NOTES ON ESPECIAL CONTINUED FRACTION EXPANSIONS AND REAL QUADRATIC NUMBER FIELDS

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1 Ozer / Kirklareli University Journal of Engineering an Science (06) NOTES ON ESPECIAL CONTINUED FRACTION EXPANSIONS AND REAL QUADRATIC NUMBER FIELDS Özen ÖZER Department of Mathematics, Faculty of Science an Arts, Kırklareli University, 3900, Kırklareli - TURKEY ozenozer39@gmail.com Abstract The primary purpose of this paper is to classify real quaratic fiels Q( ) which inclue the form of specific continue fraction expansion of integral basis element w for arbitrary perio length l = l() where,3(mo4) is a square free positive integers. Furthermore, the present paper eals with etermining new certain parametric formulas of funamental unit t u an Yokoi s -invariants n, m for such real quaratic fiels. All results are also supporte by several numerical tabular forms. Key Wors: Quaratic Fiels, Continue Fractions, Funamental Units. 00 AMS Subject Classification: R, A55, R7. Özet Bu makalenin asıl amacı,,3(mo4) kare çarpansız pozitif tamsayılar olmak üzere keyfi l = l() periyo uzunluğu için tamlık taban elemanı olan w nin özel bir sürekli kesre açılımınaki formu içeren Q( ) reel kuaratik sayı cisimlerini sınıflanırmaktır. olan Ayrıca bu çalışma, ilgili reel kuaratik sayı cisimleri için Yokoi nin -invaryantları n, m ile t u temel biriminin kesin parametrik formüllerinin belirlenmesi ile ilgilenmekteir. Tüm sonuçlar bir takım nümerik tablolar ile e esteklenmekteir. Anahtar Kelimeler: Kuaratik Cisimler, Sürekli Kesirler, Temel Birimler. Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 74

2 Ozer / Kirklareli University Journal of Engineering an Science (06) INTRODUCTION Quaratic fiels have applications in ifferent areas of mathematics such as quaratic forms, algebraic geometry, iophantine equations, algebraic number theory, an even cryptography. The Unit Theorem for real quaratic fiels says that every unit in the integer ring of a quaratic fiel is given in terms of the funamental unit of the quaratic fiel. Thus, etermining the funamental units of quaratic fiels is of great importance. Let k = Q( ) be a real quaratic number fiel where > 0 is a positive square free integer. Integral basis element is enote by w = [ a 0 ; a, a,, a l(), a 0 ] an l() is the perio length in simple continue fraction expansion of algebraic integer w for,3(mo4). The funamental unit of real quaratic number fiel is also enote by t u where N ( ) = ( ) l(). Furthermore, Yokoi s invariants are expresse by m = u t an n = t u where x represents the greatest integer not greater than x. The sequence {Y n } is also special sequence which will be efine in Section. By using coefficients of funamental unit H.Yokoi efine two significant invariants such as m, n for class number problem an the solutions of Pell equation in [9]. by using In [7], Tomita escribe explicitly form of the funamental units of the real quaratic fiels Fibonacci sequence an continue fraction. He also gave some results for the continue fraction expansion of w where (mo4) for l() = 3 in [6]. Determining of some certain funamental units t u of k = Q( ) was stuie by R.Sasaki an R.A.Mollin ([4], []). Moreover, please see [3],[5] an [8] for more etails about continue fraction expansions. We will investigate the continue fraction expansions which have partial quotients elements as 5s (except the last igit of the perio, which is always for w ) with a given perio length. Although there are infinitely many values of having all 5s in the symmetric part of perio of integral basis element, we will classify them accoring to Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 75

