/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 8, No. 3, 2015, 343-356
ISSN 1307-5543 – www.ejpam.com

An Algorithm for Explicit Form of Fundamental Units of Certain

Real Quadratic Fields and Period Eight

Özen Özer1,∗, Ayten Pekin 2

1 Kırklareli University Faculty of Art and Sciences Department of Mathematics, Kırklareli, Turkey
2 Department of Mathematics, Faculty of Sciences, Istanbul University, Istanbul, Turkey

Abstract. In this paper, we have given an explicit formulation to determine the form of the fundamental
units of certain real quadratic number fields. This new algorithm for such quadratic fields is first in the
literature and it gives us a more practical way to calculate the fundamental unit. Where, the period in
the continued fraction expansion of the quadratic irrational number of the certain real quadratic fields
is equal to 8.

2010 Mathematics Subject Classifications: 11A55, 11R27
Key Words and Phrases: real quadratic field, continued fraction, fundamental unit

1. Introduction and Notation

Determination of the fundamental units of quadratic fields has a great importance at many
branches in number theory. Although the fundamental units of real quadratic fields of Richaut-
Degert type are well-known, explicit form of the fundamental units are not known very well
and these determinations were very limited except for these type an K. therefore, Tomita has
described explicitly the form of the fundamental units of the real quadratic fields Q(

p
d) such

that d is a square-free positive integer congruent to 1 modulo 4 and the period kd in the

continued fraction expansion of the quadratic irrational number ωd = (
1+
p

d
2 ) in Q(

p
d) is

equal to 3 and 4, 5 respectively in [4] and [5]. Later, explicit form of the fundamental units
for all real quadratic fields Q(

p
d) such that the period kd in the continued fraction expansion

of the quadratic irrational number ωd is equal to 6, has been described in [3].
In this paper, we will deal with all real quadratic fields Q(

p
d) such that d is a square free

positive integer congruent to 1 modulo 4 and the period kd in the continued fraction expansion

of the quadratic irrational number ωd = (
1+
p

d
2 ) in Q(

p
d) is equal to 8 and describe explicitly

Td , Ud in the fundamental unit ǫd = (
Td+Ud

p
d

2 )> 1 of Q(
p

d) and d itself by using at most five
parameters appearing in the continued fraction expansion of ωd .

∗Corresponding author.

Email addresses: ozenozer@klu.edu.tr (Ö. Özer), aypekin@istanbul.edu.tr (A. Pekin)

http://www.ejpam.com 343 c© 2015 EJPAM All rights reserved.



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 344

Let I(d) be the set of all quadratic irrational numbers in Q(
p

d). For an element ξ of I(d)

if ξ > 1, −1 < ξ
′
< 0 then ξ is called reduced, where ξ

′
is the conjugate of ξ with respect to

Q. More information on reduced irrational numbers may be found in [2]. We denote by R(d)

the set of all reduced quadratic irrational numbers in I(d). It is well known that if an element
ξ of I(d) is in R(d) then the continued fractional expansion of ξ is purely periodic. Moreover,
the denominator of its modular automorphism is equal to fundamental unit ǫd of Q(

p
d) and

the norm of ǫd is (−1)kd [3]. In this paper [x] means the greatest integer less than or equal to
x and continued fraction with period k is generally denoted by [a0, a1, a2, . . . , ak].

2. Preliminaries and Lemmas

In this section some of the important required preliminaries and lemmas are given.
Now, for any square-free positive integer d, we can put d = a2+b with a, b ∈ Z , 0< b ≤ 2a.

Here, since
p

d − 1< a <
p

d the integers a and b are uniquely determined by d.
Let d be a square-free positive integer congruent to 1 modulo 4, then we consider the

following two cases:

Case 1. If a is even, then b = 4ℓ+ 1 with l ∈ Z , ℓ≥ 0.

Case 2. If a is odd, then b = 4ℓ with l ∈ Z , ℓ≥ 1.

Let denote by D the set of all positive square-free integers and by Dt
k the set of all positive

square-free integer d such that d ≡ k(8) and b ≡ t(8). Hence, we have
Dt

k = {d ∈ Z | d ≡ k(8), b ≡ t(8)}. Then, we get some remarks as follows:

Remark 1. d can be congruent to 1 or 5 modulo 8 since d is congruent to 1 modulo 4.

In the case of d ≡ 1(8), b can be congruent to 0, 1 or 5 modulo 8. Therefore, the set of all

positive square-free integers congruent to 1 modulo 8 is equal D0
1∪D1

1∪D5
1. Thus the set of all

positive square free integers congruent to1 modulo 8 is the union of D0
1, D1

1, D5
1.

In the case of d ≡ 5(8), b can be congruent to 1, 4 or 5 modulo 8. So the set of all positive

square-free integers congruent to 5 modulo 8 is equal to D1
5 ∪ D4

5 ∪ D5
5.

Remark 2. Let d be a square-free positive integer congruent to 1 modulo 4, then:

• If a is even; b can only be congruent to 1 or 5 modulo 8 since b ≡ 1(mod4) when a is even.

Then, d belongs to D1
5 ∪ D5

5 ∪ D5
1 ∪ D1

1 in the case of a is even.

• If a is odd; b can be only be congruent to 0 or 4 modulo 8 since b ≡ 0(mod4) when a is

odd. Then, d belongs to D0
1 ∪ D4

5 in the case of a is odd.

