ON A DISTRIBUTION PROPERTY OF THE RESIDUAL ORDER OF a (mod pq)

Transcription

1 Annales Univ. Sci. Budapest., Sect. Comp. 41 (2013) ON A DISTRIBUTION PROPERTY OF THE RESIDUAL ORDER OF a (mod pq) Leo Murata (Tokyo, Japan) Koji Chinen (Higashi-Osaka, Japan) Dedicated to Professors Zoltán Daróczy and Imre Kátai on their 75th anniversary Communicated by Pavel Varbanets (Received June 04, 2013; accepted June 24, 2013) Abstract. In the authors previous papers, for a positive integer a and a prime number p such that (a, p) = 1, the distribution of the residual orders D a(p) of a (mod p) was considered. Under the generalized Riemann hypothesis (GRH), the authors determined the natural density of the primes p satisfying D a(p) l (mod k) when k is a prime power, and proved the existence of an algorithm for computing the density when k is a composite integer other than prime powers. In this paper, we consider the distribution of the residual orders D a(pq) of a (mod pq) where p and q are distinct primes. Under GRH and some slight restriction to the base a, we determine the natural density of the prime pairs (p, q) satisfying D a(pq) l (mod 4) for l = 0, 1, 2, Introduction Let p be an odd prime number and Z/pZ be the multiplicative group of all invertible residue classes modulo p. We define ( ) the multiplicative order of the residue D a (p) := class a (mod p) in the group Z/pZ. Key words and phrases: Residual order, Artin s conjecture for primitive root Mathematics Subject Classification: 11N05, 11N25, 11R18.

2 188 L. Murata and K. Chinen In the papers [1] [3] and [7], we are interested in a distribution property of D a (p) with p varies and introduced the set, for an arbitrary residue class l (mod k), Q a (x; k, l) := {p x ; p : prime, D a (p) l (mod k)}. We studied about the existence of the natural density of Q a (x; k, l): 1 a (k, l) = lim x π(x) Q a(x; k, l), where π(x) is the number of primes up to x. We proved that 1. We can prove the existence of the density a (k, l), for any residue class l (mod k), 2. and in this proof, except for some special residue classes, we need Generalized Riemann Hypothesis. 3. We can calculate the exact value of the density a (k, l). Our typical example is: Theorem 1.1. Let a be a natural number and we decompose a into a = a 2 0a 1, a 1 is square-free. If a 1 is odd, then we have (1.1) (1.2) a (4, 0) = a (4, 2) = 1 3, a (4, 1) = a (4, 3) = 1 6, where we obtain (1.1) unconditionally and (1.2) under GRH (see [1] and [7] for details). Here GRH means Hypothesis 1.2. (Generalized Riemann Hypothesis) For any positive integers k and m, we assume that the Riemann Hypothesis holds for the Dedekind zeta function ζ K (s) for the field K = Q(ζ m, a 1/k ) where ζ m = exp(2πi/m). Now let p and q be two distinct prime numbers and let us consider the same distribution property of the residue class of a, but in the multiplicative group Z/pqZ. The group Z/pZ is a simple cyclic group, but Z/pqZ is no more cyclic. We are interested in the order of the residue class itself, namely we define ( ) the multiplicative order of the residue D a (pq) := class a (mod pq) in the group Z/pqZ = a (mod pq).

3 On a distribution property of the residual order of a (mod pq) 189 and R a (x; k, l) := {(p, q) ; p, q : odd primes, p x, q x, D a (pq) l (mod k)}. The purpose of the paper is 1. to prove the existence of the natural density Γ a (4, l) for l = 0, 1, 2, 3, 2. and to calculate the explicit value of Γ a (4, l), where the natural density means Our main theorem will be Γ a (k, l) := lim x π(x) 2 R a (x; k, l). Theorem 1.3. Let a be a natural number and we decompose a into a = a 2 0a 1, a 1 is square-free. If a 1 is odd, then we have (I) the natural densities Γ a (4, l), l = 0, 1, 2, 3 exist (II) and the standard distribution is (Γ a (4, 0), Γ a (4, 1), Γ a (4, 2), Γ a (4, 3)) = ( ) 5 9, 1 18, 1 3, Moreover, for the existence of Γ a (4, 1) and Γ a (4, 3), we need GRH, whereas the existence of Γ a (4, 0) and Γ a (4, 2) is unconditional. This theorem shows that, despite the change of group structure, as to the multiplicative order of a residue class, we can prove a similar result to Theorem 1.1. It is quite likely that there exists the density Γ a (k, l) for any residue class l (mod k), but we cannot verify this so far. As we describe in the next section, we can prove unconditionally from Theorem 1.1 that, Γ a (4, 0) = 5 9, Γ a(4, 2) = 1 3. So the main interest of Theorem 1.3 is the equi-distribution property Γ a (4, 1) = Γ a (4, 3) = 1 18, and to prove this we need GRH and difficult considerations so far. Throughout this paper, we use the following notations: π(x; k, l) is the number of prime numbers up to x which are congruent to l mod k, ϕ(n) means Euler s totient function, µ(n) is the Möbius function, ω(n) means the number of distinct prime factors of n, a 1, a 2,, a i means the least common multiple of integers a 1, a 2,, a i and (a, b) is the largest common divisor of integers a and b.

