Number Theory Thesis Analysis
Number Theory Thesis Analysis
A Thesis
presented to
University of Missouri
In Partial Fulfillment
Master of Arts
by
Aaron Yeager
December 2012
The undersigned, appointed by the Dean of the Graduate School, have examined the
thesis entitled
a candidate for the degree of Master of Arts and hereby certify that in their opinion
it is worthy of acceptance.
To begin, I would like to thank my friends and family members for their support,
today. I must also give credit to the numerous excellent math teachers, instructors,
and professors I have had throughout my collegiate career. I would like to thank
Ryan Alvarado and Kevin Brewster for their aid with LaTex commands that made
my thesis possible. I would like to thank my co-authors of the papers I written while
at the University of Missouri: William Banks, Ahmet M. Güloğlu, Zhenyu Guo, and
Roger Baker. I would like to thank Youssef Saab, Konstantin Makarov, and Jan
Segert for being on my master’s committee. On a final note, I would like to express
my gratitude to William Banks for being my advisor and for his assistance on my
thesis.
ii
Contents
ACKNOWLEDGEMENTS ii
Preface iv
0.1 Chapters I and II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv
0.2 Chapters III, IV, and V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
CHAPTERS 1
VITA 77
iii
RESULTS IN ANALYTIC AND ALGEBRAIC NUMBER THEORY
Aaron Yeager
Preface
This work is a compilation of five papers that contain some related topics and
techniques. These papers were written during the last two years at the University of
Missouri. The papers are divided as chapters, each having its own introduction and
bibliography. Although the topics are connected and similar techniques are used, the
The first two chapters are exposition topics in Analytic and Algebraic Number Theory.
I began working on the first chapter while I attended a course in Analytic Number
Theory during Fall 2011. The work in the second chapter started when I took a
The first chapter gives a brief introduction to sums over finite fields. The chapter
also discusses Gauss sums, generalized Gauss sums, exponential sums, and Jacobi
sums. If we let F be a finite field with q elements, it is shown that the modulus of the
√
Gauss sum and the modulus of the Jocobi sum is equal to q. Using these theorems
the chapter concludes by showing that if p is a prime such that p ≡ 1 (mod 4), then
The second chapter proves some useful results concerning the Dedekind zeta func-
tion. The chapter starts with some basic properties of this function. An auxiliary
iv
function is then defined and shown to have a functional equation. Using these results
for the auxiliary function, the Dedekind zeta function is shown to have a factorization
in the right half of the complex plane that has the gamma function and sums regard-
ing the incomplete gamma function as some of its factors. Given this factorization
and using properties of the gamma function and the incomplete gamma function as
well as using techniques from real and complex analysis, the function is proven to
be meormorphic on the whole complex plane, shown to have a simple pole, and the
residue is exhibited. The function is also shown to have a functional equation that
has a factor of the gamma function. The chapter concludes by giving the evaluation
of the function at zero and showing that this function is zero on the negative integers.
The last three chapters were written with co-authors. The chapters are an expan-
sion of a series of new research papers in mathematics with topics from Analytic
and Algebraic Number Theory. The papers that resemble chapters III and IV have
been accepted by journals and the paper that represents chapter V will be submit-
ted shortly. The papers were written in Spring 2011, Fall 2011, and Spring 2012
respectively.
My advisor is the co-author of the third chapter, which has been accepted by Col-
numbers. Specifically, let α, β ∈ R be fixed with α > 1, and suppose that α is irra-
tional and of finite type. We show that there are infinitely many Carmichael num-
bers composed solely of primes from the non-homogeneous Beatty sequence Bα,β =
(bαn + βc)∞
n=1 . The result is proved by appealing to a construction of infinitely many
v
Carmicheal numbers given by Alford, Granville, and Pomerance. To use their con-
such as exploiting the type of α, using the Balog and Perelli theorem, and by apply-
concludes with heuristic evidence via Dickson’s conjecture to support our conjecture
The fourth chapter was co-authored by my advisor and Ahmet M. Güloğlu and
numbers. We show that for any finite Galois extension K of the rational numbers Q,
there are infinitely many Carmichael numbers composed solely of primes for which
the associated class of Frobenius automorphisms coincides with any given conjugacy
class of Gal(K|Q). The result has three corollaries: for any algebraic number field K,
there are infinitely many Carmichael numbers which are composed solely of primes
that split completely in K; for every natural number n, there are infinitely many
Carmichael numbers of the form a2 + nb2 with a, b integers; and there are infinitely
To obtain our main result we begin by examining zeros of the Dedekind zeta function
in a region to get the existence of a proper integral ideal. We use this result along
with the Chebotarev Density Theorem to yield a lower bound on a counting function
that detects primes in our setting. Finally after this bound is proven, we show how
vi
The fifth chapter is to be submitted in the near future. It is co-authored by
my advisor, Roger Baker, and Zhenyu Guo. In the chapter we prove a new result
regarding Piatetski-Shapiro primes, primes from sequences of the form (bnc c)n∈N
where c > 1 and c 6∈ N, in relation to almost primes. We show that for any fixed
c ∈ (1, 77
76
) there are infinitely many primes of the form p = bnc c, where n is a natural
number with at most eight prime factors (counted with multiplicity). To achieve
sums with the saw-tooth function. Eventually we have to bound type I and type II
exponential sums. One of our main tools in the desired bounds on these sums is the
use of exponential pairs. Furthermore, we show how to find the optimal exponential
pairs for our setting. Using these techniques and appealing to a weighted sieve we
refinement of our argument that shows the same result is true when c ∈ (1, 77
76
).
vii
Chapter 1
1.1 Introduction
The contents of this chapter are a result of studies from [1] and [2]. The main object
of study is a sum that has a character convolved with an exponential function. Some
ax
where e( ax
p
) = e2πi p and ( xp ) is the Legendre symbol modulo p, or Kloosterman sums
∗
X ax + bx
S(a, b; p) = e ,
p
χ (mod p)
where the weight ∗ means that the sum can be extended to all x which are not poles for
the function in the sum and where x is the inverse in Z/pZ of the invertible element
x ∈ (Z/pZ)× . The quadratic Gauss sums are used for many results. Among these
results are the Hasse-Davenport relation and quadratic reciprocity. The Kloosterman
sums can also be used to show many results. One of the most notable results is that
for a1 , ..., a4 ≥ 1 positive integers for n large enough, depending on the a0i s, there
exists at least one integral solution (x1 , ..., x4 ) ∈ Z4 to the diophantine equation
1
provided there is no congruence obstruction.
The Gauss and Kloosterman sums can be generalized. That is, the Legendre char-
acter can be replaced with an arbitrary multiplicative character and the exponential
function can be replaced with an arbitrary arithmetic function. Sums of this type are
called Complete Sums. We will address these sums in the next section.
To begin our study of these sums let p be a prime number and consider the field
Fp = Z/pZ. Since this field has p elements, the Galois theory is easy to describe. For
Conversely, any finite field F with q elements is isomorphic to a unique field Fpd , so
we have the cardinality relation |Fn | = q n . Thus the extension Fn /F is Galois. Set
x ∈ F ⇐⇒ σ(x) = x ⇐⇒ xq = x.
n
x ∈ Fn ⇐⇒ σ n (x) = x ⇐⇒ xq = x.
n
Given this we can conclude that Fn is the splitting field for xq − x ∈ F[x].
2
1.2 Generalized Gauss Sums
The quadratic Gauss Sum can be generalized by using a multiplicative character and
X
τ (χ, ψ) = χ(x)ψ(x).
x∈F×
q
From this definition, our first theorem regards the size of the modulus of τ (χ, ψ).
√
|τ (χ, ψ)| = q.
Proof. If we consider the modulus squared, expanding the product of the Gauss sum
X X
|τ (χ, ψ)|2 = χ(u) ψ(y(u − 1)),
u∈F×
q y∈F×
q
where the inner sum above is purely additive. By the orthogonality relations of the
3
for all α of the additive characters on Fq . Using this and by the orthogonality relations
= q.
Taking the square root of both sides of the above yields the result.
As with the generalized Gauss Sums, let Fq be a field with q elements. Let χ and φ
X X
J(χ, φ) = χ(x)φ(1 − x) = χ(x)φ(y).
x∈Fq x+y=1
The values of such a sum occur in relation to local zeta-functions for diagonal forms.
