw2
65
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x
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y
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z
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28
(
x
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y
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)
w
2,
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(
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y
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+
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z
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w
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xy
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w
2,
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,
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