We prove that for all positive integers r R

(1)

The ternary Goldba h problem in arithmeti progressions

by

Jianya Liu and Tao Zhan (Jinan)

ForalargeoddintegerN and apositiveintegerr;deneb=(b

1

;b

2

;b

3 )

and

B(N;r)=fb2N 3

:1b

j

r;(b

j

;r)=1and b

1 +b

2 +b

3

N (modr)g:

It isknownthat

#B(N;r)=r 2

Y

pjr

pjN

(p 1)(p 2)

p 2

Y

pjr

p-N p

2

3p+3

p 2

:

Let "> 0 be arbitrary and R = N 1=8 "

: We prove that for all positive

integers r R ; with at most O(Rlog A

N) ex eptions, the Diophantine

equation

N =p

1 +p

2 +p

3

;

p

j

b

j

(mod r); j=1;2;3;

with prime variables is solvable whenever b 2 B(N;r); where A > 0 is

arbitrary.

1. Introdu tion and statement of results. For given odd integers

N weshall be on ernedwith thesolubilityof theequation

(1:1) N =p

1 +p

2 +p

3

inprimevariablesp

j

;thisisknownastheternaryGoldba h problem. Hardy

andLittlewood[HL℄provedin1923 thatsubje ttothegeneralizedRiemann

hypothesis (GRH hereafter) thenumberJ(N) of solutionsof (1.1) satises

1991 Mathemati sSubje tClassi ation: 11P32,11L07.

Key wordsandphrases: ternary Goldba hproblem,exponentialsumoverprimesin

arithmeti progressions,mean-valuetheorem.

Proje t supported by the Trans-Century Training Programme Foundation for the

TalentsbytheStateEdu ationCommissionandtheNationalNaturalS ien eFoundation

ofChina.

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an asymptoti formula

(1:2) J(N)=(N)

N 2

2log 3

N

(1+o(1)):

Here(N) isthesingularseries,andone has(N)1foroddN. In 1937

Vinogradov [Vi℄ obtained for the rst time a nontrivial estimate of expo-

nentialsumsoverprimes, andmanaged to establish(1.2) un onditionally.

Sin e 1923, many authors have onsidered the orresponding problems

withrestri tive onditionsposedon thethree primevariablesin(1.1). One

ofthesegeneralizationswasgivenbyRadema her[R℄in1926. Forapositive

integer r;deneb=(b

1

;b

2

;b

3 )and

(1:3) B(N;r)

=fb2N 3

:1b

j

r; (b

j

;r)=1and b

1 +b

2 +b

3

N (modr)g:

Then,a ording to Liuand Tsang[LT℄,

(1:4) #B(N;r)=r 2

Y

pjr

pjN

(p 1)(p 2)

p 2

Y

pjr

p-N p

2

3p+3

p 2

:

Radema her[R℄showed,subje ttoGRH,thatifr isaxedpositiveinteger,

and J(N;r;b)the numberof solutionsof theequation

(1:5)

N =p

1 +p

2 +p

3

;

p

j

b

j

(mod r); j=1;2;3;

thenwehave, forodd N and all b2B(N;r);

(1:6) J(N;r;b)=(N;r) N

2

2log 3

N

(1+o(1));

and thesingularseries(N;r) satises

(N;r)= C(r)

r 2

Y

pjr p

3

(p 1) 3

+1 Y

pjN

p-r

(p 1)((p 1) 2

1)

(p 1) 3

+1 (1:7)

Y

p>2

1+ 1

(p 1) 3

1;

wherep>2throughout,C(r)=2foroddr,andC(r)=8forevenr. Follow-

ing thework of Vinogradov [Vi℄, several authors establishedRadema her's

result un onditionally;seefor exampleAyoub [A℄ and Zulauf[Zu℄.

The arguments of [A℄ and [Zu℄ with some minormodi ationsa tually

give (1.6) forall r log A

N; whereA >0 is arbitrary. A natural problem

is whether (1.6) is stilltrue for largerr. The purposeof the present paper

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all positive modulir N 1=8 "

and all b 2B(N;r):Pre iselyspeaking, we

have thefollowing

Theorem 1.Let N bea xed large odd integer,">0 arbitrarily small

and

R=N 1=8 "

:

Let also A >0 be arbitrary. For a positive integer r; dene B(N;r) as in

(1.3). Then for all positive integers r R ; with at most O(Rlog A

N)

ex eptions, the Diophantine equation (1.5) with prime variables is solvable

whenever b2B(N;r); and the number of solutions is given by (1.7).

The above resultis a onsequen e ofthefollowingmean-valuetheorem.

Theorem 2.Let N bea xed large odd integer,">0 arbitrarily small

and

R=N 1=8 "

:

Let also A >0 be arbitrary. For a positive integer r; dene B(N;r) as in

(1.3). Then

(1:8)

X

rR

r max

b2B(N;r)

X

N=n

1 +n

2 +n

3

n

j

b

j (modr)

(n

1 )(n

2 )(n

3

) (N;r) N

2

N 2

log A

N;

where(n) denotes the von Mangoldt fun tion.

Remark. If ther's in thetheorems are restri tedto primes, then the

exponent1=8 anbeimprovedto3=20:Thisimprovementisusefulinstudy-

ingtheternaryGoldba hproblemwiththethreeprimesummandsrestri ted

to a thin subset of primes. This problemhas been investigated in another

paper[Li℄.

Sin ethederivationofTheorem1fromTheorem2isimmediate,wegive

ithere.

Proof of Theorem 1. LetE(R )bethesetofpositiveintegersrR

forwhi h

max

b2B(N;r)

X

N=n1+n2+n3

n

j

b

j (modr)

(n

1 )(n

2 )(n

3

) r(N;r) N

2

>

r

' 3

(r)

N

2

logN :

Then one dedu esfromTheorem 2that

X

r2E(R) r

2

' 3

(r)

log A

N

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#E(R )= X

r2E(R) 1R

X

r2E(R) r

' 3

(r)

Rlog A

N:

Sin e

(1:9)

r

' 3

(r)

(N;r) r

' 3

(r)

;

one sees that (1.6) is true for all r 62 E(R ) and all b 2 B(N;r): This

ompletes theproof ofTheorem1.

Now itremainsto establish Theorem2.

