Journal de Théorie des Nombres de Bordeaux

Chad T. Davis

Blair K. Spearman

Jeewon Yoo

Chad T. DAVIS, Blair K. SPEARMAN† et Jeewon YOO
Cubic polynomials defining monogenic fields with the same discriminant
Tome 30, no 3 (2018), p. 991-996.
<http://jtnb.cedram.org/item?id=JTNB_2018__30_3_991_0>
© Société Arithmétique de Bordeaux, 2018, tous droits réservés.
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Journal de Théorie des Nombres
de Bordeaux 30 (2018), 991–996
Cubic polynomials defining monogenic fields with
the same discriminant
par Chad T. DAVIS, Blair K. SPEARMAN† et Jeewon YOO
Résumé. Un corps de nombres K est dit monogène si son anneau des entiers
vérifie OK = Z[θ] pour un certain θ ∈ OK . La monogénéité d’un corps de
nombres n’est pas toujours assurée. En outre, il est rare que deux corps de
nombres aient le même discriminant. Donc, trouver des corps avec ces deux
propriétés est un problème intéressant. Dans cet article, nous montrons qu’il
existe une infinité de triplets de polynômes définissant des corps cubiques
monogènes distincts de même discriminant.
Abstract. Let K be a number field with ring of integers OK . K is said
to be monogenic if OK = Z[θ] for some θ ∈ OK . Monogeneity of a number
field is not always guaranteed. Furthermore, it is rare for two number fields
to have the same discriminant, thus finding fields with these two properties is
an interesting problem. In this paper we show that there exist infinitely many
triples of polynomials defining distinct monogenic cubic fields with the same
discriminant.
1. Introduction
Let K be a number field with ring of integers OK and discriminant
d(K). K is called monogenic if there exists an element θ ∈ OK with OK =
Z[θ]. The properties of monogeneity and sharing the same discriminant are
uncommon (see for instance [4], and [5, p. 80, Remark 2.4]) thus, finding
fields that have both properties is an interesting problem. The following
three polynomials provide an example of this phenomenon (see [2, p. 94,
Exercise 21])
p1(X) = X3 − 18X − 6,
p2(X) = X3 − 36X − 78,
p3(X) = X3 − 54X − 150.
Manuscrit reçu le 4 octobre 2017, révisé le 14 octobre 2017, accepté le 29 novembre 2017.
2010 Mathematics Subject Classification. 11R16, 11R29.
Mots-clefs. Cubic field, monogenic, discriminant.
The authors would like to thank Prof. Dr. Cornelius Greither and Prof. Qing Liu for their
helpful comments and suggestions on the first draft of this paper.
992 Chad T. Davis, Blair K. Spearman†, Jeewon Yoo
Note that the coefficients of these polynomials are in arithmetic progression.
The purpose of this paper is to give three infinite families of polynomials
f1(X), f2(X), and f3(X), defining cubic monogenic fields over Q of the
same discriminant. In particular we prove the following
Theorem 1. There exists infinitely many pairs of relatively prime integers
(k, e) satisfying k ≡ ±1 (mod 3), e ≡ ±1 (mod 3), and
(1.1) k(3k4 − 6k2e2 − e4)
is squarefree. For each such pair (k, e), the polynomials
(1.2)
f1(X) = X3 − 9k(k + e)X − 3k(3k2 + 6ke+ e2),
f2(X) = X3 − 9k2X − 3k(3k2 + e2),
f3(X) = X3 − 9k(k − e)X − 3k(3k2 − 6ke+ e2),
define distinct monogenic cubic fields with the same discriminant. Further-
more, the set of integers defined in equation (1.1) is infinite.
2. Preliminaries
In this section, we provide some preliminary results. Throughout this
section, k and e will be as in the statement of Theorem 1.
Lemma 1. The polynomials f1, f2, and f3 of equation (1.2) are irreducible
over Q and each have the same polynomial discriminant.
Proof. Since 3 - k, e it is clear that 3 exactly divides the constant coefficients
of f1, f2, and f3, hence they are 3-Eisenstein. Verifying that they have the
same discriminant is strictly computational. 
Lemma 2. Let f1, f2, and f3 be the polynomials from equation (1.2). Let
θi be a root of fi, and set Ki = Q(θi) for i ∈ {1, 2, 3}. Then K1, K2, and
K3 are monogenic.
Proof. If the discriminant of fi is equal to d(Ki) for each i ∈ {1, 2, 3} then
each Ki is monogenic. Thus, it suffices to show that for each prime p, the
exact power of p in the discriminant of fi is equal to the exact power of p
dividing d(Ki). This can be determined using a result due to Llorente and
Nart ([3, Theorem 2]), or alternatively, by Tables A, B, and C on p. 4–7
of [1]. The discriminant of each fi, which we denote by ∆, is equal to
∆ = 35k2(3k4 − 6k2e2 − e4).
Following the notation of [3] and [1], let vp(x) denote the exact power
of a prime p dividing an integer x and let sp = vp(∆). We show that
sp = vp(d(Ki)), i = 1, 2, 3, for all primes p. We give the proof for the field
K1 and note that the other cases are done similarly. Let a1 and b1 denote
the coefficients on X and the constant coefficient of f1 respectively. We
break into cases when p = 2, p = 3, and p > 3.
Monogenic cubic fields with the same discriminant 993
Case 1: p = 2. The assumption that k(3k4 − 6k2e2 − e4) is squarefree im-
plies that k and e have opposite parity. If k is even, then k ≡ 2 (mod 4)
lest equation (1.1) is divisible by a square. In this case, we have a1 ≡ 0
(mod 2) and b1 ≡ 2 (mod 4). Then line 2 of Table A in [1] implies that
s2 = v2(d(K1)) = 2 as desired. If k is odd and e is even, then b1 ≡ 1
(mod 2), so that line 1 of Table A of [1] implies that s2 = v2(d(K1)) = 0 as
desired.
Case 2: p = 3. Since neither k nor e is divisible by 3, it is easily verified
that a1 ≡ 0 (mod 9) and b1 ≡ 0 (mod 3) but b1 6≡ 0 (mod 9). Thus line 3
of Table B of [1] implies that s3 = v3(d(K1)) = 5 as desired.
Case 3: p > 3. First suppose that p does not divide k nor e. Then
(2.1)
a1 ≡ −9k(k + e) (mod p)
b1 ≡ −3k(3k2 + 6ke+ e2) (mod p)
∆ ≡ −4a31 + 27b21 (mod p).
If k ≡ −e (mod p), then from equation (2.1) we have
