The document analyzes a homogeneous cubic equation with six unknowns of the form αxy(x+y) + βzw(z+w) = (α + β)XY(X+Y). It presents three approaches to finding integral solutions to this equation. The first approach assumes specific forms for the unknowns and reduces the equation to a form involving two new variables, which are then related to special numbers to obtain the integral solutions. Several properties relating the solutions to other numbers are presented. The second approach uses a linear transformation and factoring to express the two variables as functions involving a parameter, yielding another set of integral solutions. More solution properties are given. The third approach similarly uses factor

1. International Journal of Engineering Science Invention
ISSN (Online): 2319 – 6734, ISSN (Print): 2319 – 6726
www.ijesi.org Volume 2 Issue 12ǁ December 2013 ǁ PP.01-09
On The Homogeneous Cubic Equation with Six Unknowns
 x y ( x  y )   z w ( z  w )  (   ) X Y ( X  Y )
M.A.Gopalan1, S.Vidhyalakshmi2, K.Lakshmi3
1, 2
, Proffessor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620002, Tamil Nadu, India
3
Lecturer, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620002, Tamil Nadu, India
ABSTRACT: The homogeneous cubic equation with six unknowns represented by the equation
 x y ( x  y )   z w ( z  w )  (    ) X Y ( X  Y ) is analyzed for its patterns of non-zero distinct integral
solutions and different methods of integral solutions are illustrated. A few relations between the solutions and
the special numbers are presented
KEYWORDS: Homogeneous cubic equation, integral solutions
M.Sc 2010 mathematics subject classification: 11D25
NOTATIONS:
T m , n -Polygonal number of rank n with size m
m
Pn - Pyramidal number of rank n with size m
P R n - Pronic number of rank n
O H n - Octahedral number of rank n
S O n -Stella octangular number of rank n
S n -Star number of rank n
J n -Jacobsthal number of rank of n
j n - Jacobsthal-Lucas number of rank n
K Y n -kynea number of rank n
C Pn , 3 - Centered Triangular pyramidal number of rank n
C Pn , 6 - Centered hexagonal pyramidal number of rank n
F 4 , n , 3 -Four Dimensional Figurative number of rank n whose generating polygon is a triangle
F 4 , n , 5 - Four Dimensional Figurative number of rank n whose generating polygon is a pentagon.
I.
INTRODUCTION
The theory of diophantine equations offers a rich variety of fascinating problems [1-3]. Particularly, in
[4-9] the cubic equations with 4 unknowns and in [10-11] cubic equations with 5 unknowns are studied for their
integral solutions. This communication concerns with an another non-zero cubic equation with six unknowns
given by  x y ( x  y )   z w ( z  w )  (    ) X Y ( X  Y ) . Infinitely many non-zero integer triples
( x , y , z ) satisfying the above equation are obtained. Various interesting properties among the values of x, y and
z are presented.
II.
METHOD OF ANALYSIS
The diophantine equation representing the cubic equation with six unknowns under consideration is
 x y ( x  y )   z w ( z  w )  (   ) X Y ( X  Y )
(1)
Assuming x  u  p , y  u  p , z  u  q , z  u  q , x  u  v , x  u  v
(2)
in (1), it reduces to the equation,
2
2
2
(3)
 p   q  (   ) v
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2. On the homogeneous cubic equation…
Again using the linear transformation
p  S  T ,q  S T
(4)
in (3), it reduces to
S
2
2
 T
2
 v
(5)
The above equation (5) is solved through different approaches and thus, one obtains
different sets of solutions to (1)
1. CaseI:   is not a square.
1.1 Approach1:
The solution to (5) can be written as
S   a
2
2
 b ,T  2ab, v    a
2
b
2
(6)
In view of (6), (4) and (2) the integral solution of (1) is obtained as


2
2
y  u  (  a  b  2  a b ) 

2
2
z  u    a  b  2 a b 

2
2
w  u  (  a  b  2  a b ) 