4 190 L. Murata and K. Chinen 2. R a (x; k, l) with l = 0 In this section, we study the set R a (x; k, 0) = {(p, q) ; p, q : odd primes, p x, q x, D a (pq) 0 (mod k)}. The next lemma is useful in decomposing our problem about (mod pq) into problems about (mod p) and (mod q) (we omit the proof). Lemma 2.1. We have (2.1) D a (pq) = D a (p), D a (q). If D a (pq) 0 (mod 4) then Lemma 2.1 shows that one of the D a (p) and D a (q) is congruent to 0 modulo 4. This means R a (x; 4, 0) = 2 {p x; D a (p) 0 (mod 4)} {p, q x; D a (p) 0 (mod 4), D a (q) 0 (mod 4)}. Then Theorem 1.1 gives Γ a (4, 0) = 5/9 directly, similarly Γ a (2, 0) = 8/9, and Γ a (4, 2) = 1/3. Remark. We can prove a little more general result. Here we take a base a N. We put a = a 2 0 a 1, a 1 : square free and let r be a prime number. When a 1 is odd, Γ a (r h, 0) = 1 (r 2 1) 2 r(2r2 r 2), if h = 1, 2r h 2r h 2 1, if h 2. r 2(h 2) (r 2 1) 2 For example, for the same a, Γ a (2, 0) = 8/9, Γ a (3, 0) = 39/ R a (x; 4, l), the case of l = 1, 3 (I) As we described in the previous section, we can calculate R a (x; 2, 0), R a (x; 4, 0) and consequently R a (x; 4, 2). But the separation of R a (x; 2, 1) into R a (x; 4, 1) and R a (x; 4, 3) is rather difficult. We begin with reducing R a (x; 4, 1) to an infinite sum of Q a (x; k, l) s. The purpose of this section is the following formulas (3.1) and (3.2).

5 On a distribution property of the residual order of a (mod pq) 191 Proposition 3.1. We have R a (x; 4, 1) = Q a (x; 4, 1) 2 + Q a (x; 4, 3) ω(d1) 1 ( Q a (x; 4D, D) Q a (x; 4D, 3D)) 2 (3.1) D<x D 1 (mod 4) D 0=1,D 1>1 D<x D 3 (mod 4) D 0=1,D 1>1 2 ω(d1) 1 ( Q a (x; 4D, D) Q a (x; 4D, 3D)) 2, where D = D 0 D 1, all the prime factors of D 0 are congruent to 1 mod 4 and all the prime factors of D 1 are congruent to 3 mod 4. We have similarly R a (x; 4, 3) = 2 Q a (x; 4, 1) Q a (x; 4, 3) 2 ω(d1) 1 ( Q a (x; 4D, D) Q a (x; 4D, 3D)) 2 + (3.2) + D<x D 1 (mod 4) D 0=1,D 1>1 D<x D 3 (mod 4) D 0=1,D 1>1 2 ω(d1) 1 ( Q a (x; 4D, D) Q a (x; 4D, 3D)) 2. Proof. Here we prove only (3.1). By a simple equality we divide the condition into two cases: (I) D a (p) 1 (mod 4) and (II) D a (p) 3 (mod 4) and First we consider (I). (3.3) p,q x D a(pq) 1 (mod 4) D a(p) 1 (mod 4) D a (q) D a (pq) = D a (p) (D a (p), D a (q)), D a (pq) 1 (mod 4) D a (q) 1 (mod 4). (D a (p), D a (q)) D a (q) 3 (mod 4). (D a (p), D a (q)) 1 = p x D a(p) 1 (mod 4) Da(q) (Da(p),Da(q)) q x 1 (mod 4) 1.

6 192 L. Murata and K. Chinen For a prime number p with D a (p) 1 (mod 4), we calculate the inner sum of (3.3). Let us introduce the number Q = (D a (p), D a (q)). Then (the inner sum of (3.3)) = = Q D a(p) Q D a(p) q x Q=(D a(p),d a(q)) D a(q)/q 1 (mod 4) Q Da(p) Q µ(q ) 1 = q x QQ D a(q) D a(q)/q 1 (mod 4) 1. Since all the numbers D a (p), Q and Q are odd, (3.4) D a (q)/q 1 (mod 4) D a (q) Q (mod 4Q). We introduce the number Q by (3.5) Q = { 1, if Q 1, (mod 4), 3, if Q 3, (mod 4). Then { Da (q) Q (mod 4Q) D a (q) 0 (mod QQ ) D a (q) Q QQ (mod 4QQ ). Consequently, (the inner sum of (3.3)) = Q D a(p) Q Da(p) Q µ(q ) q x D a(q) Q QQ (mod 4QQ ) 1.

