Integer Solution of the Homogeneous Bi-Quadratic Diophantine Equation with Five Unknowns $(x-y)(x^3-y^3)=(z^2-w^2)p^2$

Vol. 11, 2017, 55-61

ISSN: 2349-0632 (P), 2349-0640 (online) Published 11 December 2017

www.researchmathsci.org

DOI: http://dx.doi.org/10.22457/jmi.v11a8

Journal of

Integer Solution of the Homogeneous Bi-Quadratic

Diophantine Equation with Five Unknowns

(x-y)(x

3

-y

3

)=(z

2

-w

2

)p

2

R.Umamaheswari1 and A.Kavitha2

1

Department of Mathematics ,Shrimati Indira Gandhi College Trichy-2,Tamilnadu, India.

1

e-mail: [email protected] 2

e-mail: [email protected]

Received 10 November 2017; accepted 10 December 2017

Abstract. The homogeneous equation with five unknown 3 3 2 2 2

) (

) )(

(x− y x − y = z −w p is analyzed for its nonzero distinct integer solutions. Employing the transformation and applying the method of factorization, different patterns of nonzero distinct integer solutions to the above bi-quadratic equation are obtained. A few interesting relations between the solutions and special number patterns namely polygonal and pyramidal numbers are presented.

Keywords: Homogeneous bi-quadratic, Bi-quadratic equation with five unknown, integer solutions.

AMS Mathematics Subject Classification (2010): 11D25

1. Introduction

Bi-quadratic Diophantine equations, homogeneous and non -homogeneous, have aroused the interest of numerous mathematicians since ambiguity as can be seen from [1-2] particularly In [3-5] bi-quadratic diophantine equations with three unknowns are considered In [6-9] bi-quadratic equation with four unknowns are considered In [10-12] bi-quadratic equation with five unknowns are considered. In this paper, another interesting bi-quadratic equation with five unknown given by

2 2 2 3 3

)

(

)

)(

(

x

−

y

x

−

y

=

z

−

w

p

is considered and five different patterns of integral solutions are illustrated. A few interesting properties between the solutions and special number patterns are exhibited.

2. Notation

( )(

)



  



 _{− −}

+ =

2 2 1 1 ,

m n n

T_mn - Polygonal number of rank n with side m.

S_n=6n

( )

n−1+1

56

n=

( )

+ - Pronic number of rank n

2 3 3 ,

n n CPn

+

= - Centered triangular pyramidal number of rank n

CP

n,6

=

n

3 -Centered hexagonal pyramidal number of rank n

G

no₂

=

2

n

−

1

- Gnomonic number ₌

(

2 2 ₋1

)

n n

SOn - Stella Octangular number of rank n

(

n

( )

n

)

n

J 2 1

3

1 _{− −}

= - Jacobsthal number of rank n.

3. Method of analysis

The Diophantine equation representing the biquadratic equation with five unknowns under consideration is.

2 2 2 3 3

) ( ) )(

(x−y x −y = z −w p (1)

The substitution of the linear transformations in (1) gives 1

-uv w , 1 uv z v, -u y

, = = + =

+ =u v

x (2)

2 2 2

3v p

u + = (3)

We solve (3) through different methods and thus obtained different patterns of solutions to (1)

3.1. Pattern 1

Equation (3) can be written as

0 B e wher

3 = − = ≠

+

B A u p

v v

u

p ₍₄₎

Equation (4) is equivalent to the system of double equations.

0 pB vA

0 pA vB

= +

= +

uB uA

(5)

Solving (5) by applying the method of cross multiplication and using (2) the corresponding non-zero integer solution to (1) are obtained as

(

)

(

)

(

)

(

)

(

)

2 2

3 3

2 2

2 2

2 2

3 A B A,

1 2 6A B A,

1 2 6A B A,

2 3A

B A,

2 3A

B A,

B p

p

AB B w

w

AB B z

z

AB B y

y

AB B x

x

+ = =

+ − = =

+ − = =

− − = =

+ − = =

(6)

Properties:

1. 5x

( ) ( ) ( )

A,1+5yA,1+4pA,1 =62A2is a perfect square 2. z

(

A +1,A

) (

−p A +1, A

)

+2t4,A -20t4,A +4 = 0

Unknowns

57

Remark:

In addition to (4), (3) may also be expressed in the form of ratios as presented below

0 B where B

A v

u p

3v ₌ + _{= ≠}

−u p

Following the procedure as presented above the corresponding non zero integer solutions to (1) are found to be as given below.