3 Ozer / Kirklareli University Journal of Engineering an Science (06) arbitrary perio length. We will also etermine the general forms of funamental units ε an coefficents of funamental units t u t, u in the terms of {Y n } as new formulizations which have been unknown yet for such real quaratic fiels. By using results, the funamental unit, continue fraction expansions an Yokoi s invariants will be calculate more easily for such Q( ).. PRELIMINARIES We nee following efinitions an lemmas which will be use in our main results for the section 3. Definition.. {Y i } is sai to be a sequence efine by the recurrence relation Y i = 5Y i + Y i with see values Y 0 = 0 an Y =. We can calculate some values of the terms of the sequence as follows: Y = 5Y + Y 0 = 5, Y 3 = 5Y + Y = 5 + = 6, Y 4 = 5Y 3 + Y = = 35, Y 5 = 5Y 4 + Y 3 = = 70, Y 6 = 5Y 5 + Y 4 = 3640, Y 7 = 890, Y 8 = 9845, Y 9 = 50966, Y 0 = 64675,Y = 37400, Y = 73580, Y 3 = , This sequence plays an important role in this paper to escribe our lemmas an main results. Lemma.. For a square free positive integer congruent to,3 moulo 4, we put w =, a 0 = into the w R = a 0 + w. Then w R(), but w R R () hols. Moreover, for the perio l = l() of w R, we get w R = [ a 0, a, a,, a l() ] an w = [a 0 ; a, a,, a l(), a 0 ]. Furthermore, let w R = w RP l +P l = [ a w R Q l +Q l 0, a, a,, a l(), w R ] be a moular automorphism of w R. Then the funamental unit of Q( ) is given by the following formula: t u ( a0 ) Ql ( ) Ql ( ) t a0. Ql ( ) Ql ( ) an u Q l ( ) where Q is etermine by Q 0 0, Q an Q i aiqi Qi, (i ). i Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 76

4 Ozer / Kirklareli University Journal of Engineering an Science (06) Proof. Proof is omitte in [6]. Lemma.. Let,3(mo4) be the square free positive integer an w has got partial constant elements repeate 5s in the case of perio l = l(). If a 0 = enote the integer part of w = for,3(mo4), then we have continue fraction expansion w = = [ a 0 ; a, a,, a l(), a l() ] = [ a 0 ; 5,5,,5, a 0 ] for quaratic irrational numbers an w R = a 0 + = a 0 + [ a 0 ; 5,5,,5, a 0 ] = [ a 0, 5,,,5 ] for reuce quaratic irrational numbers. Furthermore, A k = a 0 Y k+ + Y k an B k = Y k+ are etermine in the continue fraction expansions where {A k } an {B k } are two sequences efine by : an A = 0, A =, Ak ak Ak Ak (for 0 k l ) B =, B = 0, Bk ak Bk Bk (for 0 k l ) A a Y 3a Y Y ( for k = l() ) l 0 l 0 l l Bl a0yl Yl ( for k = l() ) where l = l() is perio length of w =. Also, C j = A j Bj is the j th convergent in the continue fraction expansion of. Moreover, in the continue fraction [b, b, b 3, b n, ] = [ a 0, 5, 5,,5, ], P j = a 0 Y j + Y j an Q j = Y j are obtaine where {P j } an {Q j } are two sequences efine by for j 0. P = 0, P 0 =, Pj bj Pj Pj Q =, Q 0 = 0, Qj bj Q j Qj Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 77

