Citation: Özen Özer. Fundamental units for real quadratic fields determined by continued fraction conditions[J]. AIMS Mathematics, 2020, 5(4): 2899-2908. doi: 10.3934/math.2020187
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Zhang and Yue [33], established some congruence relations on the coefficients of the fundamental units for the real quadratic fields Q(√d) while class number is odd. Similarly [28], Williams and Buck compared the lengths of the continued fraction expansions of √d and 12(1+√d) and worked on Eisenstein's problem. Tomita and Yamamuro [27] obtained new sharpener lower bound for ϵd and determined continued fraction expansion of integral basis element using the Fibonacci integer sequence. Tomita also described explicitly ϵd by determining the continued fraction expansion of ωd where the period length is equal to 3 in the reference [26]. Chakraborty et al. proved a new alternative class number formula for real quadratic fields with discriminant △≡5(mod8) [4]. In the reference [9], the author Jeangho investigated a real quadratic integer ξ with fixed norm in the real quadratic field of Q(ξ) and proved real equation's solvability as well as the quadratic progressions for fundamental units. He also demonstrated that prime ideals lying over fixed norm are principal by considering fixed norm equals to −1 or prime.
In recent years, Benamar and co-authors gave the lower bound of the number of special polynomials with a stable period continued fraction expansion in the reference [3]. Badziahin and Shallit [2] considered the some real transcendental numbers σ with the explicit type of continued fraction expansions. Tomita and Kawamoto [10] showed a relation between the real quadratic fields of class number one and a mysterious behavior of the simple continued fraction expansion of certain quadratic irrationals. Besides, some authors expressed important results for several kinds of continued fractions and the real quadratic number fields in their papers. So, one may consider [1,5,6,7,8,12,13,14,23,24,25] references for getting more information about the continued fraction expansions and the fundamental units.
The authors Kim and Ryu [11] worked on the special class circular units ϵk of Q(√pq) and investigated unit group of such real quadratic field. They also proved class number of quartic fields Q(√−p,√−q) by using Sinnott's index formula. In generally, papers include specifying a very short period length but McLaughlin [13] took the very reasonable step of restricting to the case where the period need not be short, but with the constraint that all. By the way, Sasaki [24] and Mollin [14] studied bounds of ϵd for real quadratic fields Q(√d) and obtained useful significant results for that. Yokoi defined several invariants to solve class number problems and solutions of the Diophantine equations in the terms of coefficients of fundamental unit in the references [29,30,31,32]. Furthermore, the author of this paper proved some special and significant results in the references [15,16,17,18,19,20,21,22].
This paper determines the fundamental unit problem. The following has been identified; Continued fraction expansions have individual elements equal to each other's and also written as γs (except the last digit of the period) for a γ positive odd integer where d is square free integer and equivalent to 1 for modulo 4. We also state the fact that, infinitely countable values have been classified and generalized as of d having all γs in the symmetric part of period length of ωd. By using our newly identified results, we have concluded that the general forms of the following structures are found out for such real quadratic number fields;
(1) Radicand d
(2) Fundamental units ϵd
(3) Coefficients of fundamental unit td,ud
(4) Yokoi's invariants nd,md
Also, additional results on fundamental units, Yokoi's invariants, continued fraction expansions and period lengths are empirically revealed and validated with this paper as well as related published papers.
In this part, we give some basic and useful informations. Throughout this paper, we fix some notations and provide them as follows:
Let k=Q(√d) be a real quadratic number field for d>1 square-free integer, the integral basis element ωd is a part of Z[ωd] and l(d) is the period length of continued fraction expansion of ϵd, the fundamental unit ϵd of real quadratic number field is denoted by ϵd=td+ud√d2>1 such that N(ϵd)=(−1)l(d) in this paper.
Note: The set I(d) contains all quadratic irrational numbers in k=Q(√d). α in I(d) is defined as reduced if α>1, 1<α′<0 (α′ is the conjugate of α with respect to Q). Also, R(d) is the set of all reduced quadratic irrational numbers in I(d). Besides, it is well known that any number α in R(d) has purely periodic continued fraction expansion and the denominator of its modular automorphism is equal to fundamental unit ϵd of k=Q(√d). Yokoi's invariants defined by H. Yokoi are determined by the coefficients of the fundamental unit ϵd=td+ ud√d2 as md=⌊u2dtd⌋ and nd=⌊tdu2d⌋ where ⌊x⌋ represents the floor of x.
