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ON THE NON-HOMOGENEOUS CUBIC EQUATION WITH FIVE

UNKNOWNS

2 2 3

xxyy  z wT

.

S.Vidhyalakshmi 1, M. A. Gopalan 2 and K. Lakshmi 3 1,2,3

Department of Mathematics,Shrimati Indira Gandhi College,Trichy, India.

ABSTRACT

We obtain infinitely many non-zero integer solutions ( , , , , )x y z w T satisfying the non-

homogeneous cubic equation with five unknowns given by x2xyy2  z w T3.

Various interesting relations between the solutions and special numbers are presented

KEYWORDS:

Non-homogeneous cubic equation, Integral solutions, Polygonal numbers, Pyramidal numbers, Centered pyramidal numbers, Four dimensional pentagonal number. MSC 2000 Mathematics subject classification: 11D25.

NOTATIONS:

,

m n

T -Polygonal number of rank nwith size m m

n

P - Pyramidal number of rank n with size m

n

SO -Stella octangular number of rank n

n

OH - Octahedral number of rank n

n

J -Jacobsthal number of rank of n

n

j - Jacobsthal-Lucas number of rank n

n

KY -keynea number of rank n

,6

n

CP - Centered hexagonal pyramidal number of rank n

4, ,5n

(2)

The theory of Diophantine equations offers a rich variety of fascinating problems. In particular, cubic equations, homogeneous and non-homogeneous have aroused the interest of numerous mathematicians since antiquity [1-3]. For illustration, one may refer [4-11] for homogeneous and non-homogeneous cubic equations with three, four and five unknowns. This paper concerns with the problem of determining non-trivial integral solution of the non- homogeneous cubic equation with five unknowns given by x2xyy2  z w T3 A few relations between the solutions and the special numbers are presented.

Initially, the following two sets of solutions in ( , , , , )x y z w T satisfy the given equation:

2 4 2 4 2

(4k , 2 , 2(4k kk ) p, 2(4kk )p k, 2 ),

2 2 4 2 4

( 2 k , 2 , 2 kk (1 ) p, 2k (1 ) p, 2k)

However we have other patterns of solutions, which are illustrated below: Method of Analysis:

The Diophantine equation representing the non- homogeneous cubic equation is given by x2xyy2  z w T3 (1)

Introduction of the transformations

x u v y,  u v z, 2uvp w, 2uvp (2) in (1) leads to

u2v2 T3 (3)

The above equation (3) is solved through different approaches and thus, one obtains different sets of solutions to (1)

Approach1:

The solution to (3) is obtained as

2 2 2 2 2 2

( ), ( ),

ua ab vb ab Tab (4) In view of (2) and (4), the corresponding values of ( , , , , )x y z w T are represented by

2 2

2 2

2 2 2

2 2 2

2 2

( )( )

( )( )

2 ( )

2 ( )

( )

x a b a b

v a b a b

z ab a b p

w ab a b p

T a b

  

  

   

   

 

(5)

(3)

1. x a( 2,a 1) y a( 2,a 1) T a( 2,a 1) 4T4,a32CPa,648(OHa)0(mod15) 2. The following expressions are nasty numbers:

(a) 3 [2 ( , )p z a bx a b2( , )y2( , )]a b . (b) x a a(2 , )y a a(2 , ) 6 SOa6PRa. 3. The following expressions are cubic integers

(a) 9[ (2 , )x a ay(2 , )a az(2 , )a aw a a(2 , )T(2 , ) 6a aPa5]. (b)9[4 (2 , ) 4 (2 , ) (2 , ) (2 , ) 36( 3 4,a)]

a

x a ay a az a aw a aSOT 4.16[ ( ,1)x ay a( ,1)w a( ,1) 4 CPa,64T3,a1] is a quintic integer 5.x a( 1, )ay a( 1, )aT a( 1, ) 8aT3,a8CPa,612(OHa)0(mod3) 6.x(24n, 22n)y(24n, 22n)2(j12nj8n)

7. 9[4 (2 , ) 4 (2 , )x a ay a az a a(2 , )w a a(2 , ) 36( SOa3T4,a)]

8. T(22n1, 22n)3KY2n3J2n1

9. z a a(2 , )w a a(2 , )x a a(2 , )y a a(2 , ) 12 CPa,6 0(mod 2) 10. y a( ,1)x a( ,1)T a( ,1)T4,a 1

Approach2:

