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QUADRATIC DIOPHANTINE EQUATION

x2 (t2 t)y2 (4t 2)x + (4t2 4t)y = 0

ARZU OZKOC & AHMET TEKCAN

Abstract. Let t 2 be an integer. In this work, we consider the number of integersolutions of Diophantine equation D : x2 (t2 t)y2 (4t 2)x + (4t2 4t)y = 0 overZ. We also derive some recurrence relations on the integer solutions (xn, yn) of D. In the

last section, we consider the same problem over finite fields Fp for primes p 5.

AMS Subject Classification 2000: 11E04, 11E16, 11D09, 11D79.

Keywords: Diophantine equation, Pell equation.

Date: 02.04.2009.

1. Introduction.

A Diophantine equation is an indeterminate polynomial equation that allows the variables

to be integers only. In more technical language, they define an algebraic curve, algebraic

surface or more general object, and ask about the lattice points on it. The word Diophan-

tine refers to the Hellenistic mathematician of the 3rd century, Diophantus of Alexandria,

who made a study of such equations and was one of the first mathematicians to introduce

symbolism into algebra. In general, the Diophantine equation is the equation given by

(1.1) ax2 + bxy + cy2 + dx + ey + f = 0.

Also the Diophantine equation

(1.2) x2 dy2 = 1which is a special case of (1.1), known the Pell equation (see [3, 14]) which is named after

the English mathematician John Pell a mathematician who searched for integer solutions to

equations of this type in the seventeenth century. The Pell equation in (1.2) has infinitely

many integer solutions (xn, yn) for n 1. The first positive integer solution (x1, y1) of this

equation is called the fundamental solution, because all other solutions can be (easily) derivedfrom it. In fact if (x1, y1) is the fundamental solution, then the n-th positive solution of it,

say (xn, yn), is defined by the equality xn + yn

d = (x1 + y1

d)n for integer n 2. Thereare several methods for finding the fundamental solution of x2 dy2 = 1. For example, thecyclic method known in India in the 12th century, and the slightly less efficient but more

regular English method 17th century, produce all solutions of x2 dy2 = 1 (see [4]). Butthe most efficient method for finding the fundamental solution is based on the simple finite

continued fraction expansion of

d (see [2, 5, 6, 10, 11]). The Pell equation was first studied by

Brahmagupta (598-670) and Bhaskara (1114-1185) (see [1]). Its complete theory was worked

out by Lagrange (1736-1813), not Pell. It is often said that Euler (1707-1783) mistakenly

attributed Brounckers (1620-1684) work on this equation to Pell. However the equation

appears in a book by Rahn (1622-1676) which was certainly written with Pells help: some

1

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2 ARZU OZKOC & AHMET TEKCAN

say entirely written by Pell. Perhaps Euler knew what he was doing in naming the equation

(for further details on Pell and Diophantine equations see [7, 8, 9, 12, 13, 15]).

2. The Diophantine Equation x2 (t2 t)y2 (4t 2)x + (4t2 4t)y = 0.

In [16, 17, 18, 19, 20]), we considered some specific Pell (also Diophantine) equations and

their integer solutions. In the present paper, we consider the integer solutions of Diophantine

equation

(2.1) D : x2 (t2 t)y2 (4t 2)x + (4t2 4t)y = 0over Z, where t 2 is an integer. Note that it is very difficult to solve D in its present form,that is, we can not determine have many integer solutions D has and what they are. So we

have to transform D into a appropriate Diophantine equation which can be easily solved. To

get this let

(2.2) T : x = u + hy = v + k

be a translation for some h and k. In this case the pair {h, k} is called the base of T anddenote it by T[h; k] = {h, k}. If we apply T to D, then we get(2.3) T(D) = D : (u + h)2 (t2 t)(v + k)2 (4t 2)(u + h) + (4t2 4t)(v + k) = 0.In (2.3), we obtain u(2h 2 4t) and v(2kt2 2kt + 4t2 + 4t). So we get h = 2t 1 andk = 2. Consequently for x = u + 2t 1 and y = v + 2, we have the Diophantine equation(2.4) D : u2 (t2 t)v2 = 1which is a Pell equation. Now we try to find all integer solutions (un, vn) of

