In this paper, we will first summarize known results concerning continued
fractions. Then we will limit our consideration to continued fractions of
quadratic numbers. The second author described periods and sometimes precise
form of continued fractions of $\sqrt{N}$, where $N$ is a natural number. In
cases where we were able to find such results in literature, we recall the
original authors, however many results seem to be new.Comment: 13 page

Balková, Lubomíra

Hrušková, Aranka

Acta Polytechnica

English

arXiv

In this paper, we will first summarize known results concerning continued fractions. Then we will limit our consideration to continued fractions of quadratic numbers. The second author describes periods and sometimes the precise form of continued fractions of ?N, where N is a natural number. In cases where we have been able to find such results in the literature, we recall the original authors, however many results seem to be new

Balková, L'ubomíra

CTU Open Journal Systems (Czech Technical University, Prague / České vysoké učení technické v Praze)

Acta Polytechnica 53(4):322–328, 2013 © Czech Technical University in Prague, 2013available online at http://ctn.cvut.cz/ap/CONTINUED FRACTIONS OF SQUARE ROOTS OF NATURALNUMBERSĽubomíra Balkováa,∗, Aranka Hruškovába Department of Mathematics, Faculty of Nuclear Sciences and Physical Engineering, Czech TechnicalUniversity in Prague, Trojanova 13, 120 00 Praha 2, Czech Republicb Gymnázium Christiana Dopplera, Zborovská 45, 150 00 Praha 5 – Smíchov, Czech Republic∗ corresponding author: lubomira.balkova@fjfi.cvut.czAbstract. In this paper, we will first summarize known results concerning continued fractions.Then we will limit our consideration to continued fractions of quadratic numbers. The second authordescribes periods and sometimes the precise form of continued fractions of√N , where N is a naturalnumber. In cases where we have been able to find such results in the literature, we recall the originalauthors, however many results seem to be new.Keywords: continued fractions, periodic continued fractions, periods of continued fractions, quadraticnumbers.1. IntroductionContinued fractions have a long history behind them— their origin may go back to the age of Euclid’s algo-rithm for the greatest common divisor, or even earlier.However, they are experiencing a revival nowadays,thanks to their applications in high-speed and high-accuracy computer arithmetics. Some of the advan-tages of continued fractions in computer arithmeticsare: faster division and multiplication than with po-sitional number representations, fast and precise eval-uation of trigonometric, logarithmic and other func-tions, precise representation of transcendental num-bers, no roundoff or truncation errors ([6], Kahan’smethod in [3, p. 179]).2. Continued fractionsIn this section we summarize some basic definitionsand results that can be found in any number theorycourse [1, 2, 4]. We use bxc to denote the integer partof a real number x.Definition 2.1. The continued fraction (expansion)of a