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(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

Techniques to solve Diophantine Equation of Degree Ten with Six 

Unknowns 

822366 R)qp(800z3456yx   

DR. N.THIRUNIRAISELVI1*, DR. M.A.GOPALAN2 

 
1
 Assistant Professor, Department of Mathematics, School of Engineering and Technology, 
Dhanalakshmi Srinivasan University, Samayapuram, Trichy- 621 112, Tamil Nadu, India. 

 
2 

Professor, Department of Mathematics, Shrimati Indira Gandhi College, Affiliated to        Bharathidasan 
University, Trichy-620 002,Tamil Nadu, India. 

Abstract: 

        This paper focuses on finding varieties of distinct integer solutions to the Diophantine 

equation of degree ten with six unknowns given by 822366 R)qp(800z3456yx   through 

employing the substitution strategy and method of factorization. A new representation for 
the factorization of integer 40 involving sides of Pythagorean triangle has been introduced. 

Key words: Higher degree Diophantine equation, Integer solution. 

2020 Mathematics Subject Classification: 11D41. 

Introduction: 

It is well-known that the subject of Diophantine equations occupies a pivotal role in 
the Number theory. There is a vast general theory for higher degree Diophantine equations 
in many variables and it is a topic for research even today. While collecting problems on 
Diophantine equations of degree ten with six unknowns, the following paper [1] has been 
noticed and the authors have presented only few sets of integer solutions. It is worth to 
mention that the equation presented in [1] has many more fascinating patterns of solutions 
in integers. 

 In this paper, the process of obtaining some more choices of solutions in integers to 
the Diophantine equation in title is illustrated. Substitution strategy and factorization 
method are applied successfully to obtain different choices of integer solutions to the 
considered equation. It is to be noted that a new representation for the factorization of 
integer 40 involving sides of Pythagorean triangle has been introduced. 

 

 

 

 

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IJO - INTERNATIONAL JOURNAL OF MATHEMATICS  

(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

Method of Analysis: 

The Diophantine Equation is 
 

822366 R)qp(800z3456yx         (1) 

 To start with, it is observed by inspection that (1) is satisfied by the following 

sextuples {x, y, z, p, q, R} = {8u4v, 4u4v,2u8v, 18u4v,-6u4v,u2v}, {8*402*(9a2+4b2)2, 

4*402*(9a2+4b2) 2, 2*404*(9a2+4b2)4, 18*402*(9a2+4b2)2, -6*402*(9a2+4b2)2, 40*(9a2+4b2)}, 

{32w2,16w2,32w2,72w2,24w2,2w}. However there are often fascinating solution patterns to 

(1) that are illustrated as follows. 

Introduction of the transformation  

 s6t6q,s6t6p,stz,t2s3y,t2s3x     (2) 

in (1) leads to  

 422 R40t4s9           (3) 

Solving (3) through different ways and using (2), one obtains many non-zero distinct integer 
solutions to (1). 

 

Pattern 1: 

The choice  

Rt 
              (4)   

in leads to  








 


9

1R10
R4s

2
22            (5) 

Let  

19R10 22               (6) 

The smallest positive integers to (6) is 

 1R,1 00    

Let  0101 h,RhR  
           (7) 

be the second solution to (6). Substituting (7) and (6) and performing some algebra, we get 

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IJO - INTERNATIONAL JOURNAL OF MATHEMATICS  

(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

 00 18R20h   

In view of (7), we get 

 001001 19R20,18R19R    

which is written in matrix form as  

 t
00

t
11 )R(M),R(     

where 









1920

1819
M

 
and t is the transpose. 

The repetition of the above process leads to the general solution ),R( nn 
 to (6) as  

 t
00

nt
nn ),R(M),R(    

Let 
~

,~  be the eigen values of the matrix M. Then, it is found that 

 10619
~

,10619~    

It is well known that 

 )I~M(~~

~

)I
~

M(~~

~
M

nn
n 















   

where I is the unit matrix of order 2. 

Thus, t
00nnnn

nnnn

t
nn ),R(

2

~~

103

)
~~(5

102

)
~~(3

2

~~

),R( 




























  

From (5), we have  

 t
00

nt
nn ),R(M),R(    

From (5), we have 

 nnn R2s   

In view of (2), the corresponding integer solutions to (1) are given by 

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IJO - INTERNATIONAL JOURNAL OF MATHEMATICS  

(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

 

nn

nnn

nnn

nn
2

n

nnn

nnn

Rt

)21(R6q

)21(R6p

R2z

)13(R2y

)13(R2x























 

where  

 

.
2

~~

103

)
~~(5

,
102

)
~~(3

2

~~
R

nnnn

n

nnnn

n


















 

Pattern 2: 

 Taking  

kR2s             (8) 

in (3), it is written as 

 )k9R10(Rt 2222            (9) 

Let 

 )k9R10( 222                     (10) 

which is satisfied by 

 k,kR 00    

To obtain the other solutions to (10), consider the corresponding pell equation 

 1R10 22   

whose general solution )~,R
~

( nn   is given by  

 

nn

nn

f
2

1~

,g
102

1
R
~






                   (11) 

Applying the lemma of Brahmagupta between ),R( 00   and )~,R
~

( nn  , the general solution 

),R( 1n1n    to (10) is given by 

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IJO - INTERNATIONAL JOURNAL OF MATHEMATICS  

(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

 ,g
102

k
f

2

k
R nn1n                    (12) 

.g
10

k5
f

2

k
nn1n    

From (8) and (9), we get  

 
1n1n1n

1n1n

Rt

Rk2s









  