Remark 3. The sets D0
1, D1

1, D5
1, D1

5, D4
5 and D5

5 are represented as follows:

D0
1 ={d ∈ D | d = a2 + 8m, a ≡ 1(mod2), 0< 4m< a}

D1
1 ={d ∈ D | d = a2 + 8m+ 1, a ≡ 0(mod4), 0≤ 4m< a}

D5
1 ={d ∈ D | d = a2 + 8m+ 5, a ≡ 2(mod4), 0≤ 4m< a− 2}



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 345

D1
5 ={d ∈ D | d = a2 + 8m+ 1, a ≡ 2(mod4), 0≤ 4m< a}

D4
5 ={d ∈ D | d = a2 + 8m+ 4, a ≡ 1(mod2), 0≤ 4m< a− 2}

D5
5 ={d ∈ D | d = a2 + 8m+ 5, a ≡ 0(mod4), 0≤ 4m< a− 2}

Now in order to prove our theorems we need the following lemmas.

Lemma 1. For a square-free positive integer d > 5 congruent to 1 modulo 4, we putωd = (
1+
p

d
2 ),

q0 = [ωd] ωR = q0 − 1 +ω. Then ωd /∈ R(d), but ωR ∈ R(d) holds. Moreover for the period

k of ωR, we get ωR = [2q0 − 1,q1, . . . ,qk−1] and ωd = [q0,q1, . . . ,qk−1, 2q0 − 1]. Furthermore,

let ωR =
(Pk−1ωR+Pk−2)

(Qk−1ωR+Qk−2)
= [2q0 − 1,q1, . . . ,qk−1,ωR] be a modular automorphism of ωR, then the

fundamental unit ǫd of Q(
p

d) is given by the following formula:

ǫd = (
Td + Ud

p
d

2
)> 1,

where Td = (2q0 − 1)Qk−1 + 2Qk−2, Ud = Qk−1, and Q i is determined by Q−1 = 0, Q0 = 1,

Q i+1 = qi+1Q i +Q i−1, (i ≥ 0).

Proof. See [5, Lemma 1].

Lemma 2. For a square-free positive integer d, we put d = a2 + b (0 < b ≤ 2a), a, b ∈ Z.

Moreover let ωi = ℓi +
1
ωi+1

(ℓi = [ωi], i ≥ 0) be the continued fraction expansion of ω = ω0

in R(d). Then each ωi is expressed in the form ωi =
a−ri+

p
d

ci
(ci , ri ∈ Z), and ℓi , ci , ri can be

obtained from the following recurrence formula:

ω0 =
a− r0 +

p
d

c0
,

2a− ri =ciℓi + ri+1,

ci+1 =ci−1 + (ri+1 − ri)ℓi (i ≥ 0), where 0≤ ri+1 < ci , c−1 =
(b+ 2ar0 − r0

2)

c0
.

Moreover for the period k ≥ 1 of ω0, we get

ℓi =ℓk−i (1≤ i ≤ k− 1),

ri =rk−i+1, ci = ck−i (1≤ i ≤ k).

Proof. See [1, Proposition 1].

Lemma 3. For a square-free positive integer d congruent to 1 modulo 4, we put ωd = (
1+
p

d
2 ),

q0 = [ωd] and ωR = q0 − 1+ [ωd].

If we put ω =ωR in Lemma 2. , then we have the following recurrence formula:

r0 =r1 = a− l0 = a− 2q0 + 1,

c0 =2, c1 = c−1 =
(b+ 2ar0 − r0

2)

c0
,

ℓ0 =2q0 − 1,ℓi = qi (1≤ i ≤ k− 1).



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 346

Proof. It can be easily proved by using Lemma 2.

3. Theorems

Theorem 1. Let d = a2 + b ≡ 1mod(4) is a square free integer for positive integers a and b

satisfying 0 < b ≤ 2a. Let the period kd of the integral basis element of ωd = (
1+
p

d
2 ) in Q(

p
d)

be 8. If a is odd then,

wd = [
a+ 1

2
, l1, l2, l3,

s1(C + l3A)− 2l2B − rA

C(r − s1l3) + B2
, l3, l2, l1, a],

where (i = 1,2,3,4), li ≥ 1. Then the coefficients Td and Ud of ǫd

(Td , Ud) = ([(Ar + s1l1)(C
2l4 + 2AC) + 2(C(Bl4 + ℓ2) + B)], C(Cl4 + 2A))

and

d = (Ar + s1l1)
2 + 4rl2 + 4s1

hold. Where A, B, C are determined by A= l1l2 + 1, B = l2l3 + 1, and C = l1 + Al3, moreover r

and s are uniquely determined with the equalities a = Ar + s1l1 and

B(Bl4 + 2l2) = s1[C(1+ l3l4) + Al3]− r(A+ Cl4).

Proof. Let a be an odd integer then d ∈ D1
0 ∪ D5

4 . Since q0 =ωd =
a+1

2 then from Lemma 3
we obtain

r0 =r1 = a− 2q0 + 1= 0,

c0 =2,

l0 =a

(1)

and from the Lemma 2: l1 = 7, l2 = l6, l3 = l5, wd = [
a+1

2 , l1, l2, l3, l4, l3, l2, l1, a] hold for
kd = 8. Furthermore we have r1 = r8, r2 = r7, r3 = r6, r4 = r5, c1 = c7, c2 = c6, c3 = c5.