real number x is the sequence of integers (an)n∈Nobtained by the following algorithmx0 = x, an = bxnc, xn+1 ={1xn−an if xn /∈ Z,0 otherwise.Note that a0 ∈ Z and an ∈ N. The algorithmproducing the continued fraction is closely related tothe Euclidean algorithm for computing the greatestcommon divisor of two integers. It is thus readily seenthat if the number x is rational, then the algorithmeventually produces zeroes, i.e. there exists N ∈ Nsuch that an = 0 for all n > N , thusx = a0 +1a1 +1a2 +1. . . +1aN−1 +1aN. (1)We write x = [a0, . . . , aN ]. On the other hand, if wewant to find an expression of the form (1) with a0 ∈ Zand an ∈ N \ {0} otherwise, then there are exactlytwo of them – the continued fraction [a0, . . . , aN ] andx = a0 +1a1 +1a2 +1. . . +1aN−1 +1(aN − 1) +11.If the number x is irrational, then the sequence ofthe so-called convergentsa0, a0+1a1, . . . , a0+1a1 +1a2 +1. . . +1an−1 +1an(2)converges to x for n→∞. On the other hand, everysequence of rational numbers of the form (2) with322vol. 53 no. 4/2013 Continued Fractions of Square Roots of Natural Numbersa0 ∈ Z and an ∈ N \ {0} converges to an irrationalnumber (and for every irrational number there is onlyone such sequence — the sequence of convergents).We write x = [a0, . . . , an, . . . ].The convergents of the continued fraction areknown to represent irrational numbers better thanany other fractions.Theorem 2.2 (Lagrange). Let x ∈ R\Q and let pnqnbe its n-th convergent (where pn and qn are coprime)and let pq with p, q ∈ Z be distinct from pnqn and suchthat 0 < q ≤ qn. Then∣∣∣x− pnqn∣∣∣ < ∣∣∣x− pq∣∣∣.It is also known how well continued fractions ap-proximate irrational numbers.Theorem 2.3. Let x ∈ R \ Q and let pnqn be its n-th convergent (where pn and qn are coprime). Theneither∣∣∣x− pnqn∣∣∣ < 12q2n or∣∣∣x− pn+1qn+1∣∣∣ < 12q2n+1 .And in a certain way, only continued fractions getvery close to irrational numbers.Theorem 2.4 (Legendre). Let x ∈ R \ Q and letpq with p, q ∈ Z satisfy |x − pq | < 12q2 . Then pq isa convergent of x.2.1. Continued fractions andcontinuantsThe convergents of continued fractions are closelyrelated to the so-called continuants Kn(x1, . . . , xn).Theorem 2.5. Let a0 ∈ R, ai > 0, i ∈ N. Then itholdsa0 +1a1 +1. . . +1an= Kn+1(a0, a1, . . . , an)Kn(a1, . . . , an),where the polynomial Kn(x1, . . . , xn) is given bythe recurrence relation K−1 = 0, K0 = 1 and forn ≥ 1 by Kn(x1, . . . , xn) = Kn−2(x1, . . . , xn−2) +xnKn−1(x1, . . . , xn−1).Corollary 2.6. Let [a0, . . . , an, . . . ] be the continuedfraction of an irrational number x. Then its n-thconvergent pnqn satisfiespn = Kn+1(a0, . . . , an), qn = Kn(a1, . . . , an).Theorem 2.7. For every n ∈ N and a1, . . . , an ∈ R,we haveKn(a1, . . . , an) = Kn(an, . . . , a1).2.2. Continued fractions of quadraticnumbersWe will call a quadratic irrational an irrational rootα of a quadratic equationAx2 +Bx+ C = 0,where A,B,C ∈ Z. The second root of the equationwill be denoted α′ and called the (algebraic) conjugateof α.In order to state the theorem describing con-tinued fractions of