EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 7, No. 1, 2014, 55-64 ISSN 1307-5543 – www.ejpam.com Explicit Form of the Fundamental Units of Certain Real Quadratic Fields Gül Karadeniz Gözeri∗, Ayten Pekin Department of Mathematics, Faculty of Science, Istanbul University, Istanbul, Turkey Abstract. In this paper, for all real quadratic fields K = Q( p d) such that d is a positive square free integer congruent to 2 or 3 modulo 4 and the period kd of the continued fraction expansion of the quadratic irrational number ωd = p d is equal to 7, we describe Td , Ud explicitly in the fundamental unit εd = ( Td+Ud p d 2 )(> 1) of Q( p d) and d itself by using five parameters appearing in the continued fraction expansion of ωd . 2010 Mathematics Subject Classifications: 11A55, 11R11, 11R27 Key Words and Phrases: Continued Fraction, Quadratic Extensions, Fundamental Unit 1. Introduction Explicit form of the fundamental units of real quadratic fieldsQ( p d) where d is congruent to 1 modulo 4 and the period kd in the continued fraction expansion of the quadratic irrational number ωd in Q( p d) is equal to 3 and 4, 5 was described in [5, 6] respectively. Later in [3], explicit form of the fundamental units of all real quadratic fields Q( p d) such that the period in the continued fraction expansion of the quadratic irrational number ωd in Q( p d) is equal to 6 was obtained. In this paper, for all real quadratic fields Q( p d) such that d is congruent to 1 modulo 4 and the period kd in the continued fraction expansion of the quadratic irrational number ωd = 1+ p d 2 is equal to 7, we described Td , Ud explicitly in the fundamental unit εd of Q( p d) and d itself by using five parameters appearing in the continued fraction expansion of ωd . In this paper, we consider all real quadratic fields Q( p d) where d ≡ 2,3(mod4) and the period kd of the continued fraction expansion ofωd = p d is equal to 7 and describe explicitly coefficients Td and Ud in the fundamental unit εd = ( Td+Ud p d 2 )(> 1) of Q( p d) and d itself by using five parameters appearing in the continued fraction expansion of ωd . ∗Corresponding author. Email addresses: gulkaradeniz@istanbul.edu.tr (G. Karadeniz Gözeri), aypekin@istanbul.edu.tr (A. Pekin) http://www.ejpam.com 55 c© 2014 EJPAM All rights reserved. G. Gözeri, A. Pekin / Eur. J. Pure Appl. Math, 7 (2014), 55-64 56 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, 7]. 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 εd of Q( p d) and the norm of εd is (−1)kd [4]. 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 In this section some of the important required preliminaries and lemmas are given. 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. In this paper we will concern with all real quadratic fields Q( p d) such that d is congruent to 2 or 3 modulo 4 and the period kd is equal to 7. Let d = a2+ b ≡ 2, 3(mod4), then we consider the following three cases: Case 1. If b is congruent to 1 modulo 4, then d can only be congruent to 2 modulo 4. And for this case it is obvious that a is odd. Case 2. If b is congruent to 2 modulo 4, then d can be congruent to 2 or 3 modulo 