2
2
X  u   a  b


2
2
Y  u   a  b

x  u   a
2
b
2
 2 ab
(7)
Properties:
1. 3 k [( x  z )  (   (   )  1)( T 6 , a  3 ( O H a )  2 C Pa , 6 )] is a nasty number
2. x ( a , a )  w ( a , a )  2 (    )( 2 T 3 , a  T 6 , a  2 T 4 , a )  0 (m o d 2 )
5
3. k    ( 6 F 4 , a , 5  2 Pa  2 T 4 , a )  4  T 3 , a  x ( a ( a  1) , 1)  1
4. 3 ( X ( a , a )  Y ( a , a ))  (1    )( S a  1 2 C Pa , 3  6 C Pa , 6  1)  0
5. 2 ( z ( a , a )  w ( a , a ))  (    2   1)[ T1 0 , a  9 ( O H ) a  6 C Pa , 6 ]  0
6. w ( 2
2n
,2
2n
)  x(2
2n
,2
2n
)  (    (    )  1) j 4
       1 .
n 1
7. x ( a , a )  X ( a , a )   (   1)( 2 T 3 , a  T 6 , a  T 4 , a )  0
1.2 Approach2:
2
Let v  a    b
Now, rewrite (5) as,
S
2
 T
2
2
 v
(8)
2
1
(9)
Also 1 can be written as
1
(   k
2
 i2k
  )(    k
2
 i2k
 )
(10)
2 2
(   k )
Substituting (10) and (8) in (9) and using the method of factorisation, define
(S  i   T ) 
(   k
2
 i2k
  )( a  i   b )
2
(   k )
2
(11)
Equating real and imaginary parts in (11) we get
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3. On the homogeneous cubic equation…


(   k )


1
2
2
2
T 
[(    k ) 2 a b  2 k ( a    b )] 
2

(   k )

S 
1
2
2
[(    k )( a
2
2
   b )  4  k a b ]
(12)
Considering (2), (4), (8) & (12) and performing some algebra, the corresponding solutions of (1) are given by


2

y  u  (   k ) ( f   g )

2

z  u  (   k ) ( f   g )

2
w  u  (   k ) ( f   g )


2 2
2
2
X  u  (   k ) ( A    B ) 
2 2
2
2 
Y  u  (   k ) ( A    B ) 
(13)
2
   B )  4  k A B 


2
2
2
g  (   k ) 2 A B  2 k ( A    B ) 

(14)
2
x  u  (   k ) ( f   g )
where
2
f  (   k ) ( A
2
Properties:
2
1 . 6 k [ x ( a ( a  1) , a ( a  1) )  {   k } { (    k
2
2
 2 k  ) (1    )  2 ( 2   k   k  k ) }
3
{ 2 4 F 4 , a , 3  2 4 Pa  4 T 3 , a } ]
is a nasty number
2
2
2
2 . x ( a , a )  y ( a , a )  z ( a , a )  w ( a , a )  2 (    k )[ 2 (    k )(1    )  4   k  2 (    )(  k  k )
+ 2 k (1    )][ 2 T 3 , a -2 C P a ,6 + S O a ]
2
2
. 3 . k [ w ( a , a )  {   k } { (    k
is a cubic integer
2
 2 k  )(1    )  2 ( 2   k  
2
2
8
   k )} { 2 Pa  S O a } ]
2 2
4. X ( a ( a  1), a ( a  1))  Y ( a ( a  1), a ( a  1))  2 (   k ) (1    )( 2 T
3, a
2
 2 C Pa , 6 )
1.3 Approach3:
1 can also be written as
1
(     i 2   )(    i 2   )
(   )
2
Following the same procedure as in approach2 we get the integral solution of (1) as
x  u  (    ) ( f1   g 1 )


y  u  (    ) ( f1   g 1 )


z  u  (    ) ( f1   g 1 )


w  u  (    ) ( f1   g 1 )

2
2
2 
X  u  (   ) ( A    B )

2
2
2
Y  u  (   ) ( A    B ) 

(15)
Where
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4. On the homogeneous cubic equation…
2
   B )  4  A B 


2
2
g 1  2 ( A    B )  2 A B (   ) 

