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Applied Mathematical Sciences, Vol. 1, 2007, no. 8, 363 - 369
The Pell Equation x2 Dy2 = 4
Ahmet Tekcan
Uludag University, Faculty of ScienceDepartment of Mathematics
Gorukle 16059, Bursa, [email protected]
Abstract. Let D = 1 be a positive non-square integer. In this paper, weobtain some formulas for the integer solutions of the Pell equation x2Dy2 =4.
Mathematics Subject Classification: Primary 11E15; Secondary 11E18,11E25
Keywords: Pell equation, solutions of the Pell equation
1. Introduction.
Suppose that D = 1 be any positive non-square integer and N be any fixedinteger. The equation
x
2
Dy
2
= N(1)is known as Pell equation (x2Dy2 = N is the Pell equation and x2 Dy2 =N is the negative Pell equation) and is named after John Pell (1611-1685),a mathematician who searched for integer solutions to equations of this typein the seventeenth century. Ironically, Pell was not the first to work on thisproblem, nor did he contribute to our knowledge for solving it. Euler (1707-1783), who brought us the -function, accidentally named the equation afterPell, and the name stuck.
The Pell equation in (1) has infinitely many integer solutions (xn, yn) forn 1. The first non-trivial solution (x1, y1) of this equation, from which allothers are easily computed, can be found using, e.g., the cyclic method [1],known in India in the 12 th century, or using the slightly less efficient butmore regular English method [1](17th century). There are other methodsto compute this so-called fundamental solution, some of which are based ona continued fraction expansion of the square root of D. Many authors suchas Kaplan and Williams [2], Lenstra [3], Matthews [4], Mollin, Poorten and
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364 Ahmet Tekcan
Williams [5], Smarandache [6], Stevenhagen [7], Tekcan [8], and the othersconsider some specific Pell equations and their integer solutions.
For N = 1, the Pell equation
x2 Dy2 = 1(2)
is known the classical Pell equation. The Pell equation x2Dy2 = 1, was first
studied by Brahmagupta (598-670) and Bhaskara (1114-1185). Its completetheory was worked out by Lagrange (1736-1813), not Pell. It is often said thatEuler (1707-1783) mistakenly attributed Brounckers (1620-1684) work on thisequation to Pell. However the equation appears in a book by Rahn (1622-1676)which was certainly written with Pells help: some say entirely written by Pell.Perhaps Euler knew what he was doing in naming the equation.
2. The Pell Equation x2 Dy2 = 4.
In this section, we consider the solutions of the Pell equation in (1) forN = 4, i.e. x2 Dy2 = 4.
Theorem 2.1. Let (x1, y1) be the fundamental solution of the Pell equationx2 Dy2 = 4, and letunvn
=
x1 Dy1y1 x1
n
10
(3)
for n 1. Then the integer solutions of the Pell equation x2 Dy2 = 4 are(xn, yn), where
(xn, yn) =
un
2n1,
vn
2n1
.(4)
Proof. We prove the theorem by induction on n. For n = 1, we have from (3),(u1, v1) = (x1, y1) and hence x
21 Dy
21 = 4 since (x1, y1) is the fundamental
solution. With the assumption that the Pell equation x2
Dy2
= 4 is satisfiedfor (xn1, yn1), i.e.
x2n1 Dy
2
n1 =u2n1 Dv
2n1
22n4= 4.(5)
Now that the Pell equation x2 Dy2 = 4 is also satisfied for (xn, yn) will beshown. By (3), a straightforward calculation shows that
unvn
=
x1 Dy1y1 x1
n
10
= x1 Dy1y1 x1
x1 Dy1y1 x1
n1
1
0
=
x1 Dy1y1 x1
un1vn1
=
x1un1 + Dy1vn1
y1un1 + x1vn1
.(6)
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The Pell Equation 365
Hence
x2nDy2
n=
u2nDv2
n
22n2
=(x1un1 + Dy1vn1)
2 D(y1un1 + x1vn1)2
22n2
= x21u2n1 + 2x1un1Dy1vn1 + D2y21v2n1
22n2
D(y21u
2n1 + 2y1un1x1vn1 + x
21v
2n1)
22n2
=x21(u
2n1 Dv
2n1)Dy
21(u
2n1 Dv
2n1)
22n2
=(x21 Dy
21)(u
2n1 Dv
2n1)
22n2
=u2n1 Dv
2n1
22n2.