4. In this case, a is even when d is congruent to 2 modulo 4 and a is odd when d is congruent to 3 modulo 4. Case 3. If b is congruent to 3 modulo 4, then d can only be congruent to 3 modulo 4. And for this case it is obvious that a is even. Lemma 1. For a square-free positive integer d congruent to 2 or 3 modulo 4, we put ωd = p d, q0 = [ωd], ωR = q0 +ωd . Then ωd /∈ R(d), but ωR ∈ R(d) holds. Moreover, for the period k of ωR, we get ωR = [2q0, q1, . . . , qk−1] and ωd = [q0, q1, . . . , qk−1, 2q0]. Furthermore, let ωR = (Pk−1ωR+Pk−2) (Qk−1ωR+Qk−2) = [2q0, q1, . . . , qk−1,ωR] be a modular automorphism of ωR, then the fundamental unit εd of Q( p d) is given by the following formula: εd = ( Td + Ud p d 2 )> 1, Td = 2q0Qk−1+ 2Qk−2, Ud = 2Qk−1 where 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 = `i + 1 ωi+1 (`i = [ωi], i ≥ 0) be the continued fraction expansion of ω = ω0 G. Gözeri, A. Pekin / Eur. J. Pure Appl. Math, 7 (2014), 55-64 57 in R(d). Then each ωi is expressed in the form ωi = a−ri+ p d ci (ci , ri ∈ Z), and `i , ci , ri can be obtained from the following recurrence formula: ω0 = a− r0+ p d c0 , 2a−ri = ci`i + ri+1, ci+1 =ci−1+ (ri+1− ri)`i (i ≥ 0), where 0≤ ri+1 < ci , c−1 = (b+ 2ar0− r0 2) c0 . Moreover for the period k ≥ 1 of ω0, we get `i =`k−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 2 or 3 modulo 4, we put ωd = p d, q0 = [ωd] and ωR = q0+ωd . If we put ω=ωR in Lemma 2 , then we have the following recurrence formula: r0 =r1 = 0, c0 =1, c1 = b, `0 =2q0,`i = qi (1≤ i ≤ k− 1). Proof. The proof follows easily from Lemma 2. 3. Main Results Theorem 1. For a positive square-free integer d congruent to 2 modulo 4, we assume kd = 7. Then, if b is congruent to 1 modulo 4, we get ωd = [a,`1,`2,`3,`3,`2,`1, 2a] for three positive integers `1, `2, `3 such that `i ≥ 1 (i = 1,2, 3) and then (Td , Ud) = (2[a(A 2+ B2) + BC + A`2], 2(A2+ B2)) and d = A2r2+ 2rD+ E hold. Moreover r and s are positive integers determined uniquely by a =Ar + `1s A2+B2− C2− `2 2 = 2rB− 2s(A+ B`3) G. Gözeri, A. Pekin / Eur. J. Pure Appl. Math, 7 (2014), 55-64 58 where A, B, C, D and E are determined uniquely as follows: A=`1`2+ 1 B =`1+ A`3 C =`2`3+ 1 D =A`1s+ `2 E =`1 2s2+ 2s+ 1 Proof. In the case of b ≡ 1(mod4), it can be easily seen that a is an odd integer since d is congruent to 2 modulo 4. We can put b = 4m+ 1 for a non-negative integer m satisfying 0 ≤ 4m < 2a. Since q0 = [ωd] = [ p d] = a and ωR = a+ p d, it follows from Lemma 3 that r0 = r1 = 0, c0 = 1, c1 = 4m+ 1 and `0 = 2a. Since kd = 7, we get `1 = `6, `2 = `5 and `3 = `4 from Lemma 2. Then we have ωd = [a,`1,`2,`3,`3,`2,`1, 2a] for three positive integers `1, `2, `3 such that `i ≥ 1 (i = 1, 2,3). From Lemma 2 we get 2a = (4m+ 1)`1+ r2 (1) since r1 = 0 and c1 = 4m+ 1. From (1), we obtain (4m+ 1)`1 + r2 ≡ 0 (mod 2). So there exists a positive integer r such that r2 = 2r − `1. By substitution of r2 in (1) we get a = 2m`1+ r. (2) Here since a is odd, r must be an odd integer, too. It follows from Lemma 2 that c2 = 1+ r2`1 and 2a = c2`2+ r3+ r2. Thus, 2a = (1+ r2`1)`2+ r3+ r2 (3) is obtained. Then (4m+ 1)`1 = (1+ r2`1)`2+ r3 (4) holds from (1) and (3). Thus