The Barban-Davenport-Halberstam for tuples of k-free numbers

The Barban-Davenport-Halberstam Theorem for tuples of k-free numbers

Dissertation

zur Erlangung des mathematisch-naturwissenschaftlichen Doktorgrades

“Doctor rerum naturalium”

der Georg-August-Universit¨ at G¨ ottingen

im Promotionsprogramm Mathematical Sciences (SMS) der Georg-August University School of Science (GAUSS)

vorgelegt von Tomos Parry aus Llanberis

G¨ ottingen 2019

Betreuungsausschuss

Prof. J¨org Br¨udern, Mathematisches Institut G¨ottingen Prof. Valentin Blomer, Mathematisches Institut Bonn

Mitglieder der Pr¨ufungskommission

Referent: Prof. J¨org Br¨udern, Mathematisches Institut G¨ottingen Koreferent: Prof. Valentin Blomer, Mathematisches Institut Bonn

Weitere Mitglieder der Pr¨ufungskommission Prof. Preda Mih˘ailescu, Mathematisches Institut G¨ottingen Prof. Thomas Schick, Mathematisches Institut G¨ottingen

Prof. Gerlind Plonka-Hoch, Institut f¨ur Numerische und Angewandte Mathematik Prof. Tatyana Krivobokova, Institut f¨ur Mathematische Stochastik

Tag der m¨undlichen Pr¨ufung: 29. Oktober 2019

i Sara

Acknowledgments

Also. Ich hab irgendwas vern¨unftiges ¨uber irgendwas schlaues geschrieben. Das liegt bestimmt nicht an mir, von daher hier einmal eine wohlverdiente Danke an die, die dazu beigetragen haben. Ganz gleich ob es viel oder wenig war.

Danke an Christoph H¨ausser, Dino Destefano und Vlad Cri¸san, f¨ur eine nette Zeit in und ausserhalb des Instituts. Vom Institut sind noch einige zu erw¨ahnen, aber besonders hervorheben m¨ochte ich Julia Brandes und vor allem Fabian Dehnhert. Ein bisschen Schade, dass ich so lange gebraucht hab fertig zu schreiben, dass ihr beide lange weg seid, aber ich weiss, eure Gesellschaft in meinen ersten Jahren hier zu sch¨atzen. Auch wenn es schon lange her ist, m¨ochte ich mich auch beim Mathestammtisch im Pirates -danke Jimmy und Ewa- und dessen Teilnehmer, unter anderem Johann Hartleb und Thomas Sicking bedanken. F¨ur vieles in meinem ersten Jahr hier auch Magdalena Kersting. Danke auch an Prof. Stefan Halverscheid. Prof. Evelina Viada bin ich besonders dankbar und ich bezweifele, ich h¨atte ohne Ihre Hilfe die Arbeit fertig schreiben k¨onnen. Danke an den DAAD f¨ur finanzielle Unterst¨utzung w¨ahrend der Arbeit.

Danke an die netten Leute vom Caf´e Balzac f¨ur die grossz¨ugige Menge an Kaffee, die mir freundlicherweise oft geschenkt wurde. Danke an Prof. Marcus du Sautoy f¨ur seine Freundlichkeit.

Danke sogar an Ars`ene Wenger :D der einen daran errinert, seinen Prinzipien treu zu bleiben, auch wenn es schwer ist.

Ben! Raul! Gabi! Aura! Danke! Thanks also to Renato!

Danke Jan-Hendrik M¨uller und Nam Rodehutskors. Jetzt kommt’s mir vor, als w¨ar’s eine Ewigkeit her, aber ich weiss noch, ich hab in der WG eine coole Zeit haben d¨urfen, und unsere -ohne Angst, dass ich ¨ubertreibe- leg- end¨are WG-Parties sind f¨ur mich wirklich sch¨one Errinerungen. Und danke Diemut, dass du mich als Mitbewohner ausgehalten hast. Das war stark.

Danke Elisa, ganz einfach f¨ur eine sch¨one Zeit.

Thanks to the awsome footballers Ocatvio Albarran and Kengo Nishi. There are probably many others from football to mention, but in particular thanks to Csaba Kib´edi.

Thanks to Rayan for general loveliness :)

Vielen Dank an Jan Sch¨utt, mit dem ich viele Schachpartien geniessen durfte, obwohl er selber fast die H¨alfte verlieren musste. Die haben mir Spass gemacht.

Danke auch an Judith - du hast eine schwierige Doktorarbeitszeit f¨ur mich wirklich sch¨on gemacht.

Marta Vega und Roberta Astolfi - absolut wunderbar.

Ringrazio molto anche a Loveleen. Avevi ragione, la strada rimane definitivamente la stessa.

Ein Vergn¨ugen ist es mir, mich bei dem anderen Verein auf den Terrassen, dem S.C. Hainberg, erkenntlich zu zeigen. Der Verein hat meine Zeit hier stark gepr¨agt und ich hab wohl mehr Personen da zu danken, als mir hier Platz zu Verf¨ugung steht. Aber sicherlich w¨urde ich gerne Arne Labitzke erw¨ahnen, der mich zum Verein gebracht hat; die Aktion Menschen (also die V.(!) Mannschaft in 2015); alle Kinder, die zwischen 2017 und 2019 bei den Bambinis mitgekickt haben; die tollen Kicker der B-Jugend 2018/19, Bernhard Marks und Denis Ische; Kevin Kahl; und Tobias Bode f¨ur seine Freundlichkeit und Freude an Fussball. Ein besonderer Dank geht an H¨annes Essers, dem ich stundenlang zuh¨oren k¨onnte. Vor allem bin ich aber der 2016-18 Truppe von Basty Titscher dankbar, die aus den folgenden wunderbaren Leuten bestanden hat: Benni Alt, Benni Uralt, High Claas, Fabi, Felix, Gerrit Green, Matze gut, Hans, Christoph Heineken, KB6, Krankenschwester Gerrit, Lasse, Jonas, Magic, Trainer Martin, Coach Mirko, Moritz, Nikolaus, Nils, Opa, Taktikfuchs Philips, Patrick, Thomas Radschunat, Thimo, Elmar Schulte, Sandro Sehner, Finanzminister Sch¨auble, und Norman Wagenhausen (gelistet nach Sch¨onheit). Als ich nach G¨ottingen kam, hab ich gedacht, ich hab am meisten an der Uni an Beigeisterung zu gewinnen, aber schliesslich hat mich diese tolle Mannschaft am meisten beeinflusst. Die kleinen Assis. Besonders starke Leistungen auf dem Platz brachten Hans Baumann, Niklas Hossbach, Lasse Bahnsen, und Benni Mecke.