7 On a distribution property of the residual order of a (mod pq) 193 Then (3.3) turns into: = p,q x D a(pq) 1 (mod 4) D a(p) 1 (mod 4) 1 = p x Q D a(p) Q Da(p) Q D a(p) 1 (mod 4) = Q<x Q:odd = Q<x Q:odd p x D a(p) 1 (mod 4) Q D a(p) Q <x Q :odd QQ <x µ(q ) Q Da(p) Q µ(q ) µ(q ) p x D a(p) 1 (mod 4) D a(p) 0 (mod Q) Q Da(p) Q We put QQ = D, then D is odd and consequently, (3.6) (3.3) = D<x D:odd µ(q ) Q D When D 1 (mod 4), then p x D a(p) 1 (mod 4) D a(p) 0 (mod D) q x D a(q) Q QQ (mod 4QQ ) q x D a(q) Q QQ (mod 4QQ ) q x D a(q) Q QQ (mod 4QQ ) 1 = 1 = 1. q x D a(q) Q D (mod 4D) 1. { Da (p) 1 (mod 4) D a (p) 0 (mod D) D a (p) D (mod 4D), and when D 3 (mod 4), then { Da (p) 1 (mod 4) D a (p) 0 (mod D) D a (p) 3D (mod 4D). Therefore (3.6) = + D<x Q D D 1 (mod 4) D<x D 3 (mod 4) µ(q ) Q a (x; 4D, D) Q a (x; 4D, Q D) + µ(q ) Q a (x; 4D, 3D) Q a (x; 4D, Q D). Q D We can obtain a similar formula for the case (II) on p.191, then we get an

8 194 L. Murata and K. Chinen important formula: 1 = p,q x D a(pq) 1 (mod 4) (3.7) where D<x Q D D 1 (mod 4) D<x Q D D 3 (mod 4) D<x Q D D 1 (mod 4) D<x D 3 (mod 4) Q = Q = µ(q ) Q a (x; 4D, D) Q a (x; 4D, Q D) µ(q ) Q a (x; 4D, 3D) Q a (x; 4D, Q D) µ(q ) Q a (x; 4D, 3D) Q a (x; 4D, Q D) µ(q ) Q a (x; 4D, D) Q a (x; 4D, Q D), Q D { 1, if Q 1 (mod 4), 3, if Q 3 (mod 4), { 3, if Q 1 (mod 4), 1, if Q 3 (mod 4). Here we need a lemma on arithmetical functions (we omit the proof): Lemma 3.2. Let D be an odd natural number and D = p e1 1 pe2 2 pet t q f1 1 qf2 2 qfs s, where p i s and q j s are distinct primes with p i 1 (mod 4) and q j 3 (mod 4), be the primary decomposition of D (e i 1, f j 1). We define { 1, if t = 0, D 0 = p e1 1 pe2 2 pet t, if t 1, { 1, if s = 0, D 1 = q f1 1 qf2 2 qfs s, if s 1 and s = ω(d 1 ). Then we have Q D Q 1 (mod 4) Q D Q 3 (mod 4) µ(q ) = µ(q ) = 1, if D 0 = D 1 = 1, 2 s 1, if D 0 = 1, D 1 > 1, 0, if D 0 > 1, { 2 s 1, if D 0 = 1, D 1 > 1, 0, if D 0 > 1. Taking Lemma 3.2 into account, we can derive Proposition 3.1 from (3.7) easily.

9 On a distribution property of the residual order of a (mod pq) R a (x; 4, l), the case of l = 1, 3 (II) evaluation of Q a (x; 4D, D) and Q a (x; 4D, 3D) Proposition 3.1 shows that, in order to calculate the natural density Γ a (4, l) of R a (x; 4, l), we need to study Q a (4D, ld), l = 1, 3. These two calculations proceed in parallel. Here we need some new notations (cf. [1] [3] and [7], we make use of the same notations as those papers). For α N, let α (mod p) denote the cyclic subgroup generated by α (mod p) in Z/pZ and [Z/pZ : α (mod p) ] denote its index. Let N α (x; n, s (mod t)) = {p x; p : prime, p s (mod t), [Z/pZ : α (mod p) ] = n}. Our result is as follows and here we describe the proof only for Q a (x; 4D, D). Proposition 4.1. We have (4.1) Q a (x; 4D, D) = N a (x; k 2 f, 1 + kd 2 f (mod kd 2 f+2 )), f 1 k:odd (4.2) Q a (x; 4D, 3D) = f 1 k:odd and clearly these are disjoint unions. Proof. We start from a simple facts: N a (x; k 2 f, 1 + 3kD 2 f (mod kd 2 f+2 )) (4.3) Q a (x; 4D, D) = Q a D(x; 4, 1) Q a (x; D, 0), Q a (x; 4D, 3D) = Q a D(x; 4, 3) Q a (x; D, 0). Now we refer to Lemma 3.1 (iii) of [1]. This result gives Q a D(x, 4, 1) = N a D(x; (4l + 1) 2 f, f (mod 2 f+2 )) f 1 l 0 N a D(x; (4l + 3) 2 f, f (mod 2 f+2 )). f 1 l 0 Then, from (4.3), we have Q a (x, 4D, D) = N a D(x; (4l + 1) 2 f, f (mod 2 f+2 )) Q a (x; D, 0) f 1 l 0 (4.4) N a D(x; (4l + 3) 2 f, f (mod 2 f+2 )) Q a (x; D, 0). f 1 l 0

