On the real quadratic fields with certain continued fraction expansions and fundamental units

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1 Int. J. Noninear Ana. App. 8 (07) No., ISSN: (eectronic) On the rea quaratic fies with certain continue fraction expansions an funamenta units Özen Özera,, Ahme Khammas b a Department of Mathematics, Facuty of Science an Arts, Kırkarei University, Kırkarei, Turkey b Department of Mathematics, A-Qura University, Makkah,955, Saui Arabia (Communicate by M. Eshaghi) Abstract The purpose of this paper is to investigate the rea quaratic number fies Q( ) which contain the specific form of the continue fractions expansions of integra basis eement where, 3(mo4) is a square free positive integer. Besies, the present paper eas with etermining the funamenta unit ) ɛ = (t + u / an n an m Yokoi s -invariants by reference to continue fraction expansion of integra basis eement where () is a perio ength. Moreover, we mention cass number for such fies. Aso, we give some numerica resuts concue in the tabes. Keywors: Quaratic Fie; Funamenta Unit; Continue Fraction Expansion; Cass Number. 00 MSC: Primary R; Seconary R7, A55, R9.. Introuction an preiminaries A quaratic fie is efine as an agebraic number fie Q( ) of egree two over Q the rationa numbers. Q( ) is cae rea quaratic fie if > 0. The cass number of a number fie is efine by the orer of the iea cass group of its ring of integers. There are infinitey many quaratic fies an a of them have cass numbers such as one, two or more but it is not even known whether there are infinitey many rea quaratic fies with cass number one or two. So, cass number probem is particuary important. There are ifferent various methos to etermine cass number. One of them is Cassica Dirichet Cass Number formua h = D.L(,χ ) efine by iscriminat, reguator an ogɛ Özen Özer Emai aresses: ozenozer39@gmai.com (Özen Özer), sakhammash@uqu.eu.sa (Ahme Khammas) Receive: September 06 Revise: December 06

2 98 Özer, Khammas Dirichet L function. Dirichet L function is etermine by Dirichet character an zeta function. Aso, iscriminat D of the quaratic fie Q( ) is equa to if is congruent to mouo 4 an is equa to 4 if is congruent to,3 mouo 4. Besies, reguator is epen on funamenta unit ɛ.moreover, in rea quaratic fies, the ring of integers of Q( ) has infinitey many units that are equa to ɛ i or ɛi, where i is an arbitrary integer an ɛ is funamenta unit. So,it is aso very significant to etermine funamenta unit for stuying on the cass number probems, unit group an the etermination of the structures of rea quaratic number fies. In [6], Kim an Ryu worke on the specia circuar unit ɛ k of k = Q( pq) an investigate unit group of such rea quaratic fie. Aso, they prove cass number of quartic fie using Sinnott s inex formua. Moin an Wiiams etermine positive square free integers with cass number h() = in [9] an cass number h() = in [0] when perio ength of the continue fraction of w is equa or ess than 5. The theorems of Friesen in [3] an Hater-Koch in [4] investigate rea quaratic fies with arge funamenta units by constructing a infinite famiy. In [9], Tomita with Yamamuro gave some resuts for funamenta unit by using Fibonacci sequence an continue fraction.tomita an Kawamato obtaine significant resuts on specia rea quaratic fies of minima type in [5]. Moreover, Tomita etermine the continue fraction expansion of integra basis eement for perio ength is 3 in [8]. Sasaki in [6] an Moin in [8] stuie on ower boun of funamenta unit for rea quaratic number fies an get certain important resuts. In [], Benamar, Chanou an Mkaouar gave the ower boun of the number of non-squares monic poynomias with the fixe perio continue fraction expansions. Baziahin an Shait [], worke on specia continue fraction expansions of the rea numbers which are transcenenta numbers. Yokoi efine severa invariants but especiay two of them are important for cass number probem an the soutions of Pe equation ([]-[4]).Zhang an Yue in [5] gave some congruence reations about x an y whie ɛ = x + y an N(ɛ ) = + in a rea quaratic fie Q( ) with o cass number. The first author obtaine some resuts on ifferent types of continue fraction expansion of w in [],[3] an [4]. Aso, we can refer to the references [7], [],[5],[7] an [0] for reaers. Let k = Q( ) be a rea quaratic number fie where > 0 is a positive square-free integer. Integra basis eement is enote by w = for, 3(mo4) where () is the perio ength in the simpe continue fraction expansion of w. The funamenta unit ɛ of rea quaratic number ) fie is enote by ɛ = (t + u / where N (ɛ ) =( ) (). For the set I() of a quaratic irrationa numbers in Q( ), we say that α in I() is reuce if α >, < α < 0 (α is the conjugate of α with respect to Q), an enote by R() the set of a reuce quaratic irrationa numbers in I(). Then, it is we known that any number α in R() is purey perioic in the continue fraction expansion an the enominator of its mouar automorphism is equa to funamenta unit ɛ of Q( )). Besies, Yokoi s invariants are expresse by n = [[ ]] [[ ]] t u an m u = t where [[x]] represents the greatest integer not greater than x. In this paper, we interest in the probem etermining the rea quaratic number fies incue continue fraction expansions which have got partia quotient eements written as 6s (except the ast igit of the perio) for the perio ength.then, we categorize them with regar to arbitrary perio ength. Aso, we etermine the genera forms of funamenta unit ɛ an the coefficents of funamenta units t, u by using this new formuization which have not been known yet. Lasty,we cacuate cass numbers for some of such fies an give resuts on the Yokoi s invariants with tabes.