Thanks to Flo, Chiara and, not to forget, Ivonne! I can’t quite imagine how my time here would have been if it wasn’t for you guys. I don’t forget your friendship especially in my first years here and later in the house. You are all wonderful people! E lo so gi`a che Leo sar`a essattamente lo stesso :)

I fwy na neb diolch i Mam a Dad. Diolch i Mam am fynd a ni ar gwylia, am fynd a fi i piano, pel droed, snwcyr a bodim arall. Diolch am fod yn neis efo pawb bob tro, ag am bob tro neud hi’n neis bod adra:) Diolch i Dad am godi’n bora i ddal pike, neud Airfix ag adeiladu castall lego. Fwy na dim diolch am dy amynadd, a ma’n neud fi’n prowd o gal bod yn fab i chdi.

Contents

Chapter 1: Number theoretical sequences in arithmetic progressions 7

Chapter 2: Proof of Theorem 1 13

2.1 - Lemmas . . . 14 2.2 - The circle method argument . . . 34 2.3 - Completion of proof . . . 49

Chapter 3: Proof of Theorem 2 58

3.1 - Lemmas . . . 58 3.2 - Proof of theorem . . . 73

Thesis Overview

Denote by π(x) the number of primes up to x and by π_x(q, a) the number of those lying in a given arithmetic progression (a+nq)_n≥0. Denote byφ(n) Euler’s phi function.

Counting prime numbers has long been a large part of number theory. Euclid already docu- mented (around 300BC) that they were infinite in number, since if not then one may be added to the product of the finitely many primes p1, ..., pR, a list of all the primes, to obtain a new number

N =p1· · ·pR+ 1

which is divisible by none of the p1, ..., pR, a contradiction. Quite some time later on, Gauss conjectured (1792/3, so in his teenage years) on the basis of his per hand calculations that up to a given x there are approximately x/logx primes, and this was confirmed a century later independently by Hadamard and Vall´ee Poussin, the method laid out by Riemann in 1859, when he connected the distribution of the primes with the complex analytic properties of his zeta functionζ(s); by Perron’s formula

X

n≤x nis a power

of a primep

logp=− 1 2πi

Z 2+i∞

2−i∞

ζ⁰(s)x^sds ζ(s)s .

Today the approximation is known to be of strength π(x) = li(x) +O

xe^−c

√logx

where

li(x) = Z x

2

du logu but there is noδ >0 for which we know

π(x) = li(x) +O x^1−δ

. (1)

In expectation that the primes be randomly distributed there should be nothing better than

“square root cancellation” and indeed Littlewood showed (1914) π(x) = li(x) + Ω

√

xlog log logx logx

.

Establishing (1) for someδ∈(0,1/2] is in the number theory community accepted as out of reach (the validity ofδ= 1/2, up to logarithms, amounts to the Riemann Hypothesis), and therefore we settle for “average type” results, where this averaging is done over arithmetic progressions.

The first result in this context is a famous theorem by Dirichlet which asserts, for given coprime

qanda, that there are infinitely many primes congruent toamoduloq; this he established with the introduction of his characters

X

n≤x n≡a(q) nis a power

of a primep

logp= 1 φ(q)

X

χ

χ(a) X

n≤x nis a power

of a primep

χ(n) logp

and the analytic properties of their correspondingL-functions

∞

X

n=1

χ(n) n^s .

In fact for any A > 0, some c = c(A) > 0 and any q ≤ (logx)^A we have the Siegel-Walfisz Theorem

π_x(q, a) = li(x) φ(q) +OA

xe^−c

√logx .

The problem of counting primes restricted to a given arithmetic progression no easier than counting all of them, the above discussion about the size of the error term still applies, however now we can ask what happens when we varyqanda. Define

Ex(q, a) = X

n≤x nis a power

of a primep

logp− x φ(q).

One of the major theorems in 20th century number theory is the Bombieri-Vinogradov Theorem (1965):

X

q≤Q

1≤a≤qmax

(a,q)=1

maxy≤x|Ey(q, a)| √

xQ(logx)⁵, for anyA >0 and √

x/(logx)^A ≤Q≤√

x. Shortly after (1966) came the Barban-Davenport- Halberstam Theorem:

X

q≤Q q

X

a=1 (a,q)=1

|Ex(q, a)|²xQlogx+ x² (logx)^A

for Q≤x. This was improved to an asymptotic formula by Montgomery and Hooley (1970s):

for anyA >0 and somec∈R X

q≤Q q

X

a=1 (a,q)=1

|Ex(q, a)|²=xQlogQ+cxQ+O

x^3/4Q^5/4+ x² (logx)^A

. (2)

These results are important because they are suggested by the truth of Ex(q, a)≈

rx q.