10 196 L. Murata and K. Chinen We remark here that, when D D a (p), then (4.5) I a D(p) = D I a (p), where I α (p) denote the index, i.e. Then we can prove I α (p) = p 1 D α (p). (4.6) N a D(x; (4l + 1) 2 f, f (mod 2 f+2 )) Q a (x; D, 0) = = N a (x; k 2 f, f (mod 2 f+2 )) Q a (x; D, 0), where k is the number defined by (we omit the proof). k = 4l + 1 D N We remark that by the fact (4.6), we succeed in unifying the sub-index into a. Now we are in a position to prove (4.1), and for this purpose it is sufficient to prove (4.7) N a (x; k 2 f, f (mod 2 f+2 )) Q a (x; D, 0) = = N a (x; k 2 f, 1 + kd 2 f (mod kd 2 f+2 )). Proof of (4.7). Let Then and p N a (x; k 2 f, f (mod 2 f+2 )) Q a (x; D, 0). I a D(p) = 2 f Dk p 1 (mod 2 f Dk). With the condition p f (mod 2 f+2 ), the Chinense remainder theorem shows p 1 + kd 2 f (mod kd 2 f+2 ). This proves in (4.7). For the another inclusion, let p N a (x; k 2 f, 1 + kd 2 f (mod kd 2 f+2 )),

11 On a distribution property of the residual order of a (mod pq) 197 then trivially p N a (x; k 2 f, f (mod 2 f+2 )). Now we put p 1 = kd 2 f + β kd 2 f with β N. Since I a (p) = k 2 f, we have D a (p) = D(1 + 4 β). This shows p Q a (x; D, 0), and we proved (4.7) as well as Proposition Behavior of Q a (x; 4D, ld) (I) existence of the natural density In this section we present an asymptotic formula for Q a (x; 4D, ld). Our method is almost on the same line as [1, Section 4], so we omit the proof. First we prepare some notations. For any integer m, we define m 0 = q m q:prime and introduce the number fields q (i.e. the core of m) K m = Q(ζ m0, a 1/m ), G k 2 f,n,d = K k 2 f (ζ n, ζ kd 2 f, a 1/k 2f n ), G k 2f,n,d = G k 2f,n,d(ζ kd 2 f+2). We take automorphisms σ 1 and σ 3 Aut(Q(ζ kd 2 f+2)/q) defined by (5.1) σ 1 : ζ kd 2 f+2 ζ kd 2 f+2 1+kD 2f σ 3 : ζ kd 2 f+2 ζ kd 2 f+2 1+3kD 2f and consider σ 1 and σ 3 Aut( G k 2f,n,d/K k 2 f ) satisfying (5.2) { σ j Gk 2 f,n,d = id. σ j Q(ζ kd 2 f+2 ) = σ j (j = 1, 3) (we use σ 1 for estimating Q a (x; 4D, D) and σ 3 for Q a (x; 4D, 3D)). We define the number c j (n) = c j (k, n, d) by { 1, if σ c j (n) = j exists, 0, if not.

12 198 L. Murata and K. Chinen G k 2f,n,d G k 2f,n,d K k 2 f Q Q(ζ kd 2 f+2) Now our result is the following: Theorem 5.1. Let a be a natural number and we decompose a into a = a 2 0a 1, a 1 is square-free. We assume GRH. (i) For any odd natural number D and l = 1, 3, we have ( ) x Q a (x; 4D, ld) = a (4D, ld)li x + O log x log log x D2 log D, where the constant implied by O-symbol is absolute. (ii) The densities a (4D, ld) (l = 1, 3) are given by the following: (5.3) (5.4) a (4D, D) = f 1 k:odd a (4D, 3D) = f 1 k:odd 2k 0 ϕ(k 0 ) 2k 0 ϕ(k 0 ) µ(d) d d 2k 0 µ(d) d d 2k 0 n=1 n=1 µ(n)c 1 (n) [ G k 2 f,n,d : Q], µ(n)c 3 (n) [ G k 2f,n,d : Q]. 6. Behavior of Q a (x; 4D, ld) (II) the case a 1 is odd We can prove the following Theorem 6.1 and it is effectively used in determining the densities Γ a (4, l) (l = 1, 3):