3 On the rea quaratic fies with certain continue fraction... 8 (07) No., Definition.. {Z i } sequence is efine by the recurrence reation Z i = 6Z i + Z i for i where Z 0 = 0 an Z =. Main resuts epen on the terms of this sequence. Definition.. Let c n = ac n + bc n be a sequence with recurrence reation of {c n } sequence where a, b are rea numbers. The poynomia is cae as a characteristic equation if it is written in the foowing form: x ax b = 0 By using the efinition, we etermine the characteristic equation as foows: r 6r = 0 for {Z k } sequence. So, each eement of sequence can be written as foows: [ ( Z k = 3 + k ( 0) 3 ) ] k 0 0 for k 0. Remark.3. If {Z n } is efine as in Definition. then, we specify the foowing equivaent. (mo4), n, 3 (mo4); Z n (mo4), n (mo4); 0 (mo4), n 0 (mo4). for n 0. Lemma.4. Let be a square-free positive integer satisfying congruent to, 3 mouo 4. If we put ω =, a 0 = [[ω ]] into the ω R = a 0 + ω where [[ω ]] represents the greatest integer not greater than ω, then we get ω / R(), but ω R R() hos. Moreover, for the perio = () of ω R, we get ω R = [a 0, a,..., a ] an ω = [a 0, a,..., a, a 0 ]. Furthermore, et ω R = (P ω R +P ) (Q ω R +Q ( = ) [a 0, a,..., a, ω R ] be a mouar automorphism of ω R, Then the funamenta unit ɛ of Q is given by the foowing formuas: an ɛ = t + u = (a 0 + )Q () + Q () t = a 0.Q () + Q (), u = Q (). where Q i is etermine by Q 0 = 0, Q = an Q i+ = a i Q i + Q i, (i ). Proof. Proof is omitte in the reference [8].

4 00 Özer, Khammas [[ ]] Lemma.5. Let be the square free positive integer congruent to, 3 mouo 4 an a 0 = enote the the integer part of integra basis eement for congruent to, 3(mo4).If we consier w which has got partia quotient eements repeate 6s in the perio ength = (), then we obtain continue fraction expansions as foows w = = [ a 0 ; a, a,..., a (), a () ] = [ a0 ; 6, 6,..., 6, a 0 ] for quaratic irrationa numbers an w R = a 0 + = [ a 0, 6,..., 6 ] for reuce quaratic irrationa numbers. Moreover, A k = a 0 Z k+ +Z k an B k = Z k+ are etermine in the continue fraction expansions where {A k } an {B k } are two sequences efine by: A = 0, A =, A k = a k A k + A k, B =, B = 0, B k = a k B k + B k, for k 0 an k < () where () is perio ength of w. We obtain A = a 0Z + 3a 0 Z + Z an B = a 0 Z + Z for k = () too. Aso, in the continue fraction w R = a 0 + = [b, b,..., b n,... ] = [a 0, 6,..., 6,... ], P k = a 0 Z k + Z k an Q k = Z k are etermine where {P k } an {Q k } are two sequences efine by: for k 0. P = 0, P 0 =, P k+ = b k+.p k + P k, Q =, Q 0 = 0, Q k+ = b k+.q k + Q k, Proof.We can prove using inuction.it is cear that assertion is true for k = 0.Assume that the resut is true for k < i an 0 < i.using the efinition of {Z i } sequence, we obtain A k+ = a k+.a k + A k = 6 (a 0 Z k+ + Z k ) + (a 0 Z k + Z k ) = a 0 (6Z k+ + Z k ) + (6Z k + Z k ) = a 0 Z k+ + Z k+ Aso, we get the foowing resut: B k+ = a k+.b k + B k = 6Z k+ + Z k = Z k+ Moreover,( since a =.a 0 ), we obtain A = a 0Z + 3a 0 Z + Z an B = a 0.Z + Z for k = (). In the same vein, for the continue fraction of a 0 + = [b, b,..., b n,... ] = [a 0, 6,..., 6,... ], we obtain that P k = a 0 Z k + Z k an Q k = Z k are etermine where {P k } an {Q k } are two sequences efine by: for k 0.This competes the proof. P = 0, P 0 =, P k+ = b k+.p k + P k, Q =, Q 0 = 0, Q k+ = b k+.q k + Q k, Theorem.6. If k 4 then with possiby ony one more vaue remaining h() = if an ony if is entry in Tabe.