Let d ∈ D1
0 , where

D1
0 ={d ∈ D|d ≡ 1(mod8), b ≡ 0(mod8)}
={d ∈ D|d = a2 + 8m, a ≡ 1(mod2), 0< 4m< a}

In this case we have b = 8m ∋ m> 0. From Lemma 2, it can easily seen that c1 = c−1 = 4m

and 2a = 4ml1 + r2 (for i = 1).
Since r2 = 2a − 4ml1 is even then there exists a positive integer r such that r2 = 2r.

Therefore
2r = 2a− 4ml1⇒ r = a− 2ml1

and so r is an odd integer. From Lemma 2, we have 2a = c2l2 + r2 + r3 for i = 2 and

2a = (2+ 2rl1) · l2 + r2 + r3 (2)



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 347

for c2 = c0 = (r2 − r1)l1, c2 = 2+ 2rl1. If we put 4m= 2rl2 + s in (1) then we obtain

4ml1 = 2l2(rl2 + rl1 + 1) + r3. (3)

Where 2l2 + r3 ≡ 0(modl1) and so there exists positive integer s such that 2l2 + r3 = sl1 then
we can obtain

r3 = sl1 − 2l2. (4)

Here if the value r3 is written in (3) then it is immediately seen that 4m = 2rl2 + s and s is
even. If we put 4m= 2rl2 + s in (1) then

2a =(2rl2 + s)l1 + 2r ⇒ 2a = 2rl1l2 + 2r + sl1

⇒ 2a = r(2l1l2 + 2) + sl1

⇒ 2a = 2r(l1l2 + 1) + sl1

hold.
If we take A = l1 · l2 + 1 then we have a = rA+ s1l1 because of s = 2s1 is even, s1 > 0,

s1 ∈ Z . On the other hand, for i = 2 we have

c3 =c1 + (r3 − r2)l2 = 4m+ (r2 − r2)l2

c4 =c2 + (r4 − r2)l3 = (2+ 2rl1) + (r4 − r3)l3

c5 =c3 + (r5 − r4)l4 = 4m+ (r3 − r2)l2 + (r5 − r4)l4) = c3,

for (c3 = c1 + (r3 − r2)l2⇒ c3 = 4m+ (r3 − r2)l2).
From Lemma 2: (i = 3),

2a =c3l3 + r3 + r4,

r3 =2a− c3l3 − r4⇒ r3 = 2a− (4m+ (r3 − r2)l2)l3 − r4

⇒ r4 = 2a− 4ml3 + (2r − r3)l2l3 − r3) for i = 4

2a =c4l4 + r5 + r4,

r4 =r5⇒ 2a = (2+ 2rl1)l4 + (r4 − r3)l3l4 + 2r4

r4 =2rA+ 2s1l1 − 2rl2l3 − 2s1l3 + 2rl2l3 − (sl1 − 2l2)(l2l3 + 1)

⇒ r4 = 2rA+ 2s1l1 − 2rl2l3 − 2s1l3 + 2rl2l3 − sl1l2l3 − sl1 + 2l2
2 l3 + 2l2

⇒ r4 = 2rA− 2rl2l3 − 2s1l3 + 2rl2l3 − 2s1l1l2l3 + 2l2
2 l3 + 2l2

r4 =2rl1l2 + 2r − 2rl2l3 − 2s1l3 + 2rl2l3 + 2s1l1l2l3 + 2l2
2 l3 + 2l2

hold. Therefore we obtain the value r4, as

r4 = 2[(r − s1l3)A+ l2B] = (r − s1l3)A+ l2B for B = l2l3 + 1,A= l1l2 + 1. (5)

Furthermore

c4 =(2+ r2l1) + (r4 − r3)l3 = (2+ 2rl1) + [(2r − sl3)A+ 2l2B − sl1 + 2l2]l3



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 348

=(2+ 2rl1) + 2rAl3 − sl3
2A+ 2l2l3B − sl1l3 + 2l3

=2rC − sl3C + 2(1+ l2l3) + 2l2l3B

=C(2r − sl3) + 2B + 2Bl2l3

=C(2r − sl3) + 2B(1+ l2l3)

and so we have

c4 = C(2r − sl3) + 2B2 (for the values A, B and C = l1 + Al3). (6)

If the equalities a = rA+ sl1, r4 = (r − s1l3)A+ l2B and c4 = C(2r − sl3) + 2B2 are written in
l4 then

l4 =
2a− 2r4

c4
=

2rA+ sl1 − 2(2r − sl3)A− 4l2B

C(2r − sl3) + 2B2

holds. By taking s = 2s1 we obtain

l4 =
s1(C + l3A)− 2l2B − rA

C(r − s1l3) + B2
. (7)

From this equation

C(r − s1l3)l4 + B2l4 =s1(C + l3A)− 2l2B − rA

⇒ B2l4 + 2l2B = s1C + s1l3A− rA− rC l4 + s1Cl3l4

⇒ B2l4 + 2l2B = s1(C + l3A+ Cl3l4)− rA+ Cl4

⇒ B2l4 + 2l2B = s1[C(1+ l3l4) + Al3]− r(A+ Cl4)

hold and this proves that r and s1 are uniquely determined by a = rA+ s1l1.
Now, let’s determine the coefficients Td and Ud of the fundamental unit

