Table Of ContentPrimes In Arithmetic Progressions And Primitive Roots
N. A. Carella
Abstract: Let x 1 be a large number, and let 1 a < q be integers such that gcd(a,q) = 1 and
≥ ≤
q = O(logc) with c > 0 constant. This note proves that the counting function for the number of
primes p p = qn+a : n 1 with a fixed primitive root u = 1,v2 has the asymptotic formula
∈ { ≥ } 6 ±
π (x,q,a) = δ(u,q,a)x/logx+ O(x/logbx), where δ(u,q,a) > 0 is the density, and b > c+1 is a
u
constant.
Mathematics Subjects Classification: 11A07, 11N37.
Keywords: Prime Number; Primitive Root; Arithmetic Progression.
7
1
0
2
n
a
J
1
1
1 Introduction
]
M
Let u,q and a be integers with gcd(a,q) = 1. The density of the subset of primes in the arithmetic
G progression p = qn + a : gcd(a,q) = 1 and ord (u) = p 1 is defined by δ(u,q,a), and the
p
{ − }
. corresponding counting function
h
t
a
π (x,q,a) = # p = qn+a x : gcd(a,q) = 1 and ord (u) = p 1 (1)
m u p
{ ≤ − }
[
for any large number x 1 are subject of this note.
≥
1
v
8
8 Theorem 1.1 Let x 1 be a large number. Let 1 a < q be integers such that gcd(a,q) = 1 and
≥ ≤
1 q = O(logcx), with c 0 constant. Then, the arithmetic progression p = qn + a : n 1 has
3 infinitely many primes p≥ 2 with a fixed primitive root u= 1,v2. In add{ition, the counting≥fun}ction
0 ≥ 6 ±
. for these primes has the asymptotic formula
1
0
x x
7 π (x,q,a) = δ(u,q,a) +O , (2)
1 u logx logbx
(cid:18) (cid:19)
:
v
i where δ(u,q,a) 0 is the density constant depending on the integers u,q,a, and b > c+1is a constant.
X ≥
r
The preliminary background results and notation are discussed in Sections two to five. Section six
a
presents a proof of Theorem 1.1. Lastly, Corollary 7.1 provides the density for the Gauss problem
about the primitive root u= 10.
2 Representation of the Characteristic Function
The order of an element in the cyclic group G = F× is defined by
p
ord (v) = min k : vk 1 modp . (3)
p
{ ≡ }
Primitive elements in this cyclic group have order p 1 = #G. The characteristic function Ψ: G
− −→
0,1 of primitive elements is one of the standard analytic tools employed to investigate the various
{ }
properties of primitive roots in cyclic groups G. Many equivalent representations of the characteristic
function Ψ of primitive elements are possible. The standard characteristic function is discussed in [11,
p. 258]. It detects a primitive element by means of the divisors of p 1.
−
Primes And Primitive Roots, Page 2
A new representation of the characteristic function for primitive elements is developed here. It detects
the order ord (u) 1 of the element u F by means of the solutions of the equation τn u = 0 in
p p
≥ ∈ −
F , where u,τ are constants, and n is a variable such that 1 n < p 1,gcd(n,p 1) = 1.
p
≤ − −
Lemma 2.1 Let p 2 be a prime, and let τ be a primitive root mod p. Let u F be a nonzero
p
≥ ∈
element, and let, ψ = 1 be a nonprincipal additive character of order ordψ =p. Then
6
1 1 if ord (u) = p 1,
Ψ(u) = ψ((τn u)k) = p − (4)
p − 0 if ordp(u) = p 1.
gcd(n,p−1)=1 0≤k≤p−1 (cid:26) 6 −
X X
Proof: As the index n 1 ranges over the integers relatively prime to p 1, the element τn F
p
≥ − ∈
ranges over the primitive roots modp. Ergo, the equation τn u= 0 has a solution if and only if the
−
fixed element u F is a primitive root. Next, replace ψ(z) = ei2πkz/p to obtain
p
∈
1 ei2π(τn−u)k/p = 1 if ordp(u) = p−1, (5)
p 0 if ord (u) = p 1.
p
gcd(n,p−1)=1 0≤k≤p−1 (cid:26) 6 −
X X
This follows from the geometric series identity wn = (wN 1)/(w 1),w = 1 applied to
0≤n≤N−1 − − 6
the inner sum. (cid:4).