5 Ozer / Kirklareli University Journal of Engineering an Science (06) Proof. We can prove by using mathematical inuction. Using the following table which k a k a A k 0 ( a 0 ) a 0 Y + Y 0 (5a 0 + ) a 0 Y + Y (6 a 0 + 5) a 0 Y 3 + Y (35 a 0 + 6) a 0 Y 4 + Y 3 (70 a ) a 0 Y 5 + Y 4 B k 0 Y 0 Y 5 Y 6 Y 3 35 Y 4 70 Y 5 Table.. (Convergent of [ a 0 ; 5,5,,5, a 0 ] for l = l()) inclues values of A k, B k an a k, we can easily say that this is true for k = 0. Now, we assume that the result true for k < i. Using the efine relations for {Y i } sequence, we obtaine (a i = 5 for i l ) A a A A = 5(a 0 Y k+ + Y k ) + (a 0 Y k + Y k ) k k k k k k k k = a 0 (5Y k+ + Y k ) + (5Y k + Y k ) = a 0 Y k+ + Y k+ B a B B = 5Y k+ + Y k = Y k+ Moreover, since a l = a 0 we get the following result : A a Y 3a Y Y l 0 l 0 l l Bl a0yl Yl ( for k = l() ) Furthermore, in the continue fraction [b, b, b 3, b n, ] = [ a 0, 5, 5,,5, ],we have following table: j b j a ( a 0 ) (0 a 0 + ) (5 a 0 + 5) (70 a 0 + 6) P j 0 a 0 Y + Y 0 a 0 Y + Y a 0 Y 3 + Y a 0 Y 4 + Y 3 Q j 0 Y 0 Y 5 Y 6 Y 3 35 Y 4 Table.. Convergent of [ a 0, 5, 5,,5, ] Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 78

6 Ozer / Kirklareli University Journal of Engineering an Science (06) The Table. completes proof. Definition.. Let c n = ac n + bc n be the recurence relation of {c n } sequence where a, b are real numbers. The polynomial is calle as a characteristic equation is written in the form of r ar b = 0 The solutions will epen on the nature of the roots of the characteristic equation for recurence relation. By using the efinition, we fin characteristic equation as r 5r = 0 for {Y k } sequence. So, we can write each element of sequence as follows: Y k = 9 k + 9 [(5 ) ( 5 9 k ) ] for k 0. Lemma.3. Let {Y k } be the sequence efine as in Definition. an Definition.. Then, we have for k. Y k > (5 ) { 9 k + 9 (5 ) k ; if k is even integer ; if k is o integer Proof. As a result of the Lemma. this proof can be obtaine easily. Remark.. Let {Y n } be the sequence efine as in Definition.. Then, we state the following: Y n { 0(mo4) ; n 0(mo6) (mo4) ; n,,5(mo6) (mo4) ; n 3(mo6) 3(mo4) ; n 4(mo6) for n 0. Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 79

7 Ozer / Kirklareli University Journal of Engineering an Science (06) MAIN THEOREMS AND RESULTS preliminaries. The followings are our main theorem an results with the notation of the Main Theorem. Let be square free positive integer an l be a positive integer satisfying that 3 l,l. Suppose that the parametrization of is = ( 5 + (δ + )Y l ) + (δ + )Y l + where δ 0 is a positive integer. Then following conitions hol: () If l (mo6) an δ is even positive integer then (mo4) () If l (mo6) an δ is even positive integer then 3(mo4) (3) If l 4(mo6) an δ is even positive integer then 3(mo4) (4) If l 5(mo6) an δ is o positive integer then (mo4) In Q( ) real quaratic fiels, we have w = [ 5+(δ+)Y l with l = l() for,3(mo4). ; 5,5, ],5, 5 + (δ + )Y l l as follows: Aitionally, we get the funamental unit ε an coefficients of funamental unit t, u ε = ( 5+(δ+)Y l + ) Y l + Y l, t = (δ + )Y l + 5Y l + Y l an u = Y l Proof. We say that Z + by using Remark. for all l 0,3(mo6). So, we will assume that 3 l,l in orer to get Z +. First of all, we shoul show that four conitions hol as the followings: () if l (mo6) hols, then Y l (mo4) an Y l 0(mo4) hol. By substituting these values into parametrization of an consiering δ is even positive integer, we obtain (mo4). Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 80