(

)

(

)

(

)

(

)

(

)

2 2

2 2 4 4

2 2 4 4

2 2

2 2

3 A B A, p

1 10

9 A B A, w

1 10

9 A B A, z

2 3B

B A,

2 3B

B A,

B p

B A B w

B A B z

AB A y

y

AB A x

x

+ = =

− −

+ = =

+ −

+ = =

+ − = =

− − = =

Properties:

1.

( )

_{, 4 0} , 4 =

− t A

A A p

2. x

( ) ( )

A,A+yA,A=4A2 is a perfect square

3. x

( ) ( )

1,B+p1,B-2PR_A-4=0

3.2. Pattern 2

Let p=p

( )

a,b =a2+3b2 ₍₇₎ where a and b are non-zero district integers

Using (3) and (7) and applying the method of factorization, define

(

)(

) (

) (

2

)

2

3 3

3

3v u i v a i b a i b

i

u+ − = + −

Equating real and imaginary parts we have

ab v

b a u

2 3 2

2

= − =

(8)

using (8) in (2), the corresponding non-zero distinct integral solutions of (1) are given by

( )

( )

( )

( )

a,b 2 6 1 1 6 2 b a,

2 3 b

a,

2 3 b

a,

3 3

3 3

2 2

2 2

− − = =

+ − = =

− − = =

+ − = =

ab a w

w

ab b a z

z

ab b a y y

ab b a x x

(9)

Thus, (7) and (9) represent the non- zero distinct integral solutions to (1)

58 2. y

(

B,B+1

) (

+ p B,B+1

)

+4t3,A−2t4,A =0

3. y

(

B,B+1

) (

+pB,B+1

)

+4t_3,A−2

( )

t_4,A2+6t_4,A=0

4. w

( )

A,1 +9p

( )

A,1 −2CP_A,6 −2t_20,A −G_AO −27 =0

3.3. Pattern 3

(3) can be written as 1 * 3

u2+ v2=p2

(10)

Write 1 as,

(

)(

)

4 3 1 3 1

1= +i −i ₍₁₁₎

Using (7) and (11) in (10) and applying the method of factorization, define

(

) (

) ( )

1 ₂ 3 3

3 a i b2 i i

u+ = + +

Equating the real and imaginary part, we have

( )

(

)

( )

(

a b ab

)

v v

ab b a u

u

2 3 2 1 b a,

6 3 2 1 b a,

2 2

2 2

+ − = =

− − = =

As our interest is to find integer solution so we replace a by 2A and b by 2B we get,

( )

( )

A B AB

v v

AB B

A u

u

4 6 2 B A,

12 6 2 B A,

2 2

2 2

+ − = =

− − = =

(12)

Using (12) in (2) the corresponding non-zero distinct integral solutions of (1) are given by

( )

( )

AB AB y

y

AB B A B A x x

16 ,

8 12 4

, 2 2

− = =

− − = =

( )

, =4 4+32 3+24 2 2−96 3+36 4+1

=z A B A A A B AB B z

( )

, =4 4+32 3 +24 2 2−96 3+36 4−1

=wA B A A B A B AB B w

( )

2 2

3

,B A B

A p

p= = +

Properties:

1.