The proof of Theorem 2 is motivated by a paper of Wolke [W℄, whi h

ontains several new ideas to study the problem under onsideration and

the ternary Goldba h problem with the prime summands restri ted to a

thin subset of prime numbers. His method a tually gave Theorem 2 for

almost all primemodulir =pN 1=11

:

Thebasi toolofourproof,as anbeexpe ted,isthe ir lemethod. On

the minor ar s, one needs a nontrivial estimate for exponential sums over

primesin arithmeti progressions to every individualand largemodulusr:

All known results of this kind are, however, nontrivialwhen the hoi e of

minor ar s is very \thin". Consequently, the major ar is mu h \larger"

than usual. By dening the major and minor ar s in this way, the minor

ar s an then be treated easily by a result of Balog and Perelli [BP℄ on

exponential sums over primes in an arithmeti progression (see Lemma 1

below). Themain diÆ ultyof theproof omes from themajorar s,where

we usethefollowingideas:

(a) The starting point is Lemma 2 in x2, where we establish a new

formulafor

X

nN

nb(modr)

(n)e(n)

interms ofDiri hlet hara ters. It plays asimilarrole astheformula

X

nN

(n)e

n

a

q +

= 1

'(q) X

modq

G(a; ) X

nN

(n)e(n)

does in the treatment of the original ternary Goldba h problem, where

G(a;)is theGaussiansum denedas

G(a;)= k

X

n=1

(n)e

an

k

:

Consequently,ageneralizationoftheGaussiansum,namelyG(b;f;m;

g

;k)

denedasin x2,o urs. We needupperestimates forG(b;f;m;

g

;k); and

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(b) Thetreatment of themajorar seventuallyredu es tothe following

form ofmean-valueestimates forexponentialsumsoverprimes:

Theorem 3.For any A>0,thereexistsa onstant E=E(A)>0 su h

that if

(1:10) 1K x

1=3

L E

; =K 3

L E

;

then

X

qK max

yx max

(a;q)=1 max

jj

X

ny

(n)e

n

a

q +

(q)

'(q) X

ny e(n)

xL A

:

Inx 4,ageneralresult(Theorem4) ontainingthistheoremisestablished.

These mean-value estimates play important roles in the proof of Theorem

2,and theexponent 1=8 resultsfromthem.

It shouldbementionedthatMaier and Pomeran e[MP℄, Balog[B℄and

Mikawa [Mi℄ studied the distribution of prime twins with one of them in

arithmeti progressions. Theirmethods an deal withthebinaryGoldba h

problemwithoneofthesummandsinarithmeti progressions,butwe annot

applythem to theproblem onsideredinthepresent paper.

We usestandardnotations in numbertheory. Inparti ular,the letterr

inthesequel standsalwaysforpositive integers,whileLforlogN ex eptin

x4whereL=logx:TheletterÆdenotesasuÆ ientlysmallpositivenumber,

whosevalue mayvaryindierent o urren es. Forexample,we an write

N Æ

L 5

N Æ

; N

Æ

N Æ

:

The expression r R means 1

2

R < r R : A Diri hlet hara ter modq

willbewritten as

q

ifne essary.

2. Outlineof the proof of Theorem 2. Let

(2:1) RN

1=8 "

;

and

(2:2) P =R

2

L 3C

; Q=NR 2

L 4C

;

the onstant C will be spe iedlater. For ea h positive integer r R , the

majorar of the ir le methodis denedas

E

1 (R )=

[

qP q

[

a=1

(a;q)=1

a

q 1

qQ

; a

q +

1

qQ

:

Sin e 2P <Q; no two majorar s interse t. The minorar is denedas

E

2 (R )=

1

;1+ 1

E

1 (R ):

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Write 2[0;1℄intheform

(2:3) =a=q+; 1aq; (a;q)=1:

It followsfrom Diri hlet's lemmaon rationalapproximationsthat

E

2

(R )=f:P <q Q; jj1=(qQ)g:

Let (n)be thevon Mangoldtfun tion,e()=e 2i

asusual,and

(2:4) S(;r;b) =

X

nN

nb(modr)

(n)e(n):

ThenthestatementofTheorem2isequivalenttothat, forarbitraryA>0;

X

rR

r max

b2B(N;r)

1

\

0

S(;r;b

1

)S(;r;b

2

)S(;r;b

3

)e( N)d (N;r) N

2

N 2

L A

:

It thussuÆ esto prove

(2:5)

X

rR

r max

b2B(N;r)

\

E1(R)

S(;r;b

1

)S(;r;b

2

)S(;r;b

3

)e( N)d

(N;r) N

2

N 2

L A

;

and

(2:6)

X

rR

r max

b2B(N;r)

\

E

2 (R)

S(;r;b

1

)S(;r;b

2

)S(;r;b

3

)e( N)d

N 2

L A

:

The estimate of S(;r;b) with (b;r) = 1 on the minor ar s is given in

thefollowinglemma.

Lemma 1. Let A > 0 be arbitrary and 2 E

2

(R ): If C is suÆ iently

large,then

(2:7) S(;r;b)

N

rlog A

N

;

uniformly for rR :

Proof. We need the following result of Balog and Perelli [BP℄: For

M N and h=(r;q);

(2:8)

X

nM

(n)e

a

q n

L 3

hN

rq 1=2

+ q

1=2

N 1=2

h 1=2

+ N

4=5

r 2=5

:

(7)

(AsimilarresultwasalsoobtainedbyLavrik[La℄.) Nowthedesiredestimate

an beeasilyderived from(2.8) via partialsummation.

We an nowgive

Proof of (2.6). Itfollows fromLemma1 thattheintegraloverE

2 (R )

is

\

E

2 (R)

S(;r;b

1

)S(;r;b

2

)S(;r;b

3

)e( N)d

max

2E

2 (R)

jS(;r;b

1 )j

1

\

0

jS(;r;b

2 )j

2

d

1=2

1

\

0

jS(;r;b

3 )j

2

d

1=2

N

2

r 2

L A+1

;

uniformly for r R : Hen e the quantity on the left-hand side of (2.6) is

N 2

L A

;whi hproves (2.6).

Theorem2nowredu esto(2.5),whi hwillbeestablishedinthefollowing

fourse tions.

The startingpoint of theproofof (2.5) is Lemma2 below, whi htrans-

forms the exponential sum S(;r;b) into hara ter sums. To state the

lemma,weneed some morenotations.

Let d;f;g;k;m be xed positive integers, and

g

a Diri hlet hara ter

mod g:Dene

(2:9) G(d;f;m;

g

;k)= k

X

n=1

(n;k)=1

nf(modd)

(n)e(mn=k):

Obviously,thisis ageneralization ofthe Gaussiansum G(m;).