quadratic irrationals, we needto recall that a continued fraction [a0, . . . , an, . . . ]is called eventually periodic if [a0, . . . , an, . . . ] =[a0, . . . , ak−1, ak, . . . , a`] starts with a preperioda0, . . . , ak−1 and then a period ak, . . . , a` is repeatedan infinite number of times. It is called purely periodicif [a0, . . . , an, . . . ] = [a0, . . . , a`], i.e., if the preperiodis empty.Theorem 2.8 (Lagrange). Let α ∈ R \ Q. Thecontinued fraction of α is eventually periodic if andonly if α is a quadratic irrational.Theorem 2.9 (Galois). Let α be a quadratic ir-rational and α′ its conjugate. The continued frac-tion of α is purely periodic if and only if α > 1 andα′ ∈ (−1, 0).Example 2.10. Let α = 1+√52 , i.e., the so-calledGolden ratio, then it is the root of x2−x− 1 = 0 andα′ = 1−√52 ∈ (−1, 0). The continued fraction of α isindeed purely periodic sinceα = 1 + −1 +√52 = 1 +11+√52= 1 + 1α,consequently α = [1].In the sequel when we restrict our consideration tosquare roots of natural numbers, we will make use ofthe following lemma from [4].Lemma 2.11. Let α be a quadratic irrational andα′ its conjugate. If α has a purely periodic continuedfraction [a0, a1, . . . , an], then −1α′ = [an, . . . , a1, a0].3. Continued fractions of√NLet us consider N ∈ N \ {0}. If N = k2 for somek ∈ N, then √N = k and the continued fraction is√N = [k]. Therefore, we limit our considerationsto N ∈ N \ {0} which is not a square in the sequel.Then there exists a unique n ∈ N \ {0} and a uniquej ∈ {1, . . . , 2n} such that N = n2 + j.The proofs of the two following theorems can befound in [4] page 15. However we repeat them heresince they follow almost immediately from the previ-ous statements and they give an insight into the formof continued fractions of quadratic numbers.323Ľ. Balková, A. Hrušková Acta PolytechnicaTheorem 3.1. For every n ∈ N \ {0} and everyj ∈ {1, . . . , 2n} the continued fraction of√n2 + jis of the form [n, a1, . . . , ar, 2n], where a1 . . . ar isa palindrome.Proof. Denote α = n +√n2 + j. Then α isa quadratic irrational greater than 1 and α′ = n −√n2 + j ∈ (−1, 0). Therefore α has by Theorem 2.9a purely periodic continued fraction, i.e., there exista1, . . . , ar ∈ N such that α = [2n, a1, . . . , ar ]. It isthus evident that√n2 + j = [n, a1, . . . , ar, 2n]. It re-mains to prove that a1 . . . ar is a palindrome. Accord-ing to Lemma 2.11 the number −1α′ has its continuedfraction equal to [ar, . . . , a1, 2n]. We obtain thus√n2 + j = n+ 1−1n−√n2+j= n+ 1−1α′= [n, ar, . . . , a1, 2n].Since the continued fraction of irrational numbers isunique and we have√n2 + j = [n, a1, . . . , ar, 2n] = [n, ar, . . . , a1, 2n],it follows that a1 = ar, a2 = ar−1 etc. Consequently,a1 . . . ar is a palindrome.Theorem 3.2. The continued fraction of the form[n, a1, . . . , ar, 2n], where a1 . . . ar is a palindrome, cor-responds to√N for a rational number N .Proof. Denote by x the number whose continued frac-tion equals [n, a1, . . . , ar, 2n], i.e.,x = n+1a1 +1. . . +1ar +12n+ (x− n).Hence by Theorem 2.5,x− n = Kr(a2, . . . , ar, x+ n)Kr+1(a1, . . . , ar, x+ n)= Kr−2(a2, . . . , ar−1) + (x+ n)Kr−1(a2, . . . , ar)Kr−1(a1, . . . , ar−1) + (x+ n)Kr(a1, . . . , ar).By Theorem 2.7 and since a1 . . . ar is a palindrome,we have Kr−1(a1, . . . , ar−1) = Kr−1(a2, . . . , ar).Consequently, we obtainx =√n2 + 2nKr−1(a1,...,ar−1)+Kr−2(a2,...,ar−1)Kr(a1,...,ar) ,where under the square root, there is certainly a ra-tional number since by their definition, continuantswith integer variables are integers.In the sequel, let us study the length of the periodand the form of the continued fraction of√N =√n2 + j in dependence on n and j, where n ∈ N \{0} and j ∈ {1, . . . , 2n}. We will prove only someobservations since the proofs are quite technical andspace-demanding. The rest of the proofs may befound in [5]. In Table 1, we have highlighted allclasses of n and j for which their continued fractionsof√N =√n2 + j have been described.Observation 3.3. The continued fraction of√N hasperiod of length 1 if and only if N = n2 +1. It holdsthen√N = [n, 2n].Proof. This observation has already been made in [7].(⇐) :√n2 + 1 = n+√n2 + 1− n1 == n+1√n2 + 1 + n= n+12n+√n2 + 1− n1,hence√N = [n, 2n].(⇒) : If the length of the period equals 1, then byTheorem 3.1 we have√N = [n, 2n].√n2 + j = n+(√n2 + j − n)= n+12n+√n2 + j − n,hence we have√n2 + j − n = 12n+√n2 + j − n,j√n2 + j + n=1√n2 + j + n,j = 1.Observation 3.4. The continued fraction of√N hasperiod of length 2 if and only if 2nj is an integer. Itholds then√N = [n, 2nj , 2n].Proof. (⇐):√n2 + j = n+ (√n2 + j − n)= n+j√n2 + j + n= n+12nj+√n2 + j − nj,= n+12nj+1√n2 + j + n,= n+12nj+12n+ (√n2 + j − n),thus√N = [n, 2nj , 2n].324vol. 53 no. 4/2013 Continued Fractions of Square Roots of Natural NumbersTable 1. All classes of n ≤ 40 (first column) and j ≤ 31 (first row) for which their continued fractions of√N =√n2 + j have been described are highlighted.(⇒): If the length of the period equals 2, then byTheorem 3.1 we have√N = [n, x, 2n].√n2 + j = n+ (√n2 + j − n)= n+1x+12n+ (√n2 + j − n),hence we have√n2 + j − n = 1x(√n2 + j + n) + 1√n2 + j + n,x = 2nj.Observation 3.5. The continued fraction of√Nhas period of length 3 if and only if j = 4ak + 1 andn = aj + k for some a, k ∈ N, a ≥ 1, k ≥ 1. It holdsthen√N = [n, 2a, 2a, 2n] and 5 ≤ j ≤ n− 1.Proof. (⇒) : If the length of the period equals 3, thenby Theorem 3.1 we have√N = [n, x, x, 2n].√n2 + j = n+1x+1x+12n+ (√n2 + j − n),hence we get j =2xn+ 1x2 + 1 . Since j is an integer, xmust be even. Furthermore, as j 6= 1 by Observa-tion 3.3, there exists a ≥ 1 such that x = 2a. Itfollows then from j =4an+ 14a2 + 1 that n = aj +j−14a .Since n is an integer, we obtain finally j = 4ak + 1and n = aj + k for some k ≥ 1. It is easy to verifythat j ≥ 5 and j ≤ n− 1.