we get `2 + `3 ≡ 0 (mod `1). There exists a positive integer t such that r3 = `1 t − `2. By substitution of r3 in (4), we get 4m= t + 2r`2− `1`2− 1. Thus if we put A= `1`2+1, then we get t−A= 4m−2r`2. Since t−A is even, we can put t−A= 2s for a positive integer s. Hence, 4m = 2s + 2r`2 is obtained. Therefore we get a = Ar + `1s from (2). On the other hand, c3 = 4m+ 1+ (r3− r2)`2 (5) is obtained from Lemma 2. By substitution of r2 = 2r − `1 and r3 = `1 t − `2 in (5), we get c3 = At − `2 2. Moreover from Lemma 2, we get 2a = c3`3+ r3+ r4. Thus r4 = (2r − `1− t`3)A+ `2(`2`3+ 1) (6) G. Gözeri, A. Pekin / Eur. J. Pure Appl. Math, 7 (2014), 55-64 59 is obtained because of (3) and c3 = At − `2 2. Furthermore, c3 = At − `2 2 and c4 = (1+ r2`1) + (r4− r3)`3 imply At − `2 2 = 1+ r2`1+ r4`3− r3`3 since c3 = c4 . Thus, At − `2 2 = (1+ `2`3) 2+ 2r(`1+ A`3)− t`3(`1+ A`3)− `1(`1+ A`3) is obtained since r2 = 2r − `1, r3 = `1 t − `2 and r4 = (2r − `1 − t`3)A+ `2(`2`3 + 1). Thus if we put B = `1 + A`3 and C = `2`3 + 1, we get A2 + B2 − C2 − `2 2 = 2rB − 2s(A+ B`3) since t − A= 2s. If we assume that the integers r and s are not uniquely determined, we get A2+ B2 = 0 which is a contradiction. Therefore, the integers r and s are uniquely determined by a = Ar + `1s and A2+ B2− C2− `2 2 = 2rB− 2s(A+ B`3). Now, since ωd = [a,`1,`2,`3,`3,`2,`1, 2a] implies Q5 = BC + A`2 and Q6 = A2 + B2 by Lemma 1, we obtain (Td , Ud) = (2[a(A 2+ B2) + BC + A`2], 2(A2+ B2)). Furthermore, if we put D = A`1s+ `2 and E = `1 2s2+2s+1, then we get d = A2r2+2rD+ E because b = 2s+ 2r`2+ 1. Thus, the theorem is proved. As an application of this theorem, we can practically determine ωd where d = 314 = 172 + 25. Since q0 = a and `0 = 2a, it follows that q0 = 17 and `0 = 34. On the other hand, we get m= 6, since b = 4m+1. From a = 2m`1+ r, we obtain `1 = 1 and r = 5. Thus we get r2 = 9 immediately. Since 2a = (1+ r2`1)`2+ r2+ r3, c3 = 4m+ 1+ (r3− r2)`2 and 2a = c3`3+ r3+ r4, we obtain `2 = 2, r3 = 5, c3 = 17, `3 = 1 and r4 = 12. Hence ωd can be determined as follows: ωd = [17, 1,2, 1,1, 2,1, 34] Moreover fundamental unit of Q( p 314) can be easily determined as εd = 886+ 50 p 314 2 since A= 3, B = 4, C = 3. Furthermore by using r3 = `1 t − `2 and t − A= 2s, we get t = 7 and s = 2. Thus D = 8 and E = 9 is obtained easily. Theorem 2. Let d = a2+b ≡ 2,3 (mod 4) be a positive square-free integer with b ≡ 2 (mod 4). If kd = 7, then we get ωd = [a,`1,`2,`3,`3,`2,`1, 2a] for the three positive integers `1, `2, `3 such that `i ≥ (i = 1,2, 3) and then (Td , Ud) = (2[a(A 2+ B2) + BC + A`2], 2(A2+ B2)) and d = A2r2+ 2rD+ E hold. Moreover, r and s are positive integers determined uniquely by a =Ar + `1s G. Gözeri, A. Pekin / Eur. J. Pure Appl. Math, 7 (2014), 55-64 60 −`2[`2+ `3(C + 1)]− 1= 2r(`2+ A`3)− 2s(B`3+ A) where A, B, C, D and E are determined uniquely as follows: A=`1`2+ 1 B =A`3+ `1 C =`2`3+ 1 D =A`1s+ `2 E =`1 2s2+ 2s. Proof. In the case of b ≡ 2 (mod 4), we put b = 4m+ 2 for a positive integer m satisfying 0 < 2m+ 1 ≤ a. Since q0 = [ωd] = [ p d] = a, it follows from Lemma 3 that r0 = r1 = 0, c0 = 1, c1 = 4m+ 2, `0 = 2a. Since kd = 7, we get `1 = `6, `2 = `5 