Forpositiveintegersr and q;let

(2:10) h=(r;q):

Then r; q andh an be writtenas

r=p 1

1 :::p

s

s r

0

; (p

j

;r

0 )=1;

q=p

1

1 :::p

s

s q

0

; (p

j

;q

0 )=1;

h=p 1

1 :::p

s

;

where

j

;

j and

j

arepositiveintegerswith

j

=min(

j

;

j

);j=1;:::;s:

Dene

(2:11) h

1

=p Æ

1

1 :::p

Æ

s

;

whereÆ

j

=

j

or0 a ordingas

j

=

j

ornot. Then h

1

jh: Write

(2:12) h =h=h :

(8)

Then

(2:13) h

1 h

2

=h; (h

1

;h

2 )=1;

r

h

1

; q

h

2

=1:

Lemma 2. Let a;q;r be positive integers, and h, h

1 , h

2

dened as in

(2.10), (2.11) and (2.12) respe tively so that (2.13) holds. Then

S

a

q

+;r;b

=

1

'(r=h

1 )'(q=h

2 )

X

modr=h1

(b) X

modq=h2

G(h;b;a;;q)

X

nN

(n)(n)e(n)+O(L 2

);

whereG(h;b;a;;q) is dened as in (2.9).

Proof. It is easilyseenthat

S

a

q

+;r;b

= q

X

=1

( ;q)=1 e

a

q

X

nN

nb(modr)

n (modq)

(n)e(n):

The inner sum is empty unless b (mod h); we an therefore add the

restri tion b (modh) to thesum over : Ontheother hand, under the

ondition b (modh); thesimultaneous ongruen es

nb (mod r); n (modq)

areequivalent to

nb (mod r=h

1

); n (modq=h

2 )

a ording to (2.13). And onsequently,

S

a

q

+;r;b

= q

X

=1

( ;q)=1

b(modh) e

a

q

X

nN

nb(modr)

n (modq)

(n)e(n)

= q

X

=1

( ;q)=1

b(modh) e

a

q

X

nN

nb(modr=h1)

n (modq=h

2 )

(n)e(n):

Introdu ingtheDiri hlet hara ters modr=h

1

and modq=h

2

,one has

S

a

q

+;r;b

=

1

'(r=h

1 )'(q=h

2 )

q

X

=1

( ;q)=1 e

a

q

X

modr=h

1

(b)

(9)

modq=h

2

( )

nN

(n)(n)e(n)+O(L 2

)

=

1

'(r=h

1 )'(q=h

2 )

X

modr=h

1

(b) X

modq=h

2

G(h;b;a;;q)

X

nN

(n)(n)e(n)+O(L 2

):

Thisprovesthelemma.

3.ThegeneralizedGaussiansumG(d;f;m;

g

;k). Letd; f; g; m; k

be xed positiveintegers, and modg a Diri hlet hara ter. The purpose

ofthisse tionistogiveupperestimatesforthesumG(d;f;m;

g

;k)dened

asin(2.9).

The mainresult ofthisse tion isthefollowing

Lemma 3. Let djk, gjk and (m;k) = (f;k) = 1: Let also modg be

indu ed by the primitive hara ter

modg

:Then

jG(d;f;m;

g

;k)jg

1=2

:

In thespe ial aseg=k;dene

(3:1) G(d;f;m;)=G(d;f;m;

k

;k)= k

X

n=1

nf(modd)

(n)e(mn=k):

Then Lemma3 isa onsequen eof thefollowing

Lemma 4.Let djk and (m;k)=(f;k)=1:Letalso modk beindu ed

by the primitive hara ter

modk

: Then

jG(d;f;m;)jk

1=2

:

Now we deriveLemma 3from Lemma4.

Proof of Lemma 3. Let 0

k

betheprin ipal hara ter modk:Then

g

0

k

isa hara ter mod k;and onsequently,

G(d;f;m;

g

;k)=G(d;f;m;

g

0

k ):

Thedesiredresultfollowsfrom Lemma4onnotingthat

g

0

k

modk isalso

indu edby theprimitive hara ter

modg

:

It remainsto prove Lemma 4. To this end, we investigate G(d;f;m;)

forsomespe ial hara tersmodkinthefollowingLemmas5{7. Theproof

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Lemma 5. Let djk;and modk be primitive. Then

(3:2) G(d;f;m;)= k

d

1

() e

mf

k

G

k

d

; m;f;

;

and onsequently,

(3:3) jG(d;f;m;)jk

1=2

:

Hereand in the sequel () isdened by

()= k

X

n=1

(n)e(n=k):

Proof. Makingthe substitutionn=jd+f;onesees that

G(d;f;m;)= k=d

X

j=1

(jd+f)e

m(jd+f)

k

(3:4)

=e

mf

k

k=d

X

j=1

(jd+f)e

mj

k=d

:

Nowweappeal to theidentity

(a)= 1

( ) k

X

n=1

(n)e

an

k

;

whi h holdsfortheprimitive hara termodk:Therefore,

k=d

X

j=1

(jd+f)e

mj

k=d

= 1

() k

X

n=1

(n)e

fn

k

k=d

X

j=1 e

njd

k

e

mj

k=d

= 1

() k

X

n=1

(n)e

fn

k

k=d

X

j=1 e

(n+m)j

k=d

:

The innersum equals k=dor 0 a ording as n+m 0 (mod k=d) ornot.

Hen e theright-handsideabove isequalto

k

d

1

( )

k

X

n=1

n m(modk=d)

(n)e

fn

k

= k

d

1

() G

k

d

; m;f;

:

Thisin ombination with(3.4) gives(3.2).

The inequality(3.3) followsfrom thewell-knownfa tthat j()j=k 1=2

andthetrivialestimatejG(k=d; m;f;)jd:This ompletestheproofof

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Lemma 6.Let djk,(m;k) =1and modk beindu ed by the primitive

hara ter

modk

:If k

satises

pjk)pjk

;

then

(3:5) jG(d;f;m;)jk

1=2

:

Proof. Fromtheassumption of thelemma,one dedu e that

G(d;f;m;)= k

X

n=1

nf(modd)

(n)e(mn =k):

The followingargument isdividedinto2 ases.

Spe ial ase. We start from the simplest ase where k = p

for some

prime p and positive integer : Sin e k

jk and djk; we an suppose that

k

=p

and d= p

; where and areintegers satisfying1 and

0 . Itis obvious thatone has eitherk

jd ordjk

:

If k

jd; thenon setting n=du+f the above sum be omes

G(d;f;m;)= k=d

X

u=1

(ud+f)e

m(ud+f)

k

=

(f)e

mf

k

k=d

X

u=1 e

mu

k=d

:

Sin e (m;k)=1;thelastsum vanishes,and onsequently,

(3:6) G(d;f;m;)=0:

If djk

;thenon making thesubstitutionn=uk

+v one has

G(d;f;m;)= k=k

X

u=1 k

X

v=1

(uk

+v)e

m(uk

+v)

k

;

where the double sums over u;v are further restri ted by the ondition

uk

+vf (modd): Therestri tion uk

+vf (mod d)isequivalent to

vf (modd): Therefore theabove quantity an be writtenas

k=k

X

u=1 e

mu

k=k

k

X

v=1

vf(modd)

(v)e

mv

k

:

The rstsum vanishesunless k=k

;hen e fork

6=k one has

(12)

While fork

=k one obtains

G(d;f;m;)= k

X

v=1

vf(modd)

(v)e

mv

k

=G(d;f;m;

);

hen e byLemma 4,

(3:8) jG(d;f;m;)jk

1=2

:

We therefore on lude from(3.6){(3.8) that (3.5) holdsfork=p

:

General ase. We now turn to general k: To this end, we rst prove

that G(d;f;m;) is multipli ative with respe t to k: Let k = k

1 k

2 with

(k

1

;k

2

) = 1: Then for modk there exist a unique ouple of hara ters

1 modk

1

and

2 modk

2

su h that =

1

2

: Therefore, on making the

substitutionn=k

2 n

1 +k

1 n

2

;one has

(3:9) G(d;f;m;)= k

1

X

n

1

=1 k

2

X

n

2

=1

1

2 (k

2 n

1 +k

1 n

2 )e

m(k

2 n

1 +k

1 n

2 )

k

1 k

2

;

wherethedoublesumsare furtherrestri tedby

(3:10) k

2 n

1 +k

1 n

2

f (modd):

Onnotingthatdjk;we setd=d

1 d

2

withd

1 jk

1 and d

2 jk

2

:It followsfrom

(k

1

;k

2

)=1that (d

1

;d

2

)=1;hen e (3.10) isequivalentto

(3:11) n

1

fk

2

(mod d

1 ); n

2

fk

1

(modd

2 );

wherek

1 andk

2

aredenedbyk

1 k

1

1 (mod d

2

) andk

2 k

2

1 (modd

1 ).

Now(3.9) be omes

(3:12) G(d;f;m;)

=

k

1

X

n

1

=1

n

1

f

k

2 (modd

1 )

1 (k

2 n

1 )e

mn

1

k

1

k

2

X

n

2

=1

n

2

f

k

1 (modd

2 )

2 (k

1 n

2 )e

mn

2

k

2

=

1 (k

2 )

2 (k

1 )G(d

1

;fk

2

;m;

1 )G(d

2

;fk

1

;m;

2 ):

Now let

k =p 1

1 p

2

2 :::p

s

be the anoni al de omposition of k; where p

j

stands for primes, and

j

positiveintegers. A ordingly,k

andd an be writtenas

k

=p 1

1 p

2

2 :::p

s

and d=p 1

1 p

2

2 :::p

s

;

where

j

and

j

are integers satisfying 1

j

and 0

j

j .

It follows thatthere are primitive hara ters

j modp

j

;j =1;:::;s; su h

that

=

:::

;and ea h

modp

j

indu es

j

modp

j

:

(13)

Making the substitution n = n

1 K

1 +n

2 K

2

+:::+n

s K

s

; where K

j is

denedbyp

j

j K

j

=k;one sees that

G(d;f;m;)= s

Y

j=1

j (K

j )G(p

j

;fK

j

;m;

j );

whereK

j

satises

K

j K

j

1 (mod p

j

); j=1;:::;s:

It followsthat

jG(d;f;m;)j s

Y

j=1 jG(p

j

;fK

j

;m;

j )j=

s

Y

j=1 p

j=2

j

=k

1=2

:

This ompletes theproof ofthe lemma.

Lemma 7. Let djk and (m;k) = (f;k) = 1: Let also 0

modk be the

prin ipal hara ter. Thenfor (d;k=d)>1;

G(d;f;m; 0

)=0;

and for (d;k=d)=1;

G(d;f;m; 0

)=

k

d

e

fmt

d

;

wheret is dened by tk=d1 (modd):

This isHilfssatz 2 ofRadema her [R℄orTheorem2.2 ofAyoub [A℄.

We an nowgive

Proof of Lemma 4. Let

(3:13) k =k

1 k

2

with (k

1

;k

2

)=1; k

jk

1

; and pjk

1

)pjk

:

Then formodk there exist a unique oupleof hara ters

1 modk

1 and

0

2 modk

2

su hthat=

1

0

2

;where 0

2 modk

2

is theprin ipal hara ter.

On noting that djk; we set d = d

1 d

2

with d

1 jk

1 and d

2 jk

2

: It therefore

follows from(3.12) that

G(d;f;m;)=

1 (k

2 )

0

2 (k

1 )G(d

1

;fk

2

;m;

1 )G(d

2

;fk

1

;m; 0

2 ):

The statement ofthelemma nowfollows from Lemmas6 and 7.

4. A mean-value estimate for exponential sums over primes.

Wolke [W℄wastherst to studythemean-valueestimate asinTheorem3.

He proved that Theorem3is truefor

1K =x 1=4

; =min(K 4

;L E

):

A tually,Theorem3wasre entlygivenbytheauthorsinanotherjointpaper

(14)

a gap in theproof of [ZL℄: the statement \h 00

() >0 for 1=2 1" on

p.365 of [ZL℄is notalwaystrue. The proof thereforeneeds orre tions.

In thisse tion we prove thefollowinggeneralresult, whi h ontains the

assertionof Theorem3. One an seefromtheproofofTheorem2 thatthis

generaltheorem is ne essary.

Theorem 4. Let z 1 be arbitrary. For any A > 0, there exists a

onstant E =E(A)>0 su h that if

(4:1) 1K z

2=3

x 1=3

L E

; =z 2

K 3

L E

;

then

X

qK max

yx max

(a;q)=1 max

jj

X

ny

(n)e

n

a

q +

(q)

'(q) X

ny e(n)

zxL A

:

We needsome lemmasto establishthisresult.

Lemma 8. Suppose that F(u) and G(u) are real fun tions dened on

[a;b℄; and G(u) and 1=F 0

(u) are monotoni .

(i) If jF 0

(u)jm and jG(u)jM;then

b

\

a

G(u)e(F(u))duM=m:

(ii) If jF 00

(u)jr and jG(u)jM;then

b

\

a

G(u)e(F(u))duM=

p

r:

For the proof of these results, see Lemmas 3.3 and 3.4 in Tit hmarsh

[T℄.

Lemma 9. Let N(;T;) be the number of zeros % = +i of the

Diri hlet L-fun tion L(s;) in the re tangle 1; T T:

Supposeq 1 and T 2: Then,for 1=21;we have

X

modq

N(;T;)(qT)

3(1 )=(2 )

(logqT) 9

:

This isTheorem12.1 inMontgomery[Mo℄.

Lemma 10. Let a

n

;n = 1;2;:::; be omplex numbers and modq a

hara ter. Then

X

qQ X

modq T

\

T0

X

nN a

n

(n)n it

2

dt(Q 2

T +N) X

nN ja

n j

2

for arbitrary Q, T

0

;and T:

(15)

Lemma 11. Let (s) bethe Riemann zeta-fun tion, and

F(s)= X

nU

(n)=n s

; G(s)= X

nU

(n)=n s

:

Then

0

(s) F(s)

=G(s)( 0

(s)) F(s)G(s)(s)

((s)G(s) 1)

0

(s) F(s)

:

This isVaughan'sidentity;for theproof,see [Va℄.