Yet more insight into the primes might be gained by looking at their arithmetical patterns, say tuples n+h₁, ..., n+h_r, for some fixed non-negative integers 0 ≤ h₁ < h₂ < · · · < h_r, such that then+h_i are all prime. But unfortunately this strategy already fails at Euclid’s Theorem,

since it isn’t (yet) known that there are infinitely many pairs of primesn, n+ 2, never mind the Hardy-Littlewood Conjecture - this states that for someSh∈R

X

n≤x n+hiall prime

1∼ Shx (logx)^r.

It is perhaps a natural question whether such motives are present when we discuss sequences other than the primes. Let k ≥2. If for a given nthere is no prime p for whichp^k|n then n is said to bek-free. An asymptotic formula of similar shape to that in (2) is indeed known for thek-free numbers, the current state of knowledge attained and summarised by Vaughan in [17].

Crucially, the corresponding questions on r-tuples of k-free numbers are accessible; for given non-negative integers 0≤h1 < h2 <· · ·< hr we call n+h1, ..., , n+hr a k-freer-tuple if the n+hi are allk-free. The asymptotic count

X

n≤x

µ_k(n+h₁)· · ·µ_k(n+h_r) =S_hx+O

x^2/(k+1)+

,

for some S_h ∈ R, is easily established (see [10]) and we look at the distribution in arithmetic progressions. In [2] twins in arithmetic progressions were investigated and it was shown easily that

X

n≤x n≡a(q)

µk(n)µk(n+ 1) =η(q, a)x+O

x^2/(k+1)+

.

for someη(q, a)∈R, and that

V(x, Q) := X

q≤Q q

X

a=1

X

n≤x n≡a(q)

µ_k(n)µ_k(n+ 1)−η(q, a)x

2

Q² x

Q ^2/k

.

In [9] the method of Vaughan (that in [17]) is followed to show that V(x, Q)Q²

x Q

^1/k+

+x^1+2/klogQ+x^3/2+1/2k+. These results are important because the same results for primes are out of reach.

As far as we can see, however, there is no recorded asymptotic formula for the variance of twins ofk-free numbers. The main result of this thesis is an asymptotic formula for the variance

X

q≤Q q

X

a=1

X

n≤x n+hi k-free

n≡a(q)

1−η_h(q, a)x

2

ofk-freer-tuples; hereηh(q, a) is a suitable main term. We will state the precise result shortly and it is the content of Chapter 2. It seems to be the first instance of an asymptotic formula for this variance. We follow closely Vaughan’s argument in [17].

The above variance is somtimes referred to as the “improper” variance. We will also look at the “proper” variance

q

X

a=1

X

n≤x n k-free n≡a(q)

1−η(q, a)x

2

of k-free numbers; that is, averaging just over the residues. An asymptotic formula was found (when averaging over the reduced classes) for squarefrees by Ramon Nunes [12], and we improve one of the error terms (when averaging over all the classes). This result is the content of Chapter 3.

We now state these results precisely.

Theorem 1. Fix non-negative integers0≤h1< h2<· · ·< hr and let

S={n∈N|n+hi ∈ {k-frees}, i= 1, ..., r}. (3) Let for q, a∈NandQ, x >0

η(q, a) =

∞

X

d1,...,dr=1 (dki,dkj)|hi−hj(1≤i,j≤r)

(q,dki)|a+hi(1≤i≤r)

µ(d1)· · ·µ(dr)

[q, d^k₁, ..., d^k_r] , Ex(q, a) = X

n≤x n∈S n≡a(q)

1−xη(q, a) (4)

and

V(x, Q) =X

q≤Q q

X

a=1

Ex(q, a)². (5)

Takec=c(r)to be any number in[1/2,1)for which we know Z ∞

−∞

|ζ(σ+it)|^rdt (1 +|t|)^3/2

converges absolutely for allσ≥c. For each primepwriteRp for the number of different residues represented by theh1, ..., hr modulop^k. If alwaysRp< p^k then for Q≤xand >0

V(x, Q) =Q² x

Q ^1/k

P

log(x/Q)

+Ok,r,h, Q² x

Q ^c(r)/k

+x^1+2/(k+1)+

!

whereP =P(r, k,h)is a polynomial of degreer−1.

As already mentioned the only theorems in this direction, that we know of, are upper bound result for twins of squarefree numbers. In this case we can takec= 1/2 since it is contained in classical results that forσ≥1/2

Z 2T T

|ζ(σ+it)|²dt_σTlogT and so our theorem then implies

V(x, Q) =x^1/2Q^3/2P

log(x/Q) +O

x^1/4Q^7/4+x^5/3+

for some linear functionP. Of course if theh_i cover a complete residue system modulo somep^k then there are nok-freer-tuples.

Theorem 2. Let k≥2 and denote byS the set of k-free numbers. Forq, a∈N define η(q, a) =

∞

X

d=1 (q,dk)|a

µ(d)

[q, d^k], Ex(q, a) = X

n≤x n∈S n≡a(q)

1−xη(q, a) (6)

and forx >0

V_x(q) =

q

X

a=1

|E_x(q, a)|². (7)

Define

C_k = 2k (1/k−1)ζ(2)

Y

p

1−2/(p^k+p^k−1) 1−p^1−1/k and

fk(q) =Ck

Y

p|q

1−2/p^k+ (q, p^k)^1/k−1/p 1−2/p^k+ 1/p . For largexand forq≤xwe have for every >0

Vx(q) =q x

q 1/k

fk(q) +Ok, x q x

q

2/(9−2/k)

+x^1+2/(k+1) q

!!

.

This is an asymptotic formula for k = 2,3,4. Averaging just over the reduced classes an asymptotic formula for Vx(q), in the squarefree case, is already established in [12] with error essentially

q x

q ^1/3

+ x

q ^23/15

.