ǫd = (
Td+Ud

p
d

2 )> 1) for d ≡ 1(mod4) and the period kd = 8. Since

Q−1=0

Q0 =1

Q i+1 =qi+1Q i +Q i−1, (i ≥ 0)

Q1 =q1Q0 +Q−1⇒Q1 = l11+ 0⇒Q1 = l1

Q2 =q2Q1 +Q0⇒Q2 = l2l1 + 1⇒Q2 = A

Q3 =q3Q2 +Q1⇒Q3 = l3A+ l1⇒Q3 = C

Q4 =q4Q3⇒Q4 = l4C + A

Q5 =q5Q4 +Q3⇒Q5 = l3(l4C + A) + C ⇒Q5 = l3l4C + C + l3A

⇒Q5 = C(l3l4 + 1) + l3A.

Q6 =ℓ2Q5 +Q4 =⇒Q6 = ℓ2[C(ℓ3ℓ4 + 1) + ℓ3A] + ℓ4C + A= Cℓ4(ℓ2ℓ3 + 1) + Cℓ2 + A(1+ ℓ2ℓ3)

holds, where if we take B = (ℓ2ℓ3 + 1)

Q6 =Cℓ4B + Cℓ2 + AB = C(ℓ4B + ℓ2) + AB



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 349

Q7 =ℓ1Q6 +Q5 =⇒Q7 = ℓ1[C(Bℓ4 + ℓ2) + AB] + C(ℓ3ℓ4 + 1) + Aℓ3

=⇒Q7 = CBℓ4ℓ1 + Cℓ1ℓ2 + ABℓ1 + Cℓ3ℓ4 + C + ℓ3A

=(Cℓ4 + A)(ℓ1ℓ2ℓ3 + ℓ1 + ℓ3) + CA

=(Cℓ4 + A)(Aℓ3 + ℓ1) + CA

and so we can obtain Q7 = C2ℓ4 + 2AC for C = (Aℓ3 + ℓ1). Therefore we can determine that

(Td , Ud) = ([(Ar + s1ℓ1)(C
2ℓ4 + 2AC) + 2(C(Bℓ4 + ℓ2) + AB)], C(Cℓ4 + 2A))

and d = (Ar + s1ℓ1)
2 + 4rℓ2 + 4s1.

Now let d ∈ D5
4 where,

D5
4 ={d ∈ D|d ≡ 5(mod8), b ≡ 4(mod8)}
={d ∈ D | d = a2 + 8m+ 4, a ≡ 1(mod2), 0≤ 4m< a− 2}

therefore b = 8m+ 4 and m > 0 hold. Besides we have the following equations from Lemma
2:

c−1 =
b

2
= 4m+ 2

c1 =c−1 + (r1 − r0)ℓ0 =⇒ c1 = c−1 =⇒ c1 = 4m+ 2

and

2a− ri =ciℓi + ri+1 =⇒ (i = 1)

2a− r1 =c1ℓ1 + r2 =⇒ 2a = (4m+ 2)ℓ1 + r2,

r2 =2a− 2(m+ 1)ℓ1 =⇒ r1 = r8,

c1 =c7, r2 = r7, c2 = c6, r3 = r6, c3 = c5, r4 = r5.

Since r2 is even number then ∃r ∋ r2 = 2r. And so r is defined as

r =

¨

odd ℓ1 even number

even ℓ1 odd number

If we take i = 2, then from Lemma 2

2a =C2ℓ2 + r2 + r3 =⇒ 2a = (2+ 2rℓ1)ℓ2 + r2 + r3

c2 =c0 + (r2 − r1)ℓ1 =⇒ c2 = 2+ 2rℓ1.
(8)

By using the value 2a = (4m+ 2)ℓ1 + r2 and (8) we can write;

(4m+ 2)ℓ1 + r2 =(2+ 2rℓ1)ℓ2 + r2 + r3

=⇒ (4m+ 2)ℓ1 = (2+ 2rℓ1)ℓ2 + r3



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 350

=⇒ (4m+ 2)ℓ1 = 2ℓ2 + 2rℓ1ℓ2 + r3

=⇒ 2ℓ2 + r3 ≡ 0(modℓ1)

=⇒ ∃s ∈ Z ∋ 2ℓ2 + r3 = sℓ1

=⇒ r3 = sℓ1 − 2ℓ2.

For the value r3 = sℓ1 − 2ℓ2 we obtain 4m + 2 = 2rℓ2 + s where s = 4m + 2 − 2rℓ2 is even
number and so there exists s1 ∈ Z+ such that s = 2s1. If we write 4m+2= 2rℓ2+ s instead of
2a = (4m+ 2)ℓ1 + r2 then we obtain 2a = 2r(ℓ1ℓ2 + 1)sℓ1 and we can write a = rA+ s1ℓ1 by
taking A= ℓ1ℓ2 + 1. In the same way, we have

r4 =2a− (4m+ 2)ℓ3 + (2r − r3)ℓ2ℓ3 − r3

=⇒ r4 = 2rA− 2rℓ2ℓ3 − 2s1ℓ3 + 2rℓ2ℓ3 − 2s1ℓ1ℓ2ℓ3 + 2ℓ22ℓ3 + 2ℓ2
=2(r − s1ℓ3)A+ ℓ2B

for A= ℓ1ℓ2 + 1 and B = ℓ2ℓ3 + 1. Furthermore

c4 =(2+ 2rℓ1) + [(2r − sℓ3)A+ 2ℓ2B − sℓ1 + 2ℓ2]ℓ3
=⇒ c4 = C(2r − 2s1ℓ3) + 2(1+ ℓ2ℓ3) + 2ℓ2ℓ3B

=2C(r − s1ℓ3) + 2B2

and

ℓ4 =
2a− 2r4

c4
=

s1(C + ℓ3A)− 2ℓ2B − rA

C(r − s1ℓ3) + B2

for C = ℓ1 + Aℓ3 and (1+ ℓ2ℓ3) = B. This is completed the proof of the theorem.