a1 ≡ 0 (mod p) and b1 ≡ −6e3 6≡ 0 (mod p)
so that p - ∆. Consequently p - d(K1) and sp = vp(d(K1)). If k 6≡ −e
(mod p), then from equation (2.1) we have a1 6≡ 0 (mod p). If b1 ≡ 0
(mod p) then ∆ 6≡ 0 (mod p) so sp = vp(d(K1)). If b1 6≡ 0 (mod p), then
recalling that 3k4 − 6k2e2 − e4 is squarefree, we see that sp = 0 or 1. From
line 5 of Table C of [1], we get that sp = vp(d(K1)) as required.
Now suppose p divides k but does not divide e. Then using the assump-
tion that equation (1.1) is squarefree, it is easily checked that vp(a1) =
vp(b1) = 1. Using line 2 of Table C of [1], we have sp = vp(d(K1)) = 2 as
desired. Finally, if p divides e but does not divide k, then p - ∆ so that
sp = vp(d(K1)) = 0 as required.
In all cases, it has been verified that vp(d(K1)) = sp = vp(∆) for all
primes p ≥ 2. Thus d(K1) = ∆ and K1 is monogenic. 
Lemma 3. Let everything be as in Lemma 2. Then K1,K2, and K3 are
distinct.
Proof. Towards a contradiction, suppose that two of the fields are not dis-
tinct. Without loss of generality, suppose that Q(θ1) = Q(θ2) (noting that
the other two cases are done similarly). Then θ1 ∈ K2. Since K2 is mono-
genic by Lemma 2, there exist a, b, c ∈ Z such that
θ1 = a+ bθ2 + cθ22.
994 Chad T. Davis, Blair K. Spearman†, Jeewon Yoo
Since the trace of θ1 is zero, it follows that a = −6ck2. Making this substi-
tution, we calculate the minimal polynomial of the above element as
F (X) = X3 + uX + v
where
u = −9k(kb2 + (3k2 + e2)bc+ 3k3c2),
v = −3k
(
(3k2+e2)b3+18k3b2c+(27k4+9k2e2)bc2+(3ke4+18k3e2+9k5)c3
)
.
Since the minimal polynomial of θ1 is f1(X), we see that the coefficients of
f1 and F must be equal. Thus, from the constant term
0 ≡ v3 + k(3k
2 + 6ke+ e2) ≡ 2ke2(b+ 2)3 (mod 3).
Since 3 does not divide k nor e, this congruence forces b ≡ 1 (mod 3).
Substituting b = 1 + 3m for some integer m into the equations for u and v
yields two congruences
0 ≡ u9 + k(k + e) ≡ 2ke(ec+ 2) (mod 3),
0 ≡ 13
(
v
3 + k(3k
2 + 6ke+ e2)
)
≡ 2k2e(ec+ 1)3 (mod 3)
which is impossible as these two congruences can not simultaneously hold.
Thus K1 and K2 must be distinct fields. 
3. Proof of Theorem
Proof. By Lemma 1, each fi is irreducible over Q and have the same poly-
nomial discriminant. Lemma 2 implies that the fields Ki are monogenic for
each i ∈ {1, 2, 3} and Lemma 3 implies that the fields Ki are all distinct.
The only thing left to verify is that there are infinitely many integer pairs
(k, e) such that the result holds. In order for this to happen, we need only
show that equation (1.1) is squarefree for infinitely many pairs of integers
(k, e). This follows from Theorem 1 on p. 950–951 of [6] with A = ±1, B =
±1,M = 3,m = 4, w = 1, r = 5, k = 2, F = k(3k4 − 6k2e2 − e4) and let-
ting x → ∞. The statement that there are infinitely many integers as in
equation (1.1) follows immediately from this argument. Finally, note that
this implies there are infinitely many field discriminants that satisfy the
property given in the theorem. 
4. Examples
Let θi be a root of fi for each i = 1, 2, 3. The following table gives some
numerical examples of polynomials f1, f2, f3 that define distinct monogenic
fields over Q of the same discriminant. Notice that when (k, e) = (2, 1) we
recover the example from [2] cited at the beginning of this paper.
Monogenic cubic fields with the same discriminant 995
(k, e) fi(X) ∆fi Integral Basis for Ki = Q(θi)
(2, 1)
f1(X) = X3 − 54X − 150
f2(X) = X3 − 36X − 78
f3(X) = X3 − 18X − 6
22356 = 22 · 35 · 23 {1, θi, θ2i }
(1, 2)
f1(X) = X3 − 27X − 57
f2(X) = X3 − 9X − 21
f3(X) = X3 + 9X + 15
−8991 = −35 · 37 {1, θi, θ2i }
(1, 4)
f1(X) = X3 − 45X − 129
f2(X) = X3 − 9X − 57
f3(X) = X3 + 27X + 15
−84807 = −35 · 349 {1, θi, θ2i }
(1, 10)
f1(X) = X3 − 99X − 489
f2(X) = X3 − 9X − 309
f3(X) = X3 + 81X − 129
−2575071 = −35 · 10597 {1, θi, θ2i }
Table 1. Integral Bases and discriminants for K = Q(θi)
defined by fi for i ∈ {1, 2, 3}.
We end with a numerical example of an extension of Theorem 1 to four
polynomials that define distinct monogenic fields with the same discrimi-
nant. Let θi be a root of fi and set Ki = Q(θi) for i ∈ {1, 2, 3, 4}.
fi(X) ∆fi Integral Basis for Ki =Q(θi)
f1(X) = X3−990X−10830
f2(X) = X3−900X−9030
f3(X) = X3−810X−7230
f4(X) = X3−720X+5370
714395700 = 22 ·35 ·52 ·29399 {1, θi, θ2i }
Table 2. Integral Bases and discriminants for K = Q(θi)
defined by fi for i ∈ {1, 2, 3, 4}.
References
[1] S. Alaca, “p-integral bases of a cubic field”, Proc. Am. Math. Soc. 126 (1998), no. 7, p. 1949-
1953.
[2] Z. I. Borevich & I. R. Shafarevich, Number Theory, Academic Press Inc., 1966.
[3] P. Llorente & E. Nart, “Effective determination of the decomposition of the rational primes
in a cubic field”, Proc. Am. Math. Soc. 87 (1983), no. 4, p. 579-585.
[4] D. C. Mayer, “How many fields share a common discriminant? (Multiplicity problem)”,
Algebra and Algebraic Number Theory, http://www.algebra.at/index_e.htm.
[5] R. A. Mollin, Algebraic Number Theory, 2nd revised ed., Discrete Mathematics and its
Applications, CRC Press, 2011.
996 Chad T. Davis, Blair K. Spearman†, Jeewon Yoo
[6] C. L. Stewart & J. Top, “On ranks of twists of elliptic curves and power-free values of
binary forms”, J. Am. Math. Soc. 8 (1995), no. 4, p. 943-973.
Chad T. Davis
3333 University Way
University of British Columbia - Okanagan
Kelowna, BC, Canada, V1V 1V7.
E-mail: chad.davis@ubc.ca
Blair K. Spearman†
Jeewon Yoo
3333 University Way
University of British Columbia - Okanagan
Kelowna, BC, Canada, V1V 1V7
E-mail: j.yoo1026@gmail.com


Cubic polynomials defining monogenic fields with the same discriminant

Rodolphe Richard

Répartition galoisienne ultramétrique d’une classe d’isogénie de courbes elliptiques : Le cas de la mauvaise réduction. Application aux hauteurs locales.