Our next proposition shows how these sums can be expressed as general Gauss
sums.
character ψ of Fq . Then
τ (χ, ψ)τ (φ, ψ)
J(χ, φ) = . (1.1)
τ (χφ, ψ)
Furthermore,
√
|J(χ, φ)| = q.
4
X X
J(χ, φ)τ (χφ, ψ) = χ(x)φ(1 − x) χφ(y)ψ(y)
x∈Fq y∈F×
q
X X
= χ(x)φ(1 − x) χ(y)φ(y)ψ(y)
x∈Fq y∈F×
q
X X
= χ(x)φ(1 − x)χ(y)φ(y)ψ(y).
x∈Fq y∈F×
q
By χ and φ being nontrivial, the above sum can be restricted to x 6∈ {0, 1}. Set
follows that τ (χφ, ψ) 6= 0. Hence dividing both sides of the above by τ (χφ, ψ) yields
equation 1.1. To obtain the other result, we use equation 1.1 along with Theorem 1
Our final result shows how a Jacobi sum can be used to answer a classical problem
in Number Theory.
Theorem 3. Let p be a prime number such that p ≡ 1 (mod 4). Then there exists
a, b ∈ Z such that
p = a2 + b2 .
Proof. By the theory of characters, since p ≡ 1 (mod 4), there exists a character χ
of order 4 on F×
p . Consider the Jacobi sum J(χ, φ), where φ is the usual Legendre
5
character. Then by definition
X
J(χ, φ) = χ(x)φ(y).
x+y=1
of order 4 on F×
p , we have χ(x) ∈ {−i, −1, 0, 1, i}. Give this observation, the Jacobi
sum J(χ, φ) can be written as J(χ, φ) = a + bi, for some a, b ∈ Z. Therefore with
p =|J(χ, φ)|2
=|a + bi|2
√
=( a2 + b2 )2
=a2 + b2 .
6
Bibliography
7
8
Chapter 2
2.1 Introduction
The contents of this chapter are results from studies of the references and working
through problems 25 (a) and (b) and 26 (a) through (f) of [3]. The lemmas and
theorems in the paper are these problems worked out. All proofs are the author’s
work with few proofs using variations of techniques in chapter 10 of [3]. Let K be
an algebraic number field. That is, K is a finite dimensional field extension of the
rational field Q. Let N (·) be the norm on the field K . For s = σ + it ∈ C, with
X
ζK (s) = N (a)−s
a
where the sum is over all integral ideals in the ring OK of algebraic integers in K. By
∞
X aK (n)
ζK (s) = ,
1
ns
where aK (n) is the number of integral ideals of K with norm exactly n. When K = Q,
ζK (s) = ζ(s). Hence why the sum bears its name. This function is of great importance
for many reasons. For instance it can be used to show various density theorems such
9
as the Cebatorev Density theorem. It can also be used to get a generalization of
a Dirichlet L-series which leads Dirichlet’s Prime Number Theorem. Some natural
questions to consider would be: What properties does this function have like the
factored to recover the original ζ function? Is it meromorphic, and if so, what are the
singularities and residues, and so on? To answer some of these questions we need to
start with how the norm acts on ideals in OK . Given that ideals in the ring OK factor
uniquely into prime ideals, and since both the norm N (·) and n−s are multiplicative,
Y
ζK (s) = (1 − N (p)−s )−1 ,
p
for σ > 1.
Let Q(x, y) = ax2 + bxy + cy 2 , where a, b, c are real numbers, and put d = b2 − 4ac.
Suppose that Q is positive-definite, which is to say that a > 0 and d < 0. For z ∈ C
X 2
√ X 2 z/
√
ϑQ (z) = e−πzn −d/(2a)
e−2πa(m+bn/(2a)) −d
.
n m
Proof. Completing the square in the definition of the quadratic form and combining
10
Q(m, n) = am2 + bmn + cn2
(bn)2
= a(m + bn/(2a))2 − + cn2
4a
2
(bn) − cn2 4a
= a(m + bn/(2a))2 −
4a
2
nd
= a(m + bn/(2a))2 − .
4a
Thus
X √
ϑQ (z) = e−2πQ(m,n)z/ −d
m,n∈Z
X 2 +bmn+cn2 )z/
√
= e−2π(am −d
m,n∈Z
XX bn 2 n2 d
√
= e−2π(a(m+ 2a ) − 4a )z/ −d
n m
X 2
√ X 2 z/
√
= e−πzn −d/(2a)
e−2πa(m+bn/(2a)) −d
.
n m
bn 2az
Proof. Applying Theorem 10.1 of [3], with choices of α = 2a
and z̃ = √
−d
, to the
X √ 2az −1 √
−2πa(m+bn/(2a))2 z/ −d 2
X 2
e = √ e(kbn/(2a))e−πk −d/(2az) .
m
−d k
√ √ √ √
−πzn2 −d/(2a) + πikbn/a = −π(n − ikb/(z −d))2 z −d/(2a) − π(kb)2 /(2az −d)
11
we see that
X 2
√ X 2
√ √ 2
√
e−πzn −d/(2a)+πikbn/a = e−π(kb) /(2az −d) e−π(n−ikb/(z −d)) z −d/(2a)
n n
√ √ √
−π(kb)2 /(2az −d)
X
−d))2 z −d/(2a)
=e e−π(n−ikb/(z .
n
−ikb 2az
Using Theorem 10.1 of [3] again, this time with α = √
z −d
and z̃ = √
−d
, we obtain
If we now combine the exponents from the exponential functions and factor them, we
see that
√ √ √ √
−π(kb)2 /(2az −d)+2πlkb/(z −d)−πl2 /(z −d/(2a)) = −2πa/(z −d)(l−kb/(2a))2 .
Putting this in the sum and combining with the previous sum shows
2az −12
z √−d −1
2
X 2
√ X 2
√
√ e−πk −d/(2az) e−2πa(l−bk/(2a) /(z −d)
−d 2a k l
X √ √
−πk2 −d/(2az)
X 2
−1
= (z) e e−2πa(l−bk/(2a) /(z −d) .
k l
Let us now consider the case when K is a complex quadratic field. Let d be the
√
discriminant of K. Then K = Q( d), where d < 0. Let w be the number of units in
Ok , and h be the class number of K. Then there are h reduced positive definite binary
quadratic forms of discriminant d, say Q1 , ..., Qh . Consider when (m, n) 6= (0, 0). As
these run over integral values, Qi (m, n) runs over the values of N (a) for ideals a ∈ Ci ,
where Ci is the ith ideal class, each value being taken w times. Hence on each Qi ,
X
ζQi (s) = w N (a)−s .
a∈Ci
12
Theorem 4. For R(z) ≥ 0, let
h
X ∞
X √
ϑK (z) = ϑQi (z) = h + w r(n)e−2πnz/ −d
i=1 n=1
P
where r(n) = rK (n) = k|n χd (k) is the number of ideals in Ok with norm equal to
Proof. By definition
X √
ϑQi (z) = e−2πQi (m,n)z/ −d
m,n∈OK
X √
=1+ e−2πQi (m,n)z/ −d
.
m,n∈OK
m,n6=0
If we sum from 1 to h, by the definition of how Qi (m, n) runs over the its values, we
see
h
X ∞
X √
ϑK (z) = ϑQi (z) = h + w r(n)e−2πnz/ −d
.
i=1 n=1
13
Proof. By Euler’s integral formula for Γ(s), if σ > 0, then
Z ∞
Γ(s) = e−x xs−1 dx.
0
√
If we make of change of variables of x = 2πnu/ −d we obtain
∞ √ √ √
Z
Γ(s) = e−2πnu/ −d
(2πnu/ −d)s−1 (2πn/ −d)du
0
√ ∞ √
Z
= (2πn/ −d)s e−2πnu/ −d s−1
u du.
0
Thus
Z ∞ √
−s −s/2 −s
n Γ(s)(−d) (2π) = e−2πnu/ −d s−1
u du.
0
Since r(n) is the number of ideals in OK with norm exactly n, for σ > 0, if we multiply
both sides of the equation above by r(n) and then sum over n we achieve,
∞
X Z ∞ √
−s/2 −s
ζK (s)Γ(s)(−d) (2π) = r(n) e−2πnu/ −d s−1
u du
n=1 0
∞ ∞ (2.1)
√
Z X
−2πnu/ −d s−1
= r(n)e u du.