The assumption

uUT v, VT (6) in (3) yields to

U2V2T (7) Taking T  t2 (8) in (7), we get

2 2 2

UtV (9) (i) Then the solution to (9) is given by

2 2 2 2

2 , , , 0

t  V   U     (OR) (10)

2 2 2 2

2 , , , 0

U  V   t     (11) From (6), (8) and (10) we get

2 2 2 2

2 2 2 2

2 2

4 ( )

4 ( )

4

u

v

T

   

   

 

  



  

  

(12)

(4)

4 2

2 4

4 4 4 4

4 4 4 4

2 2

8 8

32 ( )

32 ( )

4 x y z p w p T                           Remark: 1

By considering (6), (8), (11) and (2), we get the corresponding integral solution to (1).as

2 2 2 2

2 2 2 2

2 2 2 2 4

2 2 2 2 4

2 2 2

( ) ( )

( ) ( )

4 ( )( )

4 ( )( )

( ) x y z p w p T                                       Properties: 1. 3 ,6

(2 , ) (2 , ) (2 , ) (2 , ) 72 0( 2)

a

x a ay a az a aw a aCPmod

2. x(2 , )a ay(2 , ) 360(a aPa5 2) 90T4,a(2Pa82T3,aT4,a)

3. 3[72(2T3,aT4,a) 36 PRaT a( 1,a1)] is a cubic integer

4. z(2 , )a aw a a(2 , ) 2 p42F4, ,5a 21CPa,614T4,a is a biquadratic integer 5. z a( ,1)w a( ,1) 16 CPa,6(T4,a1)40

(ii) Now, rewrite (9) as,

U2t2 1*V2 (13) Also 1 can be written as

1 ( ) ( ) i n i n (14) Let Va2b2 (15) Substituting (14) and (15) in (13) and using the method of factorisation, define,

2

(Uit)i a ibn(  ) (16) Equating real and imaginary parts in (16) we get

2 2

2 2

cos ( ) 2 sin

2 2

2 cos sin ( )

2 2

n n

U a b ab

n n

t ab a b

               (17)

(5)

2 2 2 2

2 2 2 2

4 2 2 2 2

4 2 2 2 2

2 2 2

[cos ( ) 2 sin ( )]

2 2

[cos ( ) 2 sin ( )]

2 2

2 ( )[cos ( ) 2 sin ]

2 2

2 ( )[cos ( ) 2 sin ]

2 2

[sin ( ) 2 cos ]

2 2

n n

x T a b ab a b

n n

y T a b ab a b

n n

z T a b a b ab p

n n

w T a b a b ab p

n n

T a b ab

                                 

(iii) 1 can also be written as

2 2 2 2

2 2 2

(( ) 2 )(( ) 2 )

1

( )

m n i mn m n i mn

m n

   

 (18)

Substituting (15) and (18) in (13) and using the method of factorisation, define,

2 2

2 2 2 2

( ) 2 )

( ) ( )

( )

m n i mn

U it a ib

m n

 

  

 (19)

Equating real and imaginary parts in (19) we get

2 2 2 2

2 2

2 2 2 2

2 2

1

{( )( ) 4 }

1

{2 ( ) 2 ( )}

U m n a b mnab

m n

t ab m n mn a b

m n                (20)

In view of (2), (6), (8) and (20), the corresponding values of x y z w T, , , , are represented as

2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2

4 2 2 4 4 4 4

4 2 2 4 4 4 4

2 2 2 2 2 2 2

( ) [( )( ) 4 ( )( )]

( ) [( )( ) 4 ( )( )]

2 ( )[( )( ) 4 ]

2 ( )[( )( ) 4 ]

( ) [2 ( ) 2 ( )]

x m n T m n A B mnAB m n A B

y m n T m n A B mnAB m n A B

z T m n A B m n mnAB p

w T m n A B m n mnAB p

T m n AB m n mn A B

       

       

     

     

      2

Remark2: By writing 1 as

2 2 2 2

2 2 2

(2 ( )(2 ( )

1

( )

mn i m n mn i m n

m n

   

and performing the same procedure as above the corresponding integral solution to (1) can be obtained

Approach3:

Equation (3) can be written as

(6)

Set2: u v T3, u v 1

Solving set1, the corresponding values of u v and, Tare given by

u4k36k23k1,v4k36k23 ,k T 2k1 (22) In view of (22) and (2), the corresponding solutions to (1) obtained from set1 are represented as shown below:

3 2

3 2 3 2

3 2 3 2

8 12 6 1

1

2(4 6 3 1)(4 6 3 )

2(4 6 3 1)(4 6 3 )

2 1

x k k k y

z k k k k k k p

w k k k k k k p

T k

   

      

      

 

Properties:

1. x a( )y a( ) 24 Pa40 (mod 2)

2.6[2 ( ) 2 ( )x ay az a( )w a( )T a( ) 4 T3,aT4,a1] is a nasty integer. 3.4[ ( )x ay a( ) 24 Pa45CPa,63CPa,10] is a cubic integer.