Before this only upper bound results are recorded (see [8] and the references therein), although these are stronger in the range where the above asymptotic formulas don’t hold and are concerned with more general sequences than the squarefrees. The relevance of our result is the improvement in the first error term, which fork= 2 seems decently small. This is obtained by a careful analysis of the integrals arising from an application of Perron’s formula. (Our second error term is weaker than in [12] but most likely can be made to be just as small for the squarefress by arguing, as in that paper, with the square sieve.)

Chapter 1

Number theoretical sequences in arithmetic progressions

By a “number theoretical sequence” we are being rather vague - we mean simply a sequence that would be of interest to a number theorist. For clarity, let’s just say a number theoretical sequence is a subset ofN.

LetN denote a number theoretical sequence and supposeN satisfies for someη:N×N→C X

n≤x n∈N n≡a(q)

1 =η(q, a)x+o(x) (8)

with x→ ∞. Of course, the question of uniformity in q and a is an important one but, as a minimum, we should require that the above is true for each fixedqandaif we are to study the average behaviour over arithmetic progressions. Define

Ex(q, a) = X

n≤x n∈N n≡a(q)

1−η(q, a)x and V(x, Q) =X

q≤Q q

X

a=1

|Ex(q, a)|².

We have already alluded to the relevance of the Barban-Davenport-Halberstam (BDH) Theorem and to the question of when N obeys a BDH type law. Hooley asks this question in [5] and answers

V(x, Q)AxQ+ x² (logx)^A for eachA >0, whenever

Ex(q, a)A

x (logx)^A andη satisfies

η(q, a) =η(q,(q, a)) ; (9)

in [7] he removes this condition on η. Of course, should a better approximation in arithmetic progressions hold, this would be reflected in the second error term. As with (the shorter proofs of) the Bombieri-Vinogradov Theorem, Hooley’s result was made possible with the large sieve:

if we define

W(α) = X

n≤x n∈N

e(αn) then the Large Sieve Inequality says

X

q≤Q q

X⁰

a=1

|W(a/q)|²x

x+Q²

; (10)

here and in what follows we write Σ^0q to mean the sum is taken over a= 1, ..., q coprime to q.

For illustrative purposes we present a weak form of his argument in the stronger setting where N satisfies

E_x(q, a)x^∆. (11)

AssumeQ≤xandQis large. By orthogonality E_x(q, a) = 1

q

q

X

b=1

e

−ab q

W(b/q)−x

q

X

N=1

η(q, N)e N b

q !

=: 1 q

q

X

b=1

e

−ab q

X_q,b so that

q

X

a=1

|Ex(q, a)|²= 1 q

q

X

b=1

|Xq,b|² and therefore

V(x, Q) = X

q≤Q

1 q

X

d|q q

X

b=1 (b,q)=d

|Xq,b|²

= X

q≤Q

1 q

X

d|q d

X⁰

b=1

|Xd,b|²

logQX

d≤Q

1 d

d

X⁰

b=1

|X_d,b|²

(logQ)²max

R≤Q



 1 R

X

R≤q<2R q

X⁰

b=1

|Xq,b|²





(logQ)²max

R≤Q

1 R

X

R≤q<2R q

X⁰

b=1



|W(b/q)|²+ x

q

X

N=1

η(q, N)e N b

q

2



=: (logQ)²max

R≤Q

ΥR

R . (12)

By (9)

q

X⁰

b=1

q

X

N=1

η(q, N)e N b

q

2

=

q

X

N,N⁰=1

η(q, N)η(q, N⁰)c_q(N−N⁰)

= X

d,d⁰|q

η(q, d)η(q, d⁰)

q/d

X⁰

N=1 q/d⁰

X⁰

N⁰=1

cq(N d−N⁰d⁰) and here theN, N⁰ sum is

q

X⁰

a=1

c_q/d(a)c_q/d⁰(−a) =µ(q/d)µ(q/d⁰)φ(q)q so that

q

X⁰

b=1

q

X

N=1

η(q, N)e N b

q

2

q X

d,d⁰|q

|η(q, d)η(q, d⁰)|. (13)

Clearly

X

n≤x n∈N n≡a(q)

1 x q so

η(q, a) = lim

x→∞

1 x

X

n≤x n∈N n≡a(q)

1 1 q therefore (13) becomes

q

X⁰

b=1

q

X

N=1

η(q, N)e N b

q

2

q⁻¹. (14)

This with (10) says

ΥRx(x+R²) +x²X

q≤R

q⁻¹xR²+x²⁺; on the other hand (11) implies

ΥR X

R≤q<2R q

X

a=1

|Ex(q, a)|²x^2∆R²

and therefore (12) says

V(x, Q)xmax

R≤Q

min

xR+x² R, x^2∆R

.

ForR≤x^1−∆ the minimum is≤x^1+∆ and for x^1−∆< R≤Qthe minimum is ≤xQ+x^1+∆, therefore

V(x, Q)x

xQ+x^1+∆

which is a weak result of BDH type.

