Shifted primes without large prime fa tors
by
R. C. Baker (Provo, Ut.) and G. Harman (Cardi )
1. Introdu tion. Letadenotea xednon-zerointeger,andletP +
(m)
denote thelargest primefa tor of aninteger m>1. Let
(x;y)=
X
a<px
P +
(p a)y 1:
Here and subsequently,theletterp is reservedfor aprime variable.
Theorem 1. For yx
we have
(1:1) (x;y)>
x
(logx) C
1
for x x
0
. Here = 0:2961; x
0
may depend on a; C
1
is an absolute
onstant.
The exponent an be repla ed by a very slightlysmaller onstant, as
willbeapparent from ourmethod. Thepreviousbest exponent is
1
2 p
e
+"=0:3032::: (">0arbitrarilysmall)
(Friedlander [8℄). Earlierresults inthisdire tion were obtained by Pomer-
an e[11℄, Balog[3℄ andFouvryand Grupp[7℄.
We notetwo orollariesof Theorem1.
Corollary 1 (Erd}os{Pomeran e). Let m
1
< m
2
< ::: denote
those positive integers m su h that the equation (n) = m has more
than m 1
solutions n. Then the sequen e (m
i
) is in nite, and satis es
lim
i!1 logm
i+1
=logm
i
=1.
Corollary 2 (Alford, Granville and Pomeran e). The number of
Carmi hael numbers x isx
(5 5 )=12
for large x.
1991 Mathemati sSubje tClassi ation: Primary11N25.
Resear hofthe rstauthorpartiallysupportedbytheNationalSe urityAgen y.
These results essentially follow from Theorem 1 taken in onjun tion
withthe argumentsin[11℄ and [1℄.
Let Q denote a number in (x 1=2
;x 11=20
). Our work depends on good
boundsfor
(x;S)= X
q2S X
xp<2x
pa(q) 1
whereS isa sequen e of theform fs
1 :::s
t :s
i 2S
i
glying in(Q;2Q℄. The
elementsof S are ountedwithmultipli ity;with thisinmind,write
jSj= X
s2S 1:
Thequantitytisabsolutelybounded. Wewriteuv(q)asanabbreviation
foruv (modq). Theorem1 follows fromboundsof theshape
(1:2) (x;S)< xL 1
X
q2S 1
(q)
; (x;S)>
0
xL 1
X
q2S 1
(q)
where L denotes logx and ; 0
are onstants not mu h greater than one.
Friedlander[8℄alsousesboundsofthisshape,withS =(x 1=2+Æ
;x 1=2+2Æ
)\Z
andÆ small,sothat ; 0
maybetaken arbitrarily loseto 1. The basi idea
of hispaperis to ount solutionsof theequation
p a=mn; m2(x 1=2+Æ
;x 1=2+2Æ
); xp<2x;
whi h an be interpreted in terms of primes in ongruen e lasses, and to
show thatthe m and nwith alargeprime fa tor annot be responsiblefor
all solutions,via theupperboundin(1.2). Thisidea insomewhatdi erent
form originatesinBalog [3℄.
Starting from the Balog{Friedlander onstru tion, we found that exi-
bility ouldbe gainedbyfollowinga similarpro edure withtheequation
p a=lmn; xp<2x
where m;n are about x 1
in size, and l is the produ t of many integer
fa tors of about x
"
in size. The best result was obtained by taking =
0:516. Theshapeofourequation wassuggestedbytheavailableingredients
for results of type (1.2), whi h are mostly in Bombieri, Friedlander and
Iwanie [4, 5℄.
Inordertosavespa e,wequotenumerousresultsandargumentsfrom[2℄.
Some of the ideas used arry over to improve by10 3
the exponent in [2℄,
Theorem3:
Theorem 2. For in nitely manyprimes p, we have
P +
(p a)>p 0:677
:
One anbemore pre isethan(1.2)inthefollowingexample(whi hwill
beapplied inx2).
Lemma 1. Let R<x 1=10 "
and QR<x 1 "
. Then
X
rR
(r;a)=1
(r) B
X
(q;a)=1 X
px
pa(qr)
1 xL 1
X
(q;a)=1 1
(qr)
xL A
:
Proof. Thisisaslightvariantof[4℄,Theorem9,andmaybeproved in
exa tlythe sameway.
Inthisandsubsequentstatements,qQisanabbreviationforQq <
2Q. Further, " is a suÆ iently small positive onstant, whi h we x on e
andforall,whileB isapositive onstant dependingat moston". We shall
frequently writeA for a suÆ ientlylarge positive onstant. If a statement
ontains A, it is true with every hoi e of A. It is to be understood that
x>x
1
where x
1
dependsat moston a;"and A.
WhenweuseU =O(V)ortheVinogradovnotationU V,theimplied
onstant willdependona;"andA. (Lemma2 isanex eptiontothisrule.)
The notationU V meansthat U V and V U. Onseveral o asions
we need a onstant that may depend on a, but does not depend on the
hoi e of " or A. This will be denoted by C
2
. Finally, we emphasize that
A;B and C
2
neednotbethesame at ea h o urren e.
2. The onstru tion. Let; 1=2<<0:55, bean absolute onstant
spe i edlaterinthisse tion. Let H betheintegersu h that
(2:1)
2 1
H
<" 2 1
H 1
:
De ne u by: u>0,
u H
=x 2 1
:
Clearly x
"=2
< u < x
"
. For ea h i = 1;:::;H let L be the set fl : lu;
(l;a)=1g. LetG be thesequen e fl
1 :::l
H :l
i
2Lg,sothat
(2:2)
X
l2G 1 x
2 1
; lx 2 1
(l2G):
Now letN denotethe numberof solutionsp;l;m;nof
(2:3) p a=lmn; l2G; mx 1
; (m;a)=1; px:
Thenumberof(p;l;m;n) ountedbyN forwhi hp ahasnoprimefa tor
x
isdenoted byN 0
. Now
N 0
N N N :
Here N
1
isthe numberof p;l;k;p
0
;nsatisfying
(2:4) p a=lp
0
kn; l2G; x
<p
0
2x 1
; k x 1
p 1
0
;
(k;a)=1; px
and N
2
is thenumberof p;l;m;p
0
;j satisfying
(2:5) p a=lmp
0
j; l2G; x
<p
0
2x 1
; mx 1
;
(m;a)=1; px:
In order to establish Theorem 1 we bound N N
1 N
2
from below. By
Lemma1,
(2:6) N = X
l2G X
mx 1
(m;a)=1 X
px
pa(lm)
1=(1+O(L 1
)) X
l2G X
mx 1
(m;a)=1 x
L (lm) :
Now
1
(lm)
= 1
(l)
!
l (m)
m
with !
l (m)=
Y
pjm
p-l
1 1
p
1
:
It is an elementaryexer ise to showthat
X
mM
(m;a)=1 m
1
!
l
(m)=G
l
log2+O( (a)M 1
logM);
where
G
l
=
(a)
a Y
p-la
1+ 1
p(p 1)
:
We note that
(2.7) G
l
>C
2
;
(2.8) N =(1+O(L
1
)) xlog2
L X
l2G G
l
(l) :
Now
(2:9) N
1
= X
l2G
X
x
<p
0
2x 1
X
jx 1
p 1
0
(j;a)=1
X
px
pa(modlp
0 j)
1:
Let () be a ertain monotoni fun tion on [ ;1 +"℄ whose de -
nition (involving multipleintegrals) we defer to x6. Let N
1
() denote the
ontributionto N
1
fromp
0
x
. Weshall showinx x4{6 that
(2:10) N
1 ()
()x
L X
l2G X
p
0
x
X
jx 1
p 1
0 x
(lp
0 j)
for 2 [ ;1 +"℄. Sin e (lp
0 j) = p
0
(lj)(1+O(L 1
)), an argument
similarto thatleading to (2.8) yields
N
1
xlog2
L
(1+O(L 1
)) X
l2G G
l
(l) X
h
hlog2
L
X
p
0
2 h
1
p
0
wherethesummation onditionon h is
x
=22 h
2x 1
:
A straightforward al ulation gives
(2:11) N
1
xlog2
L
1
\
()
d+"
X
l2G G
l
(l) :
Nowde ne N
2
() asthe ontributionto N
2
fromp
0
x
in(2.5).