Peter Bundschuh

Keijo Väänänen

Peter BUNDSCHUH et Keijo VÄÄNÄNEN
Note on the Stern-Brocot sequence, some relatives, and their generating
power series
Tome 30, no 1 (2018), p. 195-202.
<http://jtnb.cedram.org/item?id=JTNB_2018__30_1_195_0>
© Société Arithmétique de Bordeaux, 2018, tous droits réservés.
L’accès aux articles de la revue « Journal de Théorie des Nom-
bres de Bordeaux » (http://jtnb.cedram.org/), implique l’accord
avec les conditions générales d’utilisation (http://jtnb.cedram.
org/legal/). Toute reproduction en tout ou partie de cet article sous
quelque forme que ce soit pour tout usage autre que l’utilisation à
fin strictement personnelle du copiste est constitutive d’une infrac-
tion pénale. Toute copie ou impression de ce fichier doit contenir la
présente mention de copyright.
cedram
Article mis en ligne dans le cadre du
Centre de diffusion des revues académiques de mathématiques
http://www.cedram.org/
Journal de Théorie des Nombres
de Bordeaux 30 (2018), 195–202
Note on the Stern-Brocot sequence, some
relatives, and their generating power series
par Peter BUNDSCHUH et Keijo VÄÄNÄNEN
Résumé. Trois variantes de la suite de Stern-Brocot sont liées
à la célèbre suite de Thue-Morse. Dans la présente note, les fonc-
tions génératrices de ces quatre suites sont considérées. Tandis
que l’une d’entre elles est connue comme étant rationnelle, l’in-
dépendance algébrique sur C(z) des trois autres est démontrée
ici. Puis, ce théorème est généralisé de sorte que les fonctions
Φ(z),Φ(−z),Ψ(z),Ψ(z2) sont aussi considérées, où Φ et Ψ in-
diquent les fonctions génératrices des suites de Rudin-Shapiro et
de Baum-Sweet, respectivement. Quelques applications arithmé-
tiques sont également données.
Abstract. Three variations on the Stern-Brocot sequence are
related to the celebrated Thue-Morse sequence. In the present
note, the generating power series of these four sequences are con-
sidered. Whereas one of these was known to define a rational func-
tion, the other three are proved here to be algebraically indepen-
dent over C(z). Then this statement is fairly generalized by includ-
ing the functions Φ(z),Φ(−z),Ψ(z),Ψ(z2), where Φ and Ψ denote
the generating power series of the Rudin-Shapiro and Baum-Sweet
sequence, respectively. Moreover, some arithmetical applications
are given.
1. Introduction and first results
In their Remark 1.3, Allouche and Mendès France [1] defined a map C
associating with two sequences a = (an)n≥0 and b = (bn)n≥0 the sequence
C(a,b) :=
 ∑
0≤2ν≤n
(
n− ν
ν
)
2
aνbn−ν