8 Ozer / Kirklareli University Journal of Engineering an Science (06) () If l (mo6) satisfies, then Y l (mo4) an Y l (mo4) satisfy. By consiering δ is even positive substituting these values into parametrization of an rearranging, we have 3(mo4). (3) If l 4(mo6) an δ is even positive integer, then Y l 3(mo4) an Y l (mo4) hol an also by substituting these values into parametrization of, then 3(mo4) hols. (4) If l 5(mo6) an δ is o positive integer then we get Y l (mo4) an Y l 3(mo4). By substituting these values into parametrization of an rearranging, we have (mo4). So, conitions are satisfie. By using Lemma. we have w R = ( 5+(δ+)Y l ) + [ 5+(δ+)Y l ; 5,5, ],5, 5 + (δ + )Y l l w R = (5 + (δ + )Y l ) w R = (5 + (δ + )Y l ) By using Lemma. an Lemma., we get w R w R = (5 + (δ + )Y l ) + Y l w R + Y l Y l w R + Y l Using Definition. an put Y l+ + Y l = 5Y l + Y l equation into the above equality, we obtain w R (5 + (δ + )Y l )w R ( + (δ + )Y l ) = 0 This implies that w R = ( 5+(δ+)Y l ) + since w R > 0. If we consier Lemma., we get = [ 5+(δ+)Y l ; 5,5,,5,5 + (δ + )Y l ] an l = l(). Hence, w = [ 5+(δ+)Y l ; 5,5,,5, 5 + (δ + )Y l ] hols. l Now, we can etermine ε, t an u using Lemma. as follows: Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 8

9 Ozer / Kirklareli University Journal of Engineering an Science (06) Q 0 = 0 = Y 0, Q = = Y, Q = a. Q + Q 0 Q = 5 = Y, Q 3 = a Q + Q = 5Y + Y =Y 3, Q 4 = Y 4, So, this implies that Q i = Y i by using mathematical inuction for i 0. If we substitute t u these values of sequence into the ( a0 ) Ql ( ) Ql ( ) an rearrange, we get ε = ( 5+(δ+)Y l + ) Y l + Y l, t = (δ + )Y l + 5Y l + Y l an u = Y l We can obtain following theorems an conclusions from Main Theorem. Theorem 3.. Let be square free positive integer an l be a positive integer satisfying that l 5(mo6), 3 l an l. Suppose that parametrization of is = ( 5 + Y l ) + Y l + Then,we have,3(mo4) an w = [ 5+Y l ; 5,5, ],5, 5 + Y l with l = l(). l Aitionally, we get the funamental unit ε, coefficients of funamental unit t, u an Yokoi s invariant m as follows: ε = ( 5+Y l + ) Y l + Y l, t = Y l + Y l+ + Y l an u = Y l m = 3 Proof. This theorem is obtaine from main theorem by taking δ = 0. Assume that l 5(mo6), 3 l an l. By using this assumption an Remark., we obtain that if l, (mo6), we have,3(mo4) an if l 4 (mo6), we get 3(mo4). By using Lemma. we get w R = ( 5 + Y l ) + [5 + Y l ; 5,5, ],5, 5 + Y l l Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 8

10 Ozer / Kirklareli University Journal of Engineering an Science (06) w R = (5 + Y l ) w R = (5 + Y l ) w R By using Lemma. an Lemma.3, we get w R = (5 + Y l ) + Y l w R + Y l Y l w R + Y l Using Definition. an put Y l+ + Y l = 5Y l + Y l equation into the above equality, we obtain w R (5 + Y l )w R ( + Y l ) = 0 This implies that w R = ( 5+Y l ) + since w R > 0. If we consier Lemma. we get = [ 5+Y l ; 5,5,,5,5 + Y l ] an l = l(). Hence, w = [ 5+Y l ; 5,5,,5, 5 + Y l ] hols. l Now, we can etermine ε, t an u using Lemma.. We obtain Q i = Y i for i 0. If we t u substitute these values of sequence into the ( a0 ) Ql ( ) Ql ( ) an rearrange, we get ε = ( 5+Y l + ) Y l + Y l, t = Y l + Y l+ + Y l an u = Y l Finally, we know that m is efine as m = u substitue t an u into the m, then we get We can t calculate m = u m = u = t Y l + Y l+ + Y l t from H.Yokoi s reference. If we 4Y l t ue to is not square free positive integer for l =. From the assumption an by consiering Y l is increasing sequence, we get, Y l 4 > 4 ( ) > 3,846 Y l +Y l+ +Y l Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 83

