1 Introduction
Let
$f_{k}(x)=\unicode[STIX]{x1D6FC}x^{k}+\cdots +\unicode[STIX]{x1D6FD}x+\unicode[STIX]{x1D6FE}$
be a polynomial of degree
$k>1$
with irrational leading coefficient. Inequalities of the form
$$\begin{eqnarray}\Vert f_{k}(p)\Vert <p^{-\unicode[STIX]{x1D70C}_{k}}\end{eqnarray}$$
for infinitely many primes
$p$
were studied by Vinogradov [Reference Vinogradov11]; see [Reference Baker1] for the strongest available results. The present paper gives a new result for
$k=2$
.
Theorem 1. Let
$\unicode[STIX]{x1D70C}_{2}=3/17-\unicode[STIX]{x1D700}=0.1764\ldots ;$
then (1.1) holds for infinitely many primes
$p$
.
The values
$\unicode[STIX]{x1D70C}_{2}=1/8-\unicode[STIX]{x1D700}$
,
$3/20$
,
$2/13=0.1538\ldots$
were given by Ghosh [Reference Ghosh6], Baker and Harman [Reference Baker and Harman2], and Harman [Reference Harman7] in the case
$\unicode[STIX]{x1D6FD}=0$
; Baker [Reference Baker1] extended the last result to general
$\unicode[STIX]{x1D6FD}$
.
As in [Reference Baker1, Reference Baker and Harman2, Reference Harman7] we use the Harman sieve. We make progress in the present paper by giving new bounds for sums of the shape
$$\begin{eqnarray}\mathop{\sum }_{\ell \leqslant L}c_{\ell }\mathop{\sum }_{R<r\leqslant 2R}a_{r}\mathop{\sum }_{N/2<sr\leqslant N}b_{s}e(\ell g(rs)),\end{eqnarray}$$
where
$g$
is the approximating polynomial to
$f_{2}$
in [Reference Baker1]. Type I sums (in which
$b_{s}\equiv 1$
) are treated in §2, and general (Type II) sums in §3. In the Type I case,
$(a_{r})_{R<r\leqslant 2R}$
is restricted to convolutions of shorter sequences; a lemma of Birch and Davenport [Reference Birch and Davenport4] on Diophantine approximation plays a key role. For Type II sums, a subsidiary task is the study of the average behavior, as
$n$
varies, of the number of solutions of
$$\begin{eqnarray}\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2}=n\quad (\ell _{1},\ell _{2}\leqslant L;\;y_{1},y_{2}\in [Y,2Y)).\end{eqnarray}$$
One sum that eludes the Harman sieve takes the form
$$\begin{eqnarray}\mathop{\sum }_{\substack{ (2N)^{1/2-9\unicode[STIX]{x1D70E}}\leqslant p<(2N)^{3/8+33\unicode[STIX]{x1D70E}/4} \\ pp^{\prime }\in (N/2,N],\Vert g(pp^{\prime })\Vert \leqslant \unicode[STIX]{x1D6FF}}}1\end{eqnarray}$$
(where
$\unicode[STIX]{x1D70C}=\frac{3}{17}$
,
$\unicode[STIX]{x1D70E}=\unicode[STIX]{x1D70C}-\frac{1}{6}$
and
$\unicode[STIX]{x1D6FF}=\frac{1}{2}\,N^{-\unicode[STIX]{x1D70C}+\unicode[STIX]{x1D700}/2}$
). We bound it above using the form of the linear sieve given by Iwaniec [Reference Iwaniec10]. This sum is treated in §5; the sums accessible via the Harman sieve are in §4. Section 6 contains the sieve decomposition of
$S({\mathcal{A}},(2N)^{1/2})$
(defined below), and the calculations leading to Theorem 1.
The following notation will be used:
$\langle a,b\rangle =\text{g.c.d.}$
of
$a$
,
$b$
.
$\Vert \unicode[STIX]{x1D703}\Vert =\min _{n\in \mathbb{Z}}|\unicode[STIX]{x1D703}-n|$
.
$|{\mathcal{E}}|=\sum _{n\in {\mathcal{E}}}1({\mathcal{E}}\subset [1,N])$
.
$\unicode[STIX]{x1D712}_{{\mathcal{E}}}=$
indicator function of
${\mathcal{E}}$
.
$C\ldots$
absolute constant, not the same at each occurrence.
$\unicode[STIX]{x1D706}\ldots$
real number with
$|\unicode[STIX]{x1D706}|\leqslant C$
, not the same at each occurrence.
$\unicode[STIX]{x1D700}\ldots$
sufficiently small positive number;
$\unicode[STIX]{x1D702}=\unicode[STIX]{x1D700}^{9}$
.
$\gg ,\ll \ldots$
indicate implied constants that may depend on
$\unicode[STIX]{x1D700}$
.
$y\sim Y\ldots$
indicates
$Y<y\leqslant 2Y$
.
$e(\unicode[STIX]{x1D703})=\text{e}^{2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D703}}$
.
$a/q\ldots$
fraction in lowest terms with
$|\unicode[STIX]{x1D6FC}-a/q|<1/q^{2}$
, with
$q$
sufficiently large;
$g(x)=(a/q)x^{2}+\unicode[STIX]{x1D6FD}x+\unicode[STIX]{x1D6FE}$
.