n≥0
.
Here, for integers u, v with v ≥ 0, the symbol
(u
v
)
2 equals to 0 or 1 depending
on whether the corresponding binomial coefficient
(u
v
)
is even or odd. In
particular, C(1,1) is the Stern-Brocot sequence (1 the obvious constant
Manuscrit reçu le 25 avril 2016, révisé le 4 septembre 2016, accepté le 12 septembre 2016.
2010 Mathematics Subject Classification. 11J81, 11J91, 11B37.
Mots-clefs. Stern-Brocot sequence, transcendence, algebraic independence, Mahler’s method.
196 Peter Bundschuh, Keijo Väänänen
sequence), which is identical with the sequence (an+1)n≥0 considered in [2,
(1)]. The authors of [1] recalled that the ±1 Thue-Morse sequence t =
(tn)n≥0 is given by t0 = 1 and, for all n ≥ 0, by t2n = tn and t2n+1 = −tn.
Then they introduced the sequences α = (αn)n≥0, β = (βn)n≥0, γ = (γn)n≥0
by
α := C(t,1), β := C(1, t), γ := C(t, t)
to which σ := C(1,1) is added, σ for Stern, and noted that these four
sequences satisfy the recurrences
σ0 = 1, σ1 = 1, σ2n = σn + σn−1, σ2n+1 = σn for any n ≥ 1,
α0 = 1, α1 = 1, α2n = αn − αn−1, α2n+1 = αn for any n ≥ 1,
β0 = 1, β1 = −1, β2n = βn − βn−1, β2n+1 = −βn for any n ≥ 1,
γ0 = 1, γ1 = −1, γ2n = γn + γn−1, γ2n+1 = −γn for any n ≥ 1.
Let τ be any of these sequences σ, α, β, γ and write τ = (τn)n≥0 with
τ0 = 1. Then one concludes inductively τ2m−1 = τm1 for any m = 0, 1, . . .
and, moreover, |τn| ≤ n for n = 1, 2, . . . Thus, the generating power series
Γτ (z) :=
∞∑
n=0
τnz
n ∈ Z[[z]]
of τ has convergence radius 1. In this situation, a classical result of Carl-
son [5] says that Γτ either represents a rational function or cannot be ana-
lytically continued beyond the unit circle.
At the beginning of Section 2, for each of the functions Γτ , τ ∈{σ, α, β, γ},
a functional equation of so-called Mahler-type
(1.1) Γτ (z) = pτ (z)Γτ (z2)
will be specified with a polynomial pτ (z) of degree 2 satisfying pτ (0) = 1
and the other coefficients being 1 or −1. Although these equations look
rather similar, the analytic behaviour of the four functions is quite different.
While Γγ(z) turns out to be rational, the other three are hypertranscen-
dental, hence transcendental over C(z) and therefore, by Carlson’s above-
mentioned theorem, have the unit circle as natural boundary. Remember
here that an analytic function is called hypertranscendental if it satisfies
no algebraic differential equation, that is, no finite collection of derivatives
of the function is algebraically dependent over C(z). By the way, the hy-
pertranscendence of Γσ was proved in [2, Theorem 1]. In the case of Γα,
the hypertranscendence is shown in the proof of [3, Corollary 3], and the
case of Γβ is similar. All these results are consequences of the general hy-
pertranscendence criterion for infinite products F (z) =
∏∞
j=0 P (z2
j ) with
a polynomial P (z), see [3, Theorem 1], namely Γτ (0) = 1 and iteration
The Stern-Brocot sequence 197
of (1.1) immediately leads to the product representation
(1.2) Γτ (z) =
∞∏
j=0
pτ (z2
j )
of Γτ (z) valid in D := {z ∈ C : |z| < 1}. Note also that in [4] we proved the
algebraic independence of Γσ(z) and its twist.
The sole transcendence over C(z) of each of the functions Γσ,Γα,Γβ will
now be generalized in a different direction. Namely, our first main result
reads as follows.
Theorem 1.1. The functions Γσ(z),Γα(z),Γβ(z) are algebraically inde-
pendent over C(z).
Using [7, Corollary 2] or [8, Theorem 4.2.1] we deduce from Theorem 1.1
the following arithmetical consequence.
Corollary 1.2. If ξ is an algebraic number with 0 < |ξ| < 1 such that every
ξ2
j (j = 0, 1, . . . ) is different from ±(1−
√
5)/2, then Γσ(ξ),Γα(ξ),Γβ(ξ) are
algebraically independent over Q.
Note that here the exceptional conditions on ξ are necessary. Namely,
if ξ2j equals to (1 −
√
5)/2 or to (
√
5 − 1)/2 for some j ≥ 0, then the
product representation (1.2) of Γα or Γβ implies Γα(ξ) = 0 or Γβ(ξ) = 0,
respectively.
The question of what happens in the direction of Corollary 1.2 for tran-
scendental ξ can be answered as follows using [11, Théorème 4].
Corollary 1.3. For transcendental ξ ∈ D, one has the estimate
trdegQQ(ξ,Γσ(ξ),Γα(ξ),Γβ(ξ)) ≥ 3
and this is best possible.
2. Proof of Theorem 1.1
To begin with, we use the recurrences for τ ∈ {σ, α, β, γ} as given above
to easily demonstrate the functional equation (1.1) for Γτ with
pσ(z) = 1 + z + z2 ,(2.1)
pα(z) = 1 + z − z2 ,(2.2)
pβ(z) = 1− z − z2 (= pα(−z)) ,(2.3)
pγ(z) = 1− z + z2 (= pσ(−z)) .(2.4)
Note that the first case, Γσ(z) = (1 + z+ z2)Γσ(z2), was explicitly checked
in [2, (7)].
198 Peter Bundschuh, Keijo Väänänen
To realize the rationality of Γγ(z), we simply note pγ(z)pσ(z) = pσ(z2)
and pσ(z) 6= 0 in D, by (2.1) and (2.4). Using (1.2) with τ = γ we obtain
Γγ(z) =
1
pσ(z)
= 1− z1− z3 = (1− z)
∑
m≥0
z3m,
and this even shows that γn equals 1,−1, 0 if n ≡ 0, 1, 2 (mod 3), respec-
tively.
The main tool in our proof of Theorem 1.1 is the subsequent criterion of
Kubota [6, Proposition 3] (see also [8, Theorem 3.5]) to be quoted here in
a very particular version, which suffices for our purposes.
Lemma 2.1. Let d ≥ 2 be an integer. Suppose that the series f1, . . . , fk ∈
C[[z]] \ {0} converge on D and satisfy the functional equations
fi(zd) = bi(z)fi(z) (i = 1, . . . , k)
with all bi ∈ C(z) \ {0} fulfilling the condition that, for no (n1, . . . , nk) ∈
Zk \ {0}, the functional equation
r(zd) = r(z)
k∏
i=1
bi(z)ni
has a solution r ∈ C(z)\{0}. Then the functions f1, . . . , fk are algebraically
independent over C(z).
We apply Lemma 2.1 with d = 2, k = 3 and Γσ,Γα,Γβ for the f1, f2, f3.
According to their functional equations (1.1) with pτ (z) as given by (2.1)–
(2.3), we have b1(z) = 1/pσ(z), b2(z) = 1/pα(z), b3(z) = 1/pβ(z) and we
must show, for any (nσ, nα, nβ) ∈ Z3 \ {0}, that the equation
(2.5) r(z) = r(z2)pσ(z)nσpα(z)nαpβ(z)nβ
has no non-trivial rational solution r.
To consider first the case nσ 6= 0, nα = nβ = 0, we iterate (2.5) j times
and obtain
r(z) = r(z2j )
j−1∏
i=0
pσ(z2
i)
nσ .
Letting j → ∞ and noting r(0) 6= 0,∞, this formula leads to r(z) =
r(0)Γσ(z)nσ , by (2.5) with τ = σ, a contradiction. Namely, for rational
r(z), the function Γσ(z) would be algebraic.
To consider the second case (nα, nβ) 6= (0, 0) of (2.5), we apply Lemma 2.2
below with d = 2 and
R(z) := pσ(z)nσpα(z)nαpβ(z)nβ(2.6)
=
(
(ζ−z)
(1
ζ
−z
))nσ (
(G−z)
( 1
G
+z
))nα (
(G+z)
( 1
G
−z
))nβ
The Stern-Brocot sequence 199
being the factor appearing in the right-hand side of (2.5). Here ζ := e2πi/3
and G := (1 +
√
5)/2 denotes the golden ratio. Clearly, we may choose
Ω = G,ω = −1/G if nα 6= 0, or Ω = −G,ω = 1/G if nα = 0 but nβ 6= 0. To
examine the last hypothesis of Lemma 2.2, let j ≥ 1. All solutions of z2j =
±1/G (or of z2j = ±G) satisfy |z| = G−1/2j (or |z| = G1/2j ), and no such z
can be a zero or pole of R(z2j−1) . . . R(z). Namely otherwise, according to
the right-hand side of (2.6), zeros or poles of R(z2k) at z inside or outside
of D satisfy z2k = ±1/G (or z2k = ±G) with some k ∈ {0, . . . , j− 1} which
is impossible. 
Lemma 2.2. Let d ≥ 2 be an integer. Let R ∈ C(z) have zeros or poles