13 On a distribution property of the residual order of a (mod pq) 199 Theorem 6.1. Let a = a 2 0a 1, a 1 is square free and a 1 is odd. Then we have for any odd natural number D, a (4D, D) = a (4D, 3D). This section is allotted to the proof of Theorem 6.1. We prove Theorem 6.1 by showing that the series expressing a (4D, D) and a (4D, 3D) (see (5.3) and (5.4)) coincide each other, more precisely, the coefficients c 1 (n) and c 3 (n) always the same. The methods are similar to those of our previous papers [1] and [2] (see [1, Proposition 4.8] and [2, Section 3]). The coefficient c j (n) actually depends on five parameters d, f, k, n and D. The cases where f 2 or f = 1 and d is even are rather simple and are treated like [1, Proposition 4.8] (see Proposition 6.2). For the case f = 1 and d is odd, we must investigate closely the behavior of σ j (see (5.1)) on certain number fields. It needs a little complicated calculation and will be dealt with in several steps (see Propositions 6.7, 6.9 and 6.10). Here we review the notation (see also (5.1) and (5.2)): f, n N; k N, k : odd; d N, d k 0 ; D N, D : odd, D > 1. From now on, we assume a = a 2 0a 1, a 1 is square free and a 1 is odd. The following proposition can be proved similarly to [1, Proposition 4.8], so we omit the proof: Proposition 6.2. (i) When f 2, we have c 1 (n) = c 3 (n). (ii) When f = 1 and d is even, we have c 1 (n) = c 3 (n) = 0. Now we proceed to the case f = 1 and d is odd. Here we follow the lines of [2, Section 3]. Let L = Q(ζ 8kD ), M = G 4k,n,d = Q(ζ n, ζ 2kd, a 1/2kn ). Then we have LM = G 4k,n,d. We put K = L M. In this case, σ j Aut(L/Q) is defined by (6.1) σ 1 : 1+2kD ζ 8kD ζ 8kD σ 3 : ζ 8kD ζ 1+6kD 8kD.

14 200 L. Murata and K. Chinen We know that σ j K = id c j (n) = 1 for K = M L (see [2, p.700]). So the procedure will be as follows: 1 We calculate the degree [K : Q]. 2 We determine the intersection field K and investigate the behavior of σ j on K. LM = G 4k,n,d = G 4k,n,d (ζ 8kD ) M = G 4k,n,d = Q(ζ n, ζ 2kd, a 1/2kn ) K = G 4k,n,d Q(ζ 8kD ) L = Q(ζ 8kD ) Q 1 Calculation of [K : Q] First we need the following (the proof is elementary and we omit it): Lemma 6.3. Let D, k, n N, D, k be odd, d k, d, n be square free and we denote the odd part of n by n. Then we have ϕ( n, kd )ϕ(kd) ϕ( n, kd, kd ) = ϕ(kd ), where d = (d, D), n = p:prime, p n, p D, p k p and D = d n. Remark. Note that (n, d ) = (n, kd ) = 1 and D D. We know Gal(LM/M) = Gal(L/K) for K = M L (see [2, Lemma 3.1]) and so [L : K] = [ G 4k,n,d : G 4k,n,d ]. From this, we can calculate [K : Q] by [L : Q] [K : Q] = [L : K] = [Q(ζ 8kD) : Q] [ G 4k,n,d : G 4k,n,d ] = [Q(ζ 8kD) : Q][G 4k,n,d : Q] [ G. 4k,n,d : Q] Note that G 4k,n,d = Q(ζ n,2kd, a 1/2kn ), G 4k,n,d = Q(ζ n,2kd,8kd, a 1/2kn ).

15 On a distribution property of the residual order of a (mod pq) 201 Using [7, Proposition 3.1], we can calculate the degrees [G 4k,n,d [ G 4k,n,d : Q]: : Q] and Proposition 6.4. The degrees [G 4k,n,d : Q] and [ G 4k,n,d : Q] are given as follows: Cases [G 4k,n,d : Q] a 1 1 (mod 4) a 1 n, kd 2knϕ( n, kd ) a 1 n, kd knϕ( n, kd ) a 1 3 (mod 4) 2knϕ( n, kd ) Cases [ G 4k,n,d : Q] a 1 1 (mod 4) a 1 n, kd, kd 8knϕ( n, kd, kd ) a 1 n, kd, kd 4knϕ( n, kd, kd ) a 1 3 (mod 4) a 1 n, kd, kd 8knϕ( n, kd, kd ) a 1 n, kd, kd 4knϕ( n, kd, kd ) Proof. It is easy from [7, Proposition 3.1]. Remark. Read the table like [G 4k,n,d : Q] = knϕ( n, kd ) if and only if a 1 1 (mod 4) and a 1 n, kd, etc. Next, since kd is odd, we have (6.2) [Q(ζ 8kD ) : Q] = 4ϕ(kD). Now we can determine the degree [K : Q] as follows: Proposition 6.5. The degrees [K : Q] are given as follows: Cases [K : Q] (I) a 1 1 (mod 4) (i) a 1 n, kd, a 1 n, kd, kd ϕ(kd ) (ii) a 1 n, kd, a 1 n, kd, kd ϕ(kd ) (iii) a 1 n, kd, a 1 n, kd, kd 2ϕ(kD ) (II) a 1 3 (mod 4) (i) a 1 n, kd, kd 2ϕ(kD ) (ii) a 1 n, kd, kd ϕ(kd ) Proof. The proof is easy from Proposition 6.4 and (6.2). Remark. Read the table like When a 1 1 (mod 4), [K : Q] = ϕ(kd ) if and only if a 1 n, kd and a 1 n, kd, kd, etc.