Applying (4), it is easily seen that u2n1 Dv
2n1 = 4.2
2n4 = 22n2. Hence weconclude that
x2nDy2
n=
u2nDv2
n
22n2
=(x1un1 + Dy1vn1)
2 D(y1un1 + x1vn1)2
22n2
=x21u
2n1 + 2x1un1Dy1vn1 + D
2y21v2n1
22n2
D(y21u2n1 + 2y1un1x1vn1 + x
21v
2n1)
22n2
=x21(u
2n1 Dv
2n1)Dy
21(u
2n1 Dv
2n1)
22n2
=(x21 Dy
21)(u
2n1 Dv
2n1)
22n2
=u2n1 Dv
2n1
22n2
=22n2
22n2
= 1.
Therefore (xn, yn) is also a solution of the Pell equation x2Dy2 = 4. Since n
is arbitrary, we get all integer solutions of the Pell equation x2Dy2 = 4.
From Theorem 2.1, the following corollary and conjecture can be given.
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366 Ahmet Tekcan
Corollary 2.2. If (x1, y1) is the fundamental solution of the Pell equationx2 Dy2 = 4, then
xn =x1xn1 + Dy1yn1
2, yn =
y1xn1 + x1yn12
and
xn xn1yn yn1 = 2y1
for n 2.
Proof. We know that un = x1un1+Dy1vn1, vn = y1un1+x1vn1 by (6), andun = 2
n1xn, vn = 2n1yn by (4). So un1 = 2
n2xn1 and vn1 = 2n2yn1.
Consequently, we get
un = x1un1 + Dy1vn1
2n1xn = x12n2xn1 + Dy12
n2yn1
2n1xn = 2n2(x1xn1 + Dy1yn1)
xn = 2n2
(x1xn1 + Dy1yn1)2n1
xn =x1xn1 + Dy1yn1
2
vn = y1un1 + x1vn1
2n1yn = y12n2xn1 + x12
n2yn1
2n1yn = 2n2(y1xn1 + x1yn1)
yn =2n2(y1xn1 + x1yn1)
2n1
yn =y1xn1 + x1yn1
2
and hence xn xn1yn yn1 = xnyn1 ynxn1
=
x1xn1 + Dy1yn1
2
yn1
y1xn1 + x1yn1
2
xn1
=x1xn1yn1 + Dy1y
2n1 y1x
2n1 x1xn1yn1
2
=y1(x
2n1 Dy
2n1)
2=
4y12
= 2y1.
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The Pell Equation 367
Conjecture 2.3. If (x1, y1) is the fundamental solution of the Pell equationx2 Dy2 = 4, then (xn, yn) satisfy the following recurrence relations
xn = (x1 1) (xn1 + xn2) xn3
yn = (x1 1)(yn1 + yn2) yn3
for n 4
3. The Negative Pell Equation x2 Dy2 = 4.
Theorem 3.1. If(x1, y1) is the fundamental solution of the negative Pell equa-tion x2 Dy2 = 4, then the other solutions are (x2n+1, y2n+1), where
(x2n+1, y2n+1) =u2n+1
22n,
v2n+1
22n
(7)
for n 0.
Proof. Theorem 3.1 can be proved as in the same way that Theorem 2.1 was
proved.
From Theorem 3.1, the following corollary and conjecture can be given.
Corollary 3.2. If (x1, y1) is the fundamental solution of the negative Pellequation x2 Dy2 = 4, then
x2n+1 =(x21 + Dy
21)x2n1 + 2Dx1y1y2n1
4,
y2n+1 =2x1y1x2n1 + (x
21 + Dy
21)y2n1
4
and x2n+1 x2n1y2n+1 y2n1 = 2x1y1
for n 1.