in 0 < |z| < 1 or in 1 < |z| < +∞, where ω and Ω are zeros or poles of
minimal and of maximal absolute value, respectively. Assume that, for each
integer j ≥ 1, not all solutions of zdj = ω (or of zdj = Ω) are zeros or poles
of the product R(zdj−1) . . . R(z). Then the functional equation
(2.7) r(z) = r(zd)R(z)
has no rational solution r 6= 0.
Proof. Note first that 0 < |ω| < 1 or |Ω| > 1 holds by hypothesis. Then
suppose, on the contrary, that (2.7) has a rational solution r(z) 6= 0. Clearly,
this r(z) has no zero or pole in |z| < |ω| (or in |z| > |Ω|). Since |ωd| < |ω|
(or |Ωd| > |Ω|) we know that ωd (or Ωd) is not a zero or pole of r(z), whence
ω (or Ω) is a zero or pole of r(z), by (2.7) and the hypothesis on R.
On iterating (2.7) we conclude that r satisfies
r(z) = r(zdj )R(zdj−1) . . . R(z)
for any j = 1, 2, . . . . For any such j, we consider all dj solutions of zdj =
ω (or of zdj = Ω), among which at least one is not a zero or pole of
R(zdj−1) . . . R(z). Thus, on every circle |z| = |ω|1/dj (or |z| = |Ω|1/dj ), we
have a zero or a pole of r contradicting the rationality of r. 
3. Interactions of the Γτ ’s with other generating power series
Beside the Γτ ’s studied before, we recall two more integer sequences and
their generating power series. First, let ϕ = (ϕn)n≥0 denote the Rudin-
Shapiro sequence, which can be recursively defined by ϕ0 = 1, and ϕ2n =
ϕn, ϕ2n+1 = (−1)nϕn for any n ≥ 0. Its generating power series
Φ(z) :=
∞∑
n=0
ϕnz
n
200 Peter Bundschuh, Keijo Väänänen
satisfies a system of two functional equations of Mahler-type, in matrix
notation written as
(3.1)
(
Φ(z)
Φ(−z)
)
=
(
1 z
1 −z
)(
Φ(z2)
Φ(−z2)
)
(see [7]). Secondly, let ψ = (ψn)n≥0 denote the Baum-Sweet sequence, which
is given by ψ0 = 1, ψn = 1 if the binary representation of n ≥ 1 contains
no block of consecutive 0’s of odd length, and ψn = 0 otherwise. The
corresponding generating power series
Ψ(z) :=
∞∑
n=0
ψnz
n
fulfills a system of functional equations similar to (3.1) (see [7]), namely
(3.2)
(
Ψ(z)
Ψ(z2)
)
=
(
z 1
1 0
)(
Ψ(z2)
Ψ(z4)
)
.
K. Nishioka [7] (see also [8, pp. 158–160]) proved that both pairs {Φ(z),
Φ(−z)} and {Ψ(z),Ψ(z2)} consist of functions which are algebraically in-
dependent over C(z). Using this and one of her results cited before our
Corollary 1.2, she obtained also the algebraic independence over Q of the
numbers Φ(ξ),Φ(−ξ) (and of Ψ(ξ),Ψ(ξ2)) for any non-zero algebraic ξ ∈ D.
In a recent paper, K. Nishioka and S. Nishioka [9] proved much stronger
results on Φ and Ψ. Namely, they obtained the algebraic independence of
all four functions Φ(z),Φ(−z),Ψ(z),Ψ(z2) not only over C(z) but over any
difference field extension of valuation ring type over C(z) under the trans-
formation z 7→ z2 (compare [9, Theorem 9]). This notion, too combersome
to be fully quoted here, has been introduced by S. Nishioka in his deep
work [10, Definition 1] on solvability of certain classes of difference equa-
tions. His Theorem 2 and Proposition 5 are the key tools of the proofs
in [9].
The Nishiokas’ just-quoted algebraic independence result implies the fol-
lowing, where we restrict ourselves to homogeneous systems of functional
equations.
Lemma 3.1. The four functions Φ(z),Φ(−z),Ψ(z),Ψ(z2) and power series
f1(z), . . . , fk(z) ∈ C[[z]] satisfying a system
(3.3) >(f1(z), . . . , fk(z)) = A(z) · >(f1(z2), . . . , fk(z2))
of functional equations with triangular A ∈ GLk(C(z)), > denoting matrix
transposition, are algebraically independent over C(z) if this already applies
to f1(z), . . . , fk(z).
Using this with k = 3, A(z) = diag(pσ(z), pα(z), pβ(z)), f1 = Γσ, f2 =
Γα, f3 = Γβ, (see (2.1)–(2.3)), Theorem 1.1 leads us to the following.
The Stern-Brocot sequence 201
Theorem 3.2. The seven functions Φ(z),Φ(−z),Ψ(z),Ψ(z2),Γσ(z),Γα(z),
Γβ(z) are algebraically independent over C(z). In addition, if ξ ∈ D is a
non-zero algebraic number such that every ξ2j (j = 0, 1, . . . ) is different
from ±(1 −
√
5)/2, then Φ(ξ),Φ(−ξ), Ψ(ξ),Ψ(ξ2),Γσ(ξ),Γα(ξ),Γβ(ξ) are
algebraically independent over Q.
Note that the arithmetical part of this assertion is again an application
of Nishioka’s algebraic independence criterion cited before Corollary 1.2.
The hypertranscendence of the Γτ ’s for τ ∈ {σ, α, β} we mentioned earlier
can also be combined with Lemma 3.1 to get the following consequence.
Theorem 3.3. For fixed τ ∈ {σ, α, β}, the functions Φ(z), Φ(−z), Ψ(z),
Ψ(z2), Γτ (z), Γ′τ (z),Γ′′τ (z), . . . are algebraically independent over C(z).
Moreover, if ξ ∈ D is a non-zero algebraic number, then the values Φ(ξ),
Φ(−ξ), Ψ(ξ), Ψ(ξ2), Γτ (ξ),Γ′τ (ξ), . . . are algebraically independent over Q
under the additional proviso that all ξ2j (j = 0, 1, . . . ) are different from
(1−
√
5)/2 in case τ = α, and from (
√
5− 1)/2 in case τ = β.
Sketch of proof. We apply Lemma 3.1 with f1(z) = Γτ (z), . . . , fk(z) =
Γ(k−1)τ (z) for arbitrary but fixed integer k ≥ 1. By taking ith derivatives
(i = 0, . . . , k− 1) of the functional equation (1.1) with corresponding (2.1),
(2.2), or (2.3), we obtain the validity of (3.3) with lower triangular matrix
A(z) having (2z)ipτ (z), i = 0, . . . , k − 1, on its main diagonal. 
References
[1] J.-P. Allouche & M. Mendès France, “Lacunary formal power series and the Stern-
Brocot sequence”, Acta Arith. 159 (2013), no. 1, p. 47-61.
[2] P. Bundschuh, “Transcendence and algebraic independence of series related to Stern’s se-
quence”, Int. J. Number Theory 8 (2012), no. 2, p. 361-376.
[3] ———, “Algebraic independence of infinite products and their derivatives”, in Number
theory and related fields, Springer Proceedings in Mathematics & Statistics, vol. 43, Springer,
2013, p. 143-156.
[4] P. Bundschuh & K. Väänänen, “Algebraic independence of the generating functions of
Stern’s sequence and of its twist”, J. Théor. Nombres Bordx. 25 (2013), no. 1, p. 43-57.
[5] F. D. Carlson, “Über Potenzreihen mit ganzzahligen Koeffizienten”, Math. Z. 9 (1921),
p. 1-13.
[6] K. K. Kubota, “On the algebraic independence of holomorphic solutions of certain func-
tional equations and their values”, Math. Ann. 227 (1977), p. 9-50.
[7] K. Nishioka, “New approach in Mahler’s method”, J. Reine Angew. Math. 407 (1990),
p. 202-219.
[8] ———, Mahler Functions and Transcendence, Lecture Notes in Mathematics, vol. 1631,
Springer, 1996, viii+185 pages.
[9] K. Nishioka & S. Nishioka, “Algebraic theory of difference equations and Mahler func-
tions”, Aequationes Math. 84 (2012), no. 3, p. 245-259.
[10] S. Nishioka, “Solvability of difference Riccati equations by elementary operations”, J. Math.
Sci., Tokyo 17 (2010), no. 2, p. 159-178.
[11] P. Philippon, “Indépendance algébrique et K-fonctions”, J. Reine Angew. Math. 497
(1998), p. 1-15.
202 Peter Bundschuh, Keijo Väänänen
Peter Bundschuh
Mathematisches Institut
Universität zu Köln
Weyertal 86–90
50931 Köln, Germany
E-mail: pb@math.uni-koeln.de
Keijo Väänänen
Department of Mathematical Sciences
University of Oulu
P. O. Box 3000
90014 Oulu, Finland
E-mail: keijo.vaananen@oulu.fi