16 202 L. Murata and K. Chinen 2 Determination of K and c j (n) Recall L = Q(ζ 8kD ), M = G 4k,n,d = Q(ζ n, ζ 2kd, a 1/2kn ). We determine the intersection field K = L M with the help of the fact that K is a subfield of the cyclotomic field L. Recall D = d n, d = (d, D) and n = p n, p D, p k p. Lemma 6.6. Q(ζ kd ) L M. Proof can be carried out in an elementary manner and we omit it. First we deal with the cases (I-i), (I-ii) and (II-ii) in Proposition 6.5 (where [K : Q] = ϕ(kd )): Proposition 6.7. If [K : Q] = ϕ(kd ), then K = Q(ζ kd ) and c 1 (n) = = c 3 (n) = 1. Proof. The fact [K : Q] = ϕ(kd ) and Lemma 6.6 implies K = Q(ζ kd ). We look into the actions of σ 1 and σ 3. Putting D = D D, we have ζ kd = ζ 8kD 8D and σ 1 (ζ kd ) = (ζ 8kD 8D ) 1+2kD = ζ 8kD 8D ζ 8kD 16kDD = ζ kd. This shows σ 1 K = id and c 1 (n) = 1. Similarly we can easily see σ 3 K = id and c 3 (n) = 1. For the cases (I-iii) and (II-i) in Proposition 6.5 (where [K : Q] = 2ϕ(kD )), we know Q(ζ kd ) K = L M and [K : Q(ζ kd )] = 2 from Proposition 6.5 and Lemma 6.6. So, to determine the field K, we must find a quadratic algebraic number α such that α L M and α Q(ζ kd ). Then we have K = Q(ζ kd, α).

17 On a distribution property of the residual order of a (mod pq) 203 L = Q(ζ 8kD ) M = Q(ζ n, ζ 2kd, a 1/2kn ) K = L M 2 Q(ζ kd ) Q ϕ(kd ) The following lemma is required: Lemma 6.8. Let m Z, D m be the discriminant of Q( m) and d = D m. Then we have Q( m) Q(ζ d ). Conversely, if Q( m) Q(ζ n ), then d n. Proof. See a suitable textbook in algebraic number theory (see also [2, Lemma 3.2]). First we consider the case (I-iii) of Proposition 6.5 (recall the conditions a 1 n, kd and a 1 n, kd, kd ). We decompose a 1 as a 1 = b 1 b 2, where b 1 is the product of all the primes which divide n, kd. The following conditions are frequently used in the subsequent discussion: b 1 n, kd, b 2 n, kd, b 2 kd, (b 2, n, kd ) = 1, (b 2, kd ) = 1. Proposition 6.9. When a 1 1 (mod 4), a 1 n, kd and a 1 n, kd, kd, we have ( ) K = Q ζ kd, ( 1) b b 2 and c 1 (n) = c 3 (n) = 1. Proof. First we consider the case b 1 1 (mod 4). Then b 2 1 (mod 4) since a 1 1 (mod 4). We can see b 1 M = G 4k,n,d = Q(ζ 2 n,kd, a 1/2kn ) because

18 204 L. Murata and K. Chinen Q( b 1 ) Q(ζ b1 ) (see Lemma 6.8) and ζ b1 = ζ 2 n,kd 2 n,kd /b 1 M (note that b 1 n, kd ). On the other hand, we see b2 = a1 b1 M since a 1 M. Moreover we know b 2 L because Q( b 2 ) Q(ζ b2 ) (see Lemma 6.8) and ζ b1 = ζ 8kD 8kD/b 2 L (note that b 2 kd). Thus we have b2 L M. We also notice that b 2 Q(ζ kd ) since (b 2, kd ) = 1. Hence we obtain K = Q(ζ kd, b 2 ). Next we consider the action of σ 1 on K. We already know σ 1 (ζ kd ) = ζ kd (see the proof of Proposition 6.7). As to b 2, we have σ 1 (ζ b2 ) = σ 1 (ζ 8kD 8 kd/b 2 ) = (ζ 8kD 1+2kD ) 8 kd/b2 = ζ b2, for b 2 kd. So, Q(ζ b2 ) is fixed by σ 1. Since b 2 Q(ζ b2 ), σ 2 b 2 = b2. Thus we have proved σ 1 K = id and c 1 (n) = 1. The case b 1 3 (mod 4) is similar: we have K = Q(ζ kd, b 2 ) and can verify σ 3 K = id, thus c 3 (n) = 1. The case (II-i) in Proposition6.5 can be dealt with similarly to the previous proposition, so we state the result without proof: Proposition Let a 1 3 (mod 4) and a 1 n, kd, kd. Then we have the following: (a) If a 1 n, kd, then ( ) K = Q ζ kd, ( 1) b b 2. (b) If a 1 n, kd, then K = Q ( ζ kd, 1 ). In both cases, c 1 (n) = c 3 (n). The results which are obtained in this section are summarized in the following theorem: Theorem If a 1 is odd, then the coefficients c j (n) in (5.3) and (5.4) always satisfy c 1 (n) = c 3 (n).

