Abstract Four different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x2 + y2 – xy = 39z2 are obtained. Some interesting relations among the special numbers and the solutions are exposed. Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers

1. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 05 Issue: 03 | Mar-2016, Available @ http://www.ijret.org 499
ON CUBIC DIOPHANTINE EQUATION
322
39 zxyyx 
P.Jayakumar1
, J.Meena2
1
Professor of Mathematics, PeriyarManiammai University, Vallam, Thanajvur -613 403, TN, India.
2
Ph.D. Scholar, Assistant Professor of Mathematics, A.V.V.M. Sri Pushpam College (Autonomous), Poondi -613 503,
Thanajvur, TN, India.
Abstract
Four different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x2
+ y2
– xy =
39z2
are obtained. Some interesting relations among the special numbers and the solutions are exposed.
Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
--------------------------------------------------------------------***----------------------------------------------------------------------
2010 Mathematics subject classification: 11D25
Symbols used:
tm,, n= )]4()2([
2
1
 mmn
pn
m
= )]5()2(3[
6
1 32
 mnmnn
Gn= 2n-1
1. INTRODUCTION
The number theory is the king of Mathematics. In particular,
the Diophantine equations have a blend of attracted
interesting problems. For a broad review of variety of
problems, one may try to see [3-12]. In this work, we are
observed another interesting four different methods of the
non-zero non-negative solutions the non - homogeneous
cubic Diophantine equation x2
+ y2
– xy = 39z3
Further,
some elegant properties among the special numbers and the
solutions are observed.
2. DESCRIPTION OF METHOD
Consider the cubic Diophantine equation
322
39 zxyyx  (1)
Take the linear transformations x = u +v,
y = u – v, u  v 0) (2)
Using (1) in (2), it gives us u2
+3v2
=39z3
(3)
If we take z = z (a, b) = a2
+3b2
= (a +i 3 b)
(a -i 3 b), (4)
where a and b non-zero non- negative different integers,
then we solve(1) through dissimilar method of solutions of
(1) which are furnished below.
2.1 Method: I
We can write 39 as
39 = (6 +i 3 ) (6-i 3 ) (5)
Using (4) and (5) in (3) and this gives
(u +i 3 v) (u -i 3 v) = (6 +i 3 ) (6-i 3 )
(a+i 3 b)3
(a-i 3 b)3
It gives us
(u+ i 3 v) = (6 + i 3 ) (a + i 3 b)3
(6)
(u - i 3 v) = (6 - i 3 ) (a - i 3 b)3
(7)
Comparing both sides of (6) or (7), we obtain
u = u(a, b) = 6a3
– 36ab2
– 6a2
b + 9b3
v = v(a, b) =a3
– 6ab2
+12a2
b – 18b3
In sight of (2), the solutions x, y are found to be
x = x (a, b) = 7a3
– 42ab2
+ 6a2
b – 9b3
(8)
y = y (a, b) = 5a3
– 30ab2
–18a2
b +27b3
(9)
Hence (4), (8) and (9) gives us two parametric the non-zero
different integral values of (1).
Observations:
1. y(b, b) – x (b, b) – 136Pb
5
+ 68t4, b= 0
2. z(a, a) -4t4, a= 0
3. x(a, 1) – 14pa
5
+ t4,a + G21a + 8 = 0
4. x(1,a) + 18Pa
5
+ 33t4,a – G31a 0 (Mod 2)
5. y(1, a) –54Pa
5
+ 57t4,a+G9a 0(Mod 3).
6. x (3b, 3b) +10 Pb
5
+ 5t4,b = 0.
7. y (2a, 2a) + 1376 Pb
5
– 688 t4,a = 0
8. z (a, 2a) – 13t4,a = 0.
2.2method: II
We also write 39 as
39 = (9 5 3)(9 5 3)
4
i i  (10)

2. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 05 Issue: 03 | Mar-2016, Available @ http://www.ijret.org 500
Using (4) and (10) in (3) and we obtain
(u +i 3 v)(u - i 3 v) = [(9+i5 3 )(9 –i5 3 )
(a + i 3 b)3
(a -i 3 b)3
]
Comparing both sides of above, we are found to be
(u +i 3 v) = 1
2
[(9 +i5 3 ) (a + i 3 b)3
] (11)
(u -i 3 v) = 1
2
[(9 -i5 3 ) (a - i 3 b)3
] (12)
Comparing both sides of (11) or (12), we obtain
u = u(a, b) = 1
2
[9 a3
– 54ab2
– 30a2
b + 45b3
]
v = v(a, b) = 1
2
[5 a3
– 30ab2
+ 18a2
b - 27b3
]
In true of (2), the values of x, y are given by
x = x (a, b) = 7a3
– 42ab2
– 6a2
b + 9b3
(13)
y = y (a, b) = 2a3
– 12ab2
– 24a2
b + 36b3
(14)
Hence (4), (13) and (14) gives us two parametric the non-
zero different integral values of (1).
Observations:
1.y(b, b) – x (b, b) + 88pb
5
- 44t4, b = 0
2. 2x (b, b) – 9y (b, b) + 164Pb
5
- 82t4, b = 0
3. x(a, 1) – 14Pa
5
+ 13t4,a + G21a0 (Mod 5)
4. y(1, b) -72Pb
5
+ t52,b+23t4,b 0 (Mod 2)
5. z(a, 2a) – t28,a - G6n– 1 = 0
6. x (a, 3a) +580 Pb
5
– 290 t4,a =0
7. z (4a, 4a) – 64 t4a, = 0
8. y (a, -a) - 4 Pb
5
+ 2t4,a = 0.
2.3 Method: III
Let us take (3) as u2
+ 3v2
= 39z3
* 1 (15)
Consider 1 as 1 = (1 3)(1 3)
4
i i  (16)
Using (4), (5) and (16) in (15) and it gives us
(u+i 3 v) (u- i 3 v) =
1
4
[(1+i 3 ) (1–i 3 )
(6+i 3 ) (6-i 3 )(a+i 3 b)3
(a–i 3 b)3
]
It gives us
(u +i 3 v) = 1
2
[(1+i 3 )(6 + i 3 )(a+ i 3 b)
(17)
(u -i 3 v) = 1
2
[ (1 -i 3 ) (6 -i 3 )(a - i 3 b)3
] (18)
Comparing both sides of (17) or (18), we obtain
u = u(a, b) = 1
2
[3 a3
– 18ab2
– 42a2
b + 63b3
]
v = v(a, b) = 1
2
[7 a3
– 42ab2
+ 6a2
b - 9b3
]
In sight of (2), the values of x, y are given by
x = x (a, b) = 5a3
– 30ab2
– 18a2
b + 27b3
(19)
y = y (a, b) = -2ab3
+ 12ab2
– 24a2
b + 36b3
(20)
Observations:
1. y(a, a) – x (a, a) - 76a2
+ 38t4, a = 0
2. x (1, b) + 57t4,b - 54Pb
5
– G9b 0 (Mod 2)
3. 2x(b, b) + 5y(b, b) – 156Pa
5
+ 78t4,b = 0
4. z (b, b) - 4t4,b = 0
5. y(1, b) – 72Pb
5
+ G12b + 48t4,b + 1 = 0.
6. x(b, 1) = 10 Pb
5
+ 23t4,b + G15,b 0 (Mod 2)
7. z(6a, 6a) – 14t t4,a = 0.
8. x(2b, 3b) – 3 Pb
5
+ 3 t4,b = 0.
2.4 Method: IV
Replace (16) by 1= (1 4 3)(1 4 3)
49
i i  (21)
Using (4), (10) and (21) in (15) and this gives us
(u+i 3 v) (u- i 3 v) 1
4 49
[(1+i4 3 )(1–i 3 )(9-i5 3 )
(9+i5 3 ) (a+i 3 b)3
(a–i 3 b)3
]
This gives us
(u +i 3 v) = 1
14
[(1 +i4 3 )(9+ i5 3 )(a + i 3 b)3
] (22)
(u -i 3 v) = 1
14
[(1 –i4 3 ) (9 –i5 3 ) (a - i 3 b)3
](23)
Comparing both sides of (22) or (23), it gives us
u = u(a, b) = 1
14
[- 51a3
+ 306ab2
– 246a2
b + 369b3
]
v = v(a, b) = 1
14
[41a3
– 246ab2
– 102a2
b + 153b3
]
In true of (2), the values of x, y are given by
x = x (a, b)= 1
7
[-5a3
+ 30ab2
– 174a2
b +261b3
(24)
y = y(a, b) = 1
7
[- 46a3
+ 276ab2
–72a2
b 108b3
] (25)

3. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 05 Issue: 03 | Mar-2016, Available @ http://www.ijret.org 501
Since our intension is to find integer solutions, taking a as
7a and b as 7b in (4), (24) and (25), the related
parametricinteger values of(1) are found as
x = x (a, b) = - 245a3
+ 1470 ab2
– 8526a2
b + 12789b3
y = y (a, b) = 2254a3
+ 13524ab2
– 3528a2
b + 5292b3
z = z (a, b) = 49a2
+ 147b2
Observations:
1.x(a, a) – y(a, a) + 24108Pa
5
– 12054 t4,a = 0
2. z (a, a) – t394,a –G97a – Pa – 1 + t4,a = 0
3. x (1, b) + 244 + G4263b – 5578Pb
5
+ 1319t4,b = 0
4. y (a, 1) –2508Pa
5
+ 5782t4,a– G6762a – 5293 = 0
5. z(a, 1) – 49t4,a  0 (Mod 7)
6. x(b, 1) + 490 Pb
5
+ 8381 t4,b – G735,b 0 (Mod 2)
7. y(b, 1) – 4508 Pb
5
+ 5782 t4,b – G6762,b – 5293 =0.
8. z(3a, 3a) – 864 t4,a = 0
3. CONCLUSION
Here we observed various process of determining infinitely
a lot of non-zero different integer values to the cubic
Diophantine equation x2
+y2
–xy =39z3
.One may try to find
non-negative integer solutions of the above equations
together with their similar observations.
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