Example 1. Let a is odd, d ∈ D5
4 and d = 869≡ 5(mod8). Since a = 29, b = 28, b = 3 · 8+ 4,

m= 3 then we can determine that ℓ1 = 4, ℓ2 = 5, ℓ3 = 1, c1 = 14 and

r2 = 2a− (4m+ 2)ℓ1 =⇒ r2 = 58− 14 · 4= 58− 56= 2=⇒ r = 1,

c2 = 10 because of ℓ1 is even.

(4m+ 2)ℓ1 = (2+ 2rℓ1) · ℓ2 + r3 =⇒ 14 · 4= 10 · 5+ r3 =⇒ r3 = 6, s = 2s1 = 4.

Therefore we obtain A = 21, B = 6, C = 25 and r4 = 18, c4 = 22, ℓ4 = 1. If it is taken

above values then the coefficients of the fundamental units of Q(
p

869) is easily determined as

Td = 49377, Ud = 1675 and so ǫd =
49377+1675

p
869

2 > 1 holds.

Theorem 2. Let d = a2 + b ≡ 1mod(4) is a square free integer for positive integers a and b

satisfying 0 < b ≤ 2a. Let the period kd of the integral basis element of ωd = (
1+
p

d
2 ) in Q(

p
d)

be 8. If a is even then,

wd = [
a

2
;ℓ,ℓ2,ℓ3,

BC + AD− 2ℓ3
ℓ23 − C D

,ℓ3,ℓ2,ℓ1, a− 1], 1≤ ℓi , (i = 2,3)



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 351

and then the coefficients Td and Ud of ǫd

Td =[(A(r + 1) + B − 2) · C + 2(C − ℓ3)](Cℓ4 + 2A) + 2Cℓ2

Ud =C(Cℓ4 + 2A)

and

d = [A(r + 1) + B − 1]2 + 2[A(r + 1) + B − 2s− 2]− 1

hold. Where A= ℓ2 + 1, B = 2s − r, C = 1+ Aℓ3 = 1+ ℓ3 + ℓ2ℓ3, D = Bℓ3 − r − 1, E = ℓ3 + 1
and r and s are uniquely determined with the equations a = A(r + 1) + B − 1 and

ℓ3(ℓ3ℓ4 + 2) = BC + AD+ 2C Dℓ4.

Proof. Let a be even and kd = 8. If d ≡ 1(mod8) then b ≡ 1(mod8) or b ≡ 5(mod8) hold.
Furthermore in the case when a is even, d can belong to D5

1 ∪ D5
5 ∪ D1

5 ∪ D1
1 . q0 = [wd] =

a
2

and from Lemma 3 we can write r0 = r1 = a − 2q0 + 1⇒ r0 = r1 = 1, c0 = 2,ℓ0 = a − 1 and
because of kd = 8 and from Lemma 2 we have ℓ1 = ℓ7, ℓ2 = ℓ6, ℓ3 = ℓ5 and r1 = r8, r2 = r7,
r3 = r6, r4 = r5 and c1 = c7, c2 = c6, c3 = c5.

We first assume that d is in D1
1∪D5

1 . Then we get b ≡ 1(mod8) and so ∃m ∈ Z+ ∋ b = 8m+1

holds. From Lemma ??? c−1 =
(8m+1+2a−1)

2 ⇒ c−1 = 4m+ a and
c1 = c−1 + (r1 − r0)ℓ0⇒ c1 = 4m+ a hold.

By taking equation 2a− ri = ciℓi + ri+1 in Lemma 2 for i = 1, we obtain

2a− r1 =c1ℓ1 + r2⇒ (r1 = 1 and c1 = 4m+ a)

⇒ 2a− 1= (4m+ a)ℓ1 + r2

⇒ 2a− 1= 4mℓ1 + aℓ1 + r2

⇒ (2− ℓ1)a = 4mℓ1 + r2 + 1> 0

⇒ 2− ℓ1 > 0

⇒ ℓ1 < 2 and ℓ1 ≥ 1

⇒ ℓ1 = 1

and so we have
wd = [

a

2
; 1,ℓ2,ℓ3,ℓ4,ℓ3,ℓ2, 1, a− 1].

Since ℓ1 = 1 then a = 4m+ 1+ r2 and if r2 = a − 4m− 1 then a, 4m are even and r2 ≥ 1 is
odd and so there exists r ≥ 0 such that r2 = 2r + 1 and we can obtain r2 < a.

If we use ci+1 = ci−1 + (ri+1 − ri)ℓi for i ≥ 0 then we obtain

c2 = c0 + (r2 − r1)ℓ1 = 2+ (r2 − 1)1= 2r + 2.