19 On a distribution property of the residual order of a (mod pq) 205 More precisely, if f 2, then c 1 (n) = c 3 (n); if f = 1 and d is even, then c 1 (n) = c 3 (n) = 0; if f = 1 and d is odd, (I) if a 1 1 (mod 4) and (i) a 1 n, kd, a 1 n, kd, kd, (ii) a 1 n, kd, a 1 n, kd, kd, then K = Q(ζ kd ), c 1 (n) = c 3 (n) = 1; (iii) a 1 n, kd, a 1 n, kd, kd, ( then K = Q ζ kd, ( 1) b b 2 ), c 1 (n) = c 3 (n) = 1; (II) if a 1 3 (mod 4) and (i-a) a 1 n, kd, a 1 n, kd, kd, ( then K = Q ζ kd, ( 1) b b 2 ), c 1 (n) = c 3 (n); (i-b) a 1 n, kd, a 1 n, kd, kd, then K = Q(ζ kd, 1), c 1 (n) = c 3 (n); (ii) a 1 n, kd, kd, then K = Q(ζ kd ), c 1 (n) = c 3 (n) = 1. We know from Theorem 6.11 that the series (5.3) and (5.4) are the same, and this completes the proof of Theorem The existence of the density Γ a (4, l) We prove the main result Theorem 1.3. Let us start from the formula (3.1). Our proof bases upon the following facts: 1 We assume GRH. When a 1 is odd, then for l = 1, 3, (7.1) Q a (x; 4, l) = 1 ( ) 6 li x + O x log x log log x ([7, Theorem 1.2]).

20 206 L. Murata and K. Chinen 2 We assume GRH. In Section 5, we proved, for any D < x, ( ) x (7.2) Q a (x; 4D, ld) = a (4D, ld) li x + O log x log log x D2 log D, where a (4D, ld) is a positive constant, the natural density of Q a (x; 4D, ld) (l = 1, 3). And if a 1 is odd, then (7.3) a (4D, D) = a (4D, 3D). 3 Furthermore here we remark that (7.4) Q a (x; 4D, ld) π(x; D, 1), in fact, if p Q a (x; 4D, ld), then p 1 is divisible by D a (p), which is divisible by D. Similarly we have (7.5) Q a (x; 4D, ld) x D. Proof of Theorem 1.3. We take y = (log log x) 1/3, z = (log x) 20/3. D<x D 1 (mod 4) D 0=1,D 1>1 = 1<D<y D 1 (mod 4) D 0=1,D 1>1 2 s 1 ( Q a (x; 4D, D) Q a (x; 4D, 3D)) 2 = + y D<z D 1 (mod 4) D 0=1,D 1>1 + z D<x D 1 (mod 4) D 0=1,D 1>1 2 s 1 ( Q a (x; 4D, D) Q a (x; 4D, 3D)) 2 = = E 1 + E 2 + E 3, say. Estimate of E 1. By making use of (7.2) and (7.3), we have

21 On a distribution property of the residual order of a (mod pq) 207 (7.6) E 1 = D<y D 1 (mod 4) D 0=1,D 1>1 2 s 1 {( a (4D, D) a (4D, 3D))li x+ ( )} 2 x +O log x log log x D2 log D = ( = 2 s 1 O D<y D 1 (mod 4) D 0=1,D 1>1 x 2 (log x) 2 (log log x) 2 D4 log 2 D D log 2 D 4 log 2 D (log x) 2 (log log x) 2 = D<y ( ) x 2 = o log 2, x x 2 ) since y = (log log x) 1/3. Estimate of E 3. Here we use the estimate (7.5): E 3 < < z<d z D<x D 1 (mod 4) D 0=1,D 1>1 log 2 x2 D D 2 = = x 2 O ( z log 2 1). 2 s 1 ( x D Since z = (log x) 20/3, we have ( ) 1 (7.7) E 3 = o log 2. x Estimate of E 2. Here we use the estimate (7.4): and in our case, D satisfies Q a (x; 4D, ld) π(x; D, 1), D < z = (log x) 20/3, ) 2 < then we have π(x; D, 1) 1 ϕ(d) π(x).

