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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, Turkeytekcan@uludag.edu.tr

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