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Applied Mathematical Sciences, Vol. 3, 2009, no. 3, 115 - 118
A Note on the Pell Equation x2 Dy2 = 4
Liqun Tao
School of Mathematical Sciences, University College DublinBelfield, Dublin 4, Ireland
Abstract
Let D be any square-free positive integer. In this note we give aproof of two conjectures in [1] on the recurrence relations of integersolutions of the Pell equation x2 Dy2 = 4.
Mathematics Subject Classifications: Primary 11E15
Keywords: Pell equation
1. Introduction
Let D be any square-free positive integer. In [1], A. Tekcan consideredthe Pell equation x2 Dy2 = 4 and proved some interesting results. Healso proposed two conjectures on the recurrence relations of integer (actuallypositive) solutions in terms of the fundamental solution of the equation.
The aim of this note is just to show that these conjectures hold and followimmediately from the main results shown in [1], which, together with theconjectures, we record as follows.
Lemma 1.1. Let (x1, y1) be the fundamental solution of the Pell equationx2 Dy2 = 4, and let
unvn
=
x1 Dy1y1 x1
n10
(1)
for n 1. Then all the positive integer solutions of the Pell equation x2Dy2 =
4 are given by (xn, yn) = (
un
2n1 ,
vn
2n1 ).
Lemma 1.2. Let (x1, y1) be the fundamental solution of the negative Pellequation x2 Dy2 = 4, and let
unvn
=
x1 Dy1y1 x1
n10
(2)
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116 Liqun Tao
for n 0. Then all the positive integer solutions of the Pell equation x2Dy2 =4 are given by (x2n+1, y2n+1) = (
u2n+122n
,v2n+1
22n).
Conjecture 1.3. Let (x1, y1) be the fundamental solution of the Pell equationx2 Dy2 = 4, then (xn, yn) satisfy the following recurrence relations
xn = (x1 1)(xn1 + xn2) xn3yn = (x1 1)(yn1 + yn2) yn3
for n 4.
Conjecture 1.4. Let (x1, y1) be the fundamental solution of the Pell equationx2 Dy2 = 4, then (x2n+1, y2n+1) satisfy the following recurrence relations
x2n+1 = (x21 + 1)(x2n1 + x2n3) x2n5
y2n+1 = (x21 + 1)(y2n1 + y2n3) y2n5
for n 3.
2. Proof of conjectures
Proof of Conjecture 1.3. Since (xn, yn) = (un
2n1 ,vn
2n1 ), we need only to showthat
unvn
=
2(x1 1)(un1 + 2un2) 8un32(x1 1)(vn1 + 2vn2) 8vn3
= 2(x1 1)
un1vn1
+ 2
un1vn2
8
un3vn3
In view of (1), it suffices to show
x1 Dy1y1 x1
3 10
=
2(x1 1)
x1 Dy1y1 x1
2
+ 2
x1 Dy1y1 x1
8
1 00 1
10
It is easy to see that
x1 Dy1y1 x1
2= x21 + Dy21 2Dx1y1
2x1y1 x21 + Dy21
and
x1 Dy1y1 x1
3=
x31 + 3Dx1y
21 3Dx
21y1 + D
2y213x21y1 + Dy
31 x
31 + 3Dx1y
21
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Pell equation 117
Using x21 Dy21 = 4, we have
2(x1 1)
x1 Dy1y1 x1
2+ 2
x1 Dy1y1 x1
8
1 00 1
10
= 2(x1 1)
x
2
1 + Dy2
12x1y1
+ 2
x1y1
8
10
=
2(x1 1)(x
21 + Dy
21 + 2x1) 8
2(x1 1)(2x1y1 + 2y1)
=
2(x1 1)(x
21 + (x
21 4) + 2x1) 8
2(x1 1)(2x1y1 + 2y1)
=
4(x1 1)(x
21 + x1 2) 8
2(x1 1)(2x1y1 + 2y1)
=
4x1(x
21 3)
4(x21 1)y1
andx1 Dy1y1 x1
310
=
x31 + 3Dx1y
21
3x21y1 + Dy31
=
x31 + 3x1(x
21 4)
3x21y1 + y1(x21 4)
=
4x1(x
21 3)
4(x21 1)y1
This completes the proof. 2
Proof of Conjecture 1.4. Since (x2n+1, y2n+1) = (u2n+1
22n ,v2n+1
22n ), we need onlyto show that
u2n+1v2n+1
=
(x21 + 1)(u2n1 + u2n3) u2n5(x21 + 1)(v2n1 + v2n3) v2n5
= (x21 + 1)
u2n1v2n1
+
u2n3v2n3
u2n5v2n5
In view of (2), it suffices to show
x1 Dy1y1 x1
6 10
=
4(x21
+ 1)
x1 Dy1y1 x1
4
+ 4
x1 Dy1y1 x1
2
64
1 00 1
10
It is easy to see that
x1 Dy1y1 x1
2=
x21 + Dy
21 2Dx1y1
2x1y1 x21 + Dy
21
x1 Dy1y1 x1
4=
(x21 + Dy
21)
2 + 4Dx21y21 4Dx1y1(x
21 + Dy
21)
4x1y1(x21 + Dy
21) (x
21 + Dy
21)
2 + 4Dx21y21
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118 Liqun Tao
and
x1 Dy1y1 x1
6=
(x2
1+ Dy2
1)3 + 12Dx2
1y21
(x21
+ Dy21
) 6Dx1y1(x2
1+ Dy2
1)2 + 8D2x3
1y31
8Dx31y31
+ 6x1y1(x2
1+ Dy2
1)2 (x2
1+ Dy2
1)3 + 12Dx2
1y21
(x1 + Dy2
1)
Using x21 Dy21 = 4, we have
4(x21 + 1)
x1 Dy1y1 x1
4+ 4
x1 Dy1y1 x1
2 64
1 00 1
10
= 4(x21 + 1)
(x21 + Dy
21)
2 + 4Dx21y21
4x1y1(x21 + Dy
21)
+ 4
x21 + Dy
21
2x1y1
64
10
=
4(x21 + 1)((x
21 + Dy
21)
2 + 4Dx21y21 + 4(x
21 + Dy
21) 64)
4(x21 + 1)(4x1y1(x21 + Dy
21) + 4 2x1y1)
=
4(x21 + 1)((x
21 + x
21 + 4)
2 + 4x21(x21 + 4) + 4(x
21 + x
21 + 4)) 64
4(x21 + 1)(4x1y1(x21 + x
21 + 4) + 8x1y1)
= 32((x21 + 1)2(x21 + 4) 2)
32x1y1(x21 + 1)(x21 + 3)
andx1 Dy1y1 x1
610
=
(x21 + Dy
21)
3 + 12Dx21y21(x
21 + Dy
21)
8Dx31y31 + 6x1y1(x
21 + Dy
21)
2
=
(x21 + x
21 + 4)
3 + 12x21(x21 + 4)(x
21 + x
21 + 4)
8x31y1(x21 + 4) + 6x1y1(x
21 + x
21 + 4)
2
=
32((x21 + 1)
2(x21 + 4) 2)32x1y1(x
21 + 1)(x
21 + 3)
This completes the proof. 2
References
[1] Ahmet Tekcan, The Pell Equation x2 Dy2 = 4, Applied MathematicalSciences, 1(8), 2007, 363-369
Received: June 8, 2008