0 n=1
I note that the exchange of the integration and the summation above is permitted by
the absolute convergence of the series. Suppose the R(z) > 0. By Cauchy’s theorem,
we may replace the path of integration by a ray from 0 that passes through z. We
R
now consider the integral from 0 to z and z to ∞ separably. Call these integrals 1
R
and 2
respectively. By reversing the steps above we have that
∞
√
Z X
−s
= (−d) s/2
(2π) r(n)n−s Γ(s, 2πnz/ −d).
2 n=1
Notice that the inner sum in (2.1) is (ϑK (u) − h)/(2w). Thus
Z Z z Z z
1 s−1 h
= ϑK (u)u du − us−1 du.
1 2w 0 2w 0
14
−hz s
The later integral above is 2ws
. From theorem 1 we know that ϑK (z) = ϑK (1/z)/z.
Changing the order of summation and integration and using the change of variables
√
of x = 2πnv/ −d we obtain
∞
(1−s)/2 s−1
X √
(−d) (2π) r(n)ns−1 Γ(1 − s, 2πn/(z −d)).
n=1
R R
Putting the integrals 1
and 2
together we obtain the result.
Proof. Isolating the Dedekind zeta function in the equation in theorem 5 gives that
for R(z) ≥ 0,
∞
1 X √
ζK (s) = r(n)n−s Γ(s, 2πnz/ −d)
Γ(s) n=1
∞
1 X √
+ (−d) (1−2s)/2
(2π)2s−1
r(n)ns−1 Γ(1 − s, 2πn/(z −d)) (2.2)
Γ(s) n=1
(2π)s hz s−1 (2π)s hz s
+ − .
Γ(s)ds/2 2w(s − 1) Γ(s)ds/2 2ws
By properties of the gamma function and the incomplete gamma function, the two
sums above represent entire functions. Hence the right side of the equation above
is analytic for all s except s = 0, 1. The fact that ζK (s) is also analytic at s = 0
15
follows from theorem 8 below. To see that ζK as a simple pole at s = 1, observe that
lims→1 |ζK (s)| = ∞ by the last term above forcing this to happen. Furthermore, by
Theorem 8. The value of the Dedekind zeta function at zero is equal to −h/(2w).
16
for all s 6= 0, 1. If we multiply the above equation by s − 1 and take the limit as s
we get
Thus equating these limits and multiplying by -1 on both sides yields the result.
Proof. Solving for ζK (s) from the functional equation in Theorem 4 yields
Γ(1 − s)
ζK (s) = ζK (1 − s) (−d)1−s (2π)2s−1 .
Γ(s)
Also by Γ(s) being analytic for positive real numbers, again for any k a positive
integer, Γ(1 − (−k)) = Γ(1 + k) is also well-defined. However since Γ(s) has poles at
the negative integers, 1/Γ(s) has simple zeros at the negative integers. Thus for all
Γ(1 + k)
ζK (−k) = ζK (1 + k) (−d)1+k (2π)−2k−1 = 0.
Γ(−k)
17
18
Bibliography
[2] G. Janusz, ‘Algebraic Number Fields,’ Academic Press, New York (1973), 117–
130.
[4] J. Neukirch, ‘Class Field Theory,’ Springer, New York (1980), 117–121.
[5] P. Samuel, ‘Algebraic Theory of Numbers,’ Dover Publications, New York (1970).
[6] A. Weil, ‘Basic Number Theory,’ Springer, New York (1974), 120–138.
19
20
Chapter 3
3.1 Introduction
Around 1910, Robert Carmichael initiated the study of composite numbers N with the
same property, which are now known as Carmichael numbers. In 1994 the existence
and Pomerance [1]. In recent years, using variants of the method of [1], several
that are composed of primes from a Beatty sequence. We recall that for fixed α, β ∈ R
by
Bα,β = bαn + βc
n∈Z
.
21
because of this versatility, the arithmetic properties of Beatty sequences have been
extensively explored in the literature; see, for example, [1, 7, 8, 9, 10, 13, 16, 17, 20,
Theorem 10. Let α, β ∈ R with α > 1, and suppose that α is irrational and of finite
type. Then, there are infinitely many Carmichael numbers composed solely of primes
A quantitative version of this result is given in §3.4; see Theorem 12. To prove
Theorem 18, we show that when α is of finite type (see §3.2.2) the set of primes
one can construct Carmichael numbers from such primes using an adaptation of the
method of [1]. To do this, we extend various results and techniques of Banks and
Shparlinski [7].
of infinite type.
For irrational numbers α of infinite type, the approach described above fails. How-
ever, assuming the validity of a certain natural extension of Dickson’s conjecture (see
Conjecture II in §3.5) we establish the above conjecture in the case that β = 1 (see
22
3.2 Preliminaries
3.2.1 General notation
The notation JtK is used to denote the distance from the real number t to the nearest
We denote by btc and {t} the greatest integer 6 t and the fractional part of t,
respectively. We also put e(t) = e2πit for all t ∈ R. As usual, we use Λ(·) and ϕ(·) to
Throughout the paper, the implied constants in symbols O, and may depend
on the parameters α, β and ε but are absolute otherwise. We recall that for functions
statement that the inequality |F | 6 C |G| holds for some constant C > 0.
Recall that the discrepancy D(M ) of a sequence of (not necessarily distinct) real
V (I, M )
D(M ) = sup − |I| , (3.1)
I⊆[0,1) M
where the supremum is taken over all intervals I contained in [0, 1), V (I, M ) denotes
the number of positive integers m 6 M such that am ∈ I, and |I| denotes the length
of the interval I.
23
By Dirichlet’s approximation theorem, one has τ > 1 for every irrational number
γ. The theorems of Khinchin [15] and of Roth [23, 24] assert that τ = 1 for almost
all real numbers (in the sense of the Lebesgue measure) and all irrational algebraic
formly distributed in [0, 1) (see, e.g., [18, Example 2.1, Chapter 1]). In the case that
γ is of finite type, the following more precise statement holds (see [18, Theorem 3.2,
Chapter 2]).
Lemma 3. Let γ be a fixed irrational number of finite type τ . Then, for every δ ∈ R
Lemma 4. Let α, β ∈ R with α > 1, and put γ = α−1 , δ = α−1 (1 − β). Then,
by
if 0 < {t} 6 α−1 ,
1
ψ(t) = (3.2)
0 otherwise.
The next statement is a simplified and weakened version of a theorem of Balog and
24
Lemma 5. For an arbitrary real number θ and coprime integers c and d with 0 6
X
Λ(n)e(θn) q −1/2 x + q 1/2 x1/2 + x4/5 (log x)3 .
n6x
n≡c (mod d)
Lemma 6. Let γ be an irrational number of finite type τ , and fix A ∈ (0, 1) and
ε > 0. For any coprime integers c and d with 0 6 c < d and any nonzero integer k
X A+2+1/τ
+ε
Λ(n)e(kγn) x 2+2/τ + x4/5 (log x)3
n6x
n≡c (mod d)
holds, where the implied constant depends only on the parameters α, β, A and ε.
A+1 τ (1 + ε) A+1
B= , C= , D= + 2ε.
1 + 1/τ 1 − ετ 1 + 1/τ
Note that B ∈ (0, 1) (since τ > 1 for an irrational γ), C ∈ (τ, 2τ ), and
A+1
D = B + 2ε > B(1 + ε) = ,
1 + 1/C
25
Let a/q be the convergent in the continued fraction expansion of kγ which has the
a 1 c0
kγ − 6 −1 D = D . (3.5)
q qc0 x qx
Since |k| 6 xA it follows that q > x−A+D/C . By (3.3) we have q > c0 x1−D for
all sufficiently large x, hence by (3.5) we see that |kγ − a/q| 6 1/x. Applying
Lemma 5 with θ = kγ, and taking into account our choice of D and the inequalities
c0 x1−D 6 q 6 c−1 D
0 x , we derive the stated bound.
For the remainder of the paper, let α, β ∈ R be fixed with α > 1, and assume that α
is irrational. The following statement provides an explicit version of [7, Theorem 5.4].
Theorem 11. If α is of finite type τ = τ (α), then for any fixed ε > 0 we have
X 1 X
Λ(n) + O x1−1/(4τ +2)+ε ,
Λ(n) = (3.6)
n6x, n∈Bα,β
α n6x
n≡c (mod d) n≡c (mod d)
Proof. Let F (x; d, a) denote the left side of (3.6), and let ψ = ψα be defined by (3.2).