Note on the Stern-Brocot sequence, some relatives, and their generating power series

Shin Hattori

On canonical subgroups of Hilbert–Blumenthal abelian varieties

Alessandro Languasco

Alessandro Zaccagnini

Short intervals asymptotic formulae for binary problems with prime powers

Nicolas Billerey

Filippo A. E. Nuccio Mortarino Majno di Capriglio

Représentations galoisiennes diédrales  et formes à multiplication complexe

Olivier Ramaré

Aled Walker

Olivier RAMARÉ et Aled WALKER
Products of primes in arithmetic progressions: a footnote in parity breaking
Tome 30, no 1 (2018), p. 219-225.
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Journal de Théorie des Nombres
de Bordeaux 30 (2018), 219–225
Products of primes in arithmetic progressions:
a footnote in parity breaking
par Olivier RAMARÉ et Aled WALKER
Résumé. Nous montrons que, étant donnés x et q 6 x1/16,
toute classe inversible a modulo q contient au moins un produit
d’exactement trois nombres premiers, chacun étant inférieur ou
égal à x1/3.
Abstract. We prove that, if x and q 6 x1/16 are two parame-
ters, then for any invertible residue class a modulo q there exists
a product of exactly three primes, each one below x1/3, that is
congruent to a modulo q.
1. Introduction and results
Xylouris’ version of Linnik’s Theorem [15] tells us that, for every mod-
ulus q and every invertible residue class a modulo q, one can find a prime
congruent to a modulo q that is below q5.18 provided q be large enough.
The proof relies on intricate techniques, and though the result is indeed
effective, no one has been able to give any explicit version of it. The aim
of this paper is to show that one can easily access a fully explicit result,
with respectable constants, provided one replaces primes by products of
three primes. Here is what we prove, by combining a simple sieve technique
together with classical additive combinatorics.
Theorem 1.1. Let x and q ≤ x1/16 be two parameters. Then for any
invertible residue class a modulo q, there exists a product of three primes,
all below x1/3, that is congruent to a modulo q.
We did not try to be optimal in our treatment but sought the simplest
argument. The main surprise is that we use sieve techniques in the form of
Brun–Titchmarsh inequality but we are not blocked by the parity principle.
The reader may argue that we use a lower bound for L(1, χ), but the bound
we employ is the weakest possible and does not rely on Siegel’s Theorem.
Manuscrit reçu le 22 mai 2016, révisé le 3 novembre 2016, accepté le 2 décembre 2016.
2010 Mathematics Subject Classification. 11N13, 11A41, 11N37, 11B13.
Mots-clefs. Primes in arithmetic progressions, Least prime quadratic residue, Linnik’s
Theorem.
The first author has been partly supported by the Indo-French Centre for the Promotion of
Advanced Research – CEFIPRA, project No 5401-1. The second author was supported by the
EPSRC Grant EP/M50659X/1.
220 Olivier Ramaré, Aled Walker
In particular, it is not strong enough to push a possible Siegel zero away
from 1, a fact known to be equivalent to the parity phenomenom (see [8]
and [11], or [12, Chapter 6] for a more complete discussion).
Our theorem is also linked with a conjecture of Erdös that says that
every invertible congruence class should contain a product of two primes
not more than q. This is discussed in work of the second author [14].
A numerically improved version is being prepared.
2. Lemmas
We begin with some crude bounds. Let us define
(2.1) f0(q) =
∏
p|q
(1− 1/√p)−1.
Lemma 2.1. For q > 2 we have f0(q) ≤ 3.32
√
q
Proof. For all primes p we have (1− 1/√p)−1 6 αp
√
p, where
αp =