5 On the rea quaratic fies with certain continue fraction... 8 (07) No., Proof. Proof is in the reference [9]. Lemma.7. f k 4 an D < then with at most one possibe exception we must have h() =. Proof. Proof is in the reference [0]. Lemma.8. For any s. an e s in, () if m is ifferent from 0 then h > s 4 σ (s ) s. (s ) s () if m = 0 (i.e. n is ifferent from 0) then h > s 4 σ exception of. og(m +) hos with one possibe exception of. (s ) s og( n +). (s ) s (3) if Q( ) is a rea quaratic fie of R.D.type, then h > s 4 σ possibe exception of. Proof. Proof is in the reference [4]. (s ) s og3 hos with one possibe. (s ) s hos with one Remark.9. In the reference [4], Yokoi gave a tabe for square free integers between = an = 499 incue Yokoi s invariants an cass numbers. Definition.0. Let = n + r, > 5 be a positive square free integer satisfying the conitions 4n 0(mor) an n < r n. In this case, the rea quaratic fie Q( ) is cae a Richau- Degert (R-D) type rea quaratic fie.. The main resuts Theorem.. Let be the square free positive integer an be a positive integer hoing that 0(mo) an >. We assume that parametrization of is = r Z + (3Z + Z ) r for r > 0 positive integer. Then the foowing conitions ho: () If 0(mo4) an r (mo4) positive integer then 3(mo4) satisfies. () If (mo4) an r (mo4) positive integer then (mo4) satisfies. Aso, we get w = rz + 3; 6, 6,..., 6, rz }{{} + 6 an = (). Besies, we obtain the foowing equaities: ( ) rz ɛ = + 3Z + Z + Z for ɛ, t an u. t = rz + 6Z + Z an u = Z

6 0 Özer, Khammas Proof. We assume that 0(mo) an >. Then, we have to investigate two cases as foows: If 0(mo4), then we get Z 0(mo4), Z (mo4). By consiering r (mo4) positive integer an substituting these equivaent an equations into the parametrization of, then we obtain 3(mo4). If (mo4), then we get Z (mo4), Z (mo4). So, we have (mo4) by using r (mo4) positive integer. We put, then we get w R = rz rz + 3; 6, 6,..., 6, rz }{{} +, w R = (rz + 6) = (rz + 6) w R w R... By using Lemma.5 an the inuction with the properties of continue fraction expansion, we obtain w R = (rz + 6) + Z w R + Z Z w R + Z, Consiering Definition. with the above equaity, we have w R (rz + 6) w R ( + rz ) = 0. This requires that w R = rz since w R > 0. If we consier Lemma.4., we get w = rz + 3; 6, 6,..., 6, rz }{{} + 6 an = (). This shows that first the part of the proof is compete. Now, we have to etermine ɛ, t an u using Lemma.4, we obtain vaues of Q k as foows: Q = = Z, Q = a.q + Q 0 Q = 6 = Z, Q 3 = a Q + Q = Z + Z = 37 = Z 3, Q 4 = 8 = Z 4,... So, this impies that Q i = Z i by using mathematica inuction for i. If we substitute these vaues of sequence into the ɛ = t +u = (a 0 + )Q () + Q () an rearrange, we have ( ) rz ɛ = + 3Z + Z + Z t = rz + 6Z + Z an u = Z for ɛ, t an u. It competes the proof of Theorem..