Note that N
1
() ountssolutions of
p a=lp
0 jn
with p x, l 2 G, p
0
x
and j;n integer variables, n running over an
interval whose endpoints are x 1
. The last senten e is true if N
1 is
repla ed by N
2
, although the interval in questionis not the same. Never-
theless, the reader will readilyverify that pre iselythe same onstant will
arisefrom thesievemethodswe employbelow, thatis,
N
2
() () X
l2G X
p
0
x
X
mM
(m;a)=1 x
L (lmp
0 )
for2[ ;1 +"℄. By aslightvariant oftheargument leadingto (2.11),
we obtain
(2:12) N
2
xlog2
L
1
\
()
d+"
X
l2G G
l
(l) :
With =0:516, we are ableto obtaina de niteupperboundless than
1=2 for
1
\
()
d:
(The orrespondingboundis outof rea h if is repla edby0.296, forany
hoi e of.) Consequently,it iseasy to see that
N 0
>"x=L
Let (p)denotethenumberofo urren esofaparti ular pinthesolu-
tions ounted byN 0
. Obviously (p) (p a) B
. Sin e
N 0
= X
px
(p) n
X
px
(p)>0 1
X
px (p)
2 o
1=2
;
we readilyobtainthelowerbound
X
px
(p)>0 1>
2x
(logx) C
1
requiredforTheorem1.
The starting point for our sieve bounds is the following \fundamental
lemma". Let P(z)= Q
p<z p:
Lemma 2. Let z 2, s 2, y = z s
. There exist two real sequen es
f
d g
djP(z)
su h that
(i)
1
=1, j
d
j1,
d
=0 for dy;
(ii) for all DjP(z),
X
djD
+
d
0;
X
djD
d
0;
(iii)for all multipli ative fun tions f(d) satisfying
Y
wp<z
p-q
1 1
f(p)
1
K
logw
logz
for someq,some K1, some >0,and all z;w with z>w>1, wehave
X
djP(z)
(d;q)=1
d
f(d)
= Y
p<z
p-q
1 1
f(p)
f1+O(s s
)g:
Theimplied onstant depends onlyon K and .
Proof. This follows from the proof of Lemma 5 of [9℄, with the error
e s
improved,asintimated there,to s s
.
We alsotake a ombinatorial identityfrom Heath-Brown [10℄.
Lemma3.Let J 1and n<2X. Let (n)bevonMangoldt'sfun tion;
then
(n)= J
X
j=1 ( 1)
j 1
J
j
(2:13)
X
m1;:::;mjX 1=J
(m
1
):::(m
j )
X
::: X
n
1 :::n
j m
1 :::m
j
=n logn
1 :
3. Results on bilinear forms. It is onvenient to assemble in this
se tion resultsfrom [4℄,[5℄ and [2℄ thatweshall employ. Forsequen es
m
(m M) and
n
(n N) the onvolution sequen e (k) = (k) is
de nedby
(k)= X
mn=k
m
n
(MN k<4MN):
Foran arithmeti fun tionf we de ne
(f;q;a)= X
na(q)
nx f(n)
1
(q) X
(n;q)=1
nx
f(n):
LetK ;L;M bepositivenumbers,KLM =x and
min(K ;L;M)>x
with =exp( 2="). For sequen es =(
k
), k K, =(
l
), l L and
=(
m
),mM,we write
(K ;L;M;Q)= X
j(;q;a)j:
Similarlyifmin(K ;L) >x
and KL=x,we write
(K ;L;Q)= X
(q;a)=1
j(;q;a)j:
We need to assume that one onvolution fa tor is well distributed in
arithmeti progressionsinthe followingsense:
(A
1
) For any d1,k 1,b6=0,(k;b)=1 we have
X
lb(k)
(l;d)=1
l
= 1
(k) X
(l;dk)=1
l
+O(kk
B
(d)L 1=2
L A
);
where kk=( P
l j
l j
2
) 1=2
.
The sequen esthat we needto onsidersatisfythe ondition
(A
2
) j
l
j (l) B
:
Some sequen esaresupportedon \almost primes"inthesensethat
(A
3
)
l
=0whenever l has a prime fa tor less than exp(L=(logL) 2
):
In thenext lemma we needthehypothesis
(A
4
) L
1 "
X
j
l j
4
X
j
l j
2
2
:
Lemma 4. Let MN = x, min(M;N) > x
, and = (
m
), m M,
=(
n
), nN. Supposesatis es (A
1 ){(A
4
)and satis es (A
2
). Then
(M;N;Q) xL A
provided that
(3:1) x
" 1
Q 2
N x 5=6 "
Q 4=3
:
Proof. This followsfrom Theorem3 of [4℄.
Lemma 5. Let KLM = x, min(K ;L;M) > x
, = (
k
), k K,
=(
l
), lL, =(
m
), mM. Suppose that ;; satisfy (A
2 ),(A
3 ),
and satis es (A
1
). Suppose furtherthat
(3.2) Q KLx
"
;
(3.3) K
2
L 3
Qx 1 "
;
(3.4) K
4
L 2
(K+L) x 2 "
:
Then
(3:5) (K ;L;M;Q) xL
A
:
Proof. This followsfrom [5℄,Theorem 3.
Lemma 6. Let KLM = x, min(K ;L;M) > x
, = (
k
), k K,
=(
l
), lL, =(
m
), mM. Suppose that ;; satisfy (A
2 ), (A
3 )
and satis es (A
1
). Then we have (3:5) provided that (3:2) holds and
KL 2
Q 2
x 2 "
; (3:6)
K 5
L 2
x 2 "
: (3:7)
Proof. This isLemma 5of [2℄,a variantof Theorem4 of[5℄.
Lemma 7.Let MN =x, min(M;N) >x
. Let (
n
), nN,and (
q ),
q Q, be sequen es satisfying (A
2
) su h that (
n
) satis es (A
1
) and (A
3 ).
Then
(3:8)
X
rR X
mM
(r;am)=1
X
(q;am)=1
q
X
nN
mna(qr)
n 1
(qr) X
nN
(n;qr)=1
n
2
k k 2
xR 1
L A
provided that
(3:9) x
"
R N x
"
minfx 1=2
Q 1=2
;x 2
Q 5
R 1
;xQ 2
R 1=2
g:
Lemma 8.Let MN =x, min(M;N)>x
. Let (
n
), nN, and (
q ),
q Q, besequen es satisfying(A
2
) su h that (
n
) satis es (A
1 ), (A
3 ) and
(A
4
). Then(3:8) holds provided that
(3:10) x
"
R N x
"
min
xR
Q 2
1=2
;
x
Q
2=5
;
x 2
Q 3
1=4
:
Proof. This followsfrom Theorem2 of [4℄.
Lemma 9. Suppose that KLM =x, that (
k
), k K, and (
m
), m
M,satisfy (A
2 ) and
(3:11)
l
=
1; Ll<L
1 ,
0; otherwise,
whereL
1
2[L;2L). Then(3:5) holds provided that
(3.12) Q KLx
"
;
(3.13) MK
4
Q x 2 "
;
(3.14) MK
2
Q 2
x 2 "
:
Proof. This followsfrom Theorem5 of [5℄.