D and then we

can retransfer all results from D to D by using the inverse of T.Theorem 2.1. Let D be the Diophantine equation in (2.4). Then

(1) The continued fraction expansion of

t2 t is

t2 t =

[1;2] if t = 2

[t 1; 2, 2t 2] if t > 2.(2) The fundamental solution of D is (u1, v1) = (2t 1, 2) .(3) Define the sequence {(un, vn)}, where

(2.5)

unvn

=

2t 1 2t2 2t

2 2t

1

n1

0

for n 1. Then (un, vn) is a solution of D.(4) The solutions (un, vn) satisfy un = (2t 1)un1 + (2t2 2t)vn1 and vn = 2un1 +

(2t 1)vn1 for n 2.(5) The solutions (un, vn) satisfy the recurrence relations un = (4t 3)(un1 + un2)

un3 and vn = (4t 3)(vn1 + vn2) vn3 for n 4.(6) The nth solution (un, vn) can be given by

(2.6)unvn

=

t 1; 2, 2t 2, , 2, 2t 2

n1 times

, 2

for n

1.

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QUADRATIC DIOPHANTINE EQUATION 3

Proof. (1) Let t = 2. Then it is easily seen that

2 = [1; 2]. Now let t > 2. Then

t2 t = t 1 + (

t2 t t + 1) = t 1 + 1

t2t+t1t1

= t 1 + 12 +

t2tt+1

t1= t 1 + 1

2 + 1t2t+t1

= t 1 + 12 + 1

2t2+(t2tt+1)

.

So

t2 t = [t 1; 2, 2t 2].(2) It is easily seen that (u1, v1) = (2t 1, 2) is the fundamental solution of D since

(2t 1)2 (t2 t)22 = 1.(3) We prove it by induction. Let n = 1. Then by (2.5), we get (u1, v1) = (2t 1, 2) which

is the fundamental solution and so is a solution of D. Let us assume that the Diophantineequation in (2.4) is satisfied for n 1, that is,

D : u2n1 (t2 t)v2n1 = 1. We want to show

that this equation is also satisfied for n. Applying (2.5), we find thatunvn

=

2t 1 2t2 2t

2 2t 1n

1

0

=

2t 1 2t2 2t

2 2t 1

2t 1 2t2 2t2 2t 1

n1 1

0

=

2t 1 2t2 2t

2 2t 1

un1vn1

=

(2t 1)un1 + (2t2 2t)vn1

2un1 + (2t 1)vn1

.(2.7)

Hence we conclude that

u2n (t2 t)v2n =

(2t 1)un1 + (2t2 2t)vn12 (t2 t) (2un1 + (2t 1)vn1)2

= u2n1 (t2 t)v2n1= 1.

So (un, vn) is also a solution of D.(4) From (2.7), we find that un = (2t1)un1+(2t22t)vn1 and vn = 2un1+(2t1)vn1

for n 2.(5) We only prove that un satisfy the recurrence relation. For n = 4, we get u1 = 2t 1,

u2 = 8t2

8t + 1, u3 = 32t

3

48t2 + 18t

1 and u4 = 128t

4

256t3 + 160t2

32t + 1. Hence

u4 = (4t 3)(u3 + u2) u1= (4t 3)(32t3 40t2 + 10t) (2t 1)= 128t4 256t3 + 160t2 32t + 1.

So un = (4t 3)(un1 + un2) un3 is satisfied for n = 4. Let us assume that this relationis satisfied for n 1, that is,

(2.8) un1 = (4t 3)(un2 + un3) un4.

Then applying the previous assertion, (2.7) and (2.8), we conclude that un = (4t 3)(un1 +un

2)

un

3 for n

4.