In [6] Hooley refines his BDH bound to an asymptotic formula on the further assumption that η(q, q) is essentially multiplicative, that is there isρ ∈R^≥ and multiplicative f :N →R^≥ for which

η(q, q) =ρf(q)

and we make a few comments as to the proof. Opening the square in the definition ofV(x, Q) we see that the difficulty lies in evaluating sums of type

S(x, Q) := X

Q<q≤x

X

n0<n≤x n≡n0(q) n,n0 ∈N

1

which with a “modulus swap” becomes X

l≤x/Q

X

Q<(n−n0)/l≤x n≡n0(l) n,n0 ∈N

1 = X

l≤x/Q

X

δ|l l

X

b=1 (b,l)=δ

X

n0 ≤x−Ql n0 ∈N n0 ≡b(l)

X

n0+Ql<n≤x n∈N n≡b(l)

1

the point being the modulus is now small. In view of (8) and (9) we expect X

X<n≤Y n∈N n≡a(q)

1≈η(q,(q, a)) (Y −X)

so that the sums overn⁰ andnabove are

≈η(l, δ) X

n0 ≤x−Ql n0 ∈N n0 ≡b(l)

x−Ql−n⁰

≈η(l, δ) Z x−Ql

0









 X

n0 ≤t n0 ∈N n0 ≡b(l)

1











dt≈ η(l, δ)²(x−Ql)² 2

and therefore

S(x, Q) ≈ 1 2

X

l≤x/Q

x−Ql²X

δ|l

η(l, δ)²

l

X

b=1 (b,l)=δ

1

=: 1 2

X

l≤x/Q

x−Ql²

M(l). (15)

An important consequence of (9) is

η(l, δ) = 1 φ(l/δ)

X

r|l/δ

µ(r)η(r, r)

(this can be shown in a few lines) which with the multiplicativity assumption means that the Dirichlet series ofM(l) is associated to an Euler product and so we can apply Perron’s formula to (15). Hooley obtains with this argument

V(x, Q) =

c+o(1)

xQ+O_A x²

(logx)^A

(16) for some constantc and whereo(1) →0 with x/Q→ ∞. An alternative approach to Hooley’s was developed by Goldston and Vaughan (see [3]) through which the variance is expressed as a binary additive problem tractable by the circle method. In this case the approach to deal with sums of type

X

q≤Q

X

n0<n≤x n,n0 ∈N n≡n0(q)

1

is to write the double sum as

X

q≤Q

X

l≤x/q

X

n,n0 ≤x n−n0=ql

n,n0 ∈N

1

and then to apply the circle method to the inner sum. In [16] a general result is obtained in this way, again under (9) of course.

A subtle point is that the constant (16) may be zero. As seen from the BDH Theorem this is not the case for the primes, but as seen from [17] this is the case for the k-free numbers.

Br¨udern has a theory for this principle (see [1]) which characterises those sequences with smaller variances as thoselimit periodic sequences. On the other hand, since we expect for the k-free numbers

E_x(q, a) x

q ^1/2k

, we should also expect that in this casec= 0 and moreover that

V(x, Q)Q² x

Q 1/k

.

In fact an asymptotic formula for the k-free variance is long known and a strong asymptotic formula is given in [17], as already mentioned. As far as strong asymptotics in arithmetic pro- gressions are concerned, the k-free numbers are indeed easier than the primes, the k-frees in a vague sense perhaps approximating the primes with decreasingk.

In summary, the variance in arithmetic progressions of both the primes and the k-frees are particularly well understood, and there is a theory for sequences in general. Once we look at r-tuples, however, we no longer have (9). In this case the situation for primes is hopeless, but we can at least address thek-frees in this thesis.

Chapter 2

Proof of Theorem 1

In [17] the evaluation of the variance is translated into a binary additive problem in k-free numbers which can be tackled with the circle method. We use the method laid out there for our proof except at the very end, worried that ther-tuples not being multiplicative would cause some problems otherwise.

Write µk for the indicator function of the k-frees and write g = gk for the unique function satisfying

µk=g ?1; (17)

specifically

g(d) =

µ(d^1/k) ifdis a kth power

0 if not (18)

since

µ_k(n) =X

d^k|n

µ(d) (19)

is a well known identity. Then from (4) η(q, a) =

∞

X

d1,...,dr=1 (di,dj)|hi−hj(1≤i,j≤r)

(q,di)|a+hi (1≤i≤r)

g(d₁)· · ·g(d_r)

[q, d1, ..., dr] . (20)

Throughout this chapter the implied constants in theOsymbol will always be understood to be dependent on k, r,hand , and may be taken to be arbitrarily small at each of its occrences.

Whenevers, σ andtappear in the same context we will always mean a complex numberswith real and imaginary partsσandt. For real positive numberscandT we often write the shorthand

Z

±T

or Z

c±iT

or Z

c±i∞

for

Z T

−T

or Z c+iT

c−iT

or

Z c+i∞

c−i∞

.

We will write statements that involve r-tuples using vectors and mean that that statement is to hold for each vector component. For example, ν ≡ d(q) would mean ν ≡ d_i(q_i) for each i = 1, ..., R, where the R, d_i, q_i would be clear from context. A sum Σ^0q will mean that the summation variables are restricted to numbers coprime to q. The R-fold divisor estimate d_R(n)n is well known, as is the (General) Chinese Remainder Theorem which says

n≡amod (q)

has exactly one solution modulo [q₁, ..., q_R] if (q_i, q_j)|a_i −a_j and has no solutions otherwise.

We will use both these facts frequently but often forget to mention where they come from. A

coprimality condition may often disappear from one line to the next with the introduction of the M¨obius function; here we are using

X

d|n

µ(d) =

1 ifn= 1 0 ifn6= 1.

2.1 - Lemmas

Our theorem rests on having a concrete expression for the Gauss sums associated with thek-frees.

The difficulties in the transition tor-tuples is contained in the following lemma.

Lemma 1. Fora, q∈Ndefine

G(q, a) =

q

X

ν=1

e aν

q

η(q, ν).

For each primepand each qa non-negative power ofpdefine g_p(q) =

1/p^k ifq|p^k 0 ifp^k+1|q,

For each primepwriteRp for the number of different residue classes modulop^k represented by the h1, ..., hr. If for all primes pwe have Rp< p^k define for q∈N

G(q) = Y

p^β||q

g_p(p^β) 1−Rp/p^k.