Let z
0
=exp(L=logL).
Lemma 10. Let MN = x. Suppose that (
q
);q Q, (Æ
r
);r R , and
(
n
), nN,satisfy (A
2
). Then
(3:15) X
qQ X
rR
(qr;a)=1
q Æ
r
X
mM X
nN
mna(qr)
m
n 1
(qr) X
mM X
nN
(mn;qr)=1
m
n
k kx 1=2 "
M 1=2
witheither of the hoi es
(3.16)
m
=1 (M m<M
1 );
m
=0 (M
1
m<2M);
(3.17)
m
=
1 for (m;P(z))=1,
0 for (m;P(z))>1 for some zz
0 ,
provided that
(3:18) M x
"
maxfQ;x 1
R 4
Q;Q 1=2
R ;x 2
Q 3
R 4
g:
Proof. This isa ombination of Theorem5and Theorem5* of [4℄.
Lemma 11.Let LMN =x. Supposethat
(3.19) LR x
1=2 "
;
1=2 "
Then
(3:21) X
rR
(r;a)=1 X
lL
(l;r)=1
X
(q;al)=1
X
mM X
nN
lmna(qr) 1
1
(qr) X
mM X
nN
(mn;qr)=1 1
xL A
:
Theanalogue of (3:21), in whi h the summation isrestri ted to l;m;n free
of prime fa tors <z,holds for ea h zz
0 .
Proof. This isa ombination of [4℄,Theorems7 and 7*.
Lemma 12. Let MN= x. Suppose that (Æ
r
);rR , (
m
);mM, and
(
n
);nN,satisfy (A
2
). Supposefurtherthat (
n
)satis es(A
1
)and (A
3 ),
q
=1 for Q<qQ
1
with Q<Q
1
2Q; QR x 1 "
and
x
"
R N x
"
(x=R ) 1=3
:
Then (3:15) holds.
Proof. This followsfrom [4℄,Theorem 6.
When we apply the lemmata of x3 below, the various onditions (A
1 ),
(A
2
) and soon are notdiÆ ult to verify, and we shall omitthe dis ussion
of this.
4. An asymptoti formula. In this se tion we sharpen Theorem 8
of [4℄a little, sothatwe an take
(4:1) ()=1+" ( <1=3 3"):
Let
1 and
2
be onstants. We suppose "is suÆ ientlysmall interms
of
1 and
2 .
Theorem 3. Suppose that
1
<1=3,
2
<1=5,
1 +
2
<29=56. Then
for any numbers
q
(q) B
;Æ
r
(r) B
,we have
X
qx
1 X
rx
2
(qr;a)=1
q Æ
r
(x;qr;a) x
(qr)
xL A
:
Here
(x;q;a)= X
nx
na(q)
(n):
We remark that Theorem 8 of [4℄ has one extra ondition on
1
;
2 ,
namely 5
1 +2
2
<2.
Proof. This follows [4℄, x15 relatively losely but we supply enough
theproblemvia Lemma3 (withJ =7) to showingthat
E = X
qQ X
rR
(qr;a)=1
q Æ
r
(qr;a) xL A
whereQ=x
1
;R=x
2
,
(q;a)=
X
m1:::mjn1:::nja(q)
(m
1
):::(m
j )
1
(q)
X
(m
1 :::m
j n
1 :::n
j
;q)=1
m
i 2M
i
;n
i 2N
i
(m
1
):::(m
j ):
Here and subsequently, P
is a summation restri ted to numbers free of
primefa tors lessthanz
0
,and we shallwrite
(4:2) M
i
=[(1 )M
i
;M
i ); N
i
=[(1 )N
i
;N
i )
with
(4:3) M
1 :::M
j N
1 :::N
j
=x; M
1
;:::;M
j
<x 1=7
and
(4:4) =L
A
:
We maysupposethat
1 +
2
>1=2 ". Asonp. 246 of[4℄,wesimplyhave
to de ompose ea h produ t M
1 :::M
j N
1 :::N
j
into blo ks in the range of
thevariablesof one ofthe lemmataofx 3.
Let M
i
=x
i
,N
i
=x
i
with
0
j
:::
1
1=7; 0
j
:::
1
;
1
+:::+
j +
1
+:::+
j
=1;
%
1
=2(
1 +
2
) 1; %
2
= 5
6 4
3 (
1 +
2 ); %
3
=
2
;
%
4
=min
1
2 +
1
2
2
1
; 2
5 (1
1 );
1
2 3
4
1
; %
5
=
1
; %
6
= 1
2 (1
2 ):
Wemaysuppose
1
+:::+
j
hasnopartialsumin[%
1 +";%
2
"℄[[%
3 +
";%
4
"℄[[%
5 +";%
6
"℄. For [%
1 +";%
2
"℄ thisfollows from Lemma 4;
for[%
3 +";%
4
"℄,from Lemma8. After verifyingthat
%
6
<min(2 5
2
1
;1 2
2
1
=2)
the result for [%
5 +";%
6
"℄ follows from Lemma 7. (We must employ
Cau hy's inequalityin onjun tionwith Lemmata7{8.)
Noti e that
% ""+maxf ; +4 1; =2+ ;3 +4 2g;
soif
1
>%
6
"thenLemma10isappli able. Consequently,wemayassume
that
1
<%
5 +":
Sin e 2(%
1
+") <%
2
", theterms of
1
+:::+
j
whi h are <%
2
"
give intotal with
<%
1 +";
of oursethese terms ontain all
1
;:::;
j
and possiblysome
i 's.
The remaining
i
must be lo ated either in I = [%
2
";%
3
+"℄ or in
J =[%
4
";%
5
+"℄. Any twonumbers 0
, 00
inI give
%
3
+"< 0
+ 00
<%
6
":
Thus 0
+ 00
must be inJ. Moreover, together withany 000
from J give
+ 000
<%
5 +%
1
+2"<%
6
"
so + 000
must be in J. From the above dis ussionit follows that we an
arrange
1
+:::+
j
asa sumof partialsums
1
+:::+
k
=1;
1
:::
k
;
ea h butat mostone lo atedin J, theex eptional one being inI. In fa t,
theex eptional one must existbe ause otherwise
k(%
4
")1; k(%
5
+")1;
giving 3<k<4. We on ludethat
(4:5) k =4;
1
;
2
;
3
2J;
4 2I:
Supposenow that
(4:6)
2 +
3
>
1 +
2 +":
We ndthatLemma 5 isappli ablewith K =x
3
,L=x
2
. We verify the
hypotheses(3.2){(3.4). Naturally(3.2) follows from (4.6). Next,
2
3 +3
2
5
3 (
1 +
2 +
3 )=
5
3 (1
4 )
5
3 (1 %
2
+")<1+
1 +
2
"
be ause
1 +
2
<29=56; thisestablishes(3.3). Next,
4
3 +3
2
7
3 (
1 +
2 +
3 )=
7
3 (1
4 )
7
3 (1 %
2
+")<2 "
be ause
1 +
2
<29=56.
It remainsto onsiderthe ase where(4.5) holdstogether with
(4:7)
2 +
3
1 +
2 +":
We have
1
>1 %
3
1
2
2"=1
1 2
2
2"and so
1 +
3
>1
1 2
2
2"+%
4
"
>
5
1 2
2
>
5 29 1
>
1 +
2 +":
Moreover, re alling (4.7), we have
1
<1 (
2 +
3 ) %
2 +"<
1
2
%
2 +"<
5
14
;
2
1 +5
3
=2(
1 +2
3 )+
3
<2(1 %
2
+")+
1 +
2
2
+"= 1
3 +
19
6 (
1 +
2
)+3"<2 ";
2
1 +
3
= 1
2 (
1 +2
3 )+
3
2
1
<
1
2 (1 %
2
+")+ 15
28
<
3
7 +
15
28
<2 2(
1 +
2 ) ";
be ause
1 +
2
< 29=56. We may now apply Lemma 6 to omplete the
proof ofTheorem3.