Furthermore we have obtain the following equalities from Lemma 2 c2 = 2r + 2,

2a = c2ℓ2 + r2 + r3⇒ 2a = (2r + 2)ℓ2 + 2r + 1+ r3

and by taking a = 4m+ 1+ r2 and a = 4m+ 2r + 2 we have

8m+ 4r + 4= (2r + 2)ℓ2 + 2r + 1+ r3 = (2r + 2)ℓ2 + r3 − 2r − 3.



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 352

Since a is even then

2a =(2r + 2)ℓ2 + (2r + 1) + r3 ≡ 0(mod4)

⇒ r3 + 2r + 1≡ 0(mod4) and ℓ2 ≡ 0(mod2)

⇒ ∃s ∈ Z+ ∋ r3 + 2r + 1= 4s, r3 ≥ 0

⇒ 4s− 2r − 1≥ 0

⇒ 4s > 2r + 1

hold, where r3 = 4s− 2r − 1 is odd number because of 4s is even and 2r + 1 is odd.
From Lemma 2. we have

c3 =c1 + (r3 − r2)ℓ2 = 4m+ a+ (4s− 2r − 1− 2r − 1)ℓ2
=a− 2r − 2+ a+ (4s− 4r − 2)ℓ2
=2a− 2r − 2+ (4s− 4r − 2)ℓ2

and if the value 2a is written instead of c3 then c3 is obtained as

c3 =(2r + 2)ℓ2 + 4s− 2r − 2+ 4sℓ2 − 4rℓ2 − 2ℓ2
=(4s− 2r)(ℓ2 + 1)− 2. (9)

By using the values c3, A= ℓ2 + 1 and B = 2s− r we have

2a =(2r + 2)ℓ2 + r3 + 2r + 1= (2r + 2)ℓ2 + 4s− 2r − 1+ 2r + 1

=2[(r + 1)ℓ2 + 2s]

a =(r + 1)ℓ2 + 2s (10)

and 2a = c3ℓ3 + r3 + r4 ise r4 = 2a− c3ℓ3 − r3

r4 =2[(r + 1)ℓ2 + 2s]− [(4s− 2r)(ℓ2 + 1)− 2]ℓ3 − (4s− 2r − 1)

=2(r + 1)ℓ2 + 4s− [(4s− 2r)(ℓ2 + 1)− 2]ℓ3 − 4s+ 2r + 1

=2(r + 1)ℓ2 + 2r + 1+ [(4s− 2r)(ℓ2 + 1)− 2]ℓ3 (11)

hold from Lemma 2. Moreover we have

a = (r + 1)ℓ2 + 2s = A(r + 1) + B − 1 (12)

for A= ℓ2 + 1, B = 2s− r and

r3 =4s− 2r − 1⇒ r3 = 2B − 1

r4 =2a− c3ℓ3 − r3⇒ r4 = 2a− 2(BA− 1)ℓ3 − 2B + 1

⇒ r4 = 2rℓ2 + ℓ2 − 2BC + 2ℓ3 + 4s+ A.

If C = 1+ Aℓ3 = 1+ ℓ3 + ℓ2ℓ3, B = 2s− r,

4s =2B + 2r ⇒ r4 = 2rℓ2 + ℓ2 − 2BC + 2ℓ3 + 2B + 2r + A



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 353

⇒ r4 = 2rA+ 2A− 2BC + 2B + 2ℓ3 − 1

hold. For ℓ1 = 1, c3 = 2BA−2= 2(BA−1). If we take c4 = (2r+2)+(2rA+2A−2BC+2ℓ3)ℓ3
then

c4 = 2r + 2+ 2rAℓ3 + 2Aℓ3 − 2BCℓ3 + 2ℓ23⇒ c4 = 2C[r + 1− Bℓ3] + 2ℓ23

hold. Where if we take r + 1− Bℓ3 = −D then we obtain c4 = 2ℓ23 − 2C D = 2(ℓ23 − C D) and
r4 = 2rA+2A−2BC+2B+2ℓ3−1 for A= ℓ2+1, B = 2s+r, C = Aℓ3+1. Finally r4 is determined
as r4 = 2rA+ 2A− 2B − 2ABℓ3 + 2B + 2ℓ3 − 1= 2ℓ3 − 2A(Bℓ3 − r − 1)− 1= 2E − 2AD− 3 for
the values D = Bℓ3 − r − 1 and E = ℓ3 + 1. On the other hand we can write

a− r4 =(r + 1)ℓ2 + 2s− 2E + 2AD+ 3

=(r + 1)ℓ2 + B + r − 2E + 2AD+ 3= r(ℓ2 + 1) + ℓ2 + B − 2E + 2AD+ 3

=B − 2E + ABℓ3 + A(Bℓ3 − r − 1) + 2

where D = Bℓ3 − r − 1,

a− r4 =B − 2E + ABℓ3 + AD+ 2= B + ABℓ3 + AD+ 2− 2ℓ3 − 2

=B + ABℓ3 + AD− 2ℓ3
=B(1+ Aℓ3) + AD− 2ℓ3 = BC + AD− 2ℓ3
⇒ a− r4 = BC + AD− 2ℓ3

holds and so we have l4 as ℓ4 =
2(a−r4)

c4
=

2(BC+AD−2ℓ3)
2(ℓ23−C D)