P
3 Estimates Of Exponential Sums
This Section provides a simple estimate for the exponential sum of interest in this analysis. The proof
of this estimate is entirely based on next theorem and elementary techniques. Some versions of the
next result for double exponential sums are also proved in [2], and [5].
Theorem 3.1 ([6, Theorem 1]) Let p 2 be a large prime, and let τ F be an element of large
p
≥ ∈
multiplicative order T = ord (τ). Let X,Y F be large subsets of cardinality #X #Y p7/8+ε
p p
⊂ ≥ ≥
Then, for any given number k > 0, the following estimate holds.
γ(x)ei2πaτxy/p #Y1−21k ·#X1−2k1+2 ·p21k+4k3+4+o(1), (6)
yX∈Y (cid:12)(cid:12)xX∈X (cid:12)(cid:12)≪ T2k1+2
(cid:12) (cid:12)
(cid:12) (cid:12)
where the coefficients γ(cid:12)(x) are complex nu(cid:12)mbers, γ(x) 1, and a 1,gcd(a,p) = 1 is an integer.
| | ≤ ≥
Lemma 3.1 Let p 2 be a large prime, and let τ be a primitive root modulo p. Then,
≥
max ei2πsτn/p p1−ε (7)
gcd(s,p−1)=1(cid:12) (cid:12)≪
(cid:12)gcd(n,p−1)=1 (cid:12)
(cid:12) X (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
for any small number ε< 1/16. (cid:12) (cid:12)
Proof: Let X = Y = n : gcd(n,p 1) = 1 F be subsets of cardinalities #X = #Y = ϕ(p 1).
p
{ − } ⊂ −
Since for each fixed y Y, the map x xy is a permutation of the subset X F , the double sum
p
∈ −→ ⊂
can be rewritten as a single exponential sum
γ(x)ei2πsτxy/p = #Y ei2πsτx/p , (8)
(cid:12) (cid:12) (cid:12) (cid:12)
yX∈Y (cid:12)xX∈X (cid:12) (cid:12)xX∈X (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Primes And Primitive Roots, Page 3
where γ(x) = 1 if x X otherwise it vanishes. As the element τ is a primitive root, it has order
∈
T = ord (τ) = p 1. Setting k = 1 in Theorem 3.1 yields
p
−
1
ei2πsτx/p = ei2πsτxy/p
#Y
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12)xX∈X (cid:12) yX∈Y (cid:12)xX∈X (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) 1 (cid:12) (cid:12)p7/8+ε
(cid:12) (cid:12) #Y(cid:12) 1/2 #X3/4 (cid:12) (9)
≪ #Y · · T1/4 !
1 p7/8+ε
ϕ(p 1)1/2 ϕ(p 1)3/4
≪ ϕ(p 1) − · − · (p 1)1/4!
− −
p7/8+ε.
≪
This is restated in the simpler notation p7/8+ε p1−ε for any number ε (0,1/16). (cid:4).
≤ ∈
A similar result appears as Theorem 6 in [3].
4 Estimate For The Error Term
The upper bound for exponential sum over subsets of elements in finite fields F stated in the last
p
Section will be used here to estimate the error term E(x) arising in the proof of Theorem 1.1.
Lemma 4.1 Let p 2 be a large prime, let ψ = 1 be an additive character, and let τ be a primitive
≥ 6
root mod p. If the element u= 0 is not a primitive root, then,
6
1 1 x1−ε
(cid:12) ψ((τn u)k)(cid:12) , (10)
(cid:12) p − (cid:12) ≪ ϕ(q)logx
(cid:12) x≤p≤2x gcd(n,p−1)=1,0<k≤p−1 (cid:12)
(cid:12) X X X (cid:12)
p≡amodq
(cid:12) (cid:12)
(cid:12) (cid:12)
where 1 a < q, gcd(cid:12)(a,q) = 1 and O(logcx) with c > 0 constan(cid:12)t, for all sufficiently large numbers
≤ (cid:12) (cid:12)
x 1 and an arbitrarily small number ε > 0.
