ON THE INTEGRAL SOLUTION OF BINARY QUADRATIC DIOPHANTINE
EQUATION 22 )1()1( yabxxa
aK. Prabhakaran, bA. Hari Ganesh and cK. Mahalakshmi
a,b Assistant Professor and cResearch Scholar
aAnnai Vailankanni Arts and Science College, Thanjvur – 613 007.
b,c Poompuhar College (Autonomous), Melaiyur – 609 107, Nagapattinam (Dt.).
Abstract:
A Diophantine equation is a polynomial equation, usually in two or more unknown, such
that only the integer solutions are studied. In this paper, we have proposed a second order
Diophantine equation 22 )1()1( yabxxa with two unknown for finding its infinite integral
solutions based on Pell’s equation.
Keywords: Diophantine Equation, Integral Solution, Pell’s Equation.
Introduction:
The homogenous and non homogenous binary quadratic Diophantine equations are being
solved in wide range on now days. Particularly, the binary quadratic non-homogeneous equations
representing hyperbolas are studied by many authors for its non - zero integral solutions [2, 3]. In
this paper, we propose a non - homogenous quadratic Diophantine equation for find its non –
zero integral solutions. Moreover, the recurrence relations for the solutions of the proposed non –
homogenous quadratic Diophantine equation are also to be presented in this paper.
Method of Analysis:
The non - homogenous quadratic Diophantine equation representing the hyperbola to be
solved for its non - zero distinct integral solution is
22 )1()1( yabxxa -------------------- (1)
The equation (1) is to be solved through the following two different cases.
Case I:
The first among the two is
22 )1()1( yabxxa -------------------- (2)
Treating (2) as quadratic in x and solving for x , we have
0)1()1( 22 yabxxa
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To eliminate the square root on the right hand side of the solution of the above equation,
assume
2222)1(4)1(2 byabxa
Now the above equation is in the form of Pellian Equation as follows:
NDyX 22
Where bxaX )1(2 , )1(4 2 aD , 2bN , 1 , ),1( aZmamb
The general solution ),( nn yx of equation (2) is given by
)1(2
)1()1(4)1(4)1(4)1(42
10
2
0
2
000
2
0
2
00
a
amyaXayXyaXayX
x
nn
n
0
2
0
2
000
2
0
2
002
)1(4)1(4)1(4)1(4)1(42
1yaXayXyaXayX
ay
nn
n
Where )1(4 2
00 ayX is the fundamental solution of 1)1(4)1(2 222 yabxa
and )1(4 2
00 ayX is the fundamental solution of 2222)1(4)1(2 byabxa
For the sake of simplicity a few solutions of (2) for 2a are presented through the following
example.
Now equation (2) becomes,
223 ybxx ------------------------------ (3)
Treating the above equation as quadratic in x and find its solution in the form of following
equation
222126 bybx
It is in the form of Pellian equation as
NDyX 22
where bxX 6 ,2bN
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The general solution ),( nn yx of equation (3) as follows:
6
3121212122
100000000 myXyXyXyX
x
nn
n
00000000 12121212
122
1yXyXyXyXy
nn
n
Where 1200 yX and 1200 yX are fundamental solutions of 1126 22 ybx
Table 1
The
Values
of n
For m = 1, b = 3, N = 9,
the solutions
),(),( , mnmn yx
For m = 2, b = 6, N = 36