Definition 2.1. Let's fix a positive odd integer γ and define {ci} sequence with the recurrence relation
ci=γci−1+ci−2 |
for i≥2 with the seed values c0=0 and c1=1.
Remark 2.1. If {ci} sequence is defined as Definition 2.1, then we state the following congruence for γ≡1(mod4):
cn≡{0(mod4), n≡0(mod6);1(mod4), n≡1,2,5 (mod6);2(mod4), n≡3(mod6);3(mod4), n≡4(mod6). |
and also, for γ≡3(mod4):
cn≡{0(mod4), n≡0(mod6);1(mod4), n≡1,4,5 (mod6);3(mod4), n≡2(mod6);2(mod4), n≡3(mod6). |
is satisfied where n≥0.
Lemma 2.1. Let d be a square-free positive integer. If we put ωd=σ−1+√dσ, a0=⌊ωd⌋ into the ωR, then we get ωd∉R(d) but ωR∈R(d). Also, let ωR=PlωR+Pl−1QlωR+Ql−1 be an image of the ωR under the particular automorphism, then the fundamental unit ϵd of Q(√d) is determined as follows:
(i) If d is congruent to 1 modulo 4, then σ=2 with
ϵd=td+ud√d2 |
and
td=(2a0−1).Ql(d)+2Ql(d)−1,ud=Ql(d) |
(ii) If d is congruent to 2 or 3 modulo 4 then σ=1 as well as
ϵd=td+ud√d2=ωR.Ql(d)+Ql(d)−1>1, |
and
td=2a0Ql(d)+2Ql(d)−1,ud=2Ql(d) |
Since the denominator of the particular automorphism is the fundamental unit of Q(√d) with norm (−1)l(d) where Qi is determined by Q0=0, Q1=1 and Q(i+1)=aiQi+Qi−1, (i≥1).
Proof: (ⅰ) Proof of the part in Lemma 2.1. was demonstrated in the reference [26,Lemma 1, pp 41].
(ⅱ) It is trivial that second part of Lemma 2.1 can be proven with the consideration of proof (ⅰ).
Our theorems and results are given as follows:
Theorem 3.1. Let d be square-free positive integer and l>1 be a positive integer.
(1) We suppose
d=(2tcl+γ)2+8tcl−1+4 |
for t>0 positive integer. In this case, we obtain that d≡1(mod4) and
ωd=[tcl+γ+12;¯γ,γ,…,γ⏟l−1,2tcl+γ] |
and l=l(d). Moreover, we obtain
td=2tcl2+γcl+2cl−1,ud=cl |
for ϵd=td+ud√d2. Conversely, d≡1(mod4) is written in the terms of {ci} sequence, γ and t if
ωd=[tcl+γ+12;¯γ,γ,…,γ⏟l−1,2tcl+γ] |
holds.
(2) Let l be divisible by 3. if we suppose
d=(tcl+γ)2+4tcl−1+4 |
for t>0 positive odd integer, then we have d≡1(mod4) and
ωd=[t2cl+γ+12;¯γ,γ,…,γ⏟l−1,tcl+γ] |
with l=l(d). Furthermore, in this case
td=tcl2+γcl+2cl−1,ud=cl |
hold for ϵd=td+ud√d2. On the contrary, d≡1(mod4) is defined in the terms of {ci} sequence, γ and t if ωd is determined as above;
ωd=[t2cl+γ+12;¯γ,γ,…,γ⏟l−1,tcl+γ]. |
Proof: (1) Let the parameterization of d be d=(2tcl+γ)2+8tcl−1+4. Since γ is positive odd integer (2tcl+γ)2 is positive odd integer. So, we get d≡1(mod4). From (ⅰ) in Lemma 2.1, we know that ωd=1+√d2, a0=⌊ωd⌋ and ωR=a0−1+ωd. By using these equations, we obtain
ωR= (2tcl +γ−1)2+[tcl+γ+12;¯γ,γ,…,γ⏟l−1,2tcl+γ] |
so, we get
ωR=(2tcl+γ)+1γ+⋯+1ωR |
By a straightforward induction argument, we obtain
ωR2−(2tcl+γ)ωR−(1+2tcl−1)=0. |
This requires that ωR=(2tcl+γ)+√d2 since ωR>0. If we consider again (ⅰ) in Lemma 2.1, we get
ωR= [tcl+γ+12;¯γ,γ,…,γ⏟l−1,2tcl+γ] |
and l=l(d).