=
BC+AD−2ℓ3
ℓ23−C D

⇒ ℓ4 = BC+AD−2ℓ3
ℓ23−C D

. Besides

s and r are uniquely determined by ℓ23ℓ4 + 2ℓ3 = BC + AD+ 2C Dℓ4 and (9).
Now, let’s determine the coefficients Td and Ud of the fundamental unit ǫd . Since

d ≡ 1(mod4) then we know that wd = [q0;q1, . . . ,qk−1, 2q0 − 1] and ǫd = (
Td+Ud

p
d

2 ) > 1.
Furthermore

Q−1 =0

Q0 =1

Q i+1 =qi+1Q i +Q i−1 (i ≥ 0)

Q1 =q1Q0 +Q−1⇒Q1 = ℓ1, ℓ1 = 1⇒Q1 = 1

Q2 =q2Q1 +Q0⇒Q2 = ℓ2 · 1+ 1, (A= ℓ2 + 1)⇒Q2 = A

Q3 =q3Q2 +Q1⇒Q3 = ℓ3A+ 1, (C = ℓ3A+ 1)⇒Q3 = C

Q4 =q4Q3 +Q2⇒Q4 = ℓ4C + A

Q5 =q5Q4 +Q3⇒Q5 = ℓ3(ℓ4C + A) + C ⇒Q5 = C(ℓ3ℓ4 + 1) + ℓ3A

Q6 =q6Q5 +Q4 = ℓ2[C(ℓ3ℓ4 + 1) + ℓ3A] + ℓ4C + A= C[ℓ4(ℓ2ℓ3 + 1) + ℓ2] + A(1+ ℓ2ℓ3)

=C[((A− 1)(E − 1) + 1)ℓ4 + ℓ2] + A[(A− 1)(E − 1) + 1] for ((1+ ℓ2ℓ3) = (A− 1)(E − 1) + 1 or

Q6 =(C − ℓ3)(Cℓ4 + A) + Cℓ2

Q7 =ℓ1Q6 +Q5 = 1Q6 +Q5 = (C − ℓ3)(Cℓ4 + A) + Cℓ2 + (Cℓ4 + A)ℓ3 + C = C[Cℓ4 + 2A]



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 354

and so we obtain that

Td =(A(r + 1) + B − 2)C(Cℓ4 + 2A) + 2(C − ℓ3)(Cℓ4 + A) + 2Cℓ2,

Ud =C(Cℓ4 + 2A)

and

d =a2 + b = [A(r + 1) + B − 1]2 + 2A(r + 1) + 2B − 4s− 5

=[A(r + 1) + B − 1]2 + 2[A(r + 1) + B − 2s− 2]− 1.

Now let d ∈ D1
5 ∪ D5

5 then b ≡ 5(mod8) and ∃m ∈ Z+ ∋ b = 8m+ 5. From Lemma 2 we
can obtain the following equalities:

c−1 =
(8m+ 5+ 2a− 1)

2
⇒ c−1 = 4m+ 4+ a

c1 =c−1 + (r1 − r0)ℓ0 (r1 − r0 = 0)⇒ c1 = 4m+ 4+ a

2a− r1 =c1ℓ1 + r2⇒ (r1 = 1, c1 = 4m+ 4+ a)

⇒ 2a− 1= (4m+ 4+ a) · ℓ1 + r2

⇒ (2− ℓ1)a = 4mℓ1 + 4ℓ1 + r2 + 1> 0

⇒ 2− ℓ1 > 0, ℓ1 ≥ 1

⇒ ℓ1 = 1

and then we get a = 4(m+ 1) + r2 + 1

r2 = a− 4(m+ 1)− 1⇒ r2 is an odd integer

so r2 < a holds and ∃r ≥ 0, r ∈ Z ∋ r2 = 2r + 1. For i ≥ 0 we have

c2 = c0 + (r2 − r1)ℓ1 = 2+ (r2 − 1)1= 2r + 2

from relation ci+1 = ci−1 + (ri+1 − ri)ℓi and 2a = c2ℓ2 + r2 + r3. 2a = (2r + 2)ℓ2 + 2r + 1+ r3

and

a =4(m+ 1) + 1+ r2⇒ a = 4m+ 2r + 6

⇒ 2(4m+ 2r + 6) = (2r + 2)ℓ2 + 2r + 1+ r3

⇒ 2a = (2r + 2)ℓ2 + r3 + 2r + 1≡ 0(mod4)

⇒ r3 + 2r + 1≡ 0(mod4) and ℓ2 ≡ 0(mod2)

⇒ ∃s ∈ Z+ ∋ r3 + 2r + 1= 4s, r3 ≥ 0

⇒ 4s− 2r − 1≥ 0,

therefore
r3 = 4s− 2r − 1 is odd. (13)



Ö. Özer, A. Pekin / Eur. J. Pure Appl. Math, 8 (2015), 343-356 355

From lemma 2 we know that c3 = c1+(r3− r2)ℓ2 = 4m+4+ a+(4s−2r −1−2r −1)ℓ2 then
we obtain

c3 =a− 2r − 2+ a+ (4s− 4r − 2)ℓ2⇒ c3 = 2a− 2r − 2+ (4s− 4r − 2)ℓ2
=(2r + 2)ℓ2 + 4s− 2r − 2+ (4sℓ2 − 4r − 2)ℓ2
=(4s− 2r)(ℓ2 + 1)− 2

hold for a = 4m+ r2 + 5⇒ 4m+ 4= a− r2 − 1= a− 2r − 2.
If we take the values A= ℓ2 + 1 and B = 2s− r then