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(1, 0)
(4, 6)
(49, 84)
(676, 1170)
(9409, 16296)
(131044, 226974)
(2, 0)
(8, 12)
(98, 168)
(1352, 2340)
(18818, 32592)
(262088, 453948)
(3, 3)
(27, 45)
(363, 627)
(5043, 8733)
(70227, 121635)
(978123, 1694157)
Table 2
The
Values
of n
For m = 3, b = 9, N = 81,
the solutions
),(),( , mnmn yx
For m = 4, b = 12 , N = 144,
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(3, 0)
(12, 18)
(147, 252)
(2028, 3510)
(28227, 48888)
(393132, 680922)
(4, 0)
(16, 24)
(196, 336)
(2704, 4680)
(37636, 65184)
(524176, 907896)
(6, 6)
(54, 90)
(726, 1254)
(10086, 17466)
(140454, 243270)
(1956246, 3388314)
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Table 3
The
Values
of n
For m = 5, b = 15, N = 225,
the solutions
),(),( , mnmn yx
For m = 6, b = 18 , N = 324,
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(5, 0)
(20, 30)
(245, 420)
(3380, 5850)
(47045, 81480)
(655220, 1134870)
(6, 0)
(24, 36)
(294, 504)
(4056, 7020)
(56454, 97776)
(786264, 1361844)
(9, 9)
(81, 135)
(1089, 1881)
(15129, 26199)
(210681, 364905)
(2934369, 5082471)
Table 4
The
Values
of n
For m = 7, b = 21, N = 441,
the solutions
),(),( , mnmn yx
For m = 8, b = 24 , N = 576,
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(7, 0)
(28, 42)
(343, 588)
(4732, 8190)
(65863, 114072)
(917308, 1588818)
(8, 0)
(32, 48)
(392, 672)
(5408, 9360)
(75272, 130368)
(1048352, 1815792)
(12, 12)
(108, 180)
(1452, 2508)
(20172, 34932)
(280908, 486540)
(3912492, 6776628)
Further the solutions satisfy the following recurrence relation:
(a) Recurrence relations for solution ),(),( , mnmn yx and .2 ,1,, ),(),( iyx
ii mnmn among the
different values of b
(i) 0214 )12,1()12,()12,1( bxxx knknkn where 0n and ,.......2 ,1 ,0k
(ii) 0214 )22,1()22,())22(,1( 1 bxxx knknkn where 0n and ,.......2 ,1 ,0k
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(b) Recurrence relations for solution ),(),( , mnmn yx and .2 ,1,, ),(),( iyx
ii mnmn
(i) 02 )32,())22(,()12,( 1 knknkn xxx where ,.......2 ,1 ,0k
(ii) ,0)12,())22(,()1,( 1 knknn xxx where ,.......2 ,1 k
(c) Recurrence relations for solution .2 ,1,, ),(),( iyxii mnmn
(i) myyxx knknknkn ))2(,())2(,())2(,())2(,( 2121 where ,.......2 ,1 k
Case II:
The second among the two is
22 )1()1( yabxxa -------------------------- (4)
Treating (4) as a quadratic in x and solving for x , we have
0)1()1( 22 yabxxa
To eliminate the square root on the right hand side of the solution of the above equation,
assume
2222)1(4)1(2 byabxa
Now the above equation is in the form of Pellian Equation as follows:
NDyX 22
Where bxaX )1(2 , )1(4 2 aD , 2bN , 1 , ),1( aZmamb
The general solution ),( nn yx of equation (4) is given by
)1(2
)1()1(4)1(4)1(4)1(42
10
2
0
2
000
2
0
2
00
a
amyaXayXyaXayX
x
nn
n
0
2
0
2
000
2
0
2
002
)1(4)1(4)1(4)1(4)1(42
1yaXayXyaXayX
ay
nn
n
Where )1(4 2
00 ayX is the fundamental solution of 1)1(4)1(2 222 yabxa
and )1(4 2
00 ayX is the fundamental solution of 2222)1(4)1(2 byabxa
For the sake of simplicity a few solutions of (2) for 2a are presented through the following
example.