Nowwe an,usingtheideaduetoZhan[Zh℄,givetheproofofTheorem4.

Proof of Theorem 4. Introdu ingtheDiri hlet hara ters,theex-

ponentialsumunder onsiderationbe omes

X

ny

(n)e

n

a

q +

= 1

'(q) X

modq X

ny

(n)(n)e(n) q

X

h=1

(h)e

ah

q

+O(L 2

);

and onsequently,

(4:2)

X

qQ max

yx max

jj

max

(a;q)=1

X

ny

(n)e

n

a

q +

(q)

'(q) X

ny e(n)

X

qK max

yx max

jj

max

(a;q)=1 1

'(q) X

modq

G(a;) X

ny

(n;)(n)e(n)

+KL

2

;

whereG(a;) isdenedasin x3,and

(n;)=

(n) for6= 0

,

(n) 1 for= 0

.

To estimate thesums on theright-hand sideof (4.2), one notes that if the

primitive hara ter modq indu es the hara ter modk; thenqjk; and

jG(a;q)j q 1=2

for(a;q) =1: We now ombine all ontributions made by

an individualprimitive hara ter, so that the rst term on the right-hand

sideof(4.2) is

X

qK max

yx max

jj

max

(a;q)=1 X

kK

qjk q

1=2

'(k) X

modq

X

ny

(n;)(n)e(n)

L X

max

yx max

jj

q 1=2

'(q) X

X

(n;)(n)e(n)

:

(16)

Hen e theassertionof thetheorem redu esto

(4:3) S :=

X

qD max

yx max

jj

X

modq

X

ny

(n;)(n)e(n)

zxD 1=2

L A 3

;

with1DK and K ;satisfying(4.1).

Theargumentleadingto(4.3)fallsnaturallyintotwo ases a ordingas

D is smallor large. For D L F

; where F is some positive onstant, one

usesthe lassi alzero-densityestimateandzero-freeregionfortheDiri hlet

L-fun tions. While for L F

< D K ; one appeals to ontour integration,

thelargesieveinequalityand Vaughan'sidentity.

Case 1. DL F

;whereF isapositive onstant tobespe iedlaterin

terms ofA: Inthis ase, itsuÆ es to prove that

(4:4) :=

X

ny

(n;)(n)e(n)zxL

2F A 3

foryx, jj and any primitive hara ter modd:

To estimate;one appeals to theSiegel{Walsztheorem ([D℄, x19):

X

nu

(n;)(n)e(n)= X

j jT u

%

+b()+O

u(loguqT) 2

T

where % = +i denotes nontrivial zeros of L(s;); b() is a onstant

depending on ; and T 2 is a parameter. Applying partial summation,

we have

= y

\

y=2

e(u)d

X

nu

(n;)(n)

(4:5)

= X

j jT y

\

y=2 u

% 1

e(u)du+O

(1+jjx) xL

2

T

:

Take T =x 2

;sothat theO-termis a eptable in(4.4). Sin e,foruy;

d

du

u+

2

logu

=+

2u

min

uy

j +2uj

y

;

and

d 2

du 2

u+

2

logu

=

2u 2

j j

y 2

;

we dedu efrom Lemma8thattheintegralontheright-handsideof(4.5)is

y

\

u 1

e

u+

2

logu

dumin

y

p

j j+1

;

y

min

uy

j +2uj

:

(17)

LetT

0

=4x; sothatforT

0

<j jx 2

and uy;

j +2ujj j 2uj j=2:

Then (4.5) be omes

X

j jx 2

min

y

p

j j+1

;

y

min

uy

j +2uj

+O(xL

2F A 3

) (4:6)

X

j jT0 x

p

j j+1 +

X

T0<j jx 2

x

j j

+O(xL

2F A 3

):

It is well known that for any modq there is a onstant

1

> 0 su h

thatL(s;)hasno zerointhe region

1

logq+log 4=5

(jtj+2)

;

ex ept the possible Siegel zero. But the Siegel zero does not exist in the

present situation,sin eqL F

:Therefore,one has

(4:7) x

1

exp

1 logx

logq+log 4=5

T

exp(

2 L

1=5

);

for some onstant

2

>0: Hen e the se ond sum on theright-hand sideof

(4.6) is a eptable.

To dealwiththe rstterm, one notes that

X

j jT

0 x

p

j j+1

xL max

T1T0 T

1=2

1

X

j jT

1 x

1

;

whi h is,on applyingLemma 9,

xL max

T

1

T

0 T

1=2

1

(logqT

1 )

9

max

1=21 (qT

1 )

(3 3)=(2 )

x

1

xL F+11

max

T

1

T

0 max

1=21 exp

(1 )L+

3 3

2 1

2

logT

1

=:xL F+11

max

T

1

T

0 max

1=21 f(T

1

;);

say. Therefore, inview of(4.6), theestimate (4.4) redu esto

(4:8) max

T

1

T

0 max

1=21 f(T

1

;)zL

3F A 20

:

Supposerst 4=51;sothat

3 3

2

1

2 :

It followsfrom (4.7) that

(4:9) max

T T

max

4=51 f(T

1

;) max

4=51

expf (1 )Lgexpf

2 L

1=5

g;

(18)

whi h isa eptable in(4.8). Nowwe turn to 3=5 4=5;whi hensures

that

3 3

2

1

2 :

Onnotingthat logT

1

L+O(1); and

max

3=54=5

( 1=2)

2

= 3

70

;

one dedu esthat

(4:10) max

T

1

T

0

max

3=54=5 f(T

1

;)

max

3=54=5 exp

(1 )L+

3 3

2 1

2

L

= max

3=54=5 exp

( 1=2)

2 L

x 3=70

;

and this is also a eptable in (4.8). Finally, we onsider 1=2 3=5:

Nowwehave

6

7

3 3

2 1

2

;

and onsequently,

max

T1T0

max

1=23=5 f(T

1

;)

max

T1T0

max

1=23=5 exp

(1 )L+

3 3

2 1

2

logx

exp

3 3

2 1

2

log x

T

1

:

Sin e T

1

T

0

xxL E

;theabove quantityis

max

1=23=5 exp

( 1=2)

2 L

exp

6

7

Eloglogx

(4:11)

L 6E=7

;

whi h isa eptable in(4.8) if E6F +2A+28:

Combining (4.9){(4.11) we get (4.8), hen e (4.4). This proves (4.3) in

Case1.