7 On the rea quaratic fies with certain continue fraction... 8 (07) No., Coroary.. Let be a square free positive integer an be a positive integer hoing that 0(mo) an >. We assume that the parametrization of is ( ) = Z 4 + Z+ + Z + 0. Then, we get, 3(mo4) an w = Z + 3; 6, 6,..., 6, Z }{{} + 6 an = (). Furthermore, we obtain the foowing equaities: ( ) Z ɛ = + 3Z + Z + Z t = Z + 6Z + Z an u = Z {, if = ; m = 3, if 4; for ɛ, t, u an m. Besies, we inicate the foowing Tabe where funamenta unit is ɛ, integra basis eement is w an Yokoi s invariant is m for < () 0. Tabe : Square -free positive integers where () is even an () 0. () m w ɛ 38 [6; 6, ] [7; 6, 6, 6, 34] [433; 6, 6, 6, 6, 6, 8664] [6439; 6, 6, 6, 6, 6, 6, 6, 3878] [6448; 6, 6, 6, 6, 6, 6, 6, 6, 6, ] Proof. This resut is obtaine by substituting r = into the Theorem. We have to prove that vaues of m are etermine as foows: {, if = ; m = 3, if 4. If we put t an u into the m an rearrange, then we obtain [[ ]] [[ u m = 4Z = t Z + 6Z + Z ]] By consiering above equaization, we obtain m = for =. From the assumption (since Z is increasing sequence) we get, ( 4 > Z ) > 3, 89 Z Z

8 04 Özer, Khammas [[ for 4. Therefore, we obtain m = we escribe m as foows: 4Z Z +6Z +Z ]] = 3 for 4 ue to efinition of m an m = {, if = ; 3, if 4; which competes the proof of Coroary.. Aso, Tabe can be obtaine as a iustrate of this coroary. Remark.3. In the Tabe, Q( ) is a R-D type rea quaratic fie for = 38 = 6 + but others aren t R-D type. By using the cassica Dirichet cass number formua, we cacuate cass number h = for Q( 38). Aso, the fie is obtaine in the tabe of [4] an the Tabe 3. of [9]. Besies, in the same tabe, we can see that other cass numbers such as Q( 377) with h = 8 an Q( ) with h = 44 are too bigger than cass number two by using Proposition 4. of [4]. Coroary.4. Let be a square free positive integer an be a positive integer satisfying 0(mo) an >. Suppose that the parametrization of is Then, we have, 3(mo4) an = 5Z 4 w = 5Z + (5Z + 5Z ) ; 6, 6,..., 6, 5Z }{{} + 6 an = (). Moreover, we obtain the foowing equaities: ɛ = ( 5Z + 3Z + Z ) + Z t = 5Z + 6Z + Z an u = Z n = for ɛ, t, u an Yokoi s invariant n. Besies, we get the foowing Tabe where funamenta unit is ɛ, integra basis eement is w an Yokoi s invariant is n for < () 0.(In this tabe, we rue out ()=4 since is not a square free positive integer). Tabe : Square -free positive integers where () is even an () 0 except () = 4. () n w ɛ 330 [8; 6, 36] [648; 6, 6, 6, 6, 6, 4396] [8943; 6, 6, 6, 6, 6, 6, 6, ] [3078; 6, 6, 6, 6, 6, 6, 6, 6, 6, 64456] Proof. In a simiar way of mentione above coroary, this coroary is obtaine by substituting r = 5 in Theorem..

9 On the rea quaratic fies with certain continue fraction... 8 (07) No., Remark.5. In the Tabe, there is no rea quaratic fie of R-D type except Q( 330). Aso,we can give an exampe on cass number as h = 4 for Q( 330). Theorem.6. Let be a square free positive integer an be a positive integer hoing that. Suppose that the parametrization of is = r Z + r(3z + Z ) + 0. for r > integer. If r 0(mo) is a positive integer then (mo4) hos. So, we have w = rz + 3; 6, 6,..., 6, 6 + rz }{{} with = (). Besies, we get funamenta unit ɛ, coefficients of funamenta unit t, u as foows: ɛ = (rz + 3) Z + Z + Z, t = (rz + 3) Z + Z an u = Z. Proof. If > arbitrary positive integer an r is even positive integer then we obtain (mo4) since = r Z + r(3z + Z ) + 0. If we put then we have w R = (rz + 3) + rz + 3; 6, 6,..., 6, 6 + rz }{{} w R = (rz + 6) By a straight forwar inuction argument, we get w R w R = (rz + 6) + Z w R + Z Z w R + Z. Moreover, by using Definition. with the above equaity, we obtain w R (rz + 6) w R ( + rz ) = 0. = (rz + 6) This requires that w R = (rz + 3) + since w R > 0. If we consier Lemma.5, we get w = rz + 3; 6, 6,..., 6, 6 + rz }{{} w R.