f1  (    )( A
2
(16)
Properties:
2
1. 6 k { 4 x ( a , a )  (    ) [ (   3  ) (1    )  2 (     ) ] [ T1 0 , a  6 T 3 , a  3 T 4 , a ] } is a nasty number
2
2. 3 y ( a , a )  (    ) [ ( 3    ) (1    )  2 (     ) ] [ T 8 , a  6 T 4 , a  4 T 5 , a ]  0 ( m o d 3 )
2
3. 7 z ( a , a )  (    ) [ ( 3    ) (1    )  2 ( 3     ) ] [ 2 T 9 , a  1 0 T 3 , a  5 T 4 , a ]  0 ( m o d 7 )
2
2
5
4. k { w ( a , a )  (    ) [ ( 3    ) (1    )  2 ( 3     ) ] [ 2 Pa  C Pa , 6 ] } is a cubic integer
5
5. X ( a ( a  1) , 1)  Y ( a ( a  1) , 1)  2 ( 6 F 4 , a , 5  2 Pa )  2    0
2n
2n
6. X ( 2 , 2 )  (1    )(3 J 4 n  1)  0
1.4 Approach4:
Rewriting (5) as v 2  S 2    T 2
Factorisation of the equation (17) gives
(17)
( v  S )( v  S )  (  T )(  T )
(18)
Considering (18) and using the method of cross multiplication the non-zero integral solution of (1) are obtained
as
2
2
x  u  m   n   2 mn
2
2
y  u  (m   n   2  m n )
2
2
z  u  m   n   2 m n
2
2
2
2
w  u  ( m   n   2 m n )
X  u  m   n 
2
2
Y  u  m   n 
Properties:
1. x ( 2 a , a )  y ( 2 a , a )  ( 4   5  ) ( 6 F 4 , a , 5  T
2
4 ,a )  0 (m o d 3)
5
2. z ( a , 2 a )  w ( a , 2 a )  2  Pa   C Pa , 6  0
3. 3( X ( a , 2 a )  Y ( a , 2 a ))  (   4  )( S a  1)  0 (m o d 6 )
4. 3 ( X ( a , a )  x ( a , a )  z ( a , a ))  (5    )( 2 T 5 , a  2 T 3 , a  T 4 , a )  0 (m o d 3 )
1.5 Approach5:
(17) can be written as a set of double equations in five different ways as shown below:
Set1: v  S   T , v  S   T
Set2: v  S    , v  S  T
2
2
Set3: v  S   T , v  S  
Set4: v  S   , v  S   T
2
2
Set5: v  S  T , v  S   
Solving each of the above sets, the corresponding values of v , S a n d T are given by
Set1: v  (   ) T1 , S  (   ) T1 , T  2 T1
2
2
Set2: v  2  1  1   1   1  2 T1  2 T1  1, S  2  1  1   1   1  2 T1  2 T1 , T  2 T1  1
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5. On the homogeneous cubic equation…
Set3: v   1T
2
  1 , S   1T
2
 1
2
2
Set4: v  2  T1   1 , S   1  2  T1 , T  2 T1
2
2
Set5: v  2 T1   1  , S  2 T1   1  , T  2 T1
In view of (4) and (2), the corresponding solutions to (1) obtained from set1 are represented as shown below:
x  u  (   3  ) T1
y  u  (   3  ) T1
z  u  ( 3    ) T1
w  u  ( 3    ) T1
X  u  (    ) T1
Y  u  (    ) T1
Properties:
1. 3 k [ x ( a )  z ( a )  4 (    )( 2 T 3 , a  2 C Pa , 6  S O a )] is a nasty number
3
2. x ( a )  y ( a )  2 ( 5   3  ) [ 6 Pa  6 T 3 , a  3 ( O H a )  2 C Pa , 3 ]
For simplicity, we exhibit below the integer solutions obtained from sets2 to 5 respectively
Set2:
x  u  f (  1 ,  1 , T1 )
y  u  f (  1 ,  1 , T1 )
z  u  g (  1 ,  1 , T1 )
w  u  g (  1 ,  1 , T1 )
2
X  u  ( 2  1  1   1   1  2 T1  2 T1  1)
2
Y  u  ( 2  1  1   1   1  2 T1  2 T1  1)
Where
f ( 1 ,  1 , k )  2  1  1   1   1  2 k
2
 2 k  ( 2  1  1) ( 2 k  1)
g ( 1 ,  1 , k )  2  1  1   1   1  2 k
2
 2 k  ( 2  1  1) ( 2 k  1)
Set3:
x  u   1T
2
  1  2  1T
y  u   1T
2
  1  2  1T
z  u   1T
2
  1  2  1T
w  u   1T
2
  1  2  1T
X  u   1T
Y  u   1T
2
2
 1
 1
Set4:
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6. On the homogeneous cubic equation…
2
x  u   1  2  T1  2  T1
2
y  u  (  1  2  T1  2  T )
2
z  u   1  2  T1  4  1T1
2
w  u  (  1  2  T1  4  1T1 )
2
X  u  2  T1   1
2
Y  u  2  T1   1
Set5:
2
x  u  2 T1   1   2  T1
2
y  u  ( 2 T1   1   2  T1 )
2
z  u  2 T1   1   4  1T1
2
w  u  ( 2 T1   1   4  1T1 )
2
X  u  2 T1   1 
2
Y  u  2 T1   1 
2. CaseII:
Choose  and  such that   is a perfect square, say, d
2
2
(5) is written as S  ( d T )  v
2.1 Approach6:
The solution of (19) can be written as
dT  a
2
2
 b , S  2ab, v  a
2
b
2
2
(19)
2
(20)
Considering (20), (4) & (2) and performing some algebra the integral solution of (1) is obtained as
2
 B )