International Journal of Applied Engineering Research ISSN 0973-4562 Volume 14, Number 4, 2019 (Special Issue) © Research India Publications. http://www.ripublication.com
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Now equation (4) becomes,
22 3ybxx
Treating the above equation as quadratic in x and find its solution in the form of following
equation
222122 bybx
It is in the form of Pellian equation as
NDyX 22
Where bxX 2 ,2bN
The general solution ),( nn yx of equation (3) as follows:
2
121212122
100000000 myXyXyXyX
x
nn
n
00000000 12121212
122
1yXyXyXyXy
nn
n
Where 1200 yX and 1200 yX are fundamental solutions of 1122 22 ybx
Table 5
The
Values
of n
For m = 1, b = 1, N = 1,
the solutions
),(),( , mnmn yx
For m = 2, b = 2, N = 4,
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(1, 0)
(4, 2)
(49, 28)
(676, 390)
(9409, 5432)
(131044, 75658)
(2, 0)
(8, 4)
(98, 56)
(1352, 780)
(18818, 10864)
(262088, 151316)
(3, 1)
(27, 15)
(363, 209)
(5043, 2911)
(70227, 40545)
(978123, 564719)
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Table 6
The
Values
of n
For m = 3, b = 3, N = 9,
the solutions
),(),( , mnmn yx
For m = 4, b = 4 , N = 16,
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(3, 0)
(12, 6)
(147, 84)
(2028, 1170)
(28227, 16296)
(393132, 226974)
(4, 0)
(16, 8)
(196, 112)
(2704, 1560)
(37636, 21728)
(524176, 302632)
(6, 2)
(54, 30)
(726, 418)
(10086, 5822)
(140454, 81090)
(1956246, 1129438)
Table 7
The
Values
of n
For m = 5, b = 5, N = 25,
the solutions
),(),( , mnmn yx
For m = 6, b = 6 , N = 36,
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(5, 0)
(20, 10)
(245, 140)
(3380, 1950)
(47045, 27160)
(655220, 378290)
(6, 0)
(24, 12)
(294, 168)
(4056, 2340)
(56454, 32592)
(786264, 453948)
(9, 3)
(81, 45)
(1089, 627 )
(15129, 8733)
(210681, 121635)
(2934369, 1694157)
Table 8
The
Values
of n
For m = 7, b = 7, N = 49,
the solutions
),(),( , mnmn yx
For m = 8, b = 8 , N = 64,
the solutions
.2 ,1,, ),(),( iyxii mnmn
0
1
2
3
4
5
(7, 0)
(28, 14)
(343, 196)
(4732, 2730)
(65863, 38024)
(917308, 529606)
(8, 0)
(32, 16)
(392, 224)
(5408, 3120)
(75272, 43456)
(1048352, 605264)
(12, 4)
(108, 60)
(1452, 836)
(20172, 11644)
(280908, 162180)
(3912492, 2258876)
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Further the solutions satisfy the following recurrence relation:
(c) Recurrence relations for solution ),(),( , mnmn yx and .2 ,1,, ),(),( iyx
ii mnmn among the
different values of b
(i) 0614 )12,1()12,()12,1( bxxx knknkn where 0n and ,.......2 ,1 ,0k
(ii) 0614 )22,1()22,())22(,1( 1 bxxx knknkn where 0n and ,.......2 ,1 ,0k
(d) Recurrence relations for solution ),(),( , mnmn yx and .2 ,1,, ),(),( iyx
ii mnmn
(i) 02 )32,())22(,()12,( 1 knknkn xxx where ,.......2 ,1 ,0k
(i) ,0)12,())22(,()1,( 1 knknn xxx where ,.......2 ,1 k
(c) Recurrence relations for solution .2 ,1,, ),(),( iyxii mnmn
(i) 0))2(,())2(,())2(,())2(,( 2121 knknknkn yyxx where ,.......2 ,1 k
Reference:
1. Carmichael, R.D., The Theory of Numbers and Diophantine Analysis, Dover
Publications, New York, 1950.
2. Gopalan, M.A., Vidhyalakahmi, S., and Devibala, S., On The Diophantine Equation
143 2 xyx , Acta Ciencia Indica,Vol. XXXIII M.No2, pp.645-646, 2007.
3. Gopalan, M.A., Devibala, S., & Vidhyalakahmi, R., Integral points on the hyperbola
532 22 yx , American Journal of Applied Mathematics and Mathematical
Sciences,Vol. I, No. 1, pp.1- 4, 2012.
4. Mollion, RA., All Solutions of the Diophatine Equations, nDYX 22, Far East
Journal of Mathematical Sciences,Vol. III, pp. 257 – 293, 1998.
5. Mordell, L.J., Diophantine Equations, Acadamic Press, London, 1969.
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