X
F (x; d, a) = Λ(n)ψ(γn + δ),
n6x
n≡c (mod d)
26
where γ = α−1 and δ = α−1 (1 − β). Note that α and γ are of the same type, that is,
τ (α) = τ (γ).
By a classical result of Vinogradov (see [27, Chapter I, Lemma 12]), for any ∆
1
such that 0 < ∆ < 8
and ∆ 6 12 min{γ, 1 − γ}, there is a real-valued function Ψ with
P
(iv) Ψ(t) = k∈Z g(k)e(kt) for all t ∈ R, where g(0) = γ, and the other Fourier
Since |I| = 4∆, it follows from the definition (3.1) and Lemma 3 that
Now let K > ∆−1 be a large real number, and let ΨK be the trigonometric
polynomial defined by
X
ΨK (t) = g(k)e(kt). (3.10)
|k|6K
27
Using (3.7) it is clear that the estimate
holds uniformly for all t ∈ R. Combining (3.11) with (3.8) and taking into account
X
Λ(n)ΨK (γn + δ) + O ∆x log x + x1−1/τ +ε + K −1 ∆−1 x .
F (x; d, a) =
n6x
n≡c (mod d)
∆ = x−A/2 and K = xA .
X X
Λ(n)e(kγn) + O x1−1/τ +ε + x1−A/2+ε .
F (x; d, a) = g(k)e(kδ)
|k|6xA n6x
n≡c (mod d)
X A+2+1/τ
F (x; d, a) = γ Λ(n) + O x 2+2/τ +ε + x4/5+ε + x1−1/τ +ε + x1−A/2+ε .
n6x
n≡c (mod d)
In this section, we outline our proof of Theorem 18. We shall be brief since our
28
closely parallels (and relies on) the construction of “ordinary” Carmichael numbers
given in [1]. Here, we discuss only those modifications that are needed to establish
Theorem 18.
Let P denote the set of all prime numbers, and set Pα,β = P ∩ Bα,β . The
underlying idea behind our proof of Theorem 18 is to show that Pα,β is sufficiently
well-distributed over arithmetic progressions so that, following the method of [1], the
primes used to form Carmichael numbers can all be drawn from Pα,β rather than P.
Unfortunately, this idea appears only to succeed in the case that α is of finite type,
X X
Λ(n) = log p + O(x1/2 )
n6x p6x
X 1 X
log p − log p 6 x1−1/(4τ +2)+ε (x > x1 (α, β, ε)).
p6x, p∈Bα,β
α p6x
p≡c (mod d) p≡c (mod d)
For any modulus d 6 (4α)−1 x1/(4τ +2)−ε the right side of this inequality does not exceed
x/(4αϕ(d)); therefore, applying [1, Theorem 2.1] and taking into account the above
inequality, we derive the following statement, which plays a role in our construction
Lemma 7. For every B ∈ 0, 4τ1+2 there exist numbers ηB > 0, x2 (B) and DB such
that for all x > x2 (B) there is a set DB (x) consisting of at most DB integers such
that
X x x
log p − 6
p6x, p∈Bα,β
αϕ(d) 2α ϕ(d)
p≡c (mod d)
29
whenever d is not divisible by any element of DB (x), 1 6 d 6 xB , and c is coprime
to d. Furthermore, every number in DB (x) exceeds log x, and all, but at most one,
exceeds xηB .
depend on the parameters α and β, but we have suppressed this from the notation
for the sake of clarity. Similarly, x3 (B) depends on α and β in the statement of
Lemma 8 below.
Lemma 8. Suppose that B ∈ 0, 4τ1+2 . There exists a number x3 (B) such that if
x > x3 (B) and L is a squarefree integer not divisible by any prime exceeding x(1−B)/2
2−DB −2
> # d | L : 1 6 d 6 xB .
α log x
Sketch of Proof. Let π(x; d, a) [resp. πα,β (x; d, a)] be the number of primes [resp.
ing Lemma 7 we can replace the lower bound [1, Equation (3.2)] with the bound
1 dx1−B
πα,β (dx1−B ; d, 1) > .
2α ϕ(d) log x
Also, since πα,β (x; d, a) never exceeds π(x; d, a), the upper bound that follows [1,
8 dx1−B
πα,β (dx1−B ; dq, 1) 6 .
q(1 − B) ϕ(d) log x
30
P
Taking into account the inequality prime q | L 1/q 6 (1 − B)/(32α), the proof is
Let π(x) be the number of primes p 6 x, and let π(x, y) be the number of those
for which p − 1 is free of prime factors exceeding y. As in [1], we denote by E the set
of numbers E in the range 0 < E < 1 for which there exist numbers x4 (E), γ(E) > 0
such that
for all x > x4 (E). With only a very slight modification to the proof of
[1, Theorem 4.1], using Lemma 8 in place of [1, Theorem 3.1], we derive the fol-
0, 4τ1+2
Theorem 12. For each E ∈ E, B ∈ and ε > 0, there is a number
x4 = x4 (α, β, E, B, ε) such that for any x > x4 , there are at least xEB−ε Carmichael
As before, we fix α, β ∈ R with α > 1, and assume that α is irrational. In this section,
aj , bj ∈ Z and aj > 1. In 1904, Leonard Dickson [12] made the following well known
Conjecture I. Suppose that there is no integer n > 1 with the property that n |
f1 (m) · · · fk (m) for all m ∈ Z. Then, there exist infinitely many m ∈ N such that all
31
Let S be the set of natural numbers m for which f1 (m), . . . , fk (m) are all prime
numbers. If S is an infinite set, then it seems reasonable to expect that for every irra-
in [0, 1); in particular, for any interval I contained in [0, 1) we expect that
This is easy to prove when k = 1 (see [22]), and this is the only case in which the
truth of Conjecture I has been established (Dirichlet’s theorem). For other cases,
numerical evidence in support of (3.12) can be acquired when the set {f1 , . . . , fk }
and the interval I have been specified explicitly. For our purposes here, however, we
Conjecture II. If S is an infinite set, then for every irrational number γ and every
interval I in [0, 1) with |I| > 0, there are infinitely many numbers m ∈ S for which
{mγ} ∈ I.
32
Theorem 13. Assume both Conjectures I and II are true, and suppose that
(i) there is no integer n > 1 such that n | f1 (m) · · · fk (m) for all m ∈ Z;
(ii) there is an integer m such that all of the numbers f1 (m), . . . , fk (m) lie in the
Then, for infinitely many m ∈ N, all of the numbers f1 (m), . . . , fk (m) are prime
Proof. For any real numbers c, d with c < d, we denote by (c, d] + Z the sumset of
(c, d] + Z = x + m : x ∈ (c, d] and m ∈ Z .
Let Ω be the collection of all such sumsets (c, d] + Z together with the empty set ∅,
and let Ω◦ be the collection of all finite unions of sets in Ω. Note that Ω is closed
under finite intersections, and Ω◦ is closed under finite intersections and finite unions.
As before, we let S denote the set of m ∈ N such that f1 (m), . . . , fk (m) are
all prime numbers. Under Conjecture I and using (i) we see that S is an infinite
set. Hence, under Conjecture II it follows that for any irrational number γ, the set
Sγ = {mγ : m ∈ S} has an infinite intersection with every set (c, d]+Z, and therefore
where
i − γbj − δ γ
ci,j = and di,j = ci,j + (i = 1, . . . , aj ).
aj aj
33
Hence, if we put
aj
k [
\
T = (ci,j , di,j ] + Z ,
j=1 i=1
then mγ ∈ T if and only if all of the numbers f1 (m), . . . , fk (m) lie in the Beatty
argument, we conclude that Sγ has an infinite intersection with T , and the result
follows.
Theorem 14. Assume both Conjectures I and II are true and β = 1. Then, there
are infinitely many Carmichael numbers composed solely of primes from the Beatty
sequence Bα,β .
satisfy condition (i) of Theorem 13, hence it suffices to show that condition (ii)
also holds when β = 1. Indeed, when this is the case, Theorem 13 implies (under
Conjectures I and II) that there are infinitely many triples (p, q, r) of primes in Bα,β
with p = 6m + 1, q = 12m + 1 and r = 18m + 1 for some m ∈ N, and for any such
Theorem 13 we have
a1 = 6, a2 = 12, a3 = 18, b1 = b2 = b3 = 1,
and therefore,
γ γ γ
c6,1 = 1 − , c12,2 = 1 − , c18,3 = 1 − , d6,1 = d12,2 = d18,3 = 1.
