The document discusses four different methods to find non-zero, non-negative solutions for the cubic diophantine equation x^2 + y^2 - xy = 39z^3. It highlights interesting relationships among special numbers and the solutions derived from these methods. The study emphasizes the significance of number theory and offers theoretical implications for further mathematical exploration.

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.
4. REFERENCES
[1]. Dickson, L.E., History of theory of numbers, Vol.11,
Chelsea publishing company, New –York (1952)
[2]. Mordell, L.J., Diophantine equation, Academic press,
London (1969) Journal of Science and
Research,Vol (3) Issue 12, 20-22 (December -14)
[3]. Jayakumar. P, Sangeetha, K “Lattice points on the cone
x2
+ 9y2
=50z2
” International Journal of Science and
Research, Vol(3), Issue 12, 20-22 (December 2014)
[4]. Jayakumar P, KanagaDhurga, C,” On Quadratic
Diopphantine equation x2
+16y2
= 20z2
” Galois J.
Maths, 1(1) (2014), 17-23.
[5]. Jayakumar. P, KanagaDhurga. C, “Lattice points on the
cone x2
+9y2
=50z2
” Diophantus J. Math, 3(2) (2014),
61-71
[6]. Jayakumar. P, Prabha. S “On Ternary
QuadraticDiophantine equationx2
+15y2
=14z2
”
ArchimedesJ. Math., 4(3) (2014), 159-164.
[7]. Jayakumar, P, Meena, J “Integral solutions of the
Ternary Quadratic Diophantine equation : x2
+7y2
=
16z2
” International Journal of Science and Technology,
Vol.4, Issue 4, 1-4, Dec 2014.
[8]. Jayakumar. P, Shankarakalidoss, G “Lattice points on
Homogenous cone x2
+9y2
=50z2
” International
journal of Science and Research, Vol (4), Issue 1,
2053-2055, January -2015.
[9]. Jayakumar. P, Shankarakalidoss. G “Integral points on
the Homogenous cone x2
+ y2
=10z2
International
Journalfor Scientific Research and Development, Vol
(2), Issue 11, 234-235, January -2015
[10].Jayakumar.P, Prabha.S “Integral points on the conex2
+25y2
=17z2
” International Journal of Science and
Research Vol(4), Issue 1, 2050-2052, January-2015.
[11].Jayakumar.P, Prabha. S, “Lattice points on the cone x2
+ 9y2
= 26z2
“International Journal of
ScienceandResearch Vol (4), Issue 1, 2050-2052,
January -2015
[12].Jayakumar. P, Sangeetha. K, “Integral solution of the
Homogeneous Biquadratic Diophantine equation with
six unknowns: (x3
-y3
)z = (W2
– P2
)R4
“International
Journal of Science and Research, Vol(3), Issue 12,
1021-1023 (December-2014)