We choose
$N$
so that
$$\begin{eqnarray}L_{1}^{1/2}N\ll q\ll L_{1}^{1/2}N\quad \text{with }L_{1}=2N^{\unicode[STIX]{x1D70C}-\unicode[STIX]{x1D700}/2}\end{eqnarray}$$
and write
$$\begin{eqnarray}L=N^{\unicode[STIX]{x1D70C}-\unicode[STIX]{x1D700}/3}.\end{eqnarray}$$
Here
$\unicode[STIX]{x1D70C}$
will ultimately be
$\frac{3}{17}$
; earlier in the paper we restrict
$\unicode[STIX]{x1D70C}$
somewhat less. We do suppose
$\unicode[STIX]{x1D70C}>\frac{1}{6}$
, and write
$$\begin{eqnarray}\unicode[STIX]{x1D70E}=\unicode[STIX]{x1D70C}-1/6.\end{eqnarray}$$
We reserve the symbols
$p,p_{1},p^{\prime },\ldots$
for prime numbers.
Let
$\unicode[STIX]{x1D6FF}=L_{1}^{-1}$
and
$$\begin{eqnarray}{\mathcal{A}}=\biggl\{n:n\sim \frac{N}{2},\Vert g(n)\Vert <\unicode[STIX]{x1D6FF}\biggr\},\qquad {\mathcal{B}}=\biggl\{n:n\sim \frac{N}{2}\biggr\}.\end{eqnarray}$$
We write
$I(m)$
for an arbitrary subinterval of
$(N/2m,N/m]$
.
2 Type I sums
The object of this section is to prove the following theorem.
Theorem 2. Let
$\frac{1}{6}<\unicode[STIX]{x1D70C}<\frac{2}{11}$
. Let
$V\geqslant 1,W\geqslant 1$
,
$$\begin{eqnarray}V^{3}W^{2}\ll N^{2-3\unicode[STIX]{x1D70C}},\qquad VW^{3}\ll N^{3-15\unicode[STIX]{x1D70C}/2},\qquad VW\ll N^{2-8\unicode[STIX]{x1D70C}}.\end{eqnarray}$$
Let
$$\begin{eqnarray}T:=\mathop{\sum }_{\ell =1}^{L}\mathop{\sum }_{v\sim V}\mathop{\sum }_{w\sim W}\biggl|\mathop{\sum }_{n\unicode[STIX]{x1D716}I(vw)}e(\ell g(vwn))\biggr|.\end{eqnarray}$$
Then
$$\begin{eqnarray}T\ll N^{1-10\unicode[STIX]{x1D702}}.\end{eqnarray}$$
We require several lemmas.
Lemma 1. Let
$1\leqslant Y\ll N^{1-2\unicode[STIX]{x1D70C}}$
. Let
${\mathcal{S}}$
be the set of
$y\in (Y,4Y]$
with
$\langle y,q\rangle \leqslant N^{\unicode[STIX]{x1D70C}}$
and
$$\begin{eqnarray}T(y):=\mathop{\sum }_{\ell =1}^{L}\biggl|\mathop{\sum }_{n\in I(y)}e(\ell g(yn))\biggr|>N^{1-10\unicode[STIX]{x1D702}}Y^{-1}.\end{eqnarray}$$
There are a set
${\mathcal{S}}^{\ast }\subset {\mathcal{S}}$
and positive numbers
$S$
,
$Z$
with the following properties:
(i) we have
(2.2)
$$\begin{eqnarray}\mathop{\sum }_{y\in {\mathcal{S}}}T(y)\ll L^{1/2}N^{10\unicode[STIX]{x1D702}}Z^{1/2}|{\mathcal{S}}^{\ast }|;\end{eqnarray}$$
(ii) for
$y\in {\mathcal{S}}^{\ast }$
, we have (2.3)for some$$\begin{eqnarray}\biggl|\frac{sa\,y^{2}}{q}-u\biggr|<Z^{-1}\end{eqnarray}$$
$s=s(y)$
in
$\mathbb{N},u=u(y)$
in
$\mathbb{Z}$
,
$\langle u,s\rangle =1$
, and (2.4)$$\begin{eqnarray}s\sim S\ll L^{2}N^{\unicode[STIX]{x1D702}};\end{eqnarray}$$
(iii)
$Z$
satisfies (2.5)$$\begin{eqnarray}(N/Y)^{2}L^{-1}N^{-\unicode[STIX]{x1D702}}\ll Z\ll LS^{-1}(N/Y)^{2}.\end{eqnarray}$$
Proof. This can readily be extracted from the proof of [Reference Baker1, Lemma 8], with
$5\unicode[STIX]{x1D702}$
in place of
$\unicode[STIX]{x1D702}$
.◻
Lemma 2. Let
$\unicode[STIX]{x1D703}$
be a real number and suppose there exist
$R$
distinct integer pairs
$x,z$
satisfying
$$\begin{eqnarray}|\unicode[STIX]{x1D703}x-z|<\unicode[STIX]{x1D701},\quad 0<|x|<X,\end{eqnarray}$$
where
$R\geqslant 24\unicode[STIX]{x1D701}X>0$
. Then all integer pairs
$x$
,
$z$
satisfying (2.6) have the same ratio
$z/x$
.