6 12 18
34
γ
We deduce that T contains the set (1 − 18 , 1] + Z, which shows that condition (ii) of
Theorem 13 is satisfied.
35
36
Bibliography
[3] A. Balog and A. Perelli, ‘Exponential sums over primes in an arithmetic progres-
[4] W. Banks, ‘Carmichael numbers with a square totient,’ Canad. Math. Bull. 52 (1)
[5] W. Banks, ‘Carmichael numbers with a totient of the form a2 + nb2 ,’ to appear
in Monat. Math.
[7] W. Banks and I. Shparlinski, ‘Prime numbers with Beatty sequences,’ Colloq.
[8] W. Banks and I. Shparlinski, ‘Non-residues and primitive roots in Beatty se-
37
[9] W. Banks and I. Shparlinski, ‘Short character sums with Beatty sequences,’
[10] A. Begunts, ‘An analogue of the Dirichlet divisor problem,’ Moscow Univ. Math.
[13] A. Fraenkel and R. Holzman, ‘Gap problems for integer part and fractional part
[14] J. Grantham, ‘There are infinitely many Perrin pseudoprimes,’ J. Number Theory
[16] T. Komatsu, ‘A certain power series associated with a Beatty sequence,’ Acta
38
[19] A. Lavrik, ‘Analytic method of estimates of trigonometric sums by the primes of
an arithmetic progression,’ (Russian) Dokl. Akad. Nauk SSSR 248 (1979), no. 5,
1059–1063.
[20] G. Lü and W. Zhai, ‘The divisor problem for the Beatty sequences,’ Acta Math.
[21] K. O’Bryant, ‘A generating function technique for Beatty sequences and other
[22] P. Ribenboim, The new book of prime number records. Springer-Verlag, New
York, 1996.
1–20.
39
40
Chapter 4
4.1 Introduction
The aim of the present work is to prove the following extension of the result mentioned
in the previous chapter, the existence of infinitely many Carmichael numbers from
Theorem 15. Let K/Q be a finite Galois extension. Then, there are infinitely many
Carmichael numbers composed solely of primes for which the associated class of Frobe-
infinitely many Carmichael numbers composed solely of primes that split completely
in K.
e Since such primes will necessarily split completely in K, we immediately obtain
Corollary 1. For any fixed algebraic number field K, there are infinitely many
Carmichael numbers which are composed solely of primes that split completely in
K.
Since prime numbers and Carmichael numbers are linked by the common property
41
given by Fermat’s Little Theorem, it is natural to ask whether certain questions about
primes can be settled for Carmichael numbers; see [2, 3]. For example, it is well known
that for every natural number n, there are infinitely many primes of the form a2 + nb2
with a, b ∈ Z (see the book [4] by Cox), so it is natural to ask whether the same
result holds for the set of Carmichael numbers. As a result of Corollary 1, we give an
Corollary 2. For any fixed natural number n, there are infinitely many Carmichael
Indeed, let Sn = {a2 +nb2 : a, b ∈ Z}, and let Kn be the ring class field associated
√ √
to the order Z[ −n ] in the imaginary quadratic field Q( −n ). According to [4,
and only if p ∈ Sn . Applying Corollary 1 with K = Kn , we see that there are infinitely
unity, we obtain:
4.2 Preliminaries
DK . We put
NK = d ∈ N : gcd(d, DK ) = 1 .
(4.1)
42
For any Galois extension M/N and any unramified prime ideal p of N , (p, M |N ) will
dφ(d)
DQd = (−1)φ(d)/2 Q φ(d)/(p−1)
, (4.2)
p|d p
discriminant
φ(d)
DKd = DK DQndK .
Furthermore, Gal(Kd /Q) ' Gal(K/Q) × Gal(Qd /Q), where the isomorphism is given
Proof. In view of (4.1) and (4.2), the discriminants DK and DQd are coprime for every
d ∈ NK . It follows from [10, Ch.3, Corollary 2.10] that K ∩ Qd = Q. The result now
follows from [10, Ch. 1, Proposition 2.11] and [5, 14.4, Corollary 22].
and depend only on the field K. All constants implied by the symbols O, and
depends on K.
43
4.3 Zeros of Dedekind zeta functions
For each d ∈ NK , let ζd (s) be the Dedekind zeta function ζKd (s) associated with the
Lemma 10. There are constants c1 , c2 > 0 depending only on K with the property
that for all T > 1 and U > 2 there exists a proper integral ideal f = f(K, U, T ) of K
such that for any d ∈ NK with d 6 U , f | dOK , where OK is the ring of integers of
Proof. We use the notation of [13, §1]. For each d ∈ NK with d 6 U , and any
Hence, it follows that dχ 6 (c2 U )nK for some constant c2 = c2 (K). Applying [13,
L = log(QT nK ) = nK log(c2 T U ),
we see that for some constant c1 = c1 (K), any Hecke L-function L(s, χ) with dχ 6 Q
has at most one zero in the region Ω(T, U ). Moreover, the remark following [13,
Theorem 1.9] asserts that there is at most one function L(s, χ∗ ) vanishing in Ω(T, U )
among all L(s, χ) associated with primitive characters χ with dχ 6 Q. If such a zero
exists, then it is a real number β∗ (which can be bounded in terms of Q). For such a
zero we have
c1 c1
β∗ > 1 − >1− .
log(c2 T U ) log c2
44
Replacing c1 by a smaller constant (which also depends only on K), we can assume
Y
ζd (s) = ζK (s) L(s, χ, Kd |K)
χ6=1
is the product of Artin L-functions, where χ runs over the irreducible characters of
Gal(Kd /K). Let Kχ be the fixed field of the kernel of χ. Then, χ is injective as a
character of Gal(Kχ /K). Hence, by [10, Ch.7, Theorem 10.6] there exists a primitive
Dirichlet character χ
e modulo the conductor fχ of the extension Kχ /K such that
Furthermore, since K ⊆ Kχ ⊆ Kd , we see by [7, 5.1.5] and the last paragraph of [7,
Using the remarks above we conclude that ζd (s) vanishes in Ω(T, U ) if and only
if L(s, χ∗ ) is a factor of ζd (s) and L(β∗ , χ∗ ) = 0. In this case, we know that fχ∗ | (d)
Lemma 11. There are constants c3 , c4 , c5 > 0 depending only on K with the property
that for all d ∈ NK , T > c3 d, and σ > 1 − 1/c5 , the number Nd (σ, T ) of zeros β + iγ
Proof. We continue to use notation of [13, §1]. As in the proof of Lemma 10, for each
d ∈ NK let H (in the notation of [13, §1]) denote the trivial subgroup of the ideal
class group I((d))/P(d) modulo (d), and note that the quantities hH and d(H) defined
45
by [13, Equation (1.1b)] satisfy the bound
for some constant c = c(K) in view of [13, Lemma 1.16]. The result now follows by
applying [13, Corollary 4.4] with Q = (cd)nK and T > c3 d, where c3 = c3 (K) is any
1/nK
constant that is large enough so that the conditions T 1 and T > n2K hH of [13,
Corollary 4.4] are met (for the latter condition, any number c3 > cn2K suffices by the
inequality above).
Our goal in this section is to provide a lower bound for the counting function of the
set
PCd = p ∈ PC : p ≡ 1 (mod d)
By [10, Ch.1, Corollary 10.4] we see that p ≡ 1 (mod d) if and only if p splits
completely in Qd if and only if (p, Qd |Q) = {1d } for p ∈ NK , where 1d denotes the
identity element of Gal(Qd |Q). It follows by the isomorphism in Lemma 9 that there
exists a conjugacy class Cd in Gal(Kd /Q) (one that corresponds to C × {1d }) with
46
and its weighted version
X
ψC (x; d, 1) = log p,
pm 6x
p, m:
(pm ,Kd |Q)=Cd
where the sum is taken over primes in NK . Our main result is the following:
Theorem 16. There are constants x1 , B > 0 depending only on K with the property
|C| y
πC (y; d, 1) > (x4/5 6 y 6 x) (4.4)
2nK φ(d) log y
c1 log y
θB (y) = . (4.6)
log(c2 y 5B )
θB (y) c1
1− >1− and y 3B 6 x3B ,
log y log(c2 x4B )
in ΩB (x).