Proof. By (7), it is easily seen that
u2n+1v2n+1
=
x1 Dy1y1 x1
2n+110
=
x1 Dy1y1 x1
2 x1 Dy1y1 x1
2n110
=
x1 Dy1y1 x1
2 u2n1v2n1
=
(x21 + Dy
21)u2n1 + 2Dx1y1v2n1
2x1y1u2n1 + (x21 + Dy
21)v2n1
.(8)
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368 Ahmet Tekcan
Note that u2n+1 = 22nx2n+1 and v2n+1 = 2
2ny2n+1. Hence u2n1 = 22n2x2n1
and v2n1 = 22n2y2n1. (7) and (8) yield that
u2n+1 = (x21 + Dy
21)u2n1 + 2Dx1y1v2n1
22nx2n+1 = (x2
1 + Dy2
1)(22n2x2n1) + 2Dx1y1(2
2n2y2n1)
22nx2n+1 = 22n2 (x21 + Dy21)x2n1 + 2Dx1y1y2n1
x2n+1 =22n2 [(x21 + Dy
21)x2n1 + 2Dx1y1y2n1]
22n
x2n+1 =(x21 + Dy
21)x2n1 + 2Dx1y1y2n1
4,
v2n+1 = 2x1y1u2n1 + (x2
1 + Dy2
1)v2n1
22ny2n+1 = 2x1y1(22n2x2n1) + (x
2
1 + Dy2
1)(22n2y2n1)
22ny2n+1 = 22n2
2x1y1x2n1 + (x
2
1 + Dy2
1)y2n1
y2n+1=
22n2 [2x1y1x2n1 + (x21 + Dy
21)y2n1]
22n
y2n+1 =2x1y1x2n1 + (x
21 + Dy
21)y2n1
4
and hence
x2n+1 x2n1y2n+1 y2n1 = x2n+1y2n1 x2n1y2n+1
=
(x21 + Dy
21)x2n1 + 2Dx1y1y2n1
4
y2n1
x2n1
2x1y1x2n1 + (x21 + Dy
21)y2n1
4
=(x21 + Dy
21)x2n1y2n1 + 2Dx1y1y
22n1
4
2x1y1x
22n1 (x
21 + Dy
21)x2n1y2n1
4
=2x1y1(x
22n1 Dy
22n1)
4
=2x1y1(4)
4
= 2x1y1.
Conjecture 3.3. If (x1, y1) is the fundamental solution of the negative Pellequation x2 Dy2 = 4, then (x2n+1, y2n+1) satisfy the following recurrence
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The Pell Equation 369
relations
x2n+1 = (x2
1 + 1) (x2n1 + x2n3) x2n5
y2n+1 = (x2
1 + 1)(y2n1 + y2n3) y2n5
for n 3
References[1] H.M. Edward, Fermats Last Theorem: A Genetic Introduction to Algebraic Number
Theory, Graduate Texts in Mathematics, Vol: 50, Springer-Verlag, 1977.[2] P. Kaplan, K.S. Williams, Pells Equations x2my2 = 1,4 and Continued Fractions,
Jour. Number Theory. 23(1986), 169182.[3] H.W. Lenstra, Solving The Pell Equation, Notices of the AMS. 49(2) (2002), 182192.[4] K. Matthews, The Diophantine Equation x2 Dy2 = N, D > 0, Expositiones Math.,
18 (2000), 323331.[5] R.A. Mollin, A.J. Poorten, H.C. Williams, Halfway to a Solution of x2 Dy2 = 3,
Journal de Theorie des Nombres Bordeaux, 6 (1994), 421457.[6] F. Smarandache, Method to Solve the Diophantine Equation ax2 by2 + c = 0, In
Collected Papers, Vol. 1. Lupton, AZ: Erhus University Press, 1996.[7] P. Stevenhagen, A Density Conjecture for the Negative Pell Equation, Computational
Algebra and Number Theory, Math. Appl. 325 (1992), 187200.[8] A. Tekcan, Pell Equation x2 Dy2 = 2 II, Bulletin of the Irish Mathematical Society
54 (2004), 7389.
Received: July 31, 2006