2
2
2
y  u  2d AB   d ( A  B )

2
2
2
z  u  2d AB   d ( A  B ) 

2
2
2
w  u  2d AB   d ( A  B )

2
2
2
X  u  d (A  B )


2
2
2
Y  u  d (A  B )

x  u  2d
2
AB   d (A
2
(21)
Properties:
1. x ( 2 a , a )  y ( 2 a , a )  ( 4 d
2
 3  d ) (T 6 , a  3T 4 , a  2 T5 , a )  0
2. 2 ( z ( 2 a , a )  w ( 2 a , a ) )  ( 4 d
2
 3  d ) ( T1 0 , a  3 T 4 , a  2 T 5 , a )  0
2
3. X ( 2 a , a )  Y ( 2 a , a )  d ( 2 T1 2 , a  9 T 4 , a  2 T 2 0 , a )
4. 6 ( x ( a , a )  y ( a , a )  z ( a , a )  w ( a , a )  X ( a , a )  Y ( a , a )  4 k ) is a nasty number
2
2n
2n
2
5. k [ X ( 2 , 2 )  d ( j 4 n  1  1)] is a cubic integer
2.2 Approach7:
(19) can be written as
v
2
 (dT )
2
 S
2
(22)
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7. On the homogeneous cubic equation…
Writing (22) as a set of double equations as
v  d T  1, v  d T  S
2
and Solving, the corresponding values of v , S a n d T are given by
2
2
v  2 S1 d
2
 2 S 1 d  1, T   ( 2 S 1 d  2 S 1 ), S  2 S 1 d  1
In view of (4) and (2), the corresponding solutions to (1) are represented as
2
x  u  2 S1d  1   ( 2 S1 d  2 S1 )
2
y  u  2 S1d  1   ( 2 S1 d  2 S1 )
2
z  u  2 S1d  1   ( 2 S1 d  2 S1 )
2
w  u  2 S1d  1   ( 2 S1 d  2 S1 )
2
2
X  u  2 S1 d
2
2
Y  u  2 S1 d
 2 S1d  1
 2 S1d  1
Properties:
1. x ( a )  y ( a )  (    ) [ d ( 6 F 4 , a , 5  3 C Pa , 6  T
2
2
4,a
2
)  3T 4 , a  T8 , a ]
5
2. X ( a )  2 d ( 6 Pa  C Pa , 6 )  2 d ( 2 T 3 , a  T 4 , a )  k  1
2.3 Approach8:
(19) can be rewritten as
2
S
2
 (dT )
2
(23)
Writing (23) as a set of double equations in 3 different ways as shown below
v
2
2
2
2
Set1: v  S  T , v  S  d
Set2: v  S  d , v  S  T
2
Set3: v  S  d T , v  S  d
Solving each of the above sets, the corresponding values of v , S a n d T are given by
Set1: v  2 ( f
2
 e )  2 ( f  e )  1, S  2 ( e
Set2: v  2 ( f
2
 e ), S  2 ( f
2
Set3: v  ( T
2
2
 1) f , S  ( T
2
2
2
 f
2
)  2 ( e  f ), T  2 e  1
2
 e ), T  2 e
 1) f
In view of (4) and (2), the corresponding solutions to (1) are represented as shown below:
Set1:
x  u  2 (e
2
 f
2
)  2 ( e  f )   ( 2 e  1)
y  u  2 (e
2
 f
2
)  2 ( e  f )   ( 2 e  1)
z  u  2 (e
2
 f
2
)  2 ( e  f )   ( 2 e  1)
2
 f
2
)  2 ( e  f )   ( 2 e  1)
w  u  2 (e
X  u  2( f
Y  u  2( f
2
2
2
 e )  2( f  e)  1
2
 e )  2( f  e)  1
Set2:
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7 | Page

8. On the homogeneous cubic equation…
x  u  2 (e
2
 f
2
)  2 e
y  u  2(e
2
 f
2
)  2 e
z  u  2 (e
2
 f
2
)  2 e
2
 f
2
)  2 e
w  u  2 (e
X  u  2(e
2
2
Y  u  2 (e
 f
 f
2
2
)
)
Set3:
x  u  (T
2
 1) f   T
y  u  (T
2
 1) f   T
z  u  (T
2
 1) f   T
w  u  (T
2
 1) f   T
X  u  (T
Y  u  (T
2
2
 1) f
 1) f
Remarkable observation:
If ( x 0 , y 0 , z 0 , w 0 , X 0 , Y 0 ) be any given integral solution of (1), then the general solution pattern is
presented in the matrix form as follows:
Odd ordered solutions:










0
(   )
4n3
(   )
-2  (    )
4n3
0
-(    )
4n3
(   )
2  (   )
4n3
0
-2 (   )
0
1
x2 n 1 


1
y 2 n 1


z 2 n 1   1
 
w 2 n 1  
1

X 2 n 1


1
 
Y2 n 1  
1
2 (   )
4n3
(   )(   )
-(    ) (    )
4n3
4n3
4n3
1
0
0
-1
0
0












u0 

v0

p0 

q0 

Even ordered solutions:
 x2 n

y2n

 z
2n

 w2n

X 2n

 Y
 2n
1



1



1
 


1



1




1
0
(   )
4n
0
-(    )
4n
0
0
(   )
4n
0
0
-(    )
4n
1
0
0
-1
0
0
0
0












u0 

v0

p0 

q0 

The values of x , y , z , w , X and Y in all the above approaches satisfy the following properties:
1. 3 ( x  y )( z  w  X  Y ) is a nasty number.
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8 | Page

9. On the homogeneous cubic equation…
2. x  y  z  w  2 ( X  Y )
2
2
3. 2[ x  y  X
2
2
2
2
Y
2
2
 (z  w) ]  (x  y)
2
4. x  y  z  w  X
2
Y
2
2
 (X Y )
2
 0
 0 (m o d 4 )
5. ( x  y )( z  w )( X  Y ) is a cubical integer
2
2
6. x y  z w  ( k  y )  ( k  w )  0
III.
CONCLUSION
Instead of (4), the substitution
p  S  T ,q  S T
in (3), reduces it to the same equation (5).Then different solutions can be obtained, using the same patterns
REFERENCES
[1]
[2]
L.E. Dickson, History of Theory of numbers, vol.2, Chelsea Publishing company, New York, 1952.
L.J. Mordell, Diophantine Equations, Academic press, London, 1969.
[3]
Carmichael.R.D, The Theory of numbers and Diophantine Analysis, New York, Dover,1959
[4]
M.A.Gopalan and S.Premalatha , Integral solutions of ( x  y )( xy  w
[5]
applied sciences,Vol.28E(No.2),197-202, 2009
M.A.Gopalan and V.pandiChelvi, Remarkable solutions on the cubic equation with four unknowns
x
[6]
 y
3
 z
3
 28 ( x  y  z ) w
2
)  2(k
2
 1) z
3
Bulletin of pure and
, Antarctica J.of maths, Vol.4, No.4, 393-401, 2010.
M.A.Gopalan and B.Sivagami, Integral solutions of homogeneous cubic equation with four unknowns
x
[7]
3
2
3
 y
3
 z
3
 3 xyz  2 ( x  y ) w
3
, Impact.J.Sci.Tech, Vol.4, No.3, 53-60, 2010.
M.A.Gopalan and S.Premalatha , On the cubic Diophantine equation with four unknowns
( x  y )( xy  w
2
2
)  2(n
3
 2 n ) z , International Journal of mathematical sciences,
Vol.9,No.1-2, Jan-June, 171-175, 2010.
[8]
M.A.Gopalan and J.KaligaRani, Integral solutions of
x
3
 y
3
 ( x  y ) xy  z
3
 w
3
 ( z  w ) zw ,
Bulletin of pure and applied sciences, Vol.29E (No.1), 169-173, 2010.
( x  y )( xy  w
2
3
)  2 ( k  1 ) z , The Global Journal of
[9]
M.A.Gopalan and S.Premalatha Integral solutions of
[10]
applied Mathematics and Mathematical sciences, Vol.3, No.1-2, 51-55, 2010
M.A.Gopalan, S.Vidhyalakshmi and T.R.Usha Rani , On the cubic equation with five unknowns
x
[11]
3
 y
3
 z
3
 w
3
2
 t ( x  y ) , Indian Journal of Science, Vol.1, No.1, 17-20, Nov 2012.
M.A.Gopalan, S.Vidhyalakshmi and T.R.Usha Rani , Integral solutions of the cubic equation with five unknowns
x
3
 y
3
 u
3
 v
3
3
 3 t , IJAMA, Vol.4 (2), 147-151, Dec 2012
.
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