Case 2. L F

< D K ; where F is a onstant to be spe ied in the

following argument. In this ase, we use Vaughan's identity to establish

(19)

Estimation of the sum of type I.We rstshowthat

0

:=

X

dD max

yx max

jj

X

modd

X

mny

mM

nN

a(m)b(n)(mn)e(mn)

(4:12)

zxD 1=2

L A 7

holdsfora(m)d(m) and b(n)d(n) withmM,nN and

(4:13) xMN x; M;N xD 1

L 2A 20

:

Let

f

1

(s;)= X

mM

a(m)(m)

m s

and f

2

(s;)= X

nN

b(n)(n)

n s

;

wheres=+itisa omplexvariable. Thenone sees that

(4:14) f

1 (s;)f

2

(s;)M 1

N 1

x 1

uniformlyfor 2 2:Applying Perron's summation formula(see e.g.

Lemma3.12 in[T℄) and thenshiftingthe ontourto the left,one gets

X

mnu

mM

nN

a(m)b(n)(m)(n)

= 1

2i 1+"+ix

2

\

1+" ix 2

f

1 (s;)f

2 (s;)

u s

s

ds+O(L)

= 1

2i n

1=2 ix 2

\

1+" ix 2

+ 1=2+ix

2

\

1=2 ix 2

+ 1+"+ix

2

\

1=2+ix 2

o

f

1 (s;)f

2 (s;)

u s

s

ds+O(L):

By(4.14), theintegralson thehorizontalpartsare learlyO(L):Therefore,

X

mnu

mM

nN

a(m)b(n)(m)(n)

= 1

2

x 2

\

x 2

f

1

2

+it;

f

2

1

2

+it;

u 1=2+it

1=2+it

dt+O(L):

Now, by partialsummation,the innersum of 0

is

y

\

y=2

e(u)d n

X

mnu

mM

nN

a(m)b(n)(m)(n) o

= 1

2

x 2

\

x 2

f

1

2

+it;

f

2

1

2

+it;

1

1=2+it y

\

y=2 u

1=2+it

e(u)dudt

(20)

= 1

2

x

\

x 2

f

1

2

+it;

f

2

1

2

+it;

1

1=2+it y

\

y=2 u

1=2

e

+ t

2

logu

dudt+O(xL);

whi h,bytheargumentleading to (4.6), isestimated as

x 1=2

\

jtjT

0

f

1

2

+it;

f

2

1

2

+it;

dt

p

jtj+1

+x 1=2

\

T

0

<jtjx 2

f

1

2

+it;

f

2

1

2

+it;

dt

jtj

+O(xL):

It thereforesuÆ es to show

(4:15)

X

dD X

modd T2

\

T2=2

f

1

2 +it;

f

2

1

2

+it;

dt

zx 1=2

D 1=2

T 1=2

2 L

A 8

for1T

2

T

0

;and

(4:16)

X

dD X

modd T3

\

T3=2

f

1

2 +it;

f

2

1

2

+it;

dt

zx 1=2

D 1=2

T

3 L

A 8

forT

0

T

3

x 2

:

The left-hand sideof(4.15) is,byCau hy'sinequalityandLemma 10,

X

dD X

modd T

2

\

T2=2

f

1

2

+it;

2

dt

1=2

(4:17)

X

dD X

modd T

2

\

T

2

=2

f

2

1

2

+it;

2

dt

1=2

(D 2

T

2 +M)

1=2

(D 2

T

2 +N)

1=2

L

fD 2

T

2 +DT

1=2

2 (M

1=2

+N 1=2

)+M 1=2

N 1=2

gL

zx 1=2

D 1=2

T 1=2

2 L

A 8

ifF 2A+20 and E2A+20:This yields(4.15).

(21)

Estimation of the sum of type II. Next we prove that

00

:=

X

dD max

yx max

jj

X

modd

X

mny

mM

nN

b(n)(mn)e(mn)

(4:18)

zxD 1=2

L A 7

holdsforb(n)d(n)with nN andM;N satisfying

(4:19) xMN x; M DL

2A+20

:

Arguing as before, one sees that it suÆ es to show (4.15) and (4.16)

subje tto (4.19). Here f

1

(s;);f

2

(s;)are thesame asbefore ex eptthat

a(m) = 1 in the denition of f

1

(s;): Sin e now M is large a ording to

(4.19), the above approa h to atta k the mean value of f

1

(s;) does not

work anymore;one thereforeneeds to treat f

1

(s;) dierently.

Let w=u+iv be a omplex variable. Then,applyingPerron's formula

and thenshiftingthelineof integration asbefore,one gets

f

1

2

+it;

= 1

2i 1+"+ix

2

\

1+" ix 2

L

1

2

+it+w;

M w

(M=2) w

w

dw+O(L)

= 1

2

x 2

\

x 2

L

1

2

+it+iv;

M iv

(M=2) iv

iv

dv+O(L)

x

2

\

x 2

1

jvj+1

L

1

2

+it+iv;

dv+O(L):

Consequently,byCau hy'sinequality,

f

1

2

+it;

2

x 2

\

x 2

1

jvj+1 dv

x 2

\

x 2

1

jvj+1

L

1

2

+it+iv;

2

dv

+L 2

L x

2

\

x 2

1

jvj+1

L

1

2

+it+iv;

2

dv+L 2

:

It followsthat

X

dD X

modd T

2

\

T

2

=2

f

1

2

+it;

2

dt

L max

T

4

x 2

1

T

4 X

dD X

modd T

2

\

T =2 T

4

\

T =2

L

1

2

+it+iv;

2

dvdt+D 2

T

2 L

2

(22)

L max

T2<T4x 2

1

T

4 T2

\

T2=2 X

dD X

modd T4+t

\

T4=2+t

L

1

2

+i;

2

d dt

+L max

T

4

T

2 1

T

4 T

4

\

T

4

=2

X

dD X

modd T

2 +v

\

T

2

=2+v

L

1

2

+i;

2

d

dv

+D 2

T

2 L

2

:

Applyingthe lassi alestimate

X

modq T

\

0

L

1

2

+it;

2

dtqT(logqT) 2

;

thequantityabove is

D 2

T

2 L

3

+D 2

T

2 L

3

+D 2

T

2 L

2

D 2

T

2 L

3

:

Hen e bythe argument leadingto (4.17), one has

X

mnu

mM

nN

a(m)b(n)(m)(n)

X

dD X

modd T2

\

T2=2

f

1

2 +it;

2

dt

1=2

X

dD X

modd T

2

\

T2=2

f

2

1

2

+it;

2

dt

1=2

(D 2

T

2 L

3

) 1=2

(D 2

T

2 +N)

1=2

L 1=2

fD 2

T

2 +DT

1=2

2 N

1=2

gL 2

D 1=2

x 1=2

T 1=2

2 L

A 8

ifE 2A+20: Thisproves (4.15) underthe onditionof (4.19).