≥
Proof: Byhypothesisu= τn foranyn 1suchthatgcd(n,p 1) = 1. Therefore, ψ((τn u)k) =
6 ≥ − 0<k≤p−1 −
1, and
− P
1
E(x) = (cid:12) ψ((τn u)k)(cid:12)
| | (cid:12) p − (cid:12)
(cid:12) x≤p≤2x gcd(n,p−1)=1,0<k≤p−1 (cid:12)
(cid:12) X X X (cid:12)
p≡amodq
(cid:12) (cid:12)
(cid:12) ϕ(p 1) (cid:12)
(cid:12) (cid:12)
(cid:12) − (cid:12)
≤ p
x≤p≤2x
X
p≡amodq
1 1 x x
+O , (11)
≤ 2ϕ(q)logx log2x
(cid:18) (cid:19)
where ϕ(n)/n 1/2 for any integer n 1. This implies that there is a nontrivial upper bound. To
≤ ≥
sharpen this upper bound, let ψ(z) = ei2πkz/p with 0 < k < p, and rearrange the triple finite sum in
the form
1
E(x) = (cid:12) ψ((τn u)k)(cid:12)
| | (cid:12) p − (cid:12)
(cid:12) x≤p≤2x 0<k≤p−1,gcd(n,p−1)=1 (cid:12)
(cid:12) X X X (cid:12)
p≡amodq
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) 1 (cid:12)
(cid:12) e−i2πuk/p ei2πk(cid:12)τn/p , (12)
≤ (cid:12)p (cid:12)
x≤p≤2x (cid:12) 0<k≤p−1 gcd(n,p−1)=1 (cid:12)
X (cid:12) X X (cid:12)
p≡amodq
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Primes And Primitive Roots, Page 4
and let
1 1
U = e−i2πuk/p and V = ei2πkτn/p. (13)
p p
p1/2 p1/2
0<k≤p−1 gcd(n,p−1)=1
X X
Now consider the Holder inequality AB A B with 1/r + 1/s = 1. In terms of the
1 r s
|| || ≤ || || · || ||
components in (13) this inequality has the explicit form
1/r 1/s
UpVp Up r Vp s . (14)
| |≤ | | | |
x≤p≤2x x≤p≤2x x≤p≤2x
X X X
p≡amodq p≡amodq p≡amodq
The absolute value of the first exponential sum U is given by
p
1 1
U = e−i2πuk/p = . (15)
p
| | (cid:12)p1/2 (cid:12) p1/2
(cid:12) 0<k≤p−1 (cid:12)
(cid:12) X (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
This follows from ei2πuk/p =(cid:12) 1 for u = 0 and s(cid:12)ummation of the geometric series. The
0<k≤p−1 − 6
corresponding r-norm has the upper bound
P
U r = U r
|| p||r | p|
x≤p≤2x
X
p≡amodq
r
1
=
p1/2
x≤p≤2x (cid:12) (cid:12)
X (cid:12) (cid:12)
p≡amodq
(cid:12) (cid:12)
1 (cid:12) (cid:12)
1
≤ xr/2
x≤p≤2x
X
p≡amodq
3 x1−r/2
. (16)
≤ ϕ(q) logx
Here thefinitesumover theprimesis estimated usingtheBrun-Titchmarshtheorem; thisresultstates
that the number of primes p = qn+1 in the interval [x,2x] satisfies the inequality
3 x
π(2x,q,a) π(x,q,a) , (17)
− ≤ ϕ(q)logx
see [10, p. 167], [8, p. 157], [12], and [20, p. 83].
The absolute value of the second exponential sum V = V (k) has the upper bound
p p
1
V = ei2πkτn p1/2−ε. (18)
p
| | (cid:12)p1/2 (cid:12) ≪
(cid:12) gcd(n,p−1)=1 (cid:12)
(cid:12) X (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
This exponential sum dependents on k; but it has a uniform, and independent of k upper bound
(cid:12) (cid:12)
max ei2πkτn/p p1−ε, (19)
1≤k≤p−1(cid:12) (cid:12) ≪
(cid:12)gcd(n,p−1)=1, (cid:12)
(cid:12) X (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
where ε > 0 is an arbitrarily small num(cid:12)ber, see Lemma ??. (cid:12)A similar application appears in [15, p.
1286].