47
Applying [8, Theorem 7.1] with the choices G = Gal(Kd |Q) and T = y 3B , and
Here, the inner sums are taken over the nontrivial zeros ρ = β + iγ of the Artin
Y
ζd (s) = L(s, χ, Kd |E).
χ∈H
b
Assuming ζd (s) has no zeros in the region ΩB (x), it follows by the functional equa-
tion of ζd (s) that every zero ρ = β + iγ of ζd (s), and thus also of each L(s, χ, Kd |E),
and thus |ρ| > θB (y)/ log y 1/ log y for every such zero. We conclude that
K
X y ρ
X 1 X 1
− y 1/2 nχ (0) y 1/2 log y,
ρ ρ ρ ρ |ρ|
K
1
ρ: β< 2
|ρ|< 12 |γ|61
|γ|61
where nχ (t) is the number of zeros β + iγ of L(s, χ, Kd |E) such that 0 < β < 1 and
nK φ(d)
nχ (t) log dχ + log(|t| + 2), (4.9)
|H|
48
where dχ = |DE |NE/Q (fχ ). Summing over all characters χ ∈ H
b and using (4.9) we
see that
!
X X yρ X 1 X d
1/2
χ(g) − y log y log dχ +
1
ρ ρ ρ
K |H|
χ∈H
b ρ: β< 2 χ∈H
b
(4.10)
|ρ|< 12
|γ|61
y 1/2+2B log2 y.
K
X X yρ
+ OK y 1/2+2B log2 y .
ZB (y) = χ(C) (4.12)
1
ρ
χ∈H
b ρ: β> 2
|γ|6y 3B
To estimate the sum in (4.12), we use ideas (and notation) from the proof of [1,
Theorem 2.1]. For each zero ρ = β + iγ in the sum, we have |y ρ | = y β and |ρ| >
1 b and write Pα for any sum over all zeros β + iγ of
+ |γ| 1 + |γ|. Fix χ ∈ H
4 σ
L(s, χ, Kd /E) with σ 6 β < α and |γ| 6 y 3B . Put τ = 1 − θB (y)/ log y, and note that
1
X yρ
=0
τ
ρ
49
range 1 − 1/c5 6 β 6 τ , it follows that
1−1/c5 ρ τ 1−1/c5 τ
X yρ X y X yρ X yβ X yβ
= + +
ρ ρ ρ ρ 1 + |γ| 1 + |γ|
1/2 1−1/c5 1/2 1−1/c5
β> 21 , |γ|6y 3B
τ τ Z β
1−1/c5
X 1 X 1
y + log y y σ dσ (4.13)
1 + |γ| 1 + |γ| 1−1/c5
1/2 1−1/c5
Z τ τ
!
X 1 X 1
y 1−1/c5 + log y yσ dσ.
ρ 1 + |γ| 1−1/c5 σ
1 + |γ|
|γ|6y 3B
Summing over all characters and using (4.9) the first term above can be bounded as
before:
3B
X X 1 X byXc X 1
χ(g) 6
ρ 1 + |γ| j=0 ρ 1 + |γ|
χ∈H
b χ∈H
b
|γ|6y 3B j6|γ|6j+1
(4.14)
by 3B c
X d log d + d log(1 + j)
y 2B log2 y.
K
j=0
j+1
Let Nχ (σ, T ) be the number of zeros β + iγ of L(s, χ, Kd |E) with β > σ and |γ| 6 T .
50
and assuming that B < 1/(4c5 ), we derive that
Z τ τ
!
X X 1
yσ χ(g) dσ
1−1/c5 σ
1 + |γ|
χ∈H
b
Z τ Z 1−1/(2c5 )
σ 4c5 B(1−σ)
y ·y dσ + y σ · y 2c5 B(1−σ) log y dσ
K 1−1/c5 1−1/c5
Z τ Z 1−1/(2c5 )
4c5 B σ(1−4c5 B) 2c5 B
=y y dσ + y log y y σ(1−2c5 B) dσ
1−1/c5 1−1/c5
τ (1−4c5 B) (1−1/(2c5 ))(1−2c5 B)
y y
y 4c5 B + y 2c5 B
(1 − 4c5 B) log y (1 − 2c5 B)
1+B−1/(2c5 )
y exp(−(1 − 4c5 B)θB (y)) y
= +
(1 − 4c5 B) log y (1 − 2c5 B)
where we have used the definition of τ in the last step. Combining this bound with
(4.12), (4.13) and (4.14), and assuming further that B 6 1/(5c5 ), we find that
|C| |C|
cy exp(− 15 θB (y)) + y −B log2 y
ψC (y; d, 1) − y 6 (4.16)
|G| |G|
Note that B depends only on K, the bound (4.16) holds, and we have
c 1
1
c exp − 6 .
30 B 6
On the other hand, from the definition (4.6) one sees that θB (y) > c1 /(6B) holds for
1
c exp(− 15 θB (y)) 6 (y > y1 ). (4.17)
6
51
Increasing the value of y1 if necessary, we also have
1
c y −B log2 y 6 (y > y1 ). (4.18)
6
5/4
Put x1 = y1 so that the condition y > y1 is satisfied whenever x4/5 6 y 6 x and
2|C|
ψC (y; d, 1) > y (x4/5 6 y 6 x) (4.19)
3|G|
choice of B and d with d 6 xB . We finish the proof by noting that |G| = nK φ(d).
In view of Theorem 16, our construction of Carmichael numbers with the property
stated in Theorem 18 follows closely that given in [1]. We shall be brief, since most
of the details are the same. Our principal tool is the following variant of [1, Theorem
3.1]:
Lemma 12. Let the constants x1 , B have the property stated in Theorem 16, and
suppose that x > x1 . If L is any squarefree number in NK that is not divisible by any
X 1 1
6 ,
q 60nK
prime q | L
1
# d | L : dk + 1 ∈ PC , dk + 1 6 x > · # d | L : d 6 xB .
6nK log x
52
Proof. We use ideas (and notation) from the proof of [1, Theorem 3.1].
Observe that the region ΩB (x) defined by (4.5) is the same as the region Ω(T, U )
Fix a prime p0 with the property that p0 | NK/Q (f), where f = f(K, xB , x3B ) is
that
1
# d | L0 : d 6 y > · # d | L : d 6 y
(y > 1) (4.21)
2
shows that ζd (s) has no zeros in ΩB (x); therefore, using the lower bound (4.4) from
Theorem 16 we have
|C| dx1−B
πC (dx1−B ; d, 1) > .
2nK φ(d) log x
On the other hand, since any prime divisor q of L does not exceed x(1−B)/2 , we have
10 dx1−B
πC (dx1−B ; dq, 1) 6 π(dx1−B ; dq, 1) 6 .
q φ(d) log x
Therefore, the number of primes p ∈ PCd with p 6 dx1−B and such that gcd((p −
1)/d, L) = 1 is at least
X
πC (dx1−B ; d, 1) − πC (dx1−B ; dq, 1)
prime q | L
!
1 X 1 dx1−B x1−B
> − 10 > .
2nK q φ(d) log x 3nK log x
prime q | L
Using this bound together with (4.21) (instead of [1, Equation (3.1)]), the proof can
be concluded in the same manner as that of [1, Theorem 3.1]; the remaining details
are omitted.
53
Theorem 17. There are constants x0 , c0 > 0 depending only on K such that for all
x > x0 , there are at least xc0 Carmichael numbers up to x that are composed solely
Proof. To prove this, we only need to modify the proof of [1, Theorem 4.1] slightly,
as follows.
Let E be the set of numbers E ∈ (0, 1) for which there exists a constant x2 > 0
where π(x, y) denotes the number of primes p 6 x such that p − 1 is free of prime
factors exceeding y.
Fix E = 3/5, which lies in the set E (see, e.g., [6]), and let x2 be a number for
which the bound (4.22) holds. Let x1 , B be numbers with the property stated in
Theorem 16, and put x3 = max{x1 , x2 }. Note that our choice of x3 depends only on
K.