A similar argument gives (4.16). This ompletes the proof of (4.18)

subje tto (4.19).

Appli ation ofVaughan'sidentity. ByLemma11,oneseesthattheinner

sum ofS in(4.3) isequal to

X

ny

(n)(mn)e(mn)=S

1 S

2 S

3

;

where

S

1

= X

mny

mU

(m)(logn)(mn)e(mn);

S

2

= X

mny

2

a(m)(mn)e(mn);

(23)

S

3

=

mny

m>U

n>U

a(m)(logn)(mn)e(mn);

and a(m)d(m): Therefore,

S= X

qD max

yx max

jmodqj X

modq jS

1 j+

X

qD max

yx max

jmodqj X

modq jS

2 j (4:20)

+ X

qD max

yx max

jmodqj X

modq jS

3 j:

Taking U =DL 2A+20

in(4.20) andE 2A+20in (4.1),we have

U 2

=D 2

L 4A+40

xD 1

L 2A 20

:

Hen eea hofthethreetermsontheright-handsideof(4.20) anbedivided

into O(L 4

) sums of the form 0

or 00

: Now, in view of the hoi e of E;

(4.12)and(4.18) arebothvalid,fromwhi hthedesiredresult(4.3)forCase

2 followsinthe standardway. This ompletes theproof of thetheorem.

5. Preparation for the major ar s. Letq;r bepositive integersand

(5:1) (q;r)=h:

For(a;q)=1and (b;r)=1;dene

(5:2) f(r;q;a;b) = 8

<

:

(q=h)

'(rq=h) e

abt

h

if(q=h;h)=1,tq=h1 (mod h),

0 if(q=h;h)>1.

AndforS(;r;b) dened by(2.4), let

E(r;q;a;b;)=S

a

q

+;r;b

f(r;q;a;b) X

nN e(n);

(5:3)

E

(r;q)= max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ)

jE(r;q;a;b;)j:

(5:4)

Thepurposeofthisse tionistoestablishthefollowingmean-valueestimate,

whi h plays animportant role inproving(2.5), hen eTheorem2.

Lemma 12. Let R , P and Q be dened as in (2.1) and (2.2), while f,

E and E

as in (5.2), (5.3) and (5.4). Then for any A>0; there exists a

onstant C >0 su h that

X

rR X

qP E

(r;q)NL A

:

ThisestimatedependsonLemma13below, Lemma3ofx3,and (4.3)of

(24)

Lemma 13. Let r and q bepositiveintegers,and h,h

1 , h

2

bedened as

in (2.10), (2.11) and (2.12) respe tively so that (2.13) holds. Then forxed

positive integers r

, q

;one has

(5:5)

X

rN1

r

jr=h

1 X

qN2

q

jq=h2

1

'(r=h

1 )'(q=h

2 )

d(r

)

r

q

log

3

N

1 log

2

N

2 :

Proof. Sin e n'(n)logn; one has

X

rN1

r

jr=h1 X

qN2

q

jq=h2

1

'(r=h

1 )'(q=h

2 )

logN

1 logN

2 X

rN1

r

jr=h1 X

qN2

q

jq=h2 h

1

r

h

2

q :

For axed pair r;h

1

;we set j

1

=r=h

1

:Toestimate thesums onthe right-

handside,oneneeds thenumberofpairsq;h

2

su h thatthequotientsq=h

2

assume thesame valuej

2

:Sin e h

2

ofthese pairs mustsatisfyh

2 jr=h

1

;the

requirednumberis obviously d(r=h

1

);whered(n) isthedivisorfun tion.

Hen e thedoublesumunder onsiderationis

X

rN1

r

jr=h1 X

qN2

q

jq=h2 h

1 h

2

rq

X

j1N1

r

jj1

X

j2N2

(j2;j1)=1

q

jj

2 d(j

1 )

j

1 j

2

d(r

)

r

q

X

j1N1 d(j

1 )

j

1 X

j2N2 1

j

2

d(r

)

r

q

log

2

N

1 logN

2 :

Thisprovesthelemma.

We an nowestablishthemainresult of thisse tion.

Proof of Lemma 12. ByLemma 2 we have

(5:6) S

a

q

+;r;b

=

1

'(r=h

1 )'(q=h

2 )

X

modr=h

1

(b) X

modq=h

2

G(h;b;a;;q)

X

nN

(n)(n)e(n)+O(L 2

)

=I+J +K+O(L 2

);

say,where I;J andK arethe sums orresponding to

(i) = 0

modr=h

1 , =

0

modq=h

2

;

(j) = 0

modr=h

1

;6= 0

modq=h

2

;

(k) 6= 0

modr=h

1

(25)

It is easilyseenthat

I =

1

'(r=h

1 )'(q=h

2 )

X

=

0

modq=h

2

G(h;b;a;;q)

X

nN

=

0

modrq=h

(n)(n)e(n)

= 1

'(rq=h)

G(h;b;a; 0

q=h2 )

X

nN

e(n)+

X

nN

=

0

modrq=h

(n)((n) 1)e(n)+O

L 2

'(rq=h)

=f(r;q;a;b) X

nN

e(n)+O

1

'(rq=h) X

nN

((n) 1)e(n)

+O(L 2

);

where we have used Lemma 5 and (5.2). Taking maxima ever , b and a;

and thensummingoverq andr;one gets

(5:7)

X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ)

I f(r;q;a;b) X

nN e(n)

X

rR 1

'(r) X

qP

max

jj1=(qQ)

X

nN

((n) 1)e(n)

+R PL 2

L X

kP

max

jj1=(kQ)

X

nN

((n) 1)e(n)

+R PL 2

:

We pro eedto estimate J:Onesees that

J =

1

'(r=h

1 )'(q=h

2 )

X

modq=h

2

G(h;b;a;;q) X

nN

0

(n)(n)e(n)

=

1

'(r=h

1 )'(q=h

2 )

X

modq=h

2

G(h;b;a;;q) X

nN

(n)(n)e(n)+O(L 2

):

Consequently,one has

X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ) jJj

X

rR 1

'(r=h

1 )

X

qP 1

'(q=h

2 )

max

jj1=(qQ) X

modq=h

2

jG(h;b;a;;q)j

X

nN

(n)(n;)e(n)

+P

3=2

L 3

:

To estimate thesumson the right-hand sideabove,one appeals to Lemma

3, whi h ensures that if a primitive hara ter modk indu es a hara ter

modq=h ; then kjq=h and jG(h;b;a;;q)j k 1=2

:We now ombine all

(26)

ontributionsmadebyan individualprimitive hara ter, whi hgives

(5:8)

X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ) jJj

X

kP

X

rR 1

'(r=h

1 )

X

qP

kjq=h

2 k

1=2

'(q=h

2 )

max

jj1=(kQ) X

modk

X

nN

(n)(n;)e(n)

+P

3=2

L 3

L 5

X

kP 1

k 1=2

max

jj1=(kQ) X

modk

X

nN

(n)(n;)e(n)

+P

3=2

L 3

;

wherewehave used Lemma13 to estimatethesums inbra es.