Primes And Primitive Roots, Page 5
In light of this estimate, the corresponding s-norm has the upper bound
V s = V s
|| p||s | p|
x≤p≤2x
X
p≡amodq
s
p1/2−ε
≪
x≤Xp≤2x (cid:12) (cid:12)
p≡amodq(cid:12) (cid:12)
(cid:12) (cid:12)
s
(2x)1/2−ε 1
≪
(cid:16) (cid:17) x≤Xp≤2x
p≡amodq
1 (2x)1+s/2−εs
≪ ϕ(q) logx
1 x1+s/2−εs
. (20)
≪ ϕ(q) logx
The finite sum over the primes in the third line is estimated using the Brun-Titchmarsh theorem; and
the last line uses 21/s+1/2−ε 2 for s 1.
≤ ≥
Now, replace the estimates (16) and (20) into (14), the Holder inequality, to reach
1/r 1/s
UpVp Up r Vp s
| | ≤ | | | |
x≤p≤2x x≤p≤2x x≤p≤2x
X X X
p≡amodq p≡amodq p≡amodq
1/r 1/s
1 x1−r/2 1 x1+s/2−εs
≪ ϕ(q) logx ϕ(q) logx
! !
1 x1−ε
(21)
≪ ϕ(q)logx
1 x1−ε
.
≪ ϕ(q)logx
Note that this result is independent of the parameter 1/r+1/s =1. (cid:4).
5 Evaluation Of The Main Term
Finitesumsandproductsovertheprimesnumbersoccuronvariousproblemsconcernedwithprimitive
roots. These sums and products often involve the normalized totient function ϕ(n)/n = (1 1/p)
p|n −
and the corresponding estimates, and the asymptotic formulas.
Q
Lemma 5.1 ([14, Lemma 5]) Let x 1 be a large number, and let ϕ(n) be the Euler totient function.
≥
If q logcx, with c 0 constant, an integer 1 a < q such that gcd(a,q) = 1, then
≤ ≥ ≤
ϕ(p 1) li(x) x
− = A +O , (22)
p 1 qϕ(q) logBx
p≤x − (cid:18) (cid:19)
X
p≡amodq
where li(x) is the logarithm integral, and b > c+1 1 is an arbitrary constant, as x , and
≥ → ∞
1 1
A = 1 1 . (23)
q
− p − p(p 1)
p|gcd(a−1,q)(cid:18) (cid:19) p∤q (cid:18) − (cid:19)
Y Y
Primes And Primitive Roots, Page 6
Related discussions for q = 2 are given in [19, Lemma 1], [13, p. 16], and, [21]. The case q = 2 is
ubiquitous in various results in Number Theory.
Lemma 5.2 Let x 1 be a large number, and let ϕ(n) be the Euler totient function. If q logcx,
≥ ≤
with c 0 constant, an integer 1 a < q such that gcd(a,q) = 1, then
≥ ≤
1 li(x) x
1 = A +O , (24)
p qϕ(q) logbx
p≤x gcd(n,p−1)=1 (cid:18) (cid:19)
X X
p≡amodq
where li(x) is the logarithm integral, and b > c+1 1 is an arbitrary constant, as x , and A is
q
≥ → ∞
defined in (23).
Proof: A routine rearrangement gives
1 ϕ(p 1)
1 = − (25)
p p
p≤x gcd(n,p−1)=1 p≤x
X X X
p≡amodq p≡amodq
ϕ(p 1) ϕ(p 1)
= − − .
p 1 − p(p 1)
p≤x − p≤x −
X X
p≡amodq p≡amodq
To proceed, apply Lemma 5.1 to reach
ϕ(p 1) ϕ(p 1) li(x) x
− − = A +O
p 1 − p(p 1) qϕ(q) logbx
p≤x − p≤x − (cid:18) (cid:19)
X X
p≡amodq p≡amodq
ϕ(p 1)
− (26)
− p(p 1)
p≤x −
X
p≡amodq
li(x) x
= A +O ,
qϕ(q) logbx
(cid:18) (cid:19)
where the second finite sum
ϕ(p 1)
− loglogx (27)
p(p 1) ≪
p≤x −
X
p≡amodq
is absorbed into the error term, b > c+1 1 is an arbitrary constant, and A is defined in (23). (cid:4).
q
≥
The logarithm integral has the asymptotic formula
x 1 x x
li(x) = dz = +O . (28)
logz logx log2x
Z2 (cid:18) (cid:19)
6 The Main Result
Given a fixed integer u= 1,v2, the precise primes counting function is defined by
6 ±
π (x,q,a) = # p x : p a modq and ord (u) = p 1 (29)
u p
{ ≤ ≡ − }
for 1 a < q and gcd(a,q) = 1. The limit
≤
π (x,q,a) A
ℓ q
δ(u,q,a) = lim = a (30)
u
x→∞ π(x,q,a) ϕ(q)
is the density of the subset of primes with a fixed primitive root u = 1,v2. The proof of Theorem
6 ±
1.1 is completed here.