22 208 L. Murata and K. Chinen Then, (7.8) E 2 < y<d<z π(x) 2 y<d { } 2 1 D log 2 ϕ(d) π(x) D log 2 D 2 (log log D)2 π(x) 2 y 0.3 = = o(π(x) 2 ). Now, combining (3.1), (7.1), (7.6) (7.8), we proved R a (x; 4, 1) = 1 18 (li x)2 + o((li (x)) 2 ), and this gives the natural density of R a (x; 4, 1), i.e. Γ a (4, 1) = 1/18. Similarly, Γ a (4, 3) = 1/ Numerical examples In this section, we show some results of numerical experiment on the density Γ a (4, l). For the computer calculation, we use the relation (2.1), that is, we first calculate the residual orders D a (p) in 0 < p x for some x, and next we use the GCD algorithm to get D a (pq). We use the data of D a (p) which are already obtained in our previous papers [2] and [3]. Here we take x up to 10 7 and show the values π(x) 2 R a (x; 4, l) for several a s. The program is written in the C language, compiled by GCC. If x = 10 7, then we have about pairs (p, q) satisfying 0 < q < p x. We did numerical experiments to calculate D a (pq) from the data (D a (p), D a (q)) by the GCD algorithm in the range above. Calculation time depends on a, but it takes approximately 20 hours for each a at 2.66GHz CPU. (i) The case a 1 1 (mod 4) We show two examples a = 13 and 20 (Tables 8.1 and 8.2). The theoretical densities are Γ a (4, 0) = 5 9 = , Γ a(4, 1) = 1 18 = , Γ a (4, 2) = 1 3 = , Γ a(4, 3) = 1 18 =

23 On a distribution property of the residual order of a (mod pq) 209 x l = 0 l = 1 l = 2 l = Table 8.1. The case a = 13 x l = 0 l = 1 l = 2 l = Table 8.2. The case a = 20 (ii) The case a 1 3 (mod 4) We show two examples a = 11 and 12 (Tables 8.3 and 8.4). The theoretical densities are Γ a (4, 0) = 5 9 = , Γ a(4, 1) = 1 18 = , Γ a (4, 2) = 1 3 = , Γ a(4, 3) = 1 18 = x l = 0 l = 1 l = 2 l = Table 8.3. The case a = 11

24 210 L. Murata and K. Chinen x l = 0 l = 1 l = 2 l = (iii) The case a 1 2 (mod 4) Table 8.4. The case a = 12 We show two examples a = 2 and 10 (Tables 8.5 and 8.6). At present, we know the theoretical densities Γ a (4, 0) and Γ a (4, 2) only: Γ 2 (4, 0) = = , Γ 2(4, 2) = = , Γ 10 (4, 0) = 5 9 = , Γ 10(4, 2) = 1 3 = The calculation of the exact theoretical density of Γ 2 (4, 1) needs much deeper considerations of the algebraic quantities which appear in our Section 5. x l = 0 l = 1 l = 2 l = Table 8.5. The case a = 2 x l = 0 l = 1 l = 2 l = Table 8.6. The case a = 10

25 On a distribution property of the residual order of a (mod pq) 211 References [1] Chinen, K. and L. Murata, On a distribution property of the residual order of a (mod p), J. Number Theory, 105 (2004), [2] Chinen, K. and L. Murata, On a distribution property of the residual order of a (mod p) III, J. Math. Soc. Japan, 58-3 (2006), [3] Chinen, K. and L. Murata, On a distribution property of the residual order of a (mod p) IV, in Number Theory: Tradition and Modernization, W. Zhang and Y. Tanigawa, eds., Developments in Math. 15, 11 22, Springer Verlag, [4] Hasse, H., Über die Dichte der Primzahlen p, für die eine vorgegebene ganzrationale Zahl a 0 von durch eine vorgegebene Primzahl l 2 teilbarer bzw. unteibarer Ordnung mod p ist, Math. Ann., 162 (1965), [5] Hooley, C., On Artin s conjecture, J. Reine Angew. Math., 225 (1967), [6] Lagarias, J.C. and A.M. Odlyzko, Effective versions of the Chebotarev density theorem, in Algebraic Number Fields (Durham, 1975), , Academic Press, London, [7] Murata, L. and K. Chinen, On a distribution property of the residual order of a (mod p) II, J. Number Theory, 105 (2004), [8] Odoni, R.W.K., A conjecture of Krishnamurthy on decimal periods and some allied problems, J. Number Theory, 13 (1981), L. Murata Department of Mathematics Faculty of Economics Meiji Gakuin University Shirokanedai, Minato-ku Tokyo, Japan leo@eco.meijigakuin.ac.jp K. Chinen Department of Mathematics School of Science and Engineering Kinki University 3-4-1, Kowakae Higashi-Osaka, Japan chinen@math.kindai.ac.jp