Proof. Birch and Davenport [Reference Birch and Davenport4]. ◻
Lemma 3. Suppose that
$s\geqslant 1$
,
$D\geqslant 1$
,
$sD^{2}<q$
,
$Z\geqslant 2$
. The number of solutions
$y\in (Y,4Y]$
, with
$\langle y,q\rangle \leqslant D$
, of the inequality
$$\begin{eqnarray}\biggl\|\frac{sa\,y^{2}}{q}\biggr\|<\frac{1}{Z}\end{eqnarray}$$
is
$$\begin{eqnarray}\ll N^{\unicode[STIX]{x1D702}}\biggl(\frac{Y+q^{1/2}}{Z^{1/2}}\biggr).\end{eqnarray}$$
Proof. [Reference Baker1, Lemma 3]. ◻
Proof of Theorem 2.
In the notation of Lemma 1, with
$Y=VW$
, let
${\mathcal{C}}$
be the set of pairs
$(v,w)$
for which
$v\sim V$
,
$w\sim W$
,
$vw\in {\mathcal{S}}^{\ast }$
. As in [Reference Baker1, proof of Lemma 8], it suffices to prove that
$$\begin{eqnarray}\mathop{\sum }_{(v,w)\in {\mathcal{C}}}T(vw)\ll N^{1-10\unicode[STIX]{x1D702}}.\end{eqnarray}$$
In view of the Type I result obtained in [Reference Baker1, Lemma 8], we may suppose that
$$\begin{eqnarray}VW>N^{1-5\unicode[STIX]{x1D70C}/2-\unicode[STIX]{x1D702}}.\end{eqnarray}$$
We note that
$$\begin{eqnarray}|{\mathcal{C}}|\ll \frac{N^{2\unicode[STIX]{x1D702}}S(VW+q^{1/2})}{Z^{1/2}}\end{eqnarray}$$
as a consequence of Lemma 3; indeed
$$\begin{eqnarray}|{\mathcal{C}}|\ll \frac{N^{2\unicode[STIX]{x1D702}}SVW}{Z^{1/2}},\end{eqnarray}$$
since
$VW>N^{1-5\unicode[STIX]{x1D70C}/2-\unicode[STIX]{x1D702}}>N^{1/2+\unicode[STIX]{x1D70C}/4}\gg q^{1/2}$
for
$\unicode[STIX]{x1D700}$
sufficiently small.
Suppose for a moment that
$$\begin{eqnarray}S<N^{1-23\unicode[STIX]{x1D702}}(VW)^{-1}L^{-1/2};\end{eqnarray}$$
then (2.2) gives (with a divisor argument)
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{(v,w)\in {\mathcal{C}}}T(vw) & \ll & \displaystyle L^{1/2}N^{11\unicode[STIX]{x1D702}}Z^{1/2}|{\mathcal{C}}|\nonumber\\ \displaystyle & \ll & \displaystyle N^{1-10\unicode[STIX]{x1D702}}\nonumber\end{eqnarray}$$
(from (2.8), (2.9)), giving (2.7). So we may suppose that
$$\begin{eqnarray}S\geqslant N^{1-23\unicode[STIX]{x1D702}}(VW)^{-1}L^{-1/2}.\end{eqnarray}$$
It now follows from (2.5) that
$$\begin{eqnarray}Z\ll \biggl(\frac{N}{VW}\biggr)^{2}L\,\frac{VWL^{1/2}}{N^{1-23\unicode[STIX]{x1D702}}}=\frac{N^{1+23\unicode[STIX]{x1D702}}}{VW}\,L^{3/2}.\end{eqnarray}$$
Now let
$$\begin{eqnarray}{\mathcal{C}}(w)=\{v\sim V:(v,w)\in {\mathcal{C}}\}\end{eqnarray}$$
and for
$K\geqslant$
1, let
$$\begin{eqnarray}{\mathcal{E}}(K)=\{w\sim W:K\leqslant |{\mathcal{C}}(w)|<2K\}.\end{eqnarray}$$
Then
$$\begin{eqnarray}\displaystyle |{\mathcal{C}}| & = & \displaystyle \mathop{\sum }_{w\sim W}|{\mathcal{C}}(w)|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \mathop{\sum }_{K=2^{m}\in [1,V]}2K|{\mathcal{E}}(K)|.\nonumber\end{eqnarray}$$
We choose
$K,1\leqslant K\leqslant V$
, so that
$$\begin{eqnarray}|{\mathcal{C}}|\ll \log N|{\mathcal{E}}(K)|K.\end{eqnarray}$$
Suppose for a moment that
$$\begin{eqnarray}L^{1/2}Z^{1/2}KW<N^{1-22\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Then arguing as above,
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{(v,w)\in {\mathcal{C}}}T(vw) & \ll & \displaystyle L^{1/2}N^{11\unicode[STIX]{x1D702}}Z^{1/2}\log N|{\mathcal{E}}(K)|K\nonumber\\ \displaystyle & \ll & \displaystyle N^{12\unicode[STIX]{x1D702}}L^{1/2}Z^{1/2}WK\ll N^{1-10\unicode[STIX]{x1D702}}.\nonumber\end{eqnarray}$$