For each primepwriteH₁, ..., H_Rp for the different residue classes modulop^k represented by the h₁, ..., h_r. For A∈NandP a power of pdefine forP|p^k

Ω_A(P) =−

R_p

X

n=1

e

−AH_n P

and forp^k+1|P define the LHS to be zero. ForQ, A∈Ndefine H(Q, A) =

ω

Y

i=1

Ω_AQ/Q

i(Q_i),

whereQ=Q1· · ·Qω withQi the distinct prime powers ofQand eachQ/Qi the inverse ofQ/Qi

modQi; ifQ= 1 the product is to be read as 1. Define ρ=Y

p

1−Rp

p^k

.

If Rp< p^k for allpthen for any a, q∈N G(q, a) =ρG

q (q, a)

H

q (q, a), a

(q, a)

.

Proof. We first set quite a bit of notation. For givend1, ..., dr writed^∗= [d1, ..., dr] andg(d) = g(d1)···g(dr), wheregis as in (17). By a sum Σ^∗over variablesd1, ..., drwe will mean always that for alli, j we have (di, dj)|hi−hj. For givenQ, d1, ..., dr∈NwithQ|d^∗and (di, dj)|hi−hj write νQ,d for the unique solution modulo [(Q, d1), ...,(Q, dr)] = (Q, d^∗) =Q to the systemν ≡ −h mod ((Q,d)). ForQ, A, N ∈NwithQ|N define

g_A,Q^∗ (N) = X^∗

d1,...,dr d∗=N

g(d)e

AνQ,d

Q

so that

∞

X

N=1 Q|N

g_A,Q^∗ (N)

N = X^∗

d1,...,dr Q|d∗

g(d) d^∗ e

Aν_Q,d Q

, (21)

and forP a non-negative power of a primepdefine g^∗_A,p(P) := X

t≥0 P|pt

g^∗_A,P(p^t)

p^t = X^∗

d1,...,dr P|d∗ dnon-negative

powers ofp

g(d) d^∗ e

AνP,d

P

. (22)

Forω, D⁰₁, ..., D⁰_r, D¹₁, ..., D_r¹, ..., D₁^ω, ..., D^ω_r ∈NwriteDj = (D₁^j, ..., D_r^j)∈N^r, writeD0· · ·Dr= (D₁⁰· · ·D^ω₁, ..., D⁰_r· · ·D^ω_r)∈N^r, and writeD^∗_j = [D₁^j, ..., D_r^j]. Now we can start the proof proper.

TakeQ, D₁⁰, ..., D^ω_r ∈N, whereQ=Q1· · ·Qωas a product of prime powers, where for 1≤j≤ω eachD^j_i contains only primes ofQj, and where eachD⁰_i is coprime toQ. SupposeQj|D^∗_j. Then we have moduloQ

ν_Q,D0···Dr ≡ ν_Q,D1···Dr

≡ Q/Q1(Q/Q1)νQ₁,D₁+· · ·+Q/Qω(Q/Qω)νQ_ω,D_ω

so that forN, A∈N, withN =N0N1· · ·Nωwhere theNiare the prime powers ofN and where (Q, N0) = 1 andQi|Ni, we have

g^∗_A,Q(N) = X^∗

d1,...,dr d∗=N

g(d1)· · ·g(dr)e

Aν_Q,d Q

=

ω

X

j=0

X^∗

Dj 0,...,Dj

r D∗

j=Nj

g D⁰₁· · ·D^ω₁

· · ·g D⁰_r· · ·D_r^ω e

AνQ,D₀···Dω

Q

=







 X^∗

D0 1,...,D0

r D∗

0=N0

g(D⁰₁)· · ·g(D_r⁰)









ω

Y

j=1









 X^∗

Dj 1,...,Dj

r D∗

j=Nj

g(D₁^j)· · ·g(D_r^j)e AQ/QjνQ_j,D_j

Qj

!











= g_1,1^∗ (N₀)

ω

Y

j=1

g^∗

AQ/Q_j,Q_j(N_j)

and therefore

∞

X

N=1 Q|N

g^∗_A,Q(N)

N = Y

p6|Q



 X

t≥0

g^∗_1,1(p^t) p^t





ω

Y

j=1







 X

t≥0 Qj|ptj

g^∗

AQ/Q_j,Q_j(p^t_j) p^t









= Y

p6|Q

g^∗_1,p(1)

ω

Y

j=1

g^∗_AQ/Q

j,p_j(Qj), (23)

where obviously thepj are the primes of Qj; ifQ= 1 this all goes through easier and gives the same result as long as we read the finite product as 1. We now evaluateg^∗_A,p(P) on non-negative powersP of a primepand on anyA∈Ncoprime top. Forda non-negative power ofpwe have from (18)

g(d) =







1 ifd= 1

−1 ifd=p^k 0 otherwise so from (22)

g^∗_A,p(P) = X

X⊆{1,...,r}

(−1)^|X| X^∗

d1,...,dr P|d∗ dnon-negative

powers of p di=pkifi∈X

di=1 ifi6∈X

1 d^∗e

AνP,d

P

= X

d P|d∗ dnon-negative

powers of p alldi=1

1 d^∗e

AνP,d

P

+

R_p

X

n=1

e

−AHn

P

X

X⊆{1,...,r}

hi≡Hn(pk)∀i∈X X6=∅

(−1)^|X| X

d P|d∗ dnon-negative

powers of p di=pkifi∈X

di=1 ifi6∈X

1

d^∗. (24)

Note that

X

N ⊆{1,...,n}

(−1)^{|N |} =

n

X

i=0

(−1)ⁱ n

i

= (1−1)ⁿ= 0 so that

X

N ⊆{1,...,n}

N 6=∅

(−1)^{|N |} =−1 and therefore

X

N ⊆{1,...,r}

hi≡Hn(pk)∀i∈N N 6=∅

(−1)^{|N |}=−1. (25)