In what follows, we often hoosea number 2[0;2 1℄ andwrite the
integers l2G as
(4:8) l=dh
whered and h runoversubsets of
(4:9) dD; hx
2 1
D 1
:
Here
(4:10) x
Dx
+"
:
In deriving(4.1) fromTheorem 3we hoose
=min(2 1;1=3 ) 3":
In (2.9), ombinep
0
and d to give avariable x
1
,
+2 1
1
1=3 3":
Writing
2
=
1
,we ombine h and j to give avariable x
2
,
2
( +2 1)=1 <1=5 ":
Finally,
=
1 +
2
=0:516<29=56:
ThusTheorem 3is appli ableto N
1
()and gives(2.10) with ()=1+".
5. The sieve pro edure. Let A q
denote the arithmeti progression
fqk+a:(x a)=q k <(2x a)=qgand let
(5:1) B
q
=fn:nx; (n;q)=1g:
LetE
d
=fn:dn2Eg and
(5:2) S(E
q
d
;z)=
X
n2E q
d
;(n;P(z))=1 1
Fix;1=3 3"+". Itis onvenient towriteQforthesequen e
Q=flp
0
j:l2G; p
0
x
; jx 1
p 1
0 g;
sothat
(5:3) N
1 ()=
X
q2Q S(A
q
;(2x) 1=2
):
We atta ktheright-hand sideby omparing
(5:4) S
Q (z)=
X
q2Q S(A
q
;z) with S 0
Q (z)=
X
q2Q 1
(q) S(B
q
;z)
and
(5:5) S
Q
(C;z)= X
q2Q X
j2C S(A
q
p
1 :::p
j
;z)
with
(5:6) S
0
Q
(C;z)= X
q2Q 1
(q) X
j2C S(B
q
p
1 :::p
j
;z):
Here C is a subset of R h
,
h
denotes L 1
(logp
1
;:::;logp
h
), and z is a
fun tionofp
j
. Wemake the onventionthat thep
i
dividingq areex luded
from theinnersummationsin(5.5), (5.6);theex luded termsare zero. We
shallprovethat
S
Q
(z)=(1+C")S 0
Q
(z); jCjC
2
; (5:7)
S
Q
(C;z)=(1+C")S 0
Q
(C;z); jCjC
2
; (5:8)
undervarying onditionson C and z.
It is now ne essaryto re all a good dealof notationfrom [2℄. We write
T =fs0; t0; s+t1g
and
A
j
=A
j (
0
;
1 )
=f(
1
;:::;
j ):
0
j
<:::<
1
<
1
;
1
+:::+
j
1g:
Given any set H inR j
with
j
2H ) 0
j ,
1
+:::+
j
<1, we say
that H partitions into D T if for every
j
2 H , there exist I;J with
I[J f1;:::;jg, I\J =;,and
X
i2I
i
; X
j2J
j
2D:
If I [J = f1;:::;jg for every
j
2 H we say that H partitions exa tly
into D. WesaythatasubsetofR j
ispolyhedral ifitistheunionofat most
C onvexpolytopes.
Lemma 13. Let G 1
be the setof (s;t) in T satisfying
(5:9) 2 1+"s(5 8)=6 ":
Let G 2
be the set of (s;t) in T satisfying
(5.10) s+t+";
(5.11) 2s+3t1+ ";
(5.12) 5s+2t2 ";
(5.13) 4s+3t2 ":
Let G 3
be the set of (s;t) satisfying(5:10), (5:12) and
(5:14) s+2t2 2 ":
Let G be the set of (s;t) for whi h either (s;t) or (s;1 s t) belongs to
G 1
[G 2
[G 3
. Let G
j
= f
j 2 R
j
:
j
partitions into Gg. Let C be a
polyhedral subsetof G
j
andsupposemin
i
(
j
2C). Then
S
Q (C;p
j
)=(1+O(L A
))S 0
Q (C;p
j ):
Proof. ThisisavariantofLemma7of[2℄,and anbeprovedinexa tly
thesame way.
Let
(5:15) =
5 8
6
2"; =
3(1 )
5
":
LetS be theset of(s;t)in T forwhi h
(5:16) s1 "; s+2t2 2 "; s+4t2 ":
In theremainder ofthisse tion, A
j
isan abbreviationfor A
j
(; ). LetS
j
bethesubsetof R j
whi hpartitionsexa tlyintoS. LetU
j
bethesetof
j
inA
j
su h that(
1
;:::;
j
;2 1+")2S
j+1 .
Fors1 we writeg(s)=exp( slogs).
Lemma 14. Let C bea polyhedral subset of S
j
. Then
(5:17) S
Q (C;x
)=
1+Cg
"
S 0
Q (C;x
):
We also have
S
Q (x
)=
1+Cg
"
S 0
Q (x
):
Here jCjC
2 .
Proof. Thisisavariant ofLemma14 of[2℄;it an beprovedinexa tly
thesame way.
Lemma 15. Let C be a polyhedral subset of U
j
. Then (5:8) holds with
Proof. ThisisavariantofLemma15of[2℄. Wegivemostofthedetails,
sin ewe willbe re y ling thisproof ina modi edform later on. It suÆ es
to dis uss(5.8). By Bu hstab's identity,
(5:18) S
Q (C;x
)=S
Q (C;x
) S
Q (C
(j+1)
;p
j+1 )
with
C (j+1)
=f
j+1 2A
j+1 :
j
2C;
j+1
<g:
Here,and hen eforth,
j
maybeused for(
1
;:::;
j
) on e
j+l
=(
1
;:::
:::;
j
;:::;
j+l
) is given.
We writeC
j+1
forthepart ofC (j+1)
with
j+1
<2 1+"
and C 0
j+1
forthe omplementarypart. Thus
S
Q (C;x
)=S
Q (C;x
) S
Q (C
0
j+1
;p
j+1 ) S
Q (C
j+1
;p
j+1 ):
Lemma14isappli abletoS
Q (C;x
),andLemma13toS
Q (C
j+1
;p
j+1
)(sin e
j+1
<).
We now applyBu hstab's identity to S
Q (C
j+1
;p
j+1
). Ifwe ontinue in
thisfashion,weobtainasequen eofsumsS
Q (C
j+1
;p
j+1 ),S
Q (C
j+2
;p
j+2 );:::
with
C
k
=f
k 2A
k :
j
2C;
j+1
+:::+
k
<2 1+"g:
We have
(5:19) S
Q (C
k
;p
k )=S
Q (C
k
;x
) S
Q (C
0
k+1
;p
k+1 ) S
Q (C
k+1
;p
k+1 )
with
C 0
k+1
=f
k+1 2A
k+1 :
k 2C
k
;
k+1
<;
j+1
+:::+
k+1
2 1+"g:
Lemma 14is appli abletoS
Q (C
k
;x
),sin e
j 2U
j ,
j+1
+:::+
k
<
2 1+" gives
k 2S
k
. Lemma13 appliesto S
Q (C
0
k+1
;p
k+1
), be ause
2 1+"
j+1
+:::+
k
2 1+"+=(5 8)=6 ":
After <[
1
℄steps, C
k+1
isempty. We now ombine themainterms to
form thesum S 0
Q (C;x
) by applyingthe Bu hstab identityto thissum.
The total of the moduli of the error terms is at most C"S 0
(C;x
); the
reasoningforthisis exa tly ason p. 69 of [2℄. This ompletes theproof of
Lemma15.