Now, we have to determine ϵd,td and ud by using (ⅰ) in Lemma 2.1. So, we get
Q1=1=c1 ,Q2=a1Q1+Q0⇒Q2=γ=c2 , |
Q3=a2 .Q2+Q1=γ.c2+c1=γ2+1=c3 ,Q4=c4,… |
So, this implies that Qi = ci using induction for ∀i ≥ 0. If we substitute these values of sequence into the ϵd=td+ud√d2=(ωR).Ql(d)+Ql(d)−1>1 and rearrange, we have td and ud as follows:
td=2tcl2+γcl+2cl−1,ud=cl |
using (ⅰ) in Lemma 2.1 for ϵd=td+ud√d2.
Conversely, assume that d≡1(mod4), ωd=[tcl+γ+12;¯γ,γ,…,γ⏟l−1,2tcl+γ] holds.
Since ωd=1+√d2 and (ⅰ) in Lemma 2.1, we obtain d=(2tcl+γ)2+8tcl−1+4 which completes the proof of (1).
(2) If we assume that l≡0(mod3) and the parametrization of d is given as follows:
d=(tcl+γ)2+4tcl−1+4 |
for t>0 positive odd integer, then we have d≡1(mod4) since cl is even integer. By substituting t2 instead of t into the case (1), we get
ωd=[t2cl+γ+12;¯γ,γ,…,γ⏟l−1,tcl+γ] |
and l=l(d). Furthermore,
td=tcl2+γcl+2cl−1,ud=cl |
hold for ϵd=td+ud√d2.
On the other hand, d=(tcl+γ)2+4tcl−1+4 is obtained by putting t2 instead of t into the case (1). So, the proof of (2) is also completed.
Corollary 3.1. Assume that d≡1(mod4). If d satisfies the conditions in Theorem 3.1, then Yokoi's invariant nd is nonzero (and hence md=0).
Proof: In the case of (1) in Theorem 3.1,
nd=⌊tdud2⌋=⌊2tcl2+γ.cl+2cl−1cl2⌋=2t+⌊γ.cl+2cl−1cl2⌋ |
Since t>0 is positive integer and {ci} is increasing sequence, we get nd≠0 for l>1. In a similar way, for the case of (2) in the Theorem 3.1, we obtain
nd=⌊tdud2⌋=t+⌊γcl+2cl−1cl2⌋ |
we have nd≠0 since t>0 and td>ud2.
Remark 3.1.
(i) Let d be a square free positive integer concerning the case of (1) in the Theorem 3.1. As a result of the theorem, ϵd, ωd and nd−md also have been calculated for γ=1,3,5, 7, 9 and t=1,2 with some values of 2≤l(d) in the published papers.
(ii) Let d be the square free positive integer concerning the case of (2) in the Theorem 3.1. As a result of the theorem, ϵd, ωd and nd−md also have been calculated for γ=1,3,5 and t=1,3 with 3≤l(d)≤12.
Proof: Readers can see all numerical results with tables in the references [15,16,17,18,19,20,21,22].
Remark 3.2. It is easily seen that the present paper has got the most general theorems for such types of real quadratic number fields since d is written by countable infinite integers t>0 and γ>0 (but γ is odd) as well as in the terms of increasing integer sequence {ci}.
Theorem 3.2. Let d be the square free positive integer and l>1 be a positive integer not divisible by 3. We assume that parameterization of d is
d=(γ+(2t+1)cl)24+(2t+1)cl−1+1 |
for t>0 positive integer. Assume that γ≡1(mod4) is any positive odd integer, then we have the followings:
(1) If l≡1(mod6) and t≡0(mod2) are positive integers, then d≡2(mod4) holds.