11 Ozer / Kirklareli University Journal of Engineering an Science (06) Y l for l 4. Therefore, we obtain m = = 3 for l 4 ue to efinition of m Y l +Y l+ +Y. l This completes the proof of Theorem 3.. Corollary 3.. Let be the square free positive integer positive integer satisfying the conitions in Theorem 3.. We state the following Table 3. where funamental unit is ε, integral basis elemant is w an Yokoi s invariant is m for 4 l() 3. (In this table, we rule out l() =, 7 since is not a square free positive integer in these perios). l() m w ε [70; 5,5,5,40 ] [49075; 5,5,5,5,5,5,5,9850 ] [3340; 5,5,5,5,5,5,5,5,5,64680 ] [ ; 5,5,5,5,5,5,5,5,5,5,5,5, ] Table 3.. Theorem 3.. Let be the square free positive integer an l be a positive integer satisfying that l 5(mo6) an l >. We assume that parametrization of is = ( 5 + 3Y l ) + 3Y l + Then,we get (mo4) an w = [ 5+3Y l ; 5,5, ],5, 5 + 3Y l an l = l(). l Moreover, we have following equalities : ε = (( 5+3Y l ) Y l + Y l ) + Y l for ε, t, u an Yokoi s invariant m. t = 3Y l + 5Y l + Y l an u = Y l m = Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 84

12 Ozer / Kirklareli University Journal of Engineering an Science (06) Proof. We can create this theorem by substituting δ = in Main Theorem. If we suppose that l 5(mo6) an l >,then we have Y l (mo4) an Y l 3(mo4). Also, if we put these equivalents into = ( 5+3Y l ) + 3Y l + then we get (mo4). By using Lemma., we have w R = (5 + 3Y l ) + Y l w R + Y l Y l w R + Y l We obtain the proof in a similar way of proof of Theorem 3.. By using Lemma., Lemma.3 an Definition., we get w R = ( 5+3Y l ) + since w R > 0. If we consier Lemma. we have w = = [ 5+3Y l ; 5,5, ],5, 5 + 3Y l an l = l(). l ε, t an u are etermine as follows using Lemma.. It is seen that Q i = Y i hols for i. If we substitute these values of sequence into the t u ( a0 ) Ql ( ) Ql ( ) an rearrange, we have ε = (( 3Y l + 5 ) Y l + Y l ) + Y l t = 3Y l + 5Y l + Y l an u = Y l. If we substitute t an u into the m an rearrange, then we get m = u t = 4Y l 3Y l + Y l + Y l+ From the assumption an since Y l is increasing sequence, we have Y l > 4 ( 3Y ) >,39 l + Y l + Y l+ where l 5(mo6), l >. Therefore, we obtain m = 5(mo6), l 5 which completes the proof of Theorem 3.. 4Y l 3Y l +Y l +Y l+ = for l Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 85

13 Ozer / Kirklareli University Journal of Engineering an Science (06) Corollary 3.. Let be the square free positive integer satisfying the conitions in Theorem 3.. We state the following Table 3. where funamental unit is ε, integral basis elemant is w an Yokoi s invariant is m for < l() 7. l() m w ε 3 5 [054; 5,5,5,5,08 ] [06504; 5,5, ] 5, [ ; 5,5, ] 5, Table 3.. Theorem 3.3. Let be square free positive integer an l be a positive integer satisfying that l 5(mo6), 3 l an l. Suppose that the parametrization of is = ( 5Y l + 5 ) + 5Y l + Then, we have,3(mo4) an w = [ 5Y l+5 ; 5,5, ] 5, 5 + 5Y l with l = l(). Aitionally, we get the funamental unit ε, coefficients of funamental unit t, u an Yokoi s invariant n as follows: ε = ( 5Y l+5 l + ) Y l + Y l, t = 5Y l + Y l+ + Y l an u = Y l n = Proof. We have this theorem for δ = by using Main Theorem. Suppose that l 5(mo6), 3 l an l. By using this assumption an Remark. we can fin some results as follows: (i) if l, (mo6), then Y l (mo4) as well as either Y l (mo4) or Y l 0(mo4) hols. if Y l (mo4) an Y l (mo4),then 3(mo4) otherwise (mo4) hols. So, we have,3(mo4). Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 86