Let
T
=f(s;t):3=7+"s1 "; 0t<(1 s)=2g;
U
j
=f
j 2A
j
(;1=2):
j
partitions exa tly into T
g:
Lemma 16.Let C beapolyhedral subsetof U
j
andlet z(p
j
)bea ontin-
uous fun tionon H with z(p )1=2. Then(5:8) holds. In parti ular,
this estimation applies to
X
q2Q
X
x 3=7+"
<p
1
<x 1 "
S(A q
p
1
;p
1 ):
Proof. Thisisavariant ofLemma18 of[2℄;it an beprovedinexa tly
thesame way.
Let !(t)denoteBu hstab's fun tion; ompare [2℄,p. 61.
Lemma 17.Let C=[1 ";1=2℄. Then
S
Q (C;p
1
) (1+C
2
")S 0
Q (C;p
1 )+xL
1
I
0 X
q2Q 1
(q) :
Here
I
0
=
\
D
!
2
1
!
1
1
2
3
3
d
1 d
2 d
3
2
1
2
3
and
D=f(
1
;
2
;
3 )62G
3
:
1
=2;1
1
2
1
;
3
min(
1
;(1
1
2 )=2)g:
Proof. This isa variant of (7.7) of [2℄, and an beproved inthe same
way.
Lemma 18. Let R bea polygonal subsetof
f(s;t):1=4tsmin(3=7;4 7 3");7 3+3"s+tg:
Then
S
Q (R ;x
)(1+C
2
")S 0
Q (R ;x
)+ x
L (I
1 +I
2 )
X
q2Q 1
(q) :
Here
I
1
=
\
3 2D
!
1
1
2
3
d
1
1 d
2
2 d
3
2
3
;
I
2
=
\
4 62G
4
j
(
1 +
2
;
3 +
4 )2R
!
1
1
2
3
d
1
1 d
2
2 d
3
3 d
4
2
4
;
D=f
3 :
j
;
3 62G
3
; (
1
;
2 +
3
)2R with
2
3
or(
1 +
2
;
3
)2R with
1
2 g:
Proof. This isa variant of Lemma 24of [2℄, and an be proved in the
We have not used in x 5 the spe ial properties of Q; we do this in the
next se tion.
6. The three-dimensional sieve. Let R 0
be a polyhedral set in R 2
su h that
(6:1) 4 7 3"
1
3
7 +";
1
2
+2"
2
7 3
2
+2"
for(
1
;
2 )2R
0
. We aim to givea good upperboundfor
X
q2Q X
2R 0
S(A q
p
1 p
2
;x
):
Asin[2℄, x8,we approa h thisindire tly bysieving thesequen es
H q
=fmwn:(logm;logw)2LR 0
; mwnx; (mwn;q)=1g;
F q
=f 2H q
: a(q)g:
The argument israther loseto thatof[2℄,x8,ex eptthat Lemma22is
displa ed by a ombination of results from x 3 above. The underlyingidea
isto approximatethequantity
V = X
rx
(r) X
l2G
X
kx 1
r 1
X
dK
d X
s
ds2F rlk
1
by
V 0
= X
rx
(r) X
l2G X
kx 1
r 1
1
(rlk) X
dK
d X
s
ds2H rlk
1:
Here (
d
) obeys (A
2 ) and
(6:2) K=x
2 1+2"
:
The followingresult plays therole ofLemma 22 of[2℄.
Lemma 19.With V;V 0
asabove, we have
V V
0
xL A
:
Proof. Ason p. 81of [2℄,itsuÆ es to prove theanalogousresultwith
H q
repla edby
fmwn:w2I; m2J; n2K ; (mwn;q)=1g;
I =[(1 )M;M); J =[(1 )W;W); K=[(1 )Y;Y);
whereMWN xand, inview of(6.1),
x
4 7 3"
M x 3=7+"
; (6:3)
(1 )=2+2" (7 3)=2+2"
We now employ(8.5) of [2℄. We needto showthat
X
rx
(r) X
l2G X
kx 1
X
dK
d X
efg=d X
zje
tjfe
(z)(t)
X
mwndzta(rlk)
em2I;wzf2J;ntg2K 1
1
(nlk)
X
(mwndzt;rlk)=1
em2I;wzf2J;ntg2K 1
xL A
:
We use the ombinatorial identityin Lemma 3, with J =7, to repla e
(r) by the right-handside of (2.13). By a furtherredu tion analogous to
thatemployed inx4,weneed to showthat, withM
i
;N
i
asin(4.2),
M
i
x =7
; M
1 :::M
j N
1 :::N
j
x
; R=[(1 )P;P); P x 1
;
we have
E :=
X
mi2Mi
(m
1
):::(m
j )
X
ni2Ni X
l2G X
k2R D(m
1 :::m
j n
1 :::n
j
lk) xL A
:
Here
D(q)= X
dK
d X
efg=d X
zje
tjfe
(z)(t)
X
mwndzta(q)
em2I;wzf2J
ntg2K 1
1
(q)
X
(mwndzt;q)=1
em2I;wzf2J
ntg2K 1
:
The portion of E with ztx
"
is easilyseen to be xL A
,so we may
on neattention to E 0
,thesubsumof E with
(6.5) zt<x
"
;
(6.6) wdzt2[(1 )L;L); m2[(1 )N;N):
Here
(6:7) L x
(7 3)=2+2 1+5"
=x
(11 5)=2+5"
inview of(6.4), (6.2), (6.5);similarly,
(6:8) N x
4 7 (2 1) 6"
=x
5 9 6"
:
It remainsto showthat, foranyof thepossibleM
1
;:::;M
j
;N
1
;:::;N
j ,
one of thelemmatainx3 yields
(6:9) E
0
xL A
:
Case 1: M
1 :::M
j N
1 :::N
j
P has a subprodu t U in [x
18 9+14"
;
x
5 9 7"
℄. We hoosev maximalinf1;:::;Hg su hthat
UL
1 :::L
v
x
5 9 7"
:
ClearlyQ=UL
1 :::L
v
satis es
(6:10) x
20 10+14"
Qx
5 9 7"
We apply Lemma 10, with this Q, and R x
Q 1
. We must verify
(3.18) with N inpla e ofM;that is,sin eQx 1=2
,
N max(Qx
"
;x
1+"+4
Q 3
;x
+"
Q 1=2
):
Thisis astraightforward onsequen eof (6.8) and (6.10).
Case2: WehaveN
1
x
5 9 7"
. Inthis ase,weapplyLemma11with
(6:11) Rx
N 1
1
x
10 5+7"
:
The onditions(3.19), (3.20) arereadilydedu ed from (6.7),(6.8), (6.11).
Case 3: Case1 and Case 2 do nothold. Sin e M
1 :::M
j N
1 :::N
j P
x 1
,thereisnosubprodu tin[x
8 4+8"
;x
10 19 15"
℄. NowM
i
x (1 )=7
,
P x 2=3
. Eitherthereisasubprodu tin[x
10 19 15"
;x
18 9+14"
℄,orall
M
i
;N
i
, and P are< x
8 4+8"
. Inthe latter ase we may form a subprod-
u t in [x
8 4+8"
;x
16 8+16"
℄ and this subprodu t must lie in [x
10 19 15"
;
x
16 8+16"
℄.
In either ase, we obtain a subprodu t x
in [x
10 19 15"
;x
18 9+14"
℄.
Taking the omplementarysubprodu tifneed be,we maysupposethat
(6:12) x
10 19 15"
x
x (1 )=2
:
At thispoint we usea simpleresultin realanalysis.