If p^k+1|P there are no terms in (24). If P|p^k the non-empty X (i.e. those with d^∗ = p^k) contribute

1 p^k

R_p

X

n=1

e

−AH_n P

X

X⊆{1,...,r}

hi≡Hn(pk)∀i∈X X6=∅

(−1)^|X|=p^−k

Ω_A(P) ifP 6= 1

−R_p ifP = 1

from (25), and the empty set term (i.e. d^∗= 1) only gives a contribution ifP = 1, in which case it contributes 1. Consequently for (A, P) = 1

g^∗_A,p(P) =







1−p^−kRp ifP = 1 p^−kΩA(P) if 16=P|p^k

0 ifp^k+1|P

=

1−p^−kRp ifP = 1 gp(P)ΩA(P) ifP 6= 1 and therefore from (23) we have for (A, Q) = 1

∞

X

N=1 Q|N

g_A,Q^∗ (N)

N = Y

p6|Q

1−p^−kR_p

! _ω Y

j=1

g_pj(Q_j)Ω_AQ/Q

j(Q_j)

! .

so from (21) X^∗

d1,...,dr Q|d∗

g(d) d^∗ e

AνQ,d

Q

= Y

p

1−Rp

p^k

! _ω Y

j=1

gp_j(Qj)Ω_AQ/Q

j(Qj) 1−Rp_j/p^k_j

= ρG(Q)H(Q, A) (26)

for any Q, A ∈ N with (A, Q) = 1. We now take Q = q/(q, a) and A = a/(q, a). Since for (di, dj)|hi−hj we have

q

X

ν=1 (q,d)|ν+h

e aν

q

= q

[(q, d1), ...,(q, dr)]

e(aνq,d/q) ifq/[(q, d1), ...,(q, dr)]

a

0 if not

= q

(q, d^∗)

e(AνQ,d/Q) ifQ|d^∗

0 if not

we see from (20) and (26) G(q, a) =q X

d1,...,dr (di,dj)|hi−hj

Q|d∗

g(d) [q, d^∗](q, d^∗)e

Aν_Q,d Q

=ρG(Q)H(Q, A).

This H(q, a) function of the previous lemma is special to the r-tuple case. We deal now with its main properties. When comparing with [17] it is perhaps useful to think of Φ_q(n) as Ramanujan’s sum.

Lemma 2. DefineH as in Lemma 1. Define forq, n∈N Φq(n) =

q

X⁰

a=1

|H(q, a)|²e an

q

, Φ^∗_q(n) =

q

X⁰

a=1

H(q, a)e an

q

and Φ(q) = Φq(0).

(i) BothΦ_q(n) andΦ^∗_q(n)are, for each n, multiplicative inq.

(ii) Suppose we have a function F(q, d) defined for q ∈ N and d|q. If for all (q, q⁰) = 1 and d|q, d⁰|q⁰ we haveF(qq⁰, dd⁰) =F(q, d)F(q⁰, d⁰)then the sum

q

X

A=1

F

q,(q, A) Φq(A) is multiplcative in q.

(iii) Forq a power of a prime and ford|q

q/d

X⁰

A=1

Φ_q(−Ad) = Φ(q)µ(q/d).

(iv) For anyq∈N

q

X

A=1

|Φq(A)| q¹⁺

and the same claim holds withΦq(A)replaced byΦ^∗_q(A).

(v) LetG(·)andρbe as in Lemma 1. For anyq∈N

q

X

a=1

η(q, a)²= ρ² q

X

d|q

Φ(d)G(d)².

(vi) For anyq, a∈N

H(q, a)q.

Proof. For givend₁, ..., d_r write always d^∗ = [d₁, ..., d_r] and g(d) = g(d₁)· · ·g(d_r), where g is as in (17). By a sum Σ^∗ over variables d₁, ..., d_r we will mean always that for all i, j we have (d_i, d_j)|h_i−h_j. ForP a power of a primepandA∈Nwe have

H(P, A) =−

Rp

X

n=1

e

−AHn

P

1 forP|p^k

0 forp^k+1|P (27)

whereH1, ..., HR_p are the distinct residues represented byh1, ..., hrmodulop^k.

(i) Takeq, q⁰∈Nwith (q, q⁰) = 1 and, for any primep, defineβ, β⁰ ≥0 throughp^β||qandp^β0||q⁰. For a primepdenote byRthe inverse of anyRmodulop^β+β0. Thenq⁰qq⁰/p^β+β0 ≡q/p^β mod (p^β) for anyp|q. A similar comment applies when consideringp|q⁰ so that for anya, a⁰∈N

H(qq⁰, aq⁰+a⁰q) = Y

p|qq⁰



−

R_p

X

n=1

e (aq⁰+a⁰q)qq⁰/p^β+β0H_n p^β+β0

!



= Y

p|q



−

R_p

X

n=1

e aq/p^βHn

p^β

!

 Y

p|q⁰



−

R_p

X

n=1

e a⁰q⁰/p^β0Hn

p^β0

!



= H(q, a)H(q⁰, a⁰); (28)

here we are assumingP|p^kbut if not the equality is trivial. Asa, a⁰respectively run over reduced residue systems modulo q, q⁰ the quantity aq⁰+a⁰q runs over a reduced residue sytem modulo qq⁰, therefore (28) implies

Φqq⁰(n) =

qq⁰

X⁰

A=1

|H(qq⁰, A)|²e An

qq⁰

=

q

X⁰

a=1 q⁰

X⁰

a⁰=1

|H(qq⁰, aq⁰+a⁰q)|²e

(aq⁰+a⁰q)n qq⁰

=

q

X⁰

a=1

|H(q, a)|²e an

q !



q⁰

X⁰

a⁰=1

|H(q⁰, a⁰)|²e a⁰n

q⁰





= Φq(n) Φq⁰(n) and a similar proof obviously works for Φ^∗_q(n).