Thus we may suppose that
$$\begin{eqnarray}K\geqslant \frac{N^{1-22\unicode[STIX]{x1D702}}}{L^{1/2}Z^{1/2}W}.\end{eqnarray}$$
For the next stage of the argument, let
$w$
be a fixed integer in
${\mathcal{E}}(K)$
. We apply Lemma 2, taking
$$\begin{eqnarray}\unicode[STIX]{x1D703}=\frac{w^{2}a}{q},\qquad X=8SV^{2},\qquad \unicode[STIX]{x1D701}=\frac{1}{Z}\end{eqnarray}$$
since (by (2.3) and the definition of
${\mathcal{C}}$
)
$$\begin{eqnarray}|(sv^{2})\unicode[STIX]{x1D703}-u|<\frac{1}{Z},\quad s=s(v,w),\;u=u(v,w)\end{eqnarray}$$
for every
$v$
in
${\mathcal{C}}(w)$
. By a divisor argument, the number of distinct
$sv^{2}$
as
$v$
varies over
${\mathcal{C}}(w)$
is
$\gg KN^{-\unicode[STIX]{x1D702}}$
. Thus in the notation of Lemma 2,
$$\begin{eqnarray}R\gg KN^{-\unicode[STIX]{x1D702}}\gg X\unicode[STIX]{x1D701}N^{\unicode[STIX]{x1D702}}=\frac{8SV^{2}N^{\unicode[STIX]{x1D702}}}{Z}.\end{eqnarray}$$
To see this,
$$\begin{eqnarray}KN^{-\unicode[STIX]{x1D702}}(8SV^{2}N^{\unicode[STIX]{x1D702}}/Z)^{-1}\gg \frac{N^{1-24\unicode[STIX]{x1D702}}}{Z^{1/2}L^{1/2}W}\frac{Z}{SV^{2}}\end{eqnarray}$$
(from (2.12))
$$\begin{eqnarray}\gg \frac{N^{1-25\unicode[STIX]{x1D702}}Z^{1/2}}{L^{5/2}V^{2}W}\gg \frac{N^{1-26\unicode[STIX]{x1D702}}(N/VW)}{L^{3}V^{2}W}\end{eqnarray}$$
$$\begin{eqnarray}\gg 1\end{eqnarray}$$
from the hypothesis of the theorem.
Accordingly, all
$u/sv^{2}$
with
$v\in {\mathcal{C}}(w)$
can be written in the form
$$\begin{eqnarray}\frac{u}{sv^{2}}=\frac{t}{r},\quad \langle t,r\rangle =1,\;t=t(w),\;r=r(w)\end{eqnarray}$$
for a certain
$t\in \mathbb{Z}$
,
$r\in \mathbb{N}$
independent of
$v$
.
We record a lower bound for
$K$
that does not contain
$Z$
. From (2.12), (2.10), we have
$$\begin{eqnarray}K\geqslant \frac{N^{1-22\unicode[STIX]{x1D702}}}{L^{1/2}W}\frac{(VW)^{1/2}}{L^{3/4}N^{1/2+12\unicode[STIX]{x1D702}}}=N^{1/2-34\unicode[STIX]{x1D702}}V^{1/2}W^{-1/2}L^{-5/4}.\end{eqnarray}$$
We now select a divisor
$z$
of
$r$
such that the set
$$\begin{eqnarray}{\mathcal{C}}(w,z)=\biggl\{v\in {\mathcal{C}}(w):\frac{r}{\langle s,r\rangle }=z\biggr\}\end{eqnarray}$$
satisfies
$$\begin{eqnarray}KN^{-\unicode[STIX]{x1D702}}\ll |{\mathcal{C}}(w,z)|\leqslant 2K.\end{eqnarray}$$
For each
$v$
in
${\mathcal{C}}(w,z)$
, we have
$$\begin{eqnarray}z\mid v^{2}.\end{eqnarray}$$
It is convenient to write
$z=bc^{2}$
where
$b$
is square-free, and define
$k=v^{2}/bc^{2}$
. Then
$$\begin{eqnarray}bc^{2}k=v^{2},\qquad k=bd^{2}~(\text{where }d\in \mathbb{N}),\qquad bcd=v.\end{eqnarray}$$
This leads to the upper bound
$$\begin{eqnarray}|{\mathcal{C}}(w,z)|\ll \frac{V}{bc},\end{eqnarray}$$
from which we infer, using (2.15), that
$$\begin{eqnarray}bc\ll \frac{VN^{\unicode[STIX]{x1D702}}}{K}.\end{eqnarray}$$
Now we re-examine our rational approximation
$$\begin{eqnarray}\biggl|\frac{sav^{2}w^{2}}{q}-u\biggr|<Z^{-1}\end{eqnarray}$$
(see (2.3) and the definition of
${\mathcal{C}}$
). We observe that
$$\begin{eqnarray}\displaystyle \biggl|r\frac{w^{2}a}{q}-t\biggr| & = & \displaystyle \frac{r}{sv^{2}}\biggl|\frac{sv^{2}w^{2}a}{q}-u\biggr|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{z}{V^{2}}\frac{1}{Z}=\frac{bc^{2}}{V^{2}}\frac{1}{Z}\ll \frac{N^{2\unicode[STIX]{x1D702}}}{K^{2}Z}\end{eqnarray}$$
for
$v\in {\mathcal{C}}(w,z)$
, using (2.16).