Lemma20.Let F;H be ontinuousreal fun tionson [0; ℄,F H. Then
[F(0);H( )℄ [
0Æ
[F(Æ);H(Æ)℄:
Proof. Let y 2[F(0);H( )℄. We must show that y 2[F(Æ);H(Æ)℄ for
some Æ 2[0; ℄. We may suppose that y >H(0). Sin e y H( ), we have
y=H(Æ) forsome Æ2[0; ℄; theresult follows.
LetÆbeanynumberin[0;2 1℄. Using(4.8){(4.10)weseethatLemma
7 is appli ablewith some R ; x
+Æ
R x
+Æ+"
. Let F;H bede ned on
[0;2 1℄ by
F(Æ)=+Æ+2";
H(Æ)=min 1
2 1
2
( Æ) ";2 5( Æ) (+Æ) ";
1 2( Æ) 1
2
(+Æ) "
=min(H
1 (Æ);H
2 (Æ);H
3 (Æ));
say. If W liesin[x F(Æ)
;x H(Æ)
℄, thenLemma7 yields(6.9).
Itisastraightforward onsequen eof(6.11)thatF(2 1)H
1
(2 1);
F(0)min(H
2 (0);H
3
(0)) andsoF H. NowLemma20yields(6.9) ifW
liesin[x F(0)
;x
H(2 1)
℄.
It is an easy dedu tionfrom (6.12) that
so(6.9) holdswhen
(6:13) x
+2"
W x
1 +"
:
We now obtaintheanalogous on lusion for
(6:14) x
1 +"
<W x
(7 3)=2+2"
;
bytakingQx
+2 1
;R x 1
inLemma7. This ompletestheproof
of Lemma19.
Lemma 21. Supposethat
r
2[0;1℄ (rx 2 1+"
). Then
X
q2Q X
rx 2 1+"
r S(F
q
r
;x
)
=
1+Cg
"
X
q2Q 1
(q) X
rx 2 1+"
r S(H
q
r
;x
)
with jCjC
2 .
Proof. FromLemma 19 itiseasy to see that
X
q2Q
X
dK
d X
s
ds2F q
1 1
(q) X
dK
d X
s
ds2H q
xL A
:
Let
%(d)= Y
pjd
3 3
p +
1
p 2
:
Then,fordK,
X
ds2H q
1=(1+O(L A
))
3
(q)
q 3
%(d)
d jH
q
j
by a variant of [2℄, (8.5). We apply Lemma 2 with z = x
, y = x
"
; om-
pare[2℄, proofof Lemma23. We have
X
q2Q X
rx 2 1+"
r S(F
q
r
;x
)
X
q2Q X
rx 2 1+"
r X
dx
"
djP(x
)
(d;q)=1
+
d X
ds2F q
1
X
q2Q 1
(q) X
rx 2 1+"
r X
dx
"
djP(x
)
+
d X
s
ds2H q
1+O(xL A
)
X
q2Q
2
(q)
q 3
X
rx 2 1+"
r X
dx
"
djP(x
)
(d;q)=1
d
%(d)
d jH
q
j
X
q2Q
2
(q)
q 3
X
rx 2 1+"
r jH
q
j Y
p<x
p6jq
1
%(p)
p
1+Cg
"
withjCjC
2
. Applyingthelowerboundsievein similarfashionto
X
q2Q 1
(q) X
rx 2 1+"
r S(H
q
;x
);
we obtain
X
q2Q X
rx 2 1+"
r S(F
q
r
;x
)
1+Cg
"
X
q2Q 1
(q) X
rx 2 1+"
r S(H
q
r
;x
)
withjCjC
2
. A lowerboundof thesame qualityis obtainedbya similar
argument.
Lemma 22. We have
X
q2Q S(F
q
;x
)=(1+C") X
q2Q 1
(q) S(H
q
;x
)
wherejCjC
2 .
Proof. This is a slight variant of the proof of Lemma 15; the role of
S
Q (C
k
;x
) is now played by
(6:15)
X
q2Q
X
k 2A
k
1 +:::+
k
<2 1+"
S(F q
p
1 :::p
k
;x
);
whi h byLemma 21 is
1+Cg
"
X
q2Q 1
(q)
X
k 2A
k
1+:::+
k
<2 1+"
S(H q
p1:::pk
;x
)
with jCj C
2
. The remainder of the proof may be arried through with
virtuallyno hange.
Lemma 23. The statement of Lemma 18 remains true if R is repla ed
by R 0
.
Proof. In view of Lemma 22 thismay be proved in exa tly the same
We now needto note thatLemmata 13,17, 18 and 23 an be enhan ed
byrepla ingG
j
byalarger setG
j
() thatdepends on; inLemma13, we
weaken the on lusionto
(6:16) S(C;p
j
)=(1+C")S 0
(C;p
j
); jCjC
2 :
Let
a()=max(2 1;1 );
b()=min
1+ 3
2
; 2 3
4
;
d()=min
+
2
;1
:
LetG 1
()betheunionofG 1
; [a();b()℄and[;d()℄. LetG
j
()bede ned
inthe same wayas G
j
, ex eptthat G 1
() repla esG 1
. The repla ement of
G
j by G
j
() is an appli ation of Lemmata 7 and 8. For any given in
[;+2 1℄, we may take
(6:17) x
R x
+"
; Qx
R 1
in Lemma 7; see (4.8){(4.10). Now (3.9) holds whenever (logN)=L lies in
[F
1 ();H
1
()℄, where
F
1
()=+2";
H
1
()=min(1=2 ( )=2;2 5( ) ;1 2( ) =2) ":
The onditionF
1
H
1
ofLemma 20is satis edprovidedthat
1 6"
and the union of the [F
1 ();H
1
()℄ taken over the permissible ontains
[+2";d() C"℄ withC C
2
. We may ignoretheterms in", inview of
theshapeof thebound(6.16),forexample.
In applyingLemma 8, we use (6.17) with in [1 ; ℄. Now
(3.10) holdswhenever(logN)=Llies in[F
2 ();H
2
()℄, where
F
2
()=+2"; H
2
()=min 1
2 +
1
2
( );
2
5 2
5
( );
1
2 3
4 ( )
":
The onditionF
2
H
2
issatis edprovidedthat
2 1+6"
and the union of [F
2 ();H
2
()℄ over the permissible ontains [a() +
6";b() "℄,leadingtotheenhan ementofLemmata13,17,18and23that
we laimed.
We are now in a position to establish the desired bound (2.10), start-
ing from the identity (5.3). For the sake of larity we shall suppress the
dependen e of sets A q
;A q
p
; et ., on q. It will be ta itly assumed that the
followingexpressionsareto besummedoverq 2Q. Wewillalso omit"for
brevity. We useBu hstab's identityto write
S(A;(2x) 1=2
)=S(A;x
)
X
1
3=7
162G1() S(A
p1
;p
1 )
X
q2Q
X
3=7
1
1 S(A
p1
;p
1 )
X
1 <
1
1=2 S(A
p1
;p
1 )
X
1
3=7
1 2G
1 ()
S(A
p
1
;p
1 )
=S
0 S
1 S
2 S
3 S
4
; say :
We treat S
0
;S
2
and S
3
in exa tly analogous fashion to S
1;0
;S
1;2 and S
1;3
in [2℄, pp. 86{88, usingthe enhan ed lemmata of x5 in pla e of the orre-
sponding lemmata in [2℄. By the de nition of G
1 ();S
4
an be evaluated
asymptoti allyinthesame senseasS
0
;S
2 .
We now turnto S
1
. In x 9 of[2℄ thepart ofthe orresponding sumwith
1
2[3(1 )=5;(31 15)=3℄[[4 7;3=7℄
is simply dis arded. It is vital for our bound on () that none of this
regionisdis arded;wemustonlydis ardsumswiththreeormorevariables.