X
|Q| > π(y 5/2 , y) − π(y 5/2 / log y) − 1 y 5/2 / log y (4.23)
q|DK
X X
log L = log q 6 log q = ϑ(y 5/2 ) 6 1.1y 5/2
q∈Q q6y 5/2
54
for all y > 0, where we have used [11] for the last inequality. Furthermore,
Y Y log y 5/2
λ(L) = pa 6 pb log p
c
6 y 5π(y)/2 6 eπ·y (4.24)
pa ||λ(L) p6y
where GL = (Z/LZ)∗ .
1+δ
Let x = ey where δ = 5ε/(8B). Since
X 1 X 1 log log y 1
6 64 6
q q 5 log y 60nK
prime q |L y 5/2 / log y<q6y 5/2
for sufficiently large y, it follows from Lemma 12 that there exists an integer k coprime
divisor d of L, satisfies
1
· # d | L : 1 6 d 6 xB .
|P| > (4.26)
6nK log x
that
u
ω(L) B ω(L)
#{d|L : 1 6 d 6 x } > >
u u
u (4.27)
cy 5/2
> .
2B log x
b 2B log x
1 c 5 log y
c
|P| > y 5/2−1−δ > x3B/5−ε/3
6nK log x 2B
55
for all sufficiently large values of y. Now take P 0 = P\Q. Since |Q| 6 y 5/2 , it follows
We may view P 0 as a subset of the group (Z/LZ)? by considering the residue class
and if
Y
Π(S) := p ≡ 1 (mod L),
p∈S
Π(S) ≡ 1 (mod k), and thus Π(S) ≡ 1 (mod kL), since (k, L) = 1. However, if p ∈ P 0
for all sufficiently large values of y, using (4.25) and (4.28). But each such Carmichael
number Π(S) so formed is such that Π(S) 6 xt . Thus for X = xt we have C(X) >
X 3B/5−ε for all sufficiently large y. But X = exp(y 1+δ exp(y 1+δ/2 )), so that C(X) >
X 3B/5−ε for all sufficiently large values of X. Since y can be uniquely determined
56
Bibliography
[2] W. Banks, ‘Carmichael numbers with a square totient’, Canad. Math. Bull. 52 (1)
[4] D.A. Cox, Primes of the form x2 + ny 2 . Fermat, class field theory, and complex
[5] D.S. Dummit and R. M. Foote, Abstract Algebra. Third edition. John Wiley &
[6] J.B. Friedlander, ‘Shifted primes without large prime factors’, in Number theory
[7] G. J., Janusz, Algebraic number fields, Pure and Applied Mathematics, Vol. 55.,
57
[8] J. C. Lagarias and A. M. Odlyzko, ‘Effective versions of the Chebotarev density
[9] H. L. Montgomery and R.C. Vaughan, ‘The large sieve’, Mathematika 20 (1973),
119–134.
[13] A. Weiss, ‘The least prime ideal’, J. Reine Angew. Math. 338 (1983), 56–94.
58
Chapter 5
5.1 Introduction
where btc denotes the integer part of any t ∈ R. Such sequences are named in honor
12
of Piatetski-Shapiro, who showed in [6] that for any number c ∈ (1, 11 ) the set
is infinite. The admissible range for c in this result has been extended many times
over the years and is currently known for all c ∈ (1, 243
205
) thanks to Rivat and Wu [8].
For any natural number r, let Nr denote the set of r-almost primes, i.e., the set
of natural numbers having at most r prime factors, counted with multiplicity. In this
59
(c)
Theorem 18. For any fixed c ∈ (1, 77
76
) the set P8 is infinite. More precisely,
x
n 6 x : n ∈ N8 and bnc c is prime
,
(log x)2
5.2 Notation
Throughout the chapter, we set γ := 1/c for a given real number c > 1.
small parameters ε, δ but are absolute otherwise. We use notation of the form m ∼ M
As is customary, we put
Throughout the chapter, we make considerable use of the sawtooth function defined
by
1 1
ψ(t) := t − btc − 2
= {t} − 2
(t ∈ R) (5.1)
as well as the well known approximation of Vaaler [10]: for any H > 1 there exist
X X 1 1
ψ(t) − ah e(th) 6 bh e(th), ah , bh . (5.2)
|h| H
0<|h|6H |h|6H
60
5.3 Proof of Theorem 18
5.3.1 Initial approach
We analyze exponential sums that are relevant for finding primes in Pr(c) with the
A := n 6 x : bnc c is prime .
Ad := n ∈ A : d | n .
to within O(1) the cardinality of Ad is equal to the number of primes p 6 xc for which
It is unnecessary to evaluate X more precisely than this; however, for any sufficiently
61
Splitting the range of d into dyadic subintervals and using partial summation in a
X X
Λ(n) ψ(−(n + 1)γ d−1 ) − ψ(−nγ d−1 ) x1−ε/2
(5.4)
d∼D1 N <n6N1
X
Λ(n) ψ(−(n + 1)γ d−1 ) − ψ(−nγ d−1 ) x1−ε/2 d−1
(5.5)
N <n6N1
D 6 x1−136c/157 (5.6)
and ε > 0 is sufficiently small. Suppose this has been done, and observe that
136c 8 8635
1− > whenever c < .
157 63 8568
X X
|Ad | − Xd−1 ,
d6D
(log X)2
thus we can apply the weighted sieve in the form [4, Ch. 5, Prop. 1] with the choices
63
R := 8, δR := 0.124820 · · · and g := .
8
Note that g < R − δR , and (if x is large enough) the inequality a < Dg holds for
all a ∈ A since αg > 1; thus, the conditions of [4, Ch. 5, Prop. 1] are met, and we
conclude that A contains at least X/ log X numbers with at most eight prime
factors. This yields the statement of the theorem for all c in the interval (1, 8635
8568
).
62
We now turn to the proof of (5.5) for all D satisfying (5.6). Let S denote the sum
on the left side of (5.5). From Vaaler’s approximation (5.2) we derive the inequality
X
|S| 6 |S1 | + S2 + S3 + 2b0 Λ(n), (5.7)
N <n6N1
where
X X
ah e(−h(n + 1)γ d−1 ) − e(−hnγ d−1 ) ,
S1 := Λ(n)
N <n6N1 0<|h|6H
X X
S2 := Λ(n) bh e(−h(n + 1)γ d−1 ),
N <n6N1 0<|h|6H
X X
S3 := Λ(n) bh e(−hnγ d−1 ).
N <n6N1 0<|h|6H
To ensure that the last term on the right side of (5.7) satisfies the bound
X
2b0 Λ(n) x1−ε/2 d−1 ,
N <n6N1
we choose
H := x−1+ε N d. (5.8)
Next, we use a partial summation argument from the book of Graham and Kolesnik [3].
Writing
X X
S1 = ah Λ(n)φh (n)e(hnγ d−1 )
0<|h|6H N <n6N1
with
X X
h−1 Λ(n)φh (n) e(hnγ d−1 ) x1−ε d−1 (d 6 D). (5.9)
0<|h|6H N <n6N1
63
the left side of (5.9) is, on integrating by parts, bounded by
X X
h−1 φn (N1 ) Λ(n) e(hnγ d−1 )
h6H N <n6N1
X Z N1 X
+ h−1 φ0h (t) Λ(n) e(hnγ d−1 ) dt
h6H N N <n6t
X X
N γ−1 d−1 Λ(n) e(hnγ d−1 )
h6H N <n6N2
for some number N2 ∈ (N, N1 ]. Therefore, it suffices to show that the bound
X X
Λ(n) e(hnγ d−1 ) x1−ε N 1−γ (5.10)
h6H N <n6N2
show that our type I and type II sums satisfy the uniform bounds
X X X
SI := a` e(h`γ mγ d−1 ) x1−2ε N 1−γ , (5.11)
H1 6h6H2 `∼L m∼M
`m∈J
X X X
SII := a` bm e(h`γ mγ d−1 ) x1−2ε N 1−γ , (5.12)
H1 6h6H2 `∼L m∼M
`m∈J
1367–1368] we need to show that (5.12) holds uniformly for all L in the range
and for such u we need to show that (5.11) holds uniformly for all M satisfying
M N 1/2 u−1/2 .
Put F := H1 N γ d−1 . For the type II sum, we apply Baker [1, Thm. 2], which
64
with
k/(2+2k)
−1/2 F
TII,1 := (H1 L) and TII,2 := M −(1+k−`)/(2+2k)
H1 L
for any exponent pair (k, `) provided that F > H1 L. For the type I sum, by Robert
with
1/4
F
TI,1 := , TI,2 := M −1/2 and TI,3 := F −1 .