Let
${\mathcal{F}}$
be the set of natural numbers
$b_{0}c_{0}^{2}m$
,
$(b_{0},c_{0},m)\in \mathbb{N}^{3}$
,
$b_{0}c_{0}<VN^{2\unicode[STIX]{x1D702}}/K$
,
$m<L^{2}N^{\unicode[STIX]{x1D702}}$
. The number of possibilities for
$b_{0}c_{0}^{2}$
here is
$\ll VN^{3\unicode[STIX]{x1D702}}/K$
, since
$b_{0}c_{0}^{2}$
is a divisor of
$(b_{0}c_{0})^{2}$
. Hence
$$\begin{eqnarray}|{\mathcal{F}}|\ll \frac{VL^{2}N^{4\unicode[STIX]{x1D702}}}{K}.\end{eqnarray}$$
All the integers
$r$
occurring in (2.17) are in
${\mathcal{F}}$
. Hence
$$\begin{eqnarray}\displaystyle |{\mathcal{E}}(K)| & = & \displaystyle \mathop{\sum }_{r\in {\mathcal{F}}}\mathop{\sum }_{w\in {\mathcal{E}}(K),r(w)=r}1\nonumber\\ \displaystyle & \ll & \displaystyle |{\mathcal{F}}|\frac{N^{2\unicode[STIX]{x1D702}}(W+q^{1/2})}{KZ^{1/2}},\end{eqnarray}$$
on bounding the number of
$w$
in
${\mathcal{E}}(K)$
with
$r(w)=r$
via Lemma 3. Combining (2.18), (2.19), and recalling (2.11),
$$\begin{eqnarray}\displaystyle |{\mathcal{E}}(K)| & \ll & \displaystyle \frac{VL^{2}N^{6\unicode[STIX]{x1D702}}}{K}\frac{(W+q^{1/2})}{KZ^{1/2}},\nonumber\\ \displaystyle \mathop{\sum }_{(v,w)\in {\mathcal{C}}}T(vw) & \ll & \displaystyle L^{1/2}N^{11\unicode[STIX]{x1D702}}Z^{1/2}|{\mathcal{C}}|\nonumber\\ \displaystyle & \ll & \displaystyle L^{1/2}N^{12\unicode[STIX]{x1D702}}Z^{1/2}|{\mathcal{E}}(K)|K\nonumber\\ \displaystyle & \ll & \displaystyle \frac{L^{5/2}N^{18\unicode[STIX]{x1D702}}V(W+q^{1/2})}{K}.\nonumber\end{eqnarray}$$
We now use the lower bound (2.14) for
$K$
and obtain
$$\begin{eqnarray}\mathop{\sum }_{(v,w)\in {\mathcal{C}}}T(vw)\ll L^{15/4}(VW)^{1/2}N^{-1/2+60\unicode[STIX]{x1D702}}(W+q^{1/2}).\end{eqnarray}$$
Now (2.7) follows on applying the bounds for
$VW^{3}$
and
$VW$
in the hypothesis of Theorem 2. This completes the proof of Theorem 2. ◻
3 Type II sums
Lemma 4. Let
$Y\geqslant 1$
. For
$n\in \mathbb{N}$
, let
$R(n)$
denote the number of quadruples
$(\ell _{1},\ell _{2},y_{1},y_{2})$
with
$1\leqslant \ell _{i}\leqslant L$
,
$y_{i}\sim Y$
such that
$$\begin{eqnarray}\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2}=n.\end{eqnarray}$$
Then
$$\begin{eqnarray}\displaystyle R(n) & \ll & \displaystyle (LY)^{1+\unicode[STIX]{x1D702}},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{n\in \mathbb{Z}}R(n)^{2} & \ll & \displaystyle L^{3+\unicode[STIX]{x1D702}}Y^{2+\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Proof. For (3.2), fix
$\ell _{2}$
and
$y_{2}$
, then the equation
$$\begin{eqnarray}\ell _{1}y_{1}^{2}=\ell _{2}y_{2}^{2}+n\end{eqnarray}$$
(with
$\ell _{1}\neq 0$
,
$y_{1}\neq 0$
) determines
$\ell _{1}$
,
$y_{1}$
up to
$O((LY)^{\unicode[STIX]{x1D702}})$
possibilities.