2a = (2r + 2)ℓ2 + r3 + 2r + 1= (2r + 2)ℓ2 + 4s− 2r − 1+ 2r + 1= 2[(r + 1)ℓ2 + 2s]

and a = (r + 1)ℓ2 + 2s. If 2a = c3ℓ3 + r3 + r4 then

r4 =2a− c3ℓ3 − r3

⇒ r4 = 2[(r + 1)ℓ2 + 2s]− [(4s− 2r)(ℓ2 + 1)− 2]ℓ3 − (4s− 2r − 1)

r4 =2(r + 1)ℓ2 + 2r + 1+ [(4s− 2r)(ℓ2 + 1)− 2]ℓ3 (14)

If a = (r + 1)ℓ2 + 2s, A= ℓ2 + 1, B = 2s− r then

a = rℓ2 + ℓ2 + 2s = rℓ2 + ℓ2 + 2s+ r − r = r(ℓ2 + 1) + ℓ2 + (2s− r) = Ar + A− 1+ B

and so
a = A(r + 1) + B − 1 (15)

holds. Since A= ℓ2 + 1, and B = 2s− r then

c3 =2BA− 2= 2(BA− 1)

r3 =4s− 2r − 1⇒ r3 = 2B − 1

r4 =2a− c3ℓ3 − r3⇒ r4 = 2a− 2(BA− 1)ℓ3 − 2B + 1

⇒ r4 = 2rℓ2 + ℓ2 − 2B(1+ Aℓ3) + 2ℓ3 + 4s+ A

and if we take C = 1+ Aℓ3 = 1+ ℓ3 + ℓ2ℓ3, D = Bℓ3 − r − 1 and E = ℓ3 + 1 then we obtain

r4 = 2rℓ2 + ℓ2 − 2BC + 2ℓ3 + 2B + 2r + A= 2ℓ3 − 2A(Bℓ3 − r − 1)− 1

=2(ℓ3 + 1)− 2AD− 3

⇒ r4 = 2(ℓ3 + 1)− 2AD− 3

⇒ r4 = 2E − 2AD− 3.

At the same way we can determine c4 as

c4 =(2r + 2) + (2rA+ 2A− 2BC + 2ℓ3)ℓ3 = 2r(1+ Aℓ3) + 2(1+ Aℓ3)− 2BCℓ3 + 2ℓ23
=− 2C(Bℓ3 − r − 1) + 2ℓ23 = 2(ℓ23 − C D).



REFERENCES 356

We know that ℓ4 =
2(a−r4)

c4
from Lemma ??? then we can determine the value of a− r4 in the

following:

a− r4 =(r + 1)ℓ2 + B + r − 2E + 2AD+ 3

⇒ a− r4 = B − 2E + ABℓ3 + A(Bℓ3 − r − 1) + 2

⇒ a− r4 = B(1+ Aℓ3) + AD− 2ℓ3
⇒ a− r4 = BC + AD− 2ℓ3.

Moreover we can easily seen that ℓ4 =
2(a−r4)

c4
=

2(BC+AD−2ℓ3)
2(ℓ23−C D)

=
BC+AD−2ℓ3
ℓ23−C D

=
BC+AD−2ℓ3
ℓ23−C D

and s

and r are uniquely determined from the relations a = A(r + 1) + B − 1 and
ℓ23ℓ4 + 2ℓ3 = BC + AD+ 2C Dℓ4.

Example 2. Let a is even, d ≡ 1(mod4). If we choose D = 501 ≡ 5(mod8) then we can

practically determine that a = 22, b = 17≡ 1(mod8), 17= 8m+ 1⇒ m= 2,

c1 = 4m+ a⇒ c1 = 8+ 22⇒ c1 = 30, ℓ1 = 1, ℓ2 = 2, ℓ3 = 4,

a = 4m+ 1+ r2⇒ 22 = 9+ r2⇒ r2 = 13, r2 = 2r + 1 = 13⇒ r = 6, c2 = 2r + 2⇒ c2 = 14,

2a = c2ℓ2 + r2 + r3 ⇒ r3 = 44 − 28 − 13 = 3, r3 + 2r + 1 = 4s ⇒ 3 + 13 = 4s ⇒ s = 4,

c3 = (4s − 2r)(ℓ2 + 1) = 2⇒ c3 = 4 · 3− 2 = 10, r4 = 2a − c3ℓ3 − r3⇒ r4 = 44− 40− 3 = 1,

A= ℓ2 + 1⇒ A= 3, B = 2s− r ⇒ B = 2, C = 1+ Aℓ3 = 1+ ℓ2 + ℓ2ℓ3⇒ C = 13,

D = −r+Bℓ3−1⇒ D = 1, E = ℓ3+1⇒ E = 5, r4 = 1, a = 22, c4 = 6⇒ ℓ4 = 7. Therefore the

fundamental unit of Q(
p

501) is obtained as ǫd =
28225+1261

p
501

2 for Td = 28225, Ud = 1261.

ACKNOWLEDGEMENTS This work was supported by the research scientific project of Kırk-
lareli University with the number KLUBAP-044.

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