Lemma 23 overs the interval [4 7;3=7℄ immediately. We shall dis over
thatasimplerole-reversalinthevariablesallowsustoapplyLemma23 for
thelowerintervalaswell.
Let I =[;3=7℄. Asin(9.2) of [2℄,
S
1
=
X
12I S(A
p1
;x
)+
X
2
2(A[C)nG
2 ();
1 2I
S(A
p1p2
;p
2 )
+ X
22G2()
12I S(A
p
1
;p
2 )+
X
2 2X
S(A
p
1 p
2
;p
2 ):
Here A;C are de nedason p. 54 of[2℄, whileX istheset of (
1
;
2 ) with
2
<min(
1
;(1
1
)=2);
1
2I; (
1
;
2
)62A[C[G
2 ():
We shallshowthat one of Lemmata 16, 18 or23 is appli ablethrough-
outX. Lemma18 overs thepartX
1
ofX with
3(1 )=5
1
4 7;
1 +
2
7 3:
Lemma 23 overs that part X
2
of X with
1
2 [4 7;3=7℄. For the
remainderof X ( ompare p. 84 of[2℄) we have
3(1 )=5 (31 15)=3; + <7 3:
Writing D
1
=Xn(X
1 [X
2
),we notethat
X
22D1 S(A
p
1 p
2
;p
2
)=jfp
1 p
2 p
3
2A:(logp
1
;logp
2
)2LD
1 gj
sin e p
1 p
3
2
> 2x for
2 2 D
1
. We may now ex hange the roles of the
variablesto give
(6:18)
X
2 2D
1 S(A
p
1 p
2
;p
2
)=(1+C") X
(
2
;
3 )2D
2 S(A
p
3 p
2
;p
2 )
where
3
=(logp
3 )L
1
,jCjC
2
. Thesum ontheright-hand sideextends
overthose p
2
;p
3
su h that
log
x
p
2 p
3
;logp
2
2LD
1 :
Sin e
2
62A[C[G
2
() for es
2
>(1 )=2;
1 +
2
>
6
5
(1 )>4=7,
we see thatLemma23 is appli able.
The reasoningon pp. 59{61, 86{88of [2℄now leadsto
S
Q ((2x)
1=2
) ()S 0
Q ((2x)
1=2
):
Here
()=1+ I
1 (X
1 [X
2 )+I
2 (X
1 [X
2 )+I
1 (D
2 )+I
2 (D
2 )
(6:19)
+E
1;3
()+E
3;4
()+E
5;1
()+C"
(jCjC
2
) withthe followingde nitions:
(i)I
1
(Z) andI
2
(Z)arede nedinthesame wayasI
1 and I
2
ofLemma
18 withR repla ed byZ and
(6.20) G
j
()removed fromthedomainofintegration ofea hj-dimensio-
nal integral,
(ii) E
m;n
() isde nedinthesame way asE
m;n
onp. 88 of [2℄,subje t
to theadditional ondition(6.20).
Now, byarguingasinx 2, we ndthat
S 0
Q ((2x)
1=2
)= X
q2Q 1
(q) X
px
1=(1+O(L 1
)) x
L X
q2Q 1
(q) :
Now(2.10) followswith () de ned asin(6.19).
Re all that (2.10) holds with () = 1 +" for 1=3 3".
A omputer al ulationyields
1
\
()
dlog
1=3
0:2961
+ 0:484
\
1=3 ()
d+C"<0:4999:
7. Shifted primes with a large prime fa tor. In order to obtain
Theorem2 we mustshowthat
(7:1)
X
x 1=2
<px 0:677
(x;p;a)logp<(1=2 3")x;
ompare [2℄,p. 43.
For 2 [1=2;0:6℄, let P =P() be the set of primes p x
. We shall
showthat
(7:2)
X
q2P S(A
q
;(2x) 1=2
)G() X
q2P 1
(q) :
Here G() is a monotoni fun tion, identi al with C
2
() in [2℄ ex ept for
2L=[25=49 ";92=175 "℄=[0:5102:::;0:5257:::℄,and
(7:3)
0:6
\
1=2
G()d<0:2391:
A ording to Fouvry [6℄,
(7:4) 1
x
X
x 3=5
p<x 0:677
(x;p;a)logp<8log
12
15 50:677
+"<0:26088:
It isnowa straightforwardmatter to dedu e(7.1) from (7.2){(7.4).
InL,thefun tionG()willbeobtainedbysubtra tingaone-dimensional
integralfromC
2
(),whileaddingmu hsmallerthree-dimensionalandfour-
dimensional integrals. This will be made pre ise below. A omputer al-
ulation shows that the one-dimensional integral, after integration over L,
yieldsa saving justin ex ess of 210 3
, whilethe orresponding lossfrom
three-dimensionaland four-dimensionalintegrals is<410 5
. Sin e
0:6
\
1=2 C
2
()d <0:241
by[2℄,(1.2), we readilyobtain(7.3).
We now establish (7.2), beginning with the observation that Lemmata
13{18hold (withthe sameproofs)ifQ isrepla ed byP.
Let 2L. LetR
0
be apolygonalregioninR 2
su h that
(7.5) max
19 7
7
;
50 19
17
+24"
1
3
7 +";
(7.6) 2
1
=3 "
2
(1
1 )=2;
(7.7)
1 +4
2
3 3 "
for(
1
;
2 )2R
0
. (For orientation,note that0:384
1
0:429.)
Lemma 24.Lemma 18 holds with Q;R repla ed by P;R .
Proof. Let
H q
0
=fmwn:(logm;logw)2LR
0
;mwnx;(mwn;q)=1g;
F q
0
=f 2H q
0
: a(q)g:
OurstrategyistoestablishtheanaloguesofLemmata21{23,withP;H q
0
;F q
0
inpla eofQ;H q
;F q
. Theonlyingredientofthisargumentforwhi hdetails
needbesuppliedisthe followingvariantof Lemma19.
Lemma 25. With K as in (6:2) we have
X
q2P
X
dK
d X
s
ds2F q
s 1
1
(q) X
dK
d X
s
ds2H q
s 1
xL A
:
Proof. As inx 6 we redu e theproof to showingthat
(7:8)
X
m
i 2M
i
(m
1
):::(m
j )
X
n
i 2N
i D
0
(m
1 :::m
j n
1 :::n
j
) xL A
withM
i
;N
i
asin(4.2),
(7:9) M
1 :::M
j N
1 :::N
j
x
; M
i
x
=7
:
Here D 0
(q)is de nedsimilarlyto D(q)inx 6, with
I =[(1 )x
1
;x
1
); J =[(1 )x
2
;x
2
); K=[(1 )Y;Y);
(
1
;
2
) satis es(7.5){(7.7); while(6.5) and
(7:10) mx
3
areadditional onditionsimposedon thevariablesinD 0
(q). We notethat
(7:11)
1
2+1 3"
3
1 :
It remainsonlytoshowthatthevariablesfallwithinrangestowhi hwe
mayapplyone ofLemmata 7,10 and 11. Let
(7:12) =max((6
1
2)=3;6 2 2
1 )+8":
Case 1: M
1 :::M
j N
1 :::N
j
has a subprodu tx
1
in [x
;x
1
2+1 4"
℄.
In this ase Lemma10 is appli able,sin e
1
<1=2 and
3
max (
1
;4 3
1
1;
1
=2)+"
inview of(7.11) and thede nitionof .
Case 2: We have N
1
x
1 2+1 4"
. In this ase, we applyLemma 11
with
R x
N 1
1
x 3 1
1 +4"
;
while,re alling (6.6),(6.5), (6.2), (7.6),
2
+2 1+3" 2
1
=2 1=2+3"
Thus
R L x
5 3=2 3
1
=2+7"
x 1=2 "
sin e
1
>(19 7)=7>(10 4)=3; while
R L 1=2
x
4 5=4 5
1
=4+6"
x
1
2+1 4"
x
3
"
;
sin e
1
>(19 7)=7>(8 3)=3.