For (3.3), we observe that
$$\begin{eqnarray}\mathop{\sum }_{n\in \mathbb{Z}}R(n)^{2}\end{eqnarray}$$
is the number of tuples
$\ell _{1}$
,
$\ell _{2}$
,
$\ell _{3}$
,
$\ell _{4}$
,
$y_{1}$
,
$y_{2}$
,
$y_{3}$
,
$y_{4}$
with
$1\leqslant \ell _{i}\leqslant L$
,
$y_{i}\sim Y$
and
$$\begin{eqnarray}\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2}=\ell _{3}y_{3}^{2}-\ell _{4}y_{4}^{2}.\end{eqnarray}$$
This may be expressed as an integral:
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{n\in \mathbb{Z}}R(n)^{2} & = & \displaystyle \int _{0}^{1}\biggl|\mathop{\sum }_{\ell \leqslant L}\mathop{\sum }_{y\sim Y}e(\ell y^{2}t)\biggr|^{2}\biggl|\mathop{\sum }_{\ell _{0}\leqslant L}\mathop{\sum }_{y_{0}\sim Y}e(\ell _{0}y_{0}^{2}t)\biggr|^{2}\,dt\nonumber\\ \displaystyle & {\leqslant} & \displaystyle LV\end{eqnarray}$$
by the Cauchy–Schwarz inequality, where
$$\begin{eqnarray}V=\mathop{\sum }_{1\leqslant \ell \leqslant L}\int _{0}^{1}\biggl|\mathop{\sum }_{y\sim Y}e(\ell y^{2}t)\biggr|^{2}\biggl|\mathop{\sum }_{\ell _{0}\leqslant L}\mathop{\sum }_{y_{0}\sim Y}e(\ell _{0}y_{0}^{2}t)\biggr|^{2}\,dt.\end{eqnarray}$$
Now
$V$
is the number of solutions of
$$\begin{eqnarray}\ell (y_{1}^{2}-y_{2}^{2})=\ell _{3}y_{3}^{2}-\ell _{4}y_{4}^{2}\end{eqnarray}$$
with
$1\leqslant \ell$
,
$\ell _{3}$
,
$\ell _{4}\leqslant L$
and
$y_{i}\sim Y(1\leqslant i\leqslant 4)$
.
We first consider
$V_{1}$
, the number of solutions of (3.5) with
$$\begin{eqnarray}\ell _{3}y_{3}^{2}=\ell _{4}y_{4}^{2}.\end{eqnarray}$$
If (3.5) and (3.6) hold, then
$y_{1}=y_{2}$
. There are
$O((LY)^{1+\unicode[STIX]{x1D702}})$
possibilities for
$\ell _{3}$
,
$\ell _{4}$
,
$y_{3}$
,
$y_{4}$
and for each of these, at most
$LY$
possibilities for
$\ell$
,
$y_{1}$
,
$y_{2}$
. Thus
$$\begin{eqnarray}V_{1}\ll (LY)^{2+\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Now consider
$V_{2}$
, the number of solutions of (3.5) for which (3.6) is violated. There are
$O(L^{2}Y^{2})$
possibilities for
$\ell _{3}$
,
$y_{3}$
,
$\ell _{4}$
,
$y_{4}$
. For each of these, there are
$O((LY)^{\unicode[STIX]{x1D702}})$
possibilities for
$\ell$
,
$y_{1}-y_{2}$
and
$y_{1}+y_{2}$
, and hence
$O((LY)^{\unicode[STIX]{x1D702}})$
possibilities for
$\ell$
,
$y_{1}$
,
$y_{2}$
. Thus
$$\begin{eqnarray}V_{2}\ll (LY)^{2+\unicode[STIX]{x1D702}},\qquad V\ll (LY)^{2+\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Theorem 3. For
$\frac{1}{6}<\unicode[STIX]{x1D70C}<\frac{1}{5}$
, and
$N^{\unicode[STIX]{x1D70C}}\ll Y\ll N^{1-4\unicode[STIX]{x1D70C}}$
,
$|c_{\ell }|\leqslant 1$
,
$|a_{x}|\leqslant 1$
,
$|b_{y}|\leqslant 1$
, we have
$$\begin{eqnarray}\mathop{\sum }_{l\leqslant L}c_{\ell }\mathop{\sum }_{y\sim Y}b_{y}\mathop{\sum }_{x\in I(y)}a_{x}e(\ell g(xy))\ll N^{1-10\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Proof. Just as in [Reference Baker1, proof of Lemma 9] we need only show that
$$\begin{eqnarray}S^{\prime }:=\mathop{\sum }_{\ell \leqslant L}c_{\ell }\mathop{\sum }_{y\sim Y}b_{y}\mathop{\sum }_{x\leqslant N/Y}a_{x}e(\ell g(xy))\ll N^{1-11\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Again arguing as in that proof,
$$\begin{eqnarray}|S^{\prime }|^{2}\leqslant \frac{N}{Y}\mathop{\sum }_{\boldsymbol{\ell }\in {\mathcal{L}}}|S(\boldsymbol{\ell })|,\end{eqnarray}$$
where
$\boldsymbol{\ell }=(\ell _{1},\ell _{2},y_{1},y_{2})$
,
${\mathcal{L}}=\{\boldsymbol{\ell }:\ell _{1},\ell _{2}\leqslant L,y_{1},y_{2}\sim Y\}$
and
$$\begin{eqnarray}S(\boldsymbol{\ell }):=\mathop{\sum }_{x\leqslant N/Y}e(\ell _{1}g(xy_{1})-\ell _{2}g(xy_{2})).\end{eqnarray}$$
The contribution to the right-hand side of (3.9) from those