1√
2−1 p = 2
1√
3−1 p = 3
1 otherwise,
and since α2 6 2.42 and α3 6 1.37, we obtain the inequalities
f0(q) 6 2.42 · 1.37 ·
√
q 6 3.32 · √q . 
We will also require a rudimentary estimate on φ(q).
Lemma 2.2. If q > 31 then φ(q) > 8.
Proof. Recall that
φ(n) = n
∏
p|n
(
1− 1
p
)
.
Therefore if φ(q) 6 8, the only prime factors of q are 2, 3, 5, 7. By per-
forming an easy case analysis on which of these primes divides q, one sees
that the only q for which φ(q) 6 8 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 24,
and 30. 
We will use the elementary theory of Dirichlet characters, referring the
reader to the excellent monograph [2] of Davenport for an introduction on
the subject. In particular we note the following easy bound.
Lemma 2.3. Let χ be a non-principal Dirichlet character modulo q. Let I
be a subset of {1, . . . , q}. We have∣∣∣∑
n∈I
χ(n)
∣∣∣ ≤ φ(q)/2 .
The same bound holds true for any finite interval instead of I.
Products of primes in arithmetic progressions 221
Proof. We know on the one hand that, we have
∑
1≤n≤q χ(n) = 0 by or-
thogonality, and on the other hand that χ(n) does not vanish only when n
belongs to the multiplicative group, say Uq, of Z/qZ. We can hence bound
|
∑
n∈I χ(n)| by the cardinal of I ∩ Uq and by the cardinal of Uq \ I ∩ Uq.
One of them is not more than φ(q)/2, proving the first part of the lemma.
When I is a finite interval, we note that the sum of the values of χ(n)
on any q consecutive integers vanishes, reducing the problem to the first
case. 
We next modify an idea of Gel’fond from [4], which is maybe more easily
read in [5].
Lemma 2.4. Let χ be a non-principal quadratic character modulo q. We
have
L(1, χ) ≥ π4φ(q) −
π
φ(q)2 .
Proof. We consider the sum S(α) =
∑
n≥1(1 ? χ)(n)e−nα for real positive
α. Since (1 ? χ)(m2) ≥ 1 for every integer m, and (1 ? χ)(n) ≥ 0 in general,
a comparison with an integral gives us
1 + S(α) ≥
∑
m≥0
e−m
2α ≥
∫ ∞
0
e−αt
2 dt = Γ(1/2)2
√
α
=
√
π
2
√
α
.
On the other hand we can expand (1 ? χ)(n) =
∑
d|n χ(d) and get
S(α) =
∑
d≥1
χ(d)
eαd − 1 =
L(1, χ)
α
−
∑
d≥1
χ(d)g(αd)
by using the non-negative non-increasing function g(x) = 1x−
1
ex−1 . We find
that, by Lemma 2.3,∑
d≥1
χ(d)g(αd) = −
∑
d≥1
χ(d)
∫ ∞
αd
g′(t) dt
= −
∫ ∞
0
∑
d≤t/α
χ(d)g′(t) dt
≥ φ(q)2
∫ ∞
0
g′(t) dt
= −φ(q)/4
since lim g(x) = 1/2 as x tends to 0 from above. By comparing both upper
and lower estimate for S(α), we reach
L(1, χ) ≥
√
πα
2 − α−
αφ(q)
4 .
We select α = π/φ(q)2. The lemma follows. 
222 Olivier Ramaré, Aled Walker
Lemma 2.5. Let q ≥ 3 be an integer and χ be a non-principal quadratic
character modulo q. Then there is a prime p at most q4 such that χ(p) = 1.
Proof. We adapt the proof of J. Pintz taken from [10]. Assume that no
primes not more than a given real number x are in the kernel of χ. We use
the notation d|q∞ to say that all the prime factors of d divides q. Then on
the one side we have∑
n≤x
(1 ? χ)(n) =
∑
d|q∞
∑
m2≤x/d,
(m,q)=1
1 ≤
∑
d|q∞
√
x
d
≤
√
xf0(q)
where f0 is the function defined in (2.1), while on the other side we can
approximate this sum by L(1, χ) as follows:∑
n≤x
(1 ? χ)(n) =
∑
d≤x
χ(d)
[x
d
]
= x
∑
d≤x
χ(d)
d
−
∑
d≤x
χ(d)
{x
d
}
.
The first summation over d is an approximation of L(1, χ) (recall Lem-
ma 2.3):
L(1, χ) =
∑
d≥1
χ(d)
d
=
∑
d≤x
χ(d)
d
+
∫ ∞
x
∑
x<d≤t
χ(d) dt/t2
=
∑
d≤x
χ(d)
d
+O∗
(φ(q)
2x
)
.
We treat the second summation in d above by Axer’s method from [1] (see
also [7, Theorem 8.1]):∣∣∣∣∣∣
∑
d≤x
χ(d)
{x
d
}∣∣∣∣∣∣ ≤
∑
d≤y
1 +
∑
m≤x/y
∣∣∣∣∣∣
∑
d:[x/d]=m
χ(d)
{x
d
}∣∣∣∣∣∣ ≤ y+ φ(q)x2y ≤
√
2φ(q)x
by selecting y =
√
φ(q)x/2, the second inequality following by Abel sum-
mation. All of this implies that
√
xL(1, χ) ≤ f0(q) +
√
2φ(q) +φ(q)/(2
√
x).
However, the previous lemma gives us a lower bound for L(1, χ) and thus
we should have
π
4φ(q) −
π
φ(q)2 ≤
f0(q)√
x
+
√
2φ(q)
x
+ φ(q)2x .
We substitute x = q4, using the upper bound for f0(q) provided by Lem-
ma 2.1. Replacing the left hand side of the above inequality by π/8φ(q),
which is permissible by Lemma 2.2, together with the bound φ(q) ≤ q, after
a short calculation we derive a contradiction for all q ≥ 45. Calculating
using the exact expressions for f0(q) and φ(q) when q ∈ {15, . . . , 45}, we
also derive a contraction. For the remaining q it is easy enough to find
Products of primes in arithmetic progressions 223
primes p 6 q4 such that p ≡ 1 modulo q. Indeed, for q = 2, . . . , 14 we may
take p = 3, 7, 5, 11, 7, 29, 17, 19, 11, 23, 37, 53, 29 respectively. 
We quote the following result from [6], which is a strong form of the
Brun–Titchmarsh inequality.
Lemma 2.6. When 1 ≤ q < x, we have∑
y<p≤y+x,
p≡a[q]
1 ≤ 2x
φ(q) log(x/q) .
for any positive y.
Lemma 2.7. We have π(x) ≥ x/(log x− 1) when x ≥ 5 393. Furthermore
the number of primes not more than x but prime to some fixed modulus q
below x is at least x/ log x, again when x ≥ 5 393.
Proof. The first inequality is taken from [3]. For the second, we simply note
that the number of prime factors of q is at most (log x)/ log 2 and that
x
log x− 1 −
log x
log 2 ≥
x
log x
when x ≥ 5 000. 
The final ingredient in the argument will be Kneser’s Theorem, which
we now recall (see [9, Theorem 4.3] or [13, Theorem 5.5]).
Lemma 2.8. Let A and B be two subsets of the finite abelian group G. Let
H be the subgroup of elements h of G that stabilizes A + B, i.e. that are
such that h+A+B = A+B. We have
|A+B| ≥ |A+H|+ |B +H| − |H| .
3. Proof of Theorem 1.1
Let us first treat the case x ≥ 1016.
Let X = x1/3. Since this parameter is at least 105, Lemma 2.7 tells
us that the number πq(X) of primes below X which are coprime to q is
at least X/ logX. The Brun–Titchmarsh inequality in the form given by
Montgomery & Vaughan, recalled in Lemma 2.6, tells us that the number