(2) If l≡2(mod6) and t≡0(mod2) are positive integers, then d≡3(mod4) holds.
(3) If l≡4(mod6) and t≡0(mod2) are positive integers, then d≡3(mod4) holds.
(4) If l≡5(mod6) and t≡1(mod2) are positive integers, then d≡2(mod4) holds.
Also, suppose that γ≡3(mod4) is any positive odd integer, then we have the followings:
(1∗) If l≡1(mod6) and t≡1(mod2) are positive integers then d≡2(mod4) holds.
(2∗) If l≡2(mod6) and t≡0(mod2) are positive integers then d≡3(mod4) holds.
(3∗) If l≡4(mod6) and t≡0(mod2) are positive integers then d≡3(mod4) holds.
(4∗) If l≡5(mod6) and t≡0(mod2) are positive integers then d≡2(mod4) holds.
Then, we obtain
ωd=[(2t+1)cl+γ2;¯γ,γ,…,γ⏟l−1,(2t+1)cl+γ] |
and l=l(d). Moreover, we have the following equalities:
ϵd=((2t+1)cl22+γcl2+cl−1)+cl√d |
td=(2t+1)cl2+γcl+2cl−1,ud=2cl, |
for ϵd, td and ud. Conversely is also true.
Proof: It is clear that d is not integral while l≡0(mod3) considering Remark 2.1. So, we assume that d is not divisible by 3 in order to get integer d. We first assume that l≡1(mod4) positive odd integer, l≡1(mod6), l>1 and t≡0(mod2) positive integer. So, we get d≡2(mod4) by substituting the equivalent results into the parameterization of d. We obtain the other cases in a similar way.
By using (ⅱ) in Lemma 2.1, we have
ωR=(2t+1)cl+γ2+[(2t+1)cl+γ2;¯γ,γ,…,γ⏟l−1,(2t+1)cl+γ], |
So, we get
ωR=((2t+1)cl+γ)+1γ+⋯+1ωR |
By a straightforward induction algorithm, we get
ωR=((2t+1)cl+γ)+cl−1ωR+cl−2clωR+cl−1 |
By rearranging the Definition 2.1 into above equality, we obtain
ωR2−((2t+1)cl +γ)ωR−(1+(2t+1)cl−1)=0 |
This requires that ωR=(2t+1)cl2+√d since ωR>0. If we consider (ⅱ) in Lemma 2.1, we get
ωd=√d=[(2t+1)cl+γ2;¯γ,γ,…,γ⏟l−1,(2t+1)cl+γ] |
and l=l(d). This shows that the part of continued fraction expansions is completed.
Now, we should determine ϵd, td and ud by using (ⅱ) in Lemma 2.1. Thereby, we have
Q1=1=c1 ,Q2=a1Q1+Q0⇒Q2=γ=c2, |
Q3=a2 .Q2+Q1=γc2+c1=γ2+1=c3,Q4=c4,… |
So, this implies that Qi = ci by using mathematical induction for ∀i ≥ 0. If we substitute these values of sequence into the ϵd=td+ud√d2=(ωR).Ql(d)+Ql(d)−1>1 and rearranged, we have td and ud using (ⅱ) in Lemma 2.1 as follows:
ϵd=((2t+1)cl22+γcl2+cl−1)+cl√d |
td=(2t+1)cl2+γcl+2cl−1,ud=2cl, |
for ϵd, td and ud which complete the proof of Theorem 3.2. Conversely, it is also true and trivial that if we consider (ⅱ) in Lemma 2.1 and the definition of ωd=√d.
Remark 3.3. Let d be the square free positive integer concerning Theorem 3.2. As a result of the theorem ωd, ϵd and nd−md also have been calculated for γ=1,3,5,7 and t=0,1,2,3 with some values of 2≤l(d).
Proof: Readers can see all numerical results with tables in the references [15,16,17,18,19,20,21,22].
I would like to thank the referee for thorough evaluation, constructive comments and recommendations to substantially improve the readability and quality of this paper.
The author declares that there is no conflict of interests.
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