Primes And Primitive Roots, Page 7
Proof: Suppose that u = 1,v2 is not a primitive root for all primes p x , with x 1 constant.
0 0
6 ± ≥ ≥
Let x > x be a large number, and q = O(logcx). Consider the sum of the characteristic function
0
over the short interval [x,2x], that is,
0= Ψ(u). (31)
x≤p≤2x
X
q≡amodq
Replacing the characteristic function, Lemma ??, and expanding the nonexistence equation (31) yield
0 = Ψ(u)
x≤p≤2x
X
p≡amodq
1
= ψ((τn u)k)
p −
x≤p≤2x gcd(n,p−1)=1,0≤k≤p−1
X X X
p≡amodq
1
= a 1 (32)
u
p
x≤p≤2x gcd(n,p−1)=1
X X
p≡amodq
1
+ ψ((τn u)k)
p −
x≤p≤2x gcd(n,p−1)=1,0<k≤p−1
X X X
p≡amodq
= M(x)+E(x),
where a 0 is a constant depending on the integers u= 1,v2 and q 2.
u
≥ 6 ± ≥
The main term M(x) is determined by a finite sum over the trivial additive character ψ = 1, and the
error term E(x) is determined by a finite sum over the nontrivial additive characters ψ = ei2πk/p =1.
6
Take b > c+1. Applying Lemma 5.1 to the main term, and Lemma 4.1 to the error term yield
0 = Ψ(u)
x≤p≤2x
X
p≡amodq
= M(x)+E(x)
li(2x) li(x) x 1 x1−ε
= a A − +O +O
u q ϕ(q) logbx ϕ(q)logx
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
x x
= δ(u,q,a) +O , (33)
logx logb
(cid:18) (cid:19)
where δ(u,q,a) = a A /ϕ(q) 0, and a 0 is a correction factor depending on ℓ.
ℓ q u
≥ ≥
But δ(u,q,a) > 0 contradicts the hypothesis (31) for all sufficiently large numbers x x . Ergo, the
0
≥
short interval [x,2x] contains primes p =qn+a such that the u is a fixed primitive root. Specifically,
the counting function is given by
x
π (x,q,a) = Ψ(u) = δ(u,q,a)li(x)+O . (34)
u logbx
p≤x (cid:18) (cid:19)
X
p≡amodq
This completes the verification. (cid:4).
The determination of the correction factor a in a primes counting problem is a complex problem,
u
some cases are discussed in [18], and [22].
The logarithm difference li(2x) li(x) = x/logx+O x/log2x is evaluated via (28) in Section 5.
−
(cid:0) (cid:1)
Primes And Primitive Roots, Page 8
7 Special Case u = 10
Some of the earliest work on this problem is concerned with the primitive root u = 10, see [1], and
[4]. A corollary of Theorem 1.1 provides the required density.
Corollary 7.1 There are infinitely many primes p 7 with the primitive root u= 10. Moreover, the
≥
counting function for these primes satisfies the lower bound
x x
π (x)= # p x :ord (10) =p 1 =δ(10) +O , (35)
10 { ≤ p − } logx log2x
(cid:18) (cid:19)
where δ(10) = δ(10,2,1) > 0 is a constant, for all large numbers x 1.
≥
Proof: This follows from Theorem 1.1 by replacing the primitive root base u= 10. That is,
x x
π (x)= Ψ(10) = δ(10) +O . (36)
10 logx log2x
p≤x (cid:18) (cid:19)
X
Let u = (sv2)k = 1 and s 2 mod 4 is squarefree, k 1. A formula for computing the density
6 ± ≡ ≥
appears in [7, p. 220]. In the decimal case u= 10, the density is
1 1 1
δ(u) = 1 µ(s) 1
− p2 p 1 p2 p 1 − p(p 1)
p|s − − p|s − − p∤k (cid:18) − (cid:19)
Y Y Y
p∤k p|k
18 1
= 1 = 0.357479581392.... (37)
19 − p(p 1)
p≥2(cid:18) − (cid:19)
Y
This completes the proof. (cid:4).
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