H1 LM 2
is equivalent to d x1−3ε and thus follows from the inequality D 6 x1−3ε which is
To guarantee that
u := x−2+6ε H1 N 2γ .
We need to check that the condition u 6 x−ε N 1/5 of (5.13) is met. For this, taking
D 6 x3−8ε N −4/5−2γ .
65
The worst case occurs when N = xc , leading to the constraint D 6 x1−4c/5−8ε , which
−1/2
M N 1/2 u−1/2 = x1−3ε H1 N 1/2−γ ,
1/4
M −1/2 x−1/2+2ε H1 N −1/4+γ/2 (5.17)
H1 x6/5−4ε N 1/5−6γ/5 .
Using (5.8) again, we see that (5.18) follows from the inequality
D 6 x11/5−5ε N −4/5−6γ/5 .
Next, using the definition of F and the relation LM N one sees that the bound
1/4
F
TI,1 = x1−3ε H1−1 N −γ (5.19)
H1 LM 2
holds whenever
Taking into account (5.8) and (5.17) this bound follows from the condition
D x19/7−7ε N −6/7−12γ/7 .
66
With N := xc we derive the constraint D 6 x1−6c/7−7ε , which is a consequence of
D 6 N γ−1/3−ε . (5.21)
If ε is sufficiently small, then (5.21) follows essentially from (5.6). Indeed, since c > 1
is easily verified; hence, for small enough ε (depending only on c) we have by (5.6):
2/3+2γ−ε γ−1/3−ε
D2/3+2γ−ε 6 x1−136c/157 6 x2−3ε ,
On the other hand, we can certainly assume that HN > x1−2ε N 1−γ , for otherwise
and we have
γ−1/3−ε γ−1/3−ε
x2−3ε x2−3ε
1+γ
γ−1/3−
D 6 6 6 N 1+γ ,
D d
67
which yields (5.21).
Using the relation LM N , the upper bound M N 2/3 (which follows from
L N 1/3 ) and the definition (5.8), we see that the bound (5.20) holds if
D 6 x1−cµ/3−2ε , (5.22)
where
ν − k − 2(1 − `)/3 3k + 2` + 4
µ := = .
ν−k k+2
57 64
With the choice (k, `) := ( 126 , 126 ) (which is BA5 ( 12 , 21 ) in the notation of Graham [2])
we have
µ 803 136
= < ,
3 927 157
and therefore (5.22) follows from (5.6) if ε is small enough. This proves (5.20).
Combining the bounds (5.15), (5.16), (5.18), (5.19) and (5.20), we obtain (5.14),
5.3.2 Refinement
Here, we extend the ideas of §5.3.1 to show that for any δ > 0, the bound (5.5) holds
for all sufficiently small ε > 0 (depending on δ) under the less stringent condition
that
380c
D 6 x1− 441 −δ . (5.23)
68
After this has been done, taking into account that
380c 8 77
1− > whenever c < ,
441 63 76
(i) The type II bound (5.12) holds for L in the range u0 L u20 for some
u0 ∈ [N 1/10 , N 1/6 ];
−1/2
(ii) For such u0 , the type I bound (5.11) holds whenever M N 1/2 u0 ;
(iii) For such u0 and any numbers am , ch ∈ C with |am | 6 1, |ch | 6 1, the type I 0
bound
X X X
SI 0 := ch am e(h`γ mγ d−1 ) x1−2ε N 1−γ
h∼H m∼M `∼L
The condition u0 6 N 1/6 follows from the inequality x−2+6ε HN 2γ 6 N 1/6 , which in
D 6 x3−7ε N −5/6−2γ .
380 5
Taking N := xc leads to D 6 x1−5c/6−7ε , which follows from (5.23). Since 441
> 6
69
Condition (ii) also follows from §5.3.1 in the case that u > N 1/10 . When u < N 1/10
when M N 9/20 . Taking into account (5.8) we see that the bound
D 6 x2−4ε N −31/40−γ .
380 31
The worst case N = xc leads to D 6 x1−31c/40−4ε , and since 441
> 40
this is implied
by (5.23) when ε is small enough. We also know that (5.19) holds whenever
Taking into account (5.8) this bound follows from the inequality
D 6 x8/3−3ε N −51/60−5γ/3 .
380 51
With N := xc we derive the constraint D 6 x1−51c/60−3ε , and as 441
> 60
this a
It remains to verify condition (iii). Rather than adapting Rivat and Sargos [7], we
However, in place of (X, H, M, N ) we use the quadruple (H1 N γ d−1 , M, H1 , L). The
70
check that the various hypotheses of [11, Thm. 2] are satisfied. Applying the theorem
SI 0 TI 0 ,1 + TI 0 ,2 + TI 0 ,3 + TI 0 ,4 + TI 0 ,5 + TI 0 ,6 + TI 0 ,7 xε ,
where
1/174 1/2
TI 0 ,1 := H1179 N 114+32γ L23 d−32 , TI 0 ,5 := H1 N L−1/2 ,
3/4
TI 0 ,2 := H1 N 1/2+γ/4 L1/2 d−1/4 , TI 0 ,6 := H1 N 1/2 L1/2 ,
5/4 1/2
TI 0 ,3 := H1 N 1/2+γ/4 d−1/4 , TI 0 ,7 := H1 N 1−γ/2 d1/2 .
TI 0 ,4 := H1 N L−1 ,
given that
1/174
TI 0 ,1 = H1179 N 114+32γ L23 d−32 x1−3ε N 1−γ (5.28)
is equivalent to
Using the first two inequalities and the upper bound on L in (5.27), it suffices to have
D147 6 x147−380c/3−701ε ,
71
which is (5.23). Similarly, the bounds
3/4
TI 0 ,2 = H1 N 1/2+γ/4 L1/2 d−1/4 x1−3ε N 1−γ , (5.29)
380 5
respectively, and these are easy consequences of (5.23) since 441
> 6
> 23 . On the
other hand, using the first two inequalities and the lower bound on L in (5.27), we
1/2
TI 0 ,5 = H1 N L−1/2 x1−3ε N 1−γ , (5.32)
D 6 x1−4c/5−7ε ,
380
which is implied by (5.23) since 441
> 54 . Next, using the first two inequalities in
5/4
TI 0 ,3 = H1 N 1/2+γ/4 d−1/4 x1−3ε N 1−γ (5.33)
follows from
D x1−3c/4−17ε/4 ,
380
which is implied by (5.23) since 441
> 34 . Similarly,
1/2
TI 0 ,7 = H1 N 1−γ/2 d1/2 x1−3ε N 1−γ (5.34)
72
is a consequence of the inequality
D x1−c/2−7ε/2 ,
380
which follows from (5.23) since 441
> 12 .
Combining the bounds (5.28), (5.29), (5.30), (5.31), (5.32), (5.33) and (5.34), we
73
74
Bibliography
(1994), 269–277.
[2] S. W. Graham, ‘An algorithm for computing optimal exponent pairs,’ J. London
[3] S. W. Graham and G. Kolesnik, Van der Corput’s method of exponential sums.
[4] G. Greaves, Sieves in Number Theory. Results in Mathematics and Related Areas
[7] J. Rivat and P. Sargos, ‘Nombres premiers de la forme bnc c,’ Canad. J. Math.
75
[8] J. Rivat and J. Wu, ‘Prime numbers of the form bnc c,’ Glasg. Math. J. 43 (2001),
no. 2, 237–254.
[10] J. D. Vaaler, ‘Some extremal problems in Fourier analysis,’ Bull. Amer. Math.
[11] J. Wu, ‘On the primitive circle problem,’ Monatsh. Math. 135 (2002), no. 1,
69–81.
76
VITA
Aaron Yeager was born on December 28, 1976 in Springfield, Missouri. In 1995
he graduated from Kickapoo High School. In 2002, after a few years spent pursuing
Aaron attended Cal State Los Angeles. In 2006 Aaron returned home to be with
family and attend Missouri State University. He graduated from MSU with a BS in
mathematics in 2008. Aaron began his graduate work at the University of Missouri
in 2008. In fall 2012, Aaron received his MA in mathematics from the University of
Missouri-Columbia.
77