14 Ozer / Kirklareli University Journal of Engineering an Science (06) (ii) if l 4 (mo6), then Y l 3(mo4) an Y l (mo4) hol. By substituting these values into parametrization of an rearranging, we have 3(mo4). Hence,,3(mo4) hols. We get w R = (5 + 5Y l ) + Y l w R + Y l Y l w R + Y l using Lemma. an Lemma. with the properties of continue fraction expansion. Using Definition. we have w R (5 + 5Y l )w R ( + 5Y l ) = 0 This implies that w R = ( 5+5S l ) + since w R > 0. Hence, w = [ 5Y l+5 ; 5,5, ] 5, 5 + 5Y l hols. l t u By using Q i = Y i for i into the ( a0 ) Ql ( ) Ql ( ) an rearrange, we obtain ε = ( 5Y l+5 + ) Y l + Y l, t = 5Y l + Y l+ + Y l an u = Y l Finally, we know that n is efine as n = t u. If we substitue t an u into the n, then we get n = t u = 5Y l + Y l+ + Y l 4Y l = Y l+ 4Y l + Y l 4Y l From the assumption, since Y l is increasing sequence, we calculate following inequality for l Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 87

15 Ozer / Kirklareli University Journal of Engineering an Science (06) Hence, we obtain n = + + Y l+ 4 4Y l completes the proof of Theorem < Y l + Y l+ + Y l 4Y l 0,50 + Y l 4Y l = for l ue to efinition of n.this Corollary 3.3. Let be the square free positive integer positive integer satisfying the conitions in Theorem 3.3. We state the following Table 3.3 where funamental unit is ε, integral basis elemant is w an an Yokoi s invariant is n for l() 3. (In the following table, we rule out l() = 4,0 since is not a square free positive integer in these perios). l() n w ε 3 [5; 5,30 ] [4755; 5,5,5,5,5,5,9450 ] [45365; 5,5,5,5,5,5,5, ] [ ; 5,5, ] 5, Table REFERENCES [] R.A. Mollin, Quaratics, Boca Rato, F.L, CRC Press, 996. [] C.D. Ols, Continue Functions,New York: Ranom House, 963. [3] O. Perron, Die Lehre von en Kettenbrüchen,New York: Chelsea, Reprint from Teubner Leipzig, 950. [4] R. Sasaki, A characterization of certain real quaratic fiels, Proc. Japan Aca, 6, Ser. A, 986, no. 3, Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 88

16 Ozer / Kirklareli University Journal of Engineering an Science (06) [5] W. Sierpinski, Elementary Theory of Numbers, Warsaw: Monografi Matematyczne, 964. [6] K.Tomita, Explicit representation of funamental units of some quaratic fiels, Proc. Japan Aca., 7, Ser. A, 995, no., [7] K. Tomita an K. Yamamuro, Lower bouns for funamental units of real quaratic fiels, Nagoya Math. J, Vol.66, 00, [8] K.S. Williams an N. Buck, Comparison of the lengths of the continue fractions of D an ( + D), Proc. Amer. Math. Soc., 0 no. 4, 994, [9] H. Yokoi, New invariants an class number problem in real quaratic fiels, Nagoya Math. J, 3, 993, Notes on Especial Continue Fraction Expansions an Real Quaratic Number Fiels 89