(⇐) : The reverse implication is only an exercise inmanipulation with square roots and integer parts. Wehave it for the reader.In order to save space in the proofs, let us introduce325Ľ. Balková, A. Hrušková Acta Polytechnican j√N` = 4 2k + 1, k ≥ 2 2n− 3 [n, 1, n−32 , 1, 2n]2k + 1, k ≥ 1 3n+12 [n, 1, 2, 1, 2n]3k + 2 5n+23 [n, 1, 4, 1, 2n]3k + 2, k ≥ 1 2n− 2 [n, 1, 2n−43 , 1, 2n]3k + 2 4n+13 [n, 1, 1, 1, 2n]5k + 2, k ≥ 1 n− 1 [n, 2, 2n−45 , 2, 2n]5k + 4 8n+35 [n, 1, 3, 1, 2n]6k + 1, k ≥ 1 5n+16 [n, 2, 2, 2, 2n]6k + 5 2n−13 [n, 3,n−12 , 3, 2n]9k + 4, k ≥ 1 n− 2 [n, 2, 2n−89 , 2, 2n]` = 5 2k + 1, k ≥ 1 4 [n, n−12 , 1, 1, n−12 , 2n]5k + 3 6n+25 [n, 1, 1, 1, 1, 2n]` = 6 2k, k ≥ 2 2n− 3 [n, 1, n2 − 1, 2, n2 − 1, 1, 2n]10k + 7 2n+15 [n, 4, 1,n−32 , 1, 4, 2n]3k + 1, k ≥ 1 2n+13 [n, 2, 1, n− 1, 1, 2, 2n]3k + 1, k ≥ 1 n+ 1 [n, 1, 1, 2n−23 , 1, 1, 2n]3k + 1, k ≥ 1 4n+23 [n, 1, 2, n, 2, 1, 2n]6k + 4 7n+26 [n, 1, 1, 2, 1, 1, 2n]7k + 3, k ≥ 1 n+ 2 [n, 1, 1, 2n−67 , 1, 1, 2n]` = 8 4k + 1, k ≥ 2 2n− 7 [n, 1, n−54 , 2, n−12 , 2, n−54 , 1, 2n]6k, k ≥ 1 4n3 [n, 1, 1, 1, n−22 , 1, 1, 1, 2n]6k + 2, k ≥ 1 2n−13 [n, 3, n−22 , 1, 4, 1, n−22 , 3, 2n]7k + 5 8n+27 [n, 1, 1, 3, n, 3, 1, 1, 2n]9k + 3, k ≥ 1 9 [n, 2n−69 , 1, 2, 2n−69 , 2, 1, 2n−69 , 2n]9k + 6, k ≥ 1 9 [n, 2n−39 , 2, 1, 2n−129 , 1, 2, 2n−39 , 2n]` = 10 6k + 3, k ≥ 1 4n3 [n, 1, 1, 1, n−12 , 6, n−12 , 1, 1, 1, 2n]9k + 6 10n+39 [n, 1, 1, 3, 1, n− 1, 1, 3, 1, 1, 2n]10k + 5, k ≥ 1 4n5 [n, 2, 1, 1, n−12 , 10, n−12 , 1, 1, 2, 2n]Table 2. Lengths ` of periods and the form of continued fractions for several classes.the following notation〈a0, a1, . . . , aN−1, aN 〉 = a0+1a1 +1. . . +1aN−1 +1aN,where ai ∈ N for i ∈ {0, 1, 2, . . . , N − 1}, but aN ∈ R.Observation 3.6. Let j = 4. If n is even, thenthe length of the period is 2 and√N =[n, 2nj , 2n].If n is odd, then the length of the period is 5 and√N =[n, n−12 , 1, 1,n−12 , 2n].Proof. If n is even, then 2nj is an integer and thestatement is a corollary of Observation 3.4. If n isodd, it holds√n2 + 4 = n+(√n2 + 4− n)=〈n,√n2 + 4 + n4〉=〈n,n− 12 ,√n2 + 4 + n− 2n〉=〈n,n− 12 , 1,√n2 + 4 + 2n〉=〈n,n− 12 , 1, 1,√n2 + 4 + n− 24〉=〈n,n− 12 , 1, 1,n− 12 , 2n+ (√n2 + 4− n)〉,thus√N = [n, n−12 , 1, 1,n−12 , 2n].Observation 3.7. For n > 1 and j = 2n − 1 thelength of the period is 4 and the continued fractionis then√N = [n, 1, n− 1, 1, 2n].Proof.√n2 + 2n− 1 = n+ (√n2 + 2n− 1− n)=〈n,√n2 + 2n− 1 + n2n− 1〉326vol. 53 no. 4/2013 Continued Fractions of Square Roots of Natural Numbers=〈n, 1,√n2 + 2n− 1 + (n− 1)2〉=〈n, 1, n− 1,√n2 + 2n− 1 + (n− 1)2n− 1〉=〈n, 1, n− 1, 1, 2n+ (√n2 + 2n− 1− n)〉,hence√N = [n, 1, n− 1, 1, 2n]Observation 3.8. For n > 3 and j = 2n− 3, eitherthe length of the period is 4 if n is odd and the con-tinued fraction is then√N =[n, 1, n−32 , 1, 2n], or thelength of the period is 6 if n is even and the continuedfraction is then√N = [n, 1, n2 − 1, 2, n2 − 1, 1, 2n].Proof. For n odd:√n2 + 2n− 3 = n+ (√n2 + 2n− 3− n)=〈n,√n2 + 2n− 3 + n2n− 3〉=〈n, 1,√n2 + 2n− 3− (n− 3)4〉=〈n, 1, n− 32 ,√n2 + 2n− 3 + (n− 3)2n− 3〉=〈n, 1, n− 32 , 1, 2n+(√n2 + 2n− 3− n)〉,thus√N =[n, 1, n−32 , 1, 2n].For n even:√n2 + 2n− 3 = n+ (√n2 + 2n− 3− n)=〈n,√n2 + 2n− 3 + n2n− 3〉=〈n, 1,√n2 + 2n− 3− (n− 3)2n− 3〉=〈n, 1, n2 − 1,√n2 + 2n− 3 + (n− 1)n− 1〉=〈n, 1, n2 − 1, 2,√n2 + 2n− 3 + (n− 1)4〉=〈n, 1, n2 − 1, 2,n2 − 1,√n2 + 2n− 3 + (n− 3)2n− 3〉=〈n, 1, n2 − 1, 2,n2 − 1, 1,2n+ (√n2 + 2n− 3− n)〉.Thus√N = [n, 1, n2 − 