and `3 = `4 by Lemma 2. Then we have ωd = [a,`1,`2,`3,`3,`2,`1, 2a] for three integers `1, `2, `3 such that `i ≥ 1 holds. From Lemma 2 we get 2a = (4m+ 2)`1+ r2 (7) since r1 = 0, c1 = 4m+ 2. From (7), we have (4m+ 2)`1 + r2 ≡ 0 (mod 2) and we can put r2 = 2r for an integer r such that r ≥ 0. Hence, it follows from (7) a = (2m+ 1)`1+ r. (8) It follows from Lemma 2 that c2 = 1+ `1r2 (9) and 2a = c2`2+ r2+ r3. (10) Then from (7), (9) and (10) we have (4m+ 2)`1 = c2`2+ r3. (11) Moreover, we can write `2 + r3 ≡ 0 (mod `1) from (9) and (11). So there exists a positive even integer t such that r3 = `1 t−`2. Since t is even, we can put t = 2s for a positive integer s. Thus r3 = 2s`1− `2 is obtained. By substitution of r3 in (11), we get 4m= 2s+ 2r`2− 2. (12) It follows from (8) and (12) that a = r(`1`2+ 1) + s`1 is written. Thus if we put A = `1`2 + 1, then we get a = Ar + s`1. On the other hand, we get 2a = c3`3 + r3 + r4 and c3 = 4m+ 2+ (r3 − r2)`2 from Lemma 2. It follows from a = Ar + `1s, 2a = c3`3 + r3 + r4, G. Gözeri, A. Pekin / Eur. J. Pure Appl. Math, 7 (2014), 55-64 61 c3 = 4m+ 2+ (r3 − r2)`2 and (12) that r4 = 2Ar − 2s`3A+ `2(`2`3 + 1). Thus, if we put C = `2`3+ 1 then we get r4 = 2Ar − 2s`3A+ C`2. (13) Moreover, c3 = 4m+ 2+ (r3− r2)`2 and c4 = 1+ r2`2+ (r4− r3)`3 imply 4m= 2r2`2+ r4`3− r3`3− r3`2− 1 (14) since c3 = c4. Then by substitution r3 = 2s`1− `2, (12) and (13) in (14) we obtain −`2 2− `2`3(C + 1)− 1= 2r(`2+ A`3)− 2s[(`1+ A`3)`3+ A]. Then if we put B = `1+ A`3 we get −`2 2− `2`3(C + 1)− 1= 2r(`2+ A`3)− 2s(B`3+ A). If we assume that the integers r and s are not determined uniquely from a = Ar + `1s and −`2 2 − `2`3(C + 1)− 1 = 2r(`2 + A`3)− 2s(B`3 + A) we get A(A+ B`3) + `1(A`3 + `2) = 0 which is a contradiction. Therefore, the integers r and s are uniquely determined. Now, since ωd = [a,`1,`2,`3,`3,`2,`1, 2a] implies Q5 = BC + A`2 and Q6 = A2 + B2 by Lemma 1, we obtain Td = 2[a(A2 + B2) + BC + A`2] and Ud = 2(A2 + B2), respectively. Moreover if we put D = A`1s+ `2 and E = `1 2s2 + 2s, then we get d = A2r2 + 2rD+ E since b = 2s+ r`2. Thus, the theorem is proved completely. As an application of this theorem, we can easily determine ωd where d = 202 = 142 + 6. Since q0 = a and `0 = 2a, we get q0 = 14 and `0 = 28. On the other hand, we get m= 1 since b = 4m+ 2. Then we get `1 = 4 and r = 2 from a = (2m+ 1)`1 + r and r < 2m+ 1. Hence r2 = 4 is obtained. It follows from 4m = 2s+ 2r`2 − 2 that `2 = 1 and s = 1. Moreover, we get r3 = 7 since r3 = 2s`1 − `2. Since c3 = 4m+ 2+ (r3 − r2)`2 we obtain c3 = 9. By using 2a = c3`3 + r3 + r4 and r4 < c3, we get `3 = 2 and r4 = 3. Hence ωd can be determined as follows: ωd = [14, 4,1, 2,2, 1,4, 28] Furthermore the fundamental unit of Q( p 202) can be easily determined as εd = 6282+ 442 p 202 2 since A= 5, B = 14, C = 3. Moreover it is easily seen that D = 21 and E = 18. Theorem 3. For a positive square-free integer d congruent to 3 modulo 4, we assume kd = 7. Then, if b is congruent to 3 modulo 4, we get ωd = [a,`1,`2,`3,`3,`2,`1, 2a] for three positive integers `1, `2, `3 such that `i ≥ 1 (i = 1,2, 3), and then (Td , Ud) = (2[a(A 2+ B2) + BC + A`2], 2(A2+ B2)) and