4. 8j6n12j4n6j2n3J2n1x(22n, 22n)y(24n, 22n)T(24n, 22n)0(mod 26)

5. z a( )w a( )y a( ) 2 p1 Remark3:

Similarly, the solutions corresponding to set2 can also be obtained. Approach4:

Substituting, Ta2b2

in (3) and writing it as a system of double equations as

3 3 ( ) ( ) u v a b u v a b

  

  

and solving we get

3 2

2 3

3 3

u a ab v a b b

 

   (26)

(7)

3

3

3 2 2 3

3 2 2 3

2 2

( )

( )

2( 3 )(3 )

2( 3 )(3 )

x a b

y a b

z a ab a b b p

w a ab a b b p

T a b

 

 

   

   

 

Properties:

1.3[ ( , )x a ay a a( , )z a a( , )w a a( , ) 8 CPa,6] is a nasty number

2. x a( ,1)y a( ,1) 4 CPa,36T4,a2T8,a 0

3. x a a( , )T a a( , )z a a( , )w a a( , ) 2 is a cubical integer

4. z a a( , )w a a( , )x a a( , )y a a( , ) 144 Pa572T4,a 0

5. x a( ,1)y a( ,1) 6 Pa44T4,a2T9,a 0

Acknowledgement:

* The financial support from the UGC, New Delhi (F.MRP-5123/14 (SERO/UGC) dated March 2014) for a part of this work is gratefully acknowledged

Conclusion:

In conclusion, one may search for different patterns of solutions to (1) and their corresponding properties.

REFERENCES

[1]. L.E. Dickson, History of Theory of numbers, vol.2, Chelsea Publishing company, New York, 1952.

[2]. 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w2)2(k21)z3 Bulletin of pure and applied sciences,Vol.28E(No.2),197-202, 2009.

[5]. M.A.Gopalan and V.pandiChelvi, Remarkable solutions on the cubic equation with four unknowns x3 y3 z3 28(xyz)w2,Antarctica J.of maths, Vol.4, No.4, 393- 401, 2010.

[6] M.A.Gopalan and B.Sivagami, Integral solutions of homogeneous cubic equation with four unknowns x3 y3 z3 3xyz2(xy)w3, Impact.J.Sci.Tech, Vol.4,

No.3, 53- 60, 2010.

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No.1-2, Jan-June, 171-175, 2010.

[8]. M.A.Gopalan and J.KaligaRani, Integral solutions of

x3 y3 (xy)xy z3 w3 (zw)zw,Bulletin of pure and applied sciences, Vol.29E (No.1), 169-173, 2010.

[9]. M.A.Gopalan and S.Premalatha Integral solutions of (xy)(xyw2)2(k1)z3, The Global Journal of applied Mathematics and Mathematical sciences, Vol.3,

No.1-2, 51-55, 2010.

[10]. M.A.Gopalan, S.Vidhyalakshmi and T.R.Usha Rani , On the cubic equation with five unknowns x3 y3 z3 w3 t2(xy), Indian Journal of Science, Vol.1, No.1, 17-20, Nov 2012.

[11]. M.A.Gopalan, S.Vidhyalakshmi and T.R.Usha Rani , Integral solutions of the cubic equation with five unknowns x3 y3 u3 v3 3t3, IJAMA, Vol.4 (2), 147-151, Dec 2012.

[12]. M.A.Gopalan, S.Vidhyalakshmi andG.Sumathi, On the Homogeneous Cubic Equation with four unknowns x3 y3 14z33w2(xy), Discovery, Vol.2(4), 17-19, 2012 [13] M.A.Gopalan, S.Vidhyalakshmi andG.Sumathi, On the Homogeneous Cubic Equation with four unknowns x3y3z3w2(xy), Diophantous J.Math, Vol.2(2), 99 -103, 2013.

References

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