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(6) Note that u1v1

= [t 1;2] = t 1 + 12

= 2t12

which is the fundamental solution. Let us

assume that (un, vn) is a solution of D, that is, u2n (t2 t)v2n = 1. Then by (2.6), we deriveun+1

vn+1 = t 1 +1

2 + 12t2+ 12+ 1

2t2+ 1

+2t2+12

= t 1 +1

2 + 1t1+t1+ 12+ 1

2t2+ 1

+2t2+12

= t 1 + 12 + 1

t1+unvn

=(2t 1)un + (2t2 2t)vn

2un + (2t 1)vn .

So (un+1, vn+1) is also a solution of D since u2n+1(t2t)v2n+1 = (2t 1)un + (2t2 2t)vn2(t2 t) (2un + (2t 1)vn)2 = u2n (t2 t)v2n = 1. This completes the proof. Example 2.1. Lett = 4. Then(u1, v1) = (7, 2) is the fundamental solution of D : u212v2 =1 and some other solutions are

u2v2 = 7 242 7 2

10 = 9728 u3v3

=

7 24

2 7

3 1

0

=

1351

390

u4v4

=

7 24

2 7

4 1

0

=

18817

5432

u5v5

=

7 24

2 7

5 1

0

=

262087

75658

u6v6

=

7 24

2 7

6 1

0

=

3650401

1053780

.

Also un = 7un1 + 24vn1 and vn = 2un1 + 7vn1 for n 2; un = 9(un1 + un2) un3and vn = 9(vn1 + vn2) vn3 for n 4. Further

u2v2

= [3; 2, 6, 2] =97

28u3v3

= [3; 2, 6, 2, 6, 2] =1351

390u4v4

= [3; 2, 6, 2, 6, 2, 6, 2] =18817

5432u5v5

= [3; 2, 6, 2, 6, 2, 6, 2, 6, 2] =262087

75658u6

v6= [3; 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2] =

3650401

1053780.

From above theorem we can give the following result.

Corollary 2.2. The base of the transformation T in (2.2) is the fundamental solution of D,that is T[h; k] = {h, k} = {u1, v1}.

Proof. We proved that (u1, v1) = (2t1, 2) is the fundamental solution of D. Also we showedthat h = 2t1 and k = 2. So the base of T is T[h; k] = {h, k} = {2t1, 2} as we claimed.

We saw as above that the Diophantine equation D could be transformed into the Dio-

phantine equation D via the transformation T. Also we showed that x = u + 2t 1 and

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y = v + 2. So we can retransfer all results from D to D by using the inverse of T. Thus wecan give the following main theorem.

Theorem 2.3. Let D be the Diophantine equation in (2.1). Then

(1) The fundamental (minimal) solution of D is (x1, y1) = (4t 2, 4).(2) Define the sequence {(xn, yn)}n1 = {(un + 2t 1, vn + 2)}, where {(un, vn)} defined

in (2.5). Then (xn, yn) is a solution of D. So it has infinitely many many integer

solutions (xn, yn) Z Z.(3) The solutions (xn, yn) satisfy

xn = (2t 1)xn1 + (2t2 2t)yn1 8t2 + 10t 2yn = 2xn1 + (2t 1)yn1 8t + 6

for n 2.(4) The solutions (xn, yn) satisfy the recurrence relations

xn = (4t 3)(xn1 + xn2) xn3 16t2 + 24t 8yn = (4t 3)(yn1 + yn2) yn3 16t + 16

for n 4.

3. The Diophantine Equation x2 (t2 t)y2 (4t 2)x + (4t2 4t)y = 0 over FiniteFields.

In this section, we will consider the integer solutions of D over finite fields Fp for primes

p 5. Let t Fp and let t2 t d(mod p). Then D becomes(3.1)

Ddp : u

2

dv2

1(mod p).

Let Ddp(Fp) = {(u, v) Fp Fp : u2 dv2 1(mod p)}. Then we can give the followingtheorem.

Theorem 3.1. Let Ddp be the Diophantine equation in (3.1). Then# Ddp(Fp) =

p 1 for d Qpp + 1 for d / Qp,

where Qp denote the set of quadratic residues.