(

2,

) (

₋ 2,

)

₊16 _{,6 −}

( )

_4, 2 ₋3 _{4, =} 0

A A

A t t

CP A

A y A A p

2. w

( ) ( )

1,B +z1,B −8

( )

t_4,A 2−32SO_A+320t_4,A−208t_3,A−72=0

3. x

(

A,A+1

) (

+y A,A+1

)

+64t_3,A+16pr_A−16t_4,A+12=0

Unknowns

59 Write 1 as,

(

)(

₂

)

7

3 4 1 3 4 1

1= +i −i

Proceeding as in Pattern: 3 the non-zero distinct integral solutions to (1) are

(

)

(

)

A B AB

y y

AB B

A x

x

182 63 21 B A,

154 105 35 B A,

2 2

2 2

− + − = =

− − = =

( )

A,B =196A4−3528 2 2−4606 3 +13818 3+1764 4+1

=z A B A B AB B

z

( )

A,B =196A4−3528 2 2−4606 3 +13818 3+1764 4−1

=w A B A B AB B

w

(

)

2 2

49 147 B

A, B A

p

p= = +

Properties:

1.

( )

2n,1 =147 ₂ +196

A

J p

2. x

(

A,A

) (

+y A,A

) (

+p A,A

)

+168t4,A=0

3.

(

) (

)

( )

4, 4, ,6

2 2

154 42

84 A , A A ,

A p t A t A CPA

x + − − −

3.5. Pattern 5

Write equation (3) as

1 *

3 2 2

2

u v

p − = (13)

Let u = a2 − 3b2 (14)

Write 1 as,

( )( )

2 3 2 3

1= + − (15)

Using (14) and (15) in (13) and applying the method of factorization, define

(

p+ 3v

) (

= a+ 3b

) (

2 2+ 3

)

Equating the positive and negative parts of the above equations, we have

(16) (17)

Substituting(13) and (17) in (2) we get

( )

( )

B AB

y y

AB A x

x

4 3 B A,

4 21 B A,

2 2

+ − = =

+ = =

(18)

( )

( )

A,B 4 12 9 1 1 9 12 4

B A,

4 3 3

4

4 3 3

4

− − −

+ = =

+ − −

+ = =

B B A B A A w

w

B B A B A A z

z

Thus, (16) and (18) represent the non-zero distinct integral solutions to (1)

Properties:

1. x

( ) ( )

A2 ,A +yA2,A −2

( )

t_4,A2−SO_A−3OH_A−4CP_A,6+3t_4,A=0

2. z

( ) ( )

A,1+wA,1+16CP_A,6−2

( )

t_4,A 2+18=0

2 2

2 2

B 3 A 4 ) , (

6 B 6 A 2 ) , (

+ + = =

+ + = =

AB B A v v

AB B

60

(

) (

)

4,A 3,A

Note:

Write 1 as

(

7 4 3

)(

7 4 3

)

1= + −

Proceeding as in the Pattern 5, the non-zero distinct integral solutions to (1) are

(

)

(

)

(

)

(

)

(

)

2 2

3 3

4

3 3

4 2 2

147 49

B A,

1 12 4

B A,

1 12 4

B A,

4 3 B A,

4 2 B A,

B A

p p

B A B A A w

w

B A B A A z

z

AB B y

y

AB A x

x

+ = =

− −

+ = =

+ −

+ = =

+ − = =

+ = =

Properties:

(1,B) (1,B) 36 _4, 2 6 _,7 35 _{,6 44,} 10 0(mod 3)

.

1 + −  − − − − =

    

A t A CP A

CP A

t p

w

(

A,A

) (

A,A

)

20_4, 0 .

2 p +y − t _A=

( ) ( )

A,1 A,1 4 2Pr 6 0 .

3x +y − t3,A+ A+ =

4. Conclusion

In this paper, we illustrated different methods of obtaining integer solutions to the

bi-quadratic equation with five unknowns

( )

x-y

( ) ( )

x3-y3 =z2-w2 p2. As bi-quadratic equations are rich in variety, one may consider the other forms of equations and search for their corresponding integer solutions.

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(

2

)

4

2 2

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(

2 2

) (

2 2

)

2 4

1 z

Unknowns

61

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4 4

)

(

)

3

( )

2 2 4

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)

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z

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)

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( )

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)(

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)

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