$\boldsymbol{\ell }$
with
$$\begin{eqnarray}\ell _{1}y_{1}^{2}=\ell _{2}y_{2}^{2}\end{eqnarray}$$
is
$$\begin{eqnarray}\ll \frac{N}{Y}\cdot LY\cdot \frac{N}{Y}^{1+\unicode[STIX]{x1D702}}\end{eqnarray}$$
(by a divisor argument)
$$\begin{eqnarray}\ll N^{2-22\unicode[STIX]{x1D702}}\end{eqnarray}$$
since
$Y\gg N^{\unicode[STIX]{x1D70C}}$
. The contribution from those
$\boldsymbol{\ell }$
with
$$\begin{eqnarray}|S(\boldsymbol{\ell })|\leqslant \frac{N}{Y}^{1-22\unicode[STIX]{x1D702}}L^{-2}\end{eqnarray}$$
is
$$\begin{eqnarray}{\leqslant}\frac{N}{Y}\cdot L^{2}Y^{2}\cdot \frac{N}{Y}^{1-22\unicode[STIX]{x1D702}}L^{-2}=N^{2-22\unicode[STIX]{x1D702}}.\end{eqnarray}$$
It remains to consider
${\mathcal{M}}$
, the set of
$\boldsymbol{\ell }$
in
${\mathcal{L}}$
with
$$\begin{eqnarray}\ell _{1}y_{1}^{2}>\ell _{2}y_{2}^{2}\end{eqnarray}$$
and
$$\begin{eqnarray}|S(\boldsymbol{\ell })|>\frac{N}{Y}^{1-22\unicode[STIX]{x1D702}}L^{-2}.\end{eqnarray}$$
We apply [Reference Baker1, Lemma 5] with
$M=1$
,
$X=NY^{-1}$
,
$P=N^{1-22\unicode[STIX]{x1D702}}/YL^{2}$
. We require
$$\begin{eqnarray}P\geqslant X^{1/2+\unicode[STIX]{x1D702}}\end{eqnarray}$$
which holds since
$$\begin{eqnarray}\displaystyle PX^{-1/2-\unicode[STIX]{x1D702}} & {\geqslant} & \displaystyle N^{1/2-23\unicode[STIX]{x1D702}}L^{-2}Y^{-1/2}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle N^{1/2-23\unicode[STIX]{x1D702}}L^{-2}N^{-1/2+2\unicode[STIX]{x1D70C}}\geqslant 1.\nonumber\end{eqnarray}$$
Thus for each
$\boldsymbol{\ell }\in {\mathcal{M}}$
there exists a natural number
$s$
,
$$\begin{eqnarray}\displaystyle & \displaystyle s=s(\boldsymbol{\ell })\leqslant L^{4}N^{\unicode[STIX]{x1D702}},\qquad \biggl|s(\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2})\frac{a}{q}-u_{2}\biggr|=\frac{1}{Z(\boldsymbol{\ell })} & \displaystyle \nonumber\\ \displaystyle & \displaystyle |s(\ell _{1}\unicode[STIX]{x1D6FD}y_{1}-\ell _{2}\unicode[STIX]{x1D6FD}y_{2})-u_{1}|=\frac{1}{W(\boldsymbol{\ell })} & \displaystyle\end{eqnarray}$$
with
$u_{1}$
,
$u_{2}\in \mathbb{Z}$
,
$\langle s,u_{1},u_{2}\rangle =1$
,
$$\begin{eqnarray}Z(\boldsymbol{\ell })\geqslant \biggl(\frac{N}{Y}\biggr)^{2}L^{-4}N^{-45\unicode[STIX]{x1D702}},\qquad W(\boldsymbol{\ell })\geqslant \frac{N}{Y}L^{-4}N^{-45\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Now
$s\leqslant N/Y$
and, writing
$\unicode[STIX]{x1D6FE}_{2}=(\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2})a/q$
,
$\unicode[STIX]{x1D6FE}_{1}=(\ell _{1}y_{1}-\ell _{2}y_{2})\unicode[STIX]{x1D6FD}$
,
$$\begin{eqnarray}|s\unicode[STIX]{x1D6FE}_{j}-u_{j}|\leqslant (2k^{2})^{-1}\biggl(\frac{N}{Y}\biggr)^{1-j}\quad (j=1,2).\end{eqnarray}$$
Thus we can appeal to [Reference Baker1, Lemma 7] with
$k=2$
and
$L$
replaced by 1. Let
$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FD}_{j}=\unicode[STIX]{x1D6FE}_{j}-\frac{u_{j}}{s},\qquad F(x)=\mathop{\sum }_{j=1}^{2}\unicode[STIX]{x1D6FD}_{j}x^{j},\qquad G(x)=\mathop{\sum }_{j=1}^{2}u_{j}x^{j}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle S(s,G)=\mathop{\sum }_{u=1}^{s}e\biggl(\frac{G(u)}{s}\biggr). & \displaystyle \nonumber\end{eqnarray}$$
We obtain
$$\begin{eqnarray}S(\boldsymbol{\ell })=\mathop{\sum }_{x\leqslant N/Y}e(\unicode[STIX]{x1D6FE}_{2}x^{2}+\unicode[STIX]{x1D6FE}_{1}x)=s^{-1}S(s,G)\int _{0}^{N/Y}e(F(z))\,dz+O(N^{3\unicode[STIX]{x1D702}}L^{2}).\end{eqnarray}$$
Now
$N^{3\unicode[STIX]{x1D702}}L^{2}$
is of smaller order than
$(N^{1-22\unicode[STIX]{x1D702}}/Y)L^{-2}$
, so that
$$\begin{eqnarray}\biggl|s^{-1}S(s,G)\int _{0}^{N/Y}e(F(z))\,dz\biggr|\gg \frac{N^{1-22\unicode[STIX]{x1D702}}}{Y}L^{-2}.\end{eqnarray}$$