1, 2, n2 − 1, 1, 2n].The following table will include all remaining casesof continued fractions of√n2 + j that we were ableto determine in terms of n and j.Observation 3.9. Let k ∈ N. Let us summarizein Table 2 the lengths ` of periods and the form ofcontinued fractions for several classes (described inan analogous way) of n and j.The next observation was made in a different waythan all previous ones. We prescribed the form of thecontinued fraction and searched for√N having sucha continued fraction.Observation 3.10. If the period of the continuedfraction of√N =√n2 + j contains p ≥ 1 ones asits palindromic part, i.e.,√N = [n, 1, . . . , 1︸ ︷︷ ︸p, 2n] thenn = kFp + Fp+12 for some k ∈ N, p + 1 6= 3`, where` ∈ N, and j = 2nFp−1+Fp−2Fp , where Fn denotes the n-th Fibonacci number given by the recurrence relationF−1 = 0, F0 = 1 and Fn = Fn−2+Fn−1 for all n ≥ 1.Proof. It is a direct consequence of the proof of The-orem 3.2 and the definition of continuants.The last observation is also of a different form thanthe previous ones since j and n depend on two pa-rameters.Observation 3.11. Let n = 4ka+ 2a, where k, a ∈N, k ≥ 1, a ≥ 1, and j = 8a. Then the continuedfraction of√N =√n2 + j equals[n,4n− j2j , 1, 1,n− 22 , 1, 1,4n− j2j , 2n].Proof. The proof may be found in [5].We also made one conjecture that turned out tobe false.Conjecture 3.12. For√N the length of the periodof the continued fraction is less than or equal to 2n.This observation was made when contemplatinga table of periods of√N for N ≤ 1000. However,in [8] it is shown that for N = 1726 with n = 41,the period of the continued fraction of√N is oflength 88 > 82 = 2n. A rougher upper bound comesfrom [7].Theorem 3.13. For√N the length of the period ofthe continued fraction is less than or equal to 2N .Let us terminate with two conjectures that havenot been proved yet.Conjecture 3.14. No element of the period of√Napart from the last one is bigger than n.Conjecture 3.15. There is no period of an oddlength for j = 4k + 3, where k ∈ N.AcknowledgementsThe first author acknowledges financial support fromCzech Science Foundation grant 13-03538S.References[1] Klazar, M.: Introduction to Number Theory,KAM-DIMATIA Series 782, 2006.[2] Masáková, Z., Pelantová, E.: Teorie čísel, skriptumČVUT, 2010.[3] Muller, J.-M.: Elementary Functions: Algorithms andImplementation, 2nd edition, Birkhäuser Boston, 2006.[4] Lauritzen, N.: Continued fractions and factoring,http://home.imf.au.dk/niels/cfracfact.pdf, 2009.327Ľ. Balková, A. Hrušková Acta Polytechnica[5] Hrušková, A.: Řetězové zlomky kvadratických čísel,SOČ práce, http://bimbo.fjfi.cvut.cz/∼soc/, 2013–2014[6] Seidensticker, R.: Continued fractions for high-speedand high-accuracy computer arithmetic, IEEESymposium on Computer Arithmetic (1983), 184–193[7] Sierpinski, W.: Elementary Theory of Numbers,Panstwowe Wydawnictwo Naukowe, Warszawa, 1964.[8] Hickerson, D. R.: Length of period of simplecontinued fraction expansion of√d, Pacific Journal ofMathematics 46(2) (1973), 429–432328

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