We now turnto K :One hasbythedenitionofK ;

K =

1

'(r=h

1 )'(q=h

2 )

X

modr

6=

0

X

modq=h

G(h;b;a;;q) X

nN

(n)(n)e(n)

1

'(r=h

1 )'(q=h

2 )

X

modr=h1

6=

0

X

modq=h2

jG(h;b;a;;q)j

X

nN

(n)(n)e(n)

:

Working analogously to theargument above,one sees that

X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ) jKj

X

k

1

2R X

k

2

P

(k1;k2)=1

X

rR

k

1 jr=h

1 1

'(r=h

1 )

X

qP

k2jq=h2 k

1=2

2

'(q=h

2 )

max

jj1=(k

2 Q)

X

1modk1

1 6=

0

1 X

2modk2

X

nN

1

2

(n)(n)e(n)

:

By Lemma13, thequantityinbra esis

d(k

1 )k

1=2

2

k

1 k

2 L

5

R Æ

:

(27)

(5:9)

rR qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ) jKj

R Æ

X

k

1

2R X

k

2

P

(k

1

;k

2 )=1

1

k

1 k

1=2

2

max

jj1=(k2Q)

X

1 modk

1

16=

0

1

X

2 modk

2

X

nN

1

2

(n)(n)e(n)

:

One thus on ludes from (5.6){(5.9)that

(5:10)

X

rR X

qP E

(r;q)

= X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ)

E(r;q;a;b) f(r;q;a;b) X

nN e(n)

X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ)

I f(r;q;a;b) X

nN e(n)

+ X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ) jJj

+ X

rR X

qP max

(a;q)=1 max

(b;r)=1 max

jj1=(qQ)

jKj+R PL 2

+P 3=2

L 3

L

9

R 1=2

X

kU

max

jj1=(UQ) X

modk

X

nN

(n)(n;)e(n)

+ R

Æ

R U 1=2

X

kUR

max

jj1=(UQ) X

modk

X

nN

(n)(n;)e(n)

+R PL 2

+P 3=2

L 3

;

where

(5:11) U P =R

2

L C

:

By (4.3) with z = 1, the rst term on the right-hand side of (5.10) is

admissibleif

(5:12) U N

1=3

L D

; 1

UQ

U 3

L D

:

1=2 Æ

(28)

sideof(5.10) isadmissibleif

(5:13) R U R 1=3 Æ

N 1=3

L D

; 1

UQ

R 1 Æ

(R U) 3

L D

:

In view of the denitions of Q and U (see (2.2) and (5.11)), the optimal

hoi e ofR satisfying(5.12) and (5.13) is

RN 1=8 "

asstatedin(2.1). Thisproves thelemma.

6. The major ar s. In thisse tionwe give

Proof of (2.5). In the ourse of the proof, the following elementary

estimate willbe used: IfA

j

=B+C,j=1;2;3;then

(6:1) A

1 A

2 A

3

=B 3

+C(A 2

1 +B

2

+A

1

B)=B 3

+O(jCjjA

1 j

2

+jCjjBj 2

):

If 2E

1

(R ),thenforj =1;2;3;

(6:2) S(;r;b

j

)=f(r;q;a;b

j )

X

nN

e(n)+O(E

(r;q)):

Applying(6.1),one has

I

1 (r):=

\

E1(R)

S(;r;b

1

)S(;r;b

2

)S(;r;b

3

)e( N)d

= X

qP q

X

a=1

(a;q)=1

f(r;q;a;b

1

)f(r;q;a;b

2

)f(r;q;a;b

3 )e

aN

q

\

jj1=(qQ)

X

nN e(n)

3

e( N)d

+O

E

(r;q)

\

jj1=(qQ)

S

a

q

+;r;b

1

2

d

+O

E

(r;q)

' 2

(rq=h)

\

jj1=(qQ)

X

nN e(n)

2

d

:

Thethirdintegralontheright-hand sideabove istriviallyN:While the

se ond integral, when summedovera; an beestimated as

q

X

a=1

\

jj1=(qQ)

S

a

q

+;r;b

1

2

d 1

\

0

S

a

q

+;r;b

1

2

d N

r :

(29)

Onusingtheestimate

X

nN

e(n)min(N;1=kk);

one sees thatthe rstintegralis

1=2

\

1=2

X

nN e(n)

3

e( N)d+O

1=2

\

1=(qQ)

3

d

=

X

n

1 +n

2 +n

3

=N

1n

j

N

1+O((qQ) 2

)= 1

2 N

2

+O(N 2

L C

):

We thushave

I

1 (r)=

1

2 N

2

+O(N 2

L C

)

X

qP q

X

a=1

(a;q)=1

f(r;q;a;b

1

)f(r;q;a;b

2

)f(r;q;a;b

3 )e

aN

q

+O

N

r X

qP E

(r;q)

:

We now onsiderthesingularseries

1

X

q=1 q

X

a=1

(a;q)=1

f(r;q;a;b

1

)f(r;q;a;b

2

)f(r;q;a;b

3 )e

aN

q

:

For (q=h;h) > 1; one has f(r;q;a;b

j

) = 0, j = 1;2;3, by (6.2), hen e the

series onverges absolutelyto 0. For(q=h;h)=1;the seriesredu es to

1

' 3

(r) 1

X

q=1

(q=h;h)=1

(q=h)

' 3

(q=h) q

X

a=1

(a;q)=1

a(b

1 +b

2 +b

3 )t

h

aN

q

:

It was proved by Radema her [R℄ that if N is odd and b 2 B(N;r); then

theaboveseries onverges absolutelyandequals(N;r)denedasin(1.7).

One thereforehas

I

1

(r) (N;r) N

2

N

2

' 2

(r)L C

+ N

r X

qP E

(r;q);

and onsequently,

(6:3)

X

rR

r max

(bj;r)=1

I

1

(r) (N;r) N

2

N 2

L A

;

(30)

A knowledgements. The authors would like to thank ProfessorPan

Chengdongfor onstanten ouragement. Also,weareindebtedtothereferee

forvaluablesuggestions.

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DepartmentofMathemati s

ShandongUniversity

Jinan,Shandong250100

P.R.China

E-mail:zhantaosdu.edu. n

Re eivedon8.6.1996

andinrevisedformon6.12.1996 (3004)

We prove that for all positive integers r  R

We prove that for all positive integers r R