In view of (2), the corresponding integer solutions to (1) are given by, 

 

)k2)(g
102

k
f

2

k
(3q

),k2)(g
102

k
f

2

k
(3p

),g10f10()gf10(
1040

k
z

),2k6)(gf10(
102

k
y

),2k6)(gf10(
102

k
x

1nnn1n

1nnn1n

nn
2

nn

4

1n

nn1n

nn1n





















 

Jointly with (12), where 

.)10619()10619(g

,)10619()10619(f

1n1n
n

1n1n
n








 

Pattern 3: 

Assume  

22 v9u9R 
                   (13) 

Consider  

)i62)(i62(40 
                   (14) 

Substituting (13), (14) in (3) and using the method of factorizations, we have 

4)v3iu3)(i62()t2is3( 
 

Equating real and imaginary parts, we have 

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IJO - INTERNATIONAL JOURNAL OF MATHEMATICS  

(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

 
)]vu(uv4v3vu18u3[3t

)]vu(uv24v2vu12u2[3s
4442244

4442243





 

In view of (2), we get 

 

)]vu(uv28vvu6u[3*6q

)]vu(uv20v5vu30u6[3*6p

)]vu(uv4v3vu18u3[3z

)]vu(uv32v4vu24u4[3y

)]vu(uv16v8vu48u8[3x

2242247

2242247

2242247

2242244

2242243











 

Pattern 4: 

 It is worth to mention that the integer 40 may also be written as  

222 )(

)],(g2i),(2)][,(g2i),(f2[
40






              (15) 

where )()2(3),(g;2)(3),(f 2222    

Substituting (13) & (15) in (3) and employing the method of factorization, we get 

 )]v,u(iG)v,u(F)][,(g2i),(f2[
)(

3
t2is3

22

4




              (16) 

where  

)vu(uv4)v,u(G

vvu6u)v,u(F
22

4224




 

Equating the real and imaginary part of (16), one obtains 

 

)]v,u(F),(g)v,u(G),(f[
)(

3
t

)]v,u(G),(g2)v,u(F),(f2[
)(

3
s

22

4

22

3











               (17) 

Replacing u by u)( 22   , v by v)( 22   in (16), we have 

 
)]v,u(F),(g)v,u(G),(f[)(3t

)]v,u(G),(g2)v,u(F),(f2[)(3s
3224

3223




               (18) 

Also, from (13), )vu()(9R 22222                   (19) 

From (2) and (18), it is seen that 

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IJO - INTERNATIONAL JOURNAL OF MATHEMATICS  

(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

 

)]v,u(G),(g2)v,u(F),(f2)v,u(F),(g3)v,u(G),(f3[)(3*6q

)]v,u(G),(g2)v,u(F),(f2)v,u(F),(g3)v,u(G),(f3[)(3*6p

)]v,u(F),(g)v,u(G),(f)][v,u(G),(g2)v,u(F),(f2[)(3z

)}]v,u(G)v,u(F){,(g2)}v,u(G)v,u(F){,(f2[)(3y

)}]v,u(G)v,u(F){,(g2)}v,u(G)v,u(F){,(f2[)(3x

3223

3223

6227

3224

3224











 

which satisfy (1) along with (19). 

Pattern 5: 

Rewrite (3) as  

1*R40t4s9 422                     (20) 

Assume the integer 1 on the RHS of (1) as 

  
)qp(

)pq2iqp)(pq2iqp(
1

22

2222




                  (21) 

Substituting (14), (21) in (20), we have 

 

)]q,p,v,u(iQ)q,p,v,u(P[
)qp(

)i62(3
)t2is3(

)]q,p(iK)q,p(J)][v,u(iG)v,u(F[
)qp(

)i62(3
)t2is3(

)v3is3(
)qp(

)pq2iqp(
)i62()t2is3(

22

4

22

4

4

22

22



















 

where  

 

)q,p(J)v,u(G)q,p(K)v,u(F)q,p,v,u(Q

)q,p(K)v,u(G)q,p(J)v,u(F)q,p,v,u(P

)vu(uv4)v,u(G;vvu6u)v,u(F

pq2)q,p(K;qp)q,p(J
224224

22









 

Equating real and imaginary part of the above equation, we have 

 

)]q,p,v,u(Q)q,p,v,u(P3[
)qp(

3
t

)]q,p,v,u(Q6)q,p,v,u(P2[
)qp(

3
s

22

4

22

3











 

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IJO - INTERNATIONAL JOURNAL OF MATHEMATICS  

(ISSN: 2992-4421 )                                                                                   DR. N.THIRUNIRAISELVI1* 

https://ijojournals.com/                                                                    Volume 07 || Issue 05|| May, 2024 || 

 

Replacing u by (p2+ q2) A, v by (p2+ q2) B in the values of s, t and R, we have 

 

)BA()qp(9R

)]q,p,B,A(Q)q,p,B,A(P3[)qp(81t

)]q,p,B,A(Q6)q,p,B,A(P2[)qp(27s

22222

322

322







 

Conclusion: 

It has been shown that higher degree Diophantine equation with multiple variables 
may be solved through strategies like substitution and factorization by reducing it to a lesser 
degree solvable Diophantine equation. One may search for other choices of higher degree 
Diophantine equations with multiple variables for obtaining their respective integer 
solutions. 

References: 

[1] J.Sivasankari, Dr.R.Anbuselvi, “Integral solutions for the Diophantine Equation of Higher   

      Degree with six Unknowns
822366 R)qp(800z3456yx  ”, Advances in   

      Nonlinear Variational Inequalities, Vol 27(1), 2024. 

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