Suppose that neither Case 1 nor Case 2 holds. We shall show that
M
1 :::M
j N
1 :::N
j
hasa subprodu tx
1
with
(7:13) "
1
=2+":
Forsupposethisisnotthe ase. ThenM
1 :::M
j N
1 :::N
j
learlyhasnosub-
produ t in[x
"
;x
℄[[x
;x
1 2+1 4"
℄=[x
"
;x
1 2+1 4"
℄. More-
over, there is no subprodu t in [x 3 1
1 +5"
;x
"
℄, and hen e none in
[x
3 1 1+5"
;x
1 2+1 4"
℄. Sin e Case 2 doesnot hold,all M
i
and N
i are
lessthanx 3 1
1 3"
. Wenow readilyobtaina ontradi tion sin e
2(3 1
1
+5")<
1
2+1 4"
from (7.5).
We shall now show (with
1
as in (7.13)) that Lemma 7 is appli able
withR=x
1
;Qx
1
and x
2
inpla e ofN. It suÆ esto showthat
(7:14) =2+2"
2
g "
where
g=min(1 2+3
1
=2;(1 +
1
)=2;2 5+4
1 ):
The left-hand inequalityin(7.14) isan easy onsequen e of
2
2
1
=3 "(38 14)=21 ":
Asfortheright-hand inequalityin(7.14), itsuÆ es to verifythat
(7:15)
1
1
2
min
1 2+
3( ")
2
;
1 +( ")
2
;2 5+4( ")
"
be auseof (7.6) and (7.13).
For larity,weseparatethe ases14=27and>14=27. If<14=27,
then
1
19 7
7
+24";
1
1
2
14 19
14
12";
max
6 2 19 7
;6 2
38 14
= 4
;
min 1
2 3
2
; 1
2
;2 4 5"
min
1
2 6
7
; 1
2 2
7
;2 16
7
5"
>
14 19
14
12" 1
1
2
;
whi h establishes(7.15). If14=27, then
1
50 19
17
+24";
1
1
2
18 25
17
12";
max
6 2
3
50 19
51
;6 2
100 38
17
= 2+4
17
;
min
1
2 3
2
; 1
2
;2 4
5"
min
1
2
3+6
17
; 1
2
+2
17
;2
8+16
17
5"
>
18 25
17
12" 1
1
2 :
Thus (7.15) holds in both ases, and the proof of Lemma 25 is omplete.
Indeed, the proof of Lemma 24 now goes through in the same fashion as
thatof Lemma 23.
Wenowturntotheestimate(7.2),whi hisour nalobje tive. Wehave
X
q2P S(A
q
;(2x) 1=2
)= X
q2P S(A
q
;x
) X
q2P
X
1
3=7+"
S(A q
p1
;p
1 )
X
q2P
X
3=7+" 11 "
S(A q
p
1
;p
1 )
X
q2P
X
1 "<
1
1=2 S(A
q
p1
;p
1 )
=T
0 T
1 T
2 T
3
; say.
We treat T
0
;T
2
, and T
3
in exa tly analogous fashion to S
1;0
;S
1;2 and S
1;3
in[2℄,pp. 86{88, usingthelemmataof x5 withQ repla ed by P.
We now turn to T
1
. Let D() be the set of
1
in [;3=7℄ for whi h
S(A q
p1
;p
1
) is simply dis arded in treating S
1;2
in [2℄. Disregarding " as
inx6,
D()= 8
>
<
>
:
[4 7;3=7℄ for25=4921=41,
[(3 3)=5;(31 15)=3℄[[4 7;3=7℄ for21=41<16=31,
[(3 3)=5;3=7℄ for16=31<11=21,
We an \salvage"theinterse tionof D() with
I()=
[(19 7)=7;3=7℄ for25=4914=27,
[(50 19)=17;3=7℄ for14=27<92=175.
Of ourse92=175 isthevalue of at whi h I()vanishes. That is,we have
X
q2P
X
1
2D()\I() S(A
q
p
1
;p
1 )
= X
q2P
X
12D()\I() S(A
q
p
1
;x
)+ X
q2P
X
12D()\I()
2 2C[D
S(A q
p
1 p
2
;x
)
+ X
q2P
X
12D()\I()
2 2B[E
S(A q
p
1 p
2
;p
2 )
X
q2P
X
12D()\I()
3
<
2
2 2C[D
S(A q
p
1 p
2 p
3
;p
3 )
= T
1;1 +T
1;2 +T
1;3 T
1;4
; say :
We an give asymptoti formulae for T
1;1
by Lemma 15, and for T
1;3 and
thatpartof T
1;4
forwhi h
3 2G
3
,byLemma 13. We applyLemma15 to
thepart ofT
1;2
with
2
2C. We thenapplyLemma 24 to thepartof T
1;2
with
2
2D;thusR
0
=f
2
2D:
1
2D()\I()g. (Itisreadilyveri ed
that(7.5){(7.7) hold.) Wesimplydis ardtheportionof T
1;4
with
3 62G
3 .
ThusG() isobtainedfrom theupperboundC
2
() of [2℄bysubtra ting
\
D()\I() d
1
1 (1
1 )
;
and adding (i) the integrals orresponding to 1
I
1
; 1
I
2
in Lemma 18
with R
0
in pla e of R ; (ii)the integrals arisingfrom the dis ardedportion
of T
1;4
. Thisestablishes(7.2), and Theorem2 follows.
Referen es
[1℄ W. R. Alford, A. Granville and C. Pomeran e, There are in nitely many
Carmi haelnumbers,Ann.ofMath.139(1994),703{722.
[2℄ R. C. Baker and G. Harman, The Brun{Tit hmarsh theorem on average, in:
Analyti NumberTheory,Vol.I,Birkhauser,Boston,1996,39{103.
[3℄ A.Balog,p+awithoutlargeprimefa tors,Sem.TheoriedesNombresBordeaux
(1983-84),expose31.
[4℄ E.Bombieri,J.FriedlanderandH.Iwanie ,Primesinarithmeti progressions
tolargemoduli,A taMath.156(1986),203{251.
[5℄ |,|, |,Primesin arithmeti progressions tolarge moduli II,Math. Ann. 277
(1987),361{393.
[6℄ E. Fouvry, Theoremede Brun{Tit hmarsh; appli ationau theoreme de Fermat,
[7℄ E. Fouvry and F. Grupp,On the swit hing prin iple in sieve theory, J.Reine
Angew.Math.370(1986),101{126.
[8℄ J. Friedlander, Shifted primes without large primefa tors, in: NumberTheory
andAppli ations,1989,Kluwer,Berlin,1990,393{401.
[9℄ J.B.FriedlanderandH.Iwanie ,OnBombieri'sasymptoti sieve,Ann.S uola
Norm.Sup.Pisa 5(1978),719{756.
[10℄ D.R.Heath-Brown,Thenumberofprimesinashortinterval,J.ReineAngew.
Math.389(1988),22{63.
[11℄ C.Pomeran e,PopularvaluesofEuler'sfun tion,Mathematika27(1980),84{89.
DepartmentofMathemati s S hoolofMathemati s
BrighamYoungUniversity UniversityofWales
Provo,Utah84602 College ofCardi
U.S.A. SenghennyddRoad
E-mail:bakermath.byu.edu Cardi CF24AG,U.K.
E-mail:harman f.a .uk
Re eivedon4.12.1996
andinrevisedformon12.5.1997 (3091)