Proof. Let d Qp and let p 1, 5(mod8). If v = 0, then u2 1(mod p) u 1(mod p).So

Ddp has two integer solutions (1, 0) and (p 1, 0). If u = 0, then dv2 1(mod p)

has two solutions v1, v2. So Ddp has two integer solutions (0, v1) and (0, v2). Now let Sp =Fp {1, p 1}. Then there are p52 points u in Sp such that u

21d

is a square. Set u21d

= c2

for some c = 0. Then v2 c2(modp) v c(modp). So Ddp has two solutions (u, c) and(u, c), that is, for each u in Sp, Ddp has two solutions. So it has 2(p52 ) = p5 solutions. Wesee as above that it has also four solutions (1, 0), (p 1, 0), (0, v1) and (0, v2). Therefore Ddphas p 5 + 4 = p 1 integer solutions. Now let p 3, 7(mod 8). Ifv = 0, then u2 1(mod p)and hence u = 1 and u = p 1. So Ddp has two integer solutions (1, 0) and (p 1, 0). Ifu = 0, then dv2 1(mod p) has no solution. So Ddp has no integer solution (0, v). LetHp = F

p {1, p 1}. Then there are p32 points u in Hp such that u

21d

is a square. Setu21d

= j2 for some j = 0. Then v2 j2(modp) v j(modp). So

Ddp has two solutions

(u, j) and (u,

j), that is, for each u in Hp, D

dp has two solutions. So it has 2(

p32 ) = p

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solutions. It has also two solutions (1, 0) and (p 1, 0). Therefore Ddp has p 3 + 2 = p 1integer solutions.

Similarly it can be shown that if d /

Qp, then D

dp has p + 1 integer solutions.

Example 3.1. Let t = 4. Then 12 Q23 and 12 / Q31. So

D1223(F23) =

(1, 0), (2, 11), (2, 12), (3, 4), (3, 19), (5, 5), (5, 18), (6, 1), (6, 22), (7, 2),

(7, 21), (16, 2), (16, 21), (17, 1), (17, 22), (18, 5), (18, 18), (20, 4),

(20, 19), (21, 11), (21, 12), (22, 0)

D1231(F31) =

(0, 7), (0, 24), (1, 0), (2, 15), (2, 16), (4, 3), (4, 28), (5, 8), (5, 23), (7, 2),

(7, 29), (10, 4), (10, 27), (11, 14), (11, 17), (13, 13), (13, 18), (18, 13),

(18, 18), (20, 14), (20, 17), (21, 4), (21, 27), (24, 2), (24, 29), (26, 8),

(26, 23), (27, 3), (27, 28), (29, 15), (29, 16), (30, 0)

.

For the Diophantine equation D, we set

D(Fp) = {(x, y) Fp Fp : x2 (t2 t)y2 (4t 2)x + (4t2 4t)y = 0(mod p)}.Then we can give the following theorem.

Theorem 3.2. Let D be the Diophantine equation in (2.1). Then

#D(Fp) =

p 1 for t2 t Qpp + 1 for t2 t / Qp.

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by2

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Number Theory, Math. Appl. 325(1992), 187200.

[16] A. Tekcan. Pell Equation x2Dy2 = 2, II. Bulletin of the Irish Mathematical Society 54(2004), 7389.[17] A. Tekcan, O. Bizim, M. Bayraktar. Solving the Pell Equation Using the Fundamental Element of

the FieldQ(

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[18] A. Tekcan. The Pell Equation x2

Dy2 =

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[19] A. Tekcan, B. Gezer, O. Bizim. On the Integer Solutions of the Pell Equation x2dy2 = 2t. Int. Jour.of Comp. and Math. Sci. 1(3)(2007), 204208.

[20] A. Tekcan. The Diophantine Equation 4y24yx1 = 0, Curves and Conics over Finite Fields. Acceptedfor publication to Mathematical Reports.

Uludag University, Faculty of Science, Department of Mathematics, Gorukle, 16059, Bursa-

TURKIYE

E-mail address : [email protected], [email protected]

URL: http://matematik.uludag.edu.tr/AhmetTekcan.htm

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