10 06 Özer, Khammas an = (). Now, we have to etermine ɛ, t an u using Lemma.4, we get Q = = Z, Q = a.q + Q 0 Q = 6 = Z Q 3 = a 6 Q + Q = 6Z + Z = 37 = Z 3, Q 4 = 8 = Z 4. So, this impies that Q i = Z i by using mathematica inuction for i 0. If we substitute these vaues of the sequence into the ɛ = t +u = (a 0 + )Q () + Q () an rearrange, we have ɛ = (rz + 3) Z + Z + Z, t = (rz + 3) Z + Z an u = Z. Coroary.7. Suppose that the parametrization of is = 4Z + (Z + + Z ) + 0. where is a square free positive integer an is a positive integer. Then, we have (mo4) an w = Z + 3; 6, 6,..., 6, 6 + 4Z }{{} with = (). Moreover, we can get funamenta unit ɛ, coefficients of funamenta unit t, u as foows: ɛ = (Z + 3) Z + Z + Z, t = (Z + 3) Z + Z an u = Z. Aso, we have Yokoi s - invariant vaue n = for. Furthermore, we have the foowing Tabe 3 where funamenta unit is ɛ, integra basis eement is w an an Yokoi s invariant is n for () 8. Tabe 3: Square -free positive integers with () 8. () n w ɛ 30 [5; 6, 30] [77; 6, 6, 54] [459; 6, 6, 6, 98] [83; 6, 6, 6, 6, 566] [739; 6, 6, 6, 6, 6, 34638] [06709; 6, 6, 6, 6, 6, 6, 6, 348] [657555; 6, 6, 6, 6, 6, 6, 6, 350] Proof. If we substitute r = in Theorem.6, this coroary is got.we emonstrate that the vaue of Yokoi s -invariant is [[ n = ]] for. t We know that n = consiering Yokoi s references. If we substitute t u an u into the n, then we get [[ ]] [[ t 4Z ]] [[ n = = + 6Z + Z 3 u 4Z = + + Z ]] Z Z =,

11 On the rea quaratic fies with certain continue fraction... 8 (07) No., since Z is increasing an 0 < 3 Z + Z < 0, 64 for. Therefore, we obtain n Z = for. This competes the proof of Coroary.7. Besies, Tabe 3 is create as numerica exampe. Remark.8. In the Tabe 3, there isn t any R-D type of rea quaratic fie excuing Q( 30) with h = which is aso obtaine in the Tabe. of [0] an the tabe of [4]. Aitionay, we can cacuate vaues of cass numbers for some rea quaratic fies as foows: Q( 5954) with cass number h =, Q( 0830), with cass number h =48, Q( 79388) with cass number h =38. Coroary.9. Let be a square free positive integer an be a positive integer satisfying. Suppose that parametrization of is Then, we have (mo4) an = 6Z + 4(Z + + Z ) + 0. w = 4Z + 3; 6, 6,..., 6, 6 + 8Z }{{} with = (). Aitionay, we get funamenta unit ɛ, coefficients of funamenta unit t, u as foows: ɛ = (4Z + 3) Z + Z + Z, t = (4Z + 3) Z + Z an u = Z. an Yokoi s - invariant vaue n = for. Besies, we inicate the Tabe 4 where funamenta unit is ɛ, integra basis eement is w an Yokoi s invariant is n for () 8 (In this tabe, we rue out ()=,3,4,6,8 since is not a square free positive integer). Tabe 4: Square -free positive integers with () = 5 or () = 7. () n w ɛ [563; 6, 6, 6, 6, 46] [345; 6, 6, 6, 6, 6, 6, 46830] Proof. The resut is got if we substitute r = 4 in Theorem.6 The vaue of Yokoi s -invariant is obtaine as n = for [[. ]] t We know that n = from H. Yokoi s references. If we substitute t u an u into the n, then we get [[ ]] [[ t 8Z ]] [[ n = = + 6Z + Z 3 u 4Z = + + Z ]] Z Z =, + Z Z since (Z ) is increasing an 0 < 3 Z < 0, 00 for. Therefore, we obtain n = for. This competes the proof of Coroary.9. For numerica exampes, we tabuate the Tabe 4. Remark.0. There isn t any R-D type of rea quaratic fie in the Tabe 4. As an iustration, we can say that the cass number is h = 34 for Q( ).

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