d = A2r2+ 2rD+ E G. Gözeri, A. Pekin / Eur. J. Pure Appl. Math, 7 (2014), 55-64 62 hold where A, B, C, D and E are determined uniquely as follows: A=`1`2+ 1 B =`1+ A`3 C =`2`3+ 1 D =A`1s+ `2 E =`1 2s2+ 2s+ 3. Moreover, r is an odd integer and s is a positive integer determined uniquely by a =Ar + `1s 3(A2+ B2)− C2− `2 2 = 2rB− 2s(A+ B`3). Proof. In the case of b ≡ 3 (mod 4), it can be easily seen that a is an even integer since d is congruent to 3 modulo 4. We can put b = 4m+ 3 for a non-negative integer m satisfying 0 ≤ 4m < 2a− 2. Since q0 = [ωd] = [ p d] = a and ωR = a+ p d, it follows from Lemma 3 that r0 = r1 = 0, c0 = 1, c1 = 4m+ 3 and `0 = 2a. Since kd = 7, we get `1 = `6, `2 = `5 and `3 = `4 from Lemma 2. Then we have ωd = [a,`1,`2,`3,`3,`2,`1, 2a] for three positive integers `1, `2, `3 such that `i ≥ 1 (i = 1, 2,3). From Lemma 2 we get 2a = (4m+ 3)`1+ r2 (15) since r1 = 0 and c1 = 4m+3. From (15), we obtain (4m+3)`1+r2 ≡ 0(mod2). So there exists a positive integer r such that r2 = 2r−3`1. By substitution of r2 in (15) we get a = 2m`1+ r. Here r is an even integer, since a is even. It follows from Lemma 2 that c2 = 1+ r2`1 and 2a = c2`2+ r3+ r2. Thus, 2a = (1+ r2`1)`2+ r3+ r2 is obtained. Then (4m+ 3)`1 = (1+ r2`1)`2+ r3 (16) holds from (15). Thus we get `2 + `3 ≡ 0 (mod `1). So there exists a positive integer t such that r3 = `1 t − `2. By substitution of r3 in (16), we get 4m = t + 2r`2 − 3(`1`2 + 1). Thus if we put A = `1`2 + 1, then we get t − 3A = 4m− 2r`2. Since t − 3A is even, we can put t − 3A = 2s for a positive integer s. Hence, 4m = 2s + 2r`2 is obtained. Therefore we get a = Ar + `1s since a = 2m`1+ r. On the other hand, c3 = 4m+ 3+ (r3− r2)`2 (17) is obtained from Lemma 2. Thus we get from (17) c3 = At − `2 2 since r2 = 2r − 3`1 and r3 = `1 t − `2. Moreover, from Lemma 2 we get 2a = c3`3+ r3+ r4. Thus r4 = (2r − 3`1 − t`3)A+ `2(`2`3 + 1) is obtained because of 2a = (1+ r2`1)`2 + r2 + r3 and c3 = At − `2 2. REFERENCES 63 Furthermore, c3 = At − `2 2 and c4 = (1+ r2`1) + (r4− r3)`3 imply At − `2 2 = 1+ r2`1+ r4`3− r3`3 since c3 = c4. Thus, At − `2 2 = (1+ `2`3) 2+ 2r(`1+ A`3)− t`3(`1+ A`3)− 3`1(`1+ A`3) is obtained since r2 = 2r − `1, r3 = `1 t − `2 and r4 = (2r − 3`1− t`3)A+ `2(`2`3+ 1). Thus, if we put B = `1 + A`3 and C = `2`3 + 1, we get 3(A2 + B2)− C2 − `2 2 = 2rB − 2s(A+ B`3) since t − 3A= 2s. If we assume that the integers r and s are not uniquely determined, we get A2+ B2 = 0 which is a contradiction. Therefore, the integers r and s are uniquely determined by a = Ar + `1s and 3(A2+ B2)− C2− `2 2 = 2rB− 2s(A+ B`3). Now, since ωd = [a,`1,`2,`3,`3,`2,`1, 2a] implies Q5 = BC + A`2 and Q6 = A2 + B2 by Lemma 1, we obtain Td = 2[a(A2+B2)+BC+A`2] and Ud = 2(A2+B2). Moreover, if we put D = A`1s+`2 and E = `1 2s2+2s+3, then we get d = A2r2+2rD+ E since b = 2s+2r`2+3. Thus, the proof is completed. 4. Conclusion In this paper, some results are presented in order to determine the fundamental units of certain quadratic fields Q( p d) with the period kd of the continued fraction expansion of the quadratic irrational number ωd is equal to 7. 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