(ii) SupposeM, M⁰∈Nwith (M, M⁰) = 1 so that F

M M⁰,(M M⁰, aM⁰+a⁰M)

=F

M,(M, a) F

M⁰,(M⁰, a⁰)

. (29)

Asaanda⁰run over complete residue system moduloM andM⁰respectively then so doaM⁰and a⁰M and the quantityaM⁰+a⁰M runs over a complete residue system moduloM M⁰. Therefore by (29) and part (i)

M M⁰

X

A=1

F

M M⁰,(M M⁰, A)

ΦM M⁰(A)

=

M

X

a=1 M⁰

X

a⁰=1

F

M M⁰,(M M⁰, aM⁰+a⁰M)

ΦM M⁰(aM⁰+a⁰M)

=

M

X

a=1 M⁰

X

a⁰=1

F

M,(M, a) F

M⁰,(M⁰, a⁰)

ΦM(aM⁰+a⁰M) ΦM⁰(a⁰M+aM⁰)

=

M

X

a=1 M⁰

X

a⁰=1

F

M,(M, aM⁰) F

M⁰,(M⁰, a⁰M)

ΦM(a) ΦM⁰(a⁰) and the claim follows.

(iii) The equality is trivial ifp^k+1|q, otherwise forq|p^k we have from (27) Φ_q(n) =

R_p

X

n,n⁰=1 q

X⁰

a=1

e

a(Hn−Hn⁰+n) q

(30)

so that

q/d

X⁰

A=1

Φq(−Ad) =

Rp

X

n,n⁰=1 q

X⁰

a=1

e

a(Hn−Hn⁰) q

^q/d X⁰

A=1

e

−aAd q

= µ(q/d)

R_p

X

n,n⁰=1 q

X⁰

a=1

e

a(Hn−Hn⁰) q

= µ(q/d)Φq(0) from (30).

(iv) The same argument as in part (ii) shows the sum in qustion to be multiplicative so it is enough to prove the bound forqa prime power. In that case (30) implies

Φ_q(n) =

R_p

X

n,n⁰=1

c_q(H_n−H_n⁰+n) so that

q

X

n=1

|Φ_q(n)|

q

X

n=1

|c_q(n)| q¹⁺

by a standard bound for Ramanujan’s sum and the proof is obviously the same for Φ^∗_q(n).

(v) From part (i)

d

X⁰

b=1

H(d, b)e −ab

d

is, for eacha∈N, multiplicative indand therefore so is η^∗(q, a) :=

q

X

b=1

G q

(q, b)

H q

(q, b), b (q, b)

e

−ab q

= X

d|q

G(d)

d

X⁰

b=1

H(d, b)e −ab

d

. As in (ii) it can then be shown that

q

X

a=1

|η^∗(q, a)|² is mutiplicative.

Supposeqis a prime power so that ford|q, p^k we have from (27)

d

X⁰

b=1

H(d, b)e −ab

d

=−

R_p

X

n=1

c_d(H_n−a).

and therefore (sinceGvanishes forp^k+1|d) η^∗(q, a) =−

R_p

X

n=1

X

d|q,p^k

G(d)cd(Hn−a)

so that

q

X

a=1

|η^∗(q, a)|² =

Rp

X

n,n⁰=1

X

d,d⁰|q

G(d)G(d⁰)

q

X

a=1

cd(Hn−a)cd⁰(a−Hn⁰)

= q

R_p

X

n,n⁰=1

X

d,d⁰|q

G(d)G(d⁰) [d, d⁰]

[d,d⁰]

X

a=1

cd(Hn−a)cd⁰(a−Hn⁰). (31) Then

[d,d⁰]

X

a=1

cd(Hn−a)cd⁰(a−Hn⁰)

=

d

X⁰

A=1 d⁰

X⁰

A⁰=1

e AH_n

d −A⁰H_n⁰ d⁰

^[d,d

0]

X

a=1

e

a(−A[d, d⁰]/d+A⁰[d, d⁰]/d⁰) [d, d⁰]

= [d, d⁰]

d

X⁰

A=1

d⁰

X⁰

A0=1

[d,d0]|−A[d,d0]/d+A0[d,d0]/d0

e AHn

d −A⁰Hn⁰

d⁰

but the only (prime power) d, d⁰ which can satisfy these summation conditions are those with d=d⁰, in which case theA, A⁰ sum becomes

d

X⁰

A=1

e

A(Hn−Hn⁰) d

=cd(Hn−Hn⁰) and so

1 [d, d⁰]

[d,d⁰]

X

a=1

c_d(H_n−a)c_d⁰(H_n⁰−a) =

cd(Hn−Hn⁰) ifd=d⁰

0 if not.

Therefore (31) says

q

X

a=1

|η^∗(q, a)|² = q

Rp

X

n,n⁰=1

X

d|q

|G(d)|²cd(Hn−Hn⁰)

= qX

d|q

|G(d)|²

q

X⁰

a=1

−

R_p

X

n=1

e aHn

d

2

= qX

d|q

Φ(d)|G(d)|². (32)

from (28). This holds initially only forqa prime power, but we’ve already said that the LHS is multiplicative and the RHS is multiplicative from part (i), so (32) holds for generalq. LetG(·,·) be as in Lemma 1. From orthogonality and then Lemma 1

η(q, a) = 1 q

q

X

b=1

G(q, b)e −ab

q

= ρη^∗(q, a) q