of primes less than X in any progression a mod q, for a prime to q, is at
most 3213X/(φ(q) logX). This implies, when compared to the total number of
primes coprime to q given by Lemma 2.7, that at least 1332φ(q) such residue
classes contain a prime. Let us call this set of classes A and apply Kneser’s
Theorem (Lemma 2.8) to the group G of invertible residues modulo q. Let
H be the stabilizer of A ·A . We divide into cases according to the index
of H.
224 Olivier Ramaré, Aled Walker
If H is equal to G then, since A ·A ·H = A ·A , we have A ·A = G
and of course A ·A ·A = G.
If H has index 2, then it is the kernel of some quadratic character χ.
Because A generates G multiplicatively, there is a point a in A such that
χ(a) = −1. By Lemma 2.5, there is another one, say a′, such that χ(a′) = 1.
Hence A · A also has a point b such that χ(b) = 1 and one, say b′, such
that χ(b′) = −1. This implies that A ·A ·H = G, i.e. A ·A = G.
WhenH is of index 3, then A ·H covers at least 2H-cosets (since 1332 >
1
3)
and is thus of cardinality at least 2φ(q)/3. Kneser’s Theorem ensures that
|A ·A | ≥ φ(q), i.e. that again A ·A = G.
When H is of index 4, then A · H covers at least 2 H-cosets (since
13
32 >
1
4) and is thus of cardinality at least φ(q)/2. By Kneser’s Theorem,
|A ·A | ≥ 2|A ·H| − |H| ≥ 34φ(q) .
When H is of index Y say, with Y at least 5, let us write |A |/φ(q) =
1/U . The set A · H is made out of at least dY/Ue cosets modulo H.
Using the same manipulation as above, Kneser’s Theorem ensures that
|A · A |/φ(q) ≥ (2dY/Ue − 1)/Y . A quick computation shows that the
minimum of (2dY/Ue − 1)/Y when Y ranges {5, 6, 7, 8, 9} is reached at
Y = 7 and has value 5/7. When Y is larger than 10, we directly check that
(2dY/Ue − 1)/Y ≥ 2U −
1
Y ≥
13
16 −
1
10 ≥
7
10 .
Combining these final two cases, we have proved that |A ·A | ≥ 710φ(q).
Let b be an arbitrary invertible residue class modulo q. The set b/A is
of cardinality at least |A | and, since 1332 is greater then
3
10 , this is strictly
larger than the size of the complementary set of A · A . Therefore these
sets have a point in common: there exist a, a1 and a2, all three in A such
that b/a = a1a2, proving our theorem in this case.
It remains to deal with x < 1016, which is done by explicit calculation.
The inclusion of this addendum was kindly suggested to us by an anony-
mous referee. Indeed, when x < 1016, the modulus q is restricted to be not
more than 10, implying that only a limited number of congruence classes
are to be looked at. We proceed by hand:
• when q = 2, we only need x ≥ 33;
• when q = 3, we only need x ≥ 73;
• when q = 4, we only need x ≥ 53;
• when q = 5, we only need x ≥ 193;
• when q = 6, we only need x ≥ 113;
• when q = 7, we only need x ≥ 293;
• when q = 8, we only need x ≥ 233;
• when q = 9, we only need x ≥ 233;
• when q = 10, we only need x ≥ 193.
Products of primes in arithmetic progressions 225
This takes care of the situation when x ≥ 293. However, when x is be-
low 293, the bound x1/16 is less than 2. This ends the proof of our theorem.
References
[1] A. Axer, “Über einige Grenzwertsätze”, Wien. Ber. 120 (1911), p. 1253-1298.
[2] H. Davenport, Multiplicative number theory, 3rd ed., Graduate Texts in Mathematics,
vol. 74, Springer, 2000, x+177 pages.
[3] P. Dusart, “Estimates of some functions over primes without R.H.”, https://arxiv.org/
abs/1002.0442, 2010.
[4] A. O. Gel’fond, “On the arithmetic equivalent of analyticity of the Dirichlet L-series on
the line Re s = 1”, Izv. Akad. Nauk SSSR. Ser. Mat. 20 (1956), p. 145-166.
[5] A. O. Gel’fond & Y. V. Linnik, Elementary methods in analytic number theory, Rand
McNally & Co., 1965, Translated by Amiel Feinstein. Revised and edited by L. J. Mordell,
242 pages.
[6] H. L. Montgomery & R. C. Vaughan, “Hilbert’s inequality”, J. Lond. Math. Soc. 8 (1974),
p. 73-82.
[7] ———,Multiplicative Number Theory: I. Classical Theory, Cambridge Studies in Advanced
Mathematics, vol. 97, Cambridge University Press, 2007, xvii+552 pages.
[8] Y. Motohashi, “A note on Siegel’s zeros”, Proc. Japan Acad., Ser. A 55 (1979), p. 190-192.
[9] M. B. Nathanson, Additive Number Theory –Inverse Problems and the Geometry of Sum-
sets, Graduate Texts in Mathematics, vol. 165, Springer, 1996, xiv+293 pages.
[10] J. Pintz, “Elementary methods in the theory of L-functions, VI. On the least prime qua-
dratic residue (modρ)”, Acta Arith. 32 (1977), p. 173-178.
[11] K. Ramachandra, A. Sankaranarayanan & K. Srinivas, “Ramanujan’s lattice point
problem, prime number theory and other remarks”, Hardy-Ramanujan J. 19 (1996), p. 2-
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[12] O. Ramaré, Arithmetical aspects of the large sieve inequality, Harish-Chandra Research
Institute Lecture Notes, vol. 1, Hindustan Book Agency, 2009, x+201 pages.
[13] T. Tao & V. H. Vu, Additive Combinatorics, Cambridge Studies in Advanced Mathematics,
vol. 105, Cambridge University Press, 2006, xviii+512 pages.
[14] A. Walker, “A multiplicative analogue of Schnirelmann’s theorem”, Bull. Lond. Math. Soc.
48 (2016), no. 6, p. 1018-1028.
[15] T. Xylouris, “On the least prime in an arithmetic progression and estimates for the zeros
of Dirichlet L-functions”, Acta Arith. 150 (2011), no. 1, p. 65-91.
Olivier Ramaré
CNRS / Institut de Mathématiques de Marseille
Aix Marseille Université, U.M.R. 7373
Site Sud, Campus de Luminy, Case 907
13288 MARSEILLE Cedex 9, France
E-mail: olivier.ramare@univ-amu.fr
URL: http://iml.univ-mrs.fr/~ramare/
Aled Walker
Mathematical Institute
University of Oxford
Andrew Wiles Building
Radcliffe Observatory Quarter
Woodstock Road
Oxford
OX2 6GG, United Kingdom
E-mail: walker@maths.ox.ac.uk
URL: https://www.maths.ox.ac.uk/people/aled.walker


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