Moreover, by standard bounds, the left-hand side of (3.12) is
$$\begin{eqnarray}\ll s^{-1/2}\min (N/Y,(sZ(\boldsymbol{\ell }))^{1/2}).\end{eqnarray}$$
We now use a standard splitting-up argument to choose a subset
$Q$
of
${\mathcal{M}}$
such that
$$\begin{eqnarray}s(\boldsymbol{\ell })\sim S,\biggl|s(\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2})\frac{a}{q}-u_{2}\biggr|<\frac{1}{Z}\quad (\boldsymbol{\ell }\in Q),\end{eqnarray}$$
where
$(N/Y)^{2}L^{-4}N^{-\unicode[STIX]{x1D702}}\leqslant Z\leqslant Z_{0}$
, with
$$\begin{eqnarray}\frac{N}{Y}=(SZ_{0})^{1/2},\end{eqnarray}$$
and moreover
$$\begin{eqnarray}S\leqslant L^{4}N^{\unicode[STIX]{x1D702}},\end{eqnarray}$$
while
$$\begin{eqnarray}\mathop{\sum }_{\boldsymbol{\ell }\in {\mathcal{M}}}|S(\boldsymbol{\ell })|\leqslant (\log N)^{2}\mathop{\sum }_{\boldsymbol{\ell }\in Q}|S(\boldsymbol{\ell })|.\end{eqnarray}$$
Compare e.g. the argument in [Reference Baker1, proof of Lemma 8]. In order to obtain (3.8) it remains to show that
$$\begin{eqnarray}\frac{N}{Y}\mathop{\sum }_{\boldsymbol{\ell }\in Q}|S(\boldsymbol{\ell })|\ll N^{2-23\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Using (3.11)–(3.14), we find that
$$\begin{eqnarray}|S(\boldsymbol{\ell })|\ll Z^{1/2}\quad (\boldsymbol{\ell }\in Q)\end{eqnarray}$$
and we must show that
$$\begin{eqnarray}Z^{1/2}|Q|\ll YN^{1-23\unicode[STIX]{x1D702}}.\end{eqnarray}$$
For each
$\boldsymbol{\ell }$
in
$Q$
there is an
$s\sim S$
with
$$\begin{eqnarray}s(\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2})\in {\mathcal{C}},\end{eqnarray}$$
where
$$\begin{eqnarray}{\mathcal{C}}=\biggl\{n:1\leqslant n\leqslant 2SLY^{2},~|na~(\text{mod}~q)|<\frac{q}{Z}\biggr\}.\end{eqnarray}$$
Clearly
$$\begin{eqnarray}|{\mathcal{C}}|\ll \biggl(\frac{SLY^{2}}{q}+1\biggr)\biggl(\frac{q}{Z}+1\biggr)=\frac{SLY^{2}}{Z}+\frac{SLY^{2}}{q}+\frac{q}{Z}+1.\end{eqnarray}$$
Given
$m\in {\mathcal{C}}$
, let
$h_{1}(m),\ldots ,h_{j}(m)$
,
$j=j(m)\ll N^{\unicode[STIX]{x1D702}}$
, be the divisors
$h$
of
$m$
with
$h\sim S$
. Each element of
$Q$
satisfies
$$\begin{eqnarray}(\ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2})h_{i}(m)=m\end{eqnarray}$$
for some
$m\in {\mathcal{C}}$
and some
$i$
,
$1\leqslant i\leqslant j(m)$
. Let
$$\begin{eqnarray}{\mathcal{K}}=\{m/h_{i}(m):1\leqslant i\leqslant j(m),~m\in {\mathcal{C}}\}.\end{eqnarray}$$
Then
$$\begin{eqnarray}|{\mathcal{K}}|\ll N^{\unicode[STIX]{x1D702}}|{\mathcal{C}}|,\end{eqnarray}$$
while
$$\begin{eqnarray}\displaystyle |Q| & = & \displaystyle \mathop{\sum }_{\substack{ \boldsymbol{\ell }\in {\mathcal{M}} \\ \ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2}\in {\mathcal{K}}}}1\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \mathop{\sum }_{n\in {\mathcal{K}}}\mathop{\sum }_{\substack{ \boldsymbol{\ell }\in {\mathcal{M}} \\ \ell _{1}y_{1}^{2}-\ell _{2}y_{2}^{2}=n}}1\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{n\in {\mathcal{K}}}R(n)\nonumber\end{eqnarray}$$
in the notation of Lemma 4. Applying Cauchy’s inequality,
$$\begin{eqnarray}\displaystyle |Q| & {\leqslant} & \displaystyle |{\mathcal{K}}|^{1/2}\biggl(\mathop{\sum }_{n\geqslant 1}\,R(n)^{2}\biggr)^{1/2}\nonumber\\ \displaystyle & \ll & \displaystyle (L^{3}Y^{2})^{1/2}|{\mathcal{C}}|^{1/2}N^{\unicode[STIX]{x1D702}},\end{eqnarray}$$
Alternatively, (3.2) yields
$$\begin{eqnarray}|Q|\leqslant \Bigl(\max _{n\in {\mathcal{K}}}R(n)\Bigr)|{\mathcal{K}}|\ll LY|{\mathcal{C}}|N^{2\unicode[STIX]{x1D702}}.\end{eqnarray}$$
We now find that, depending on the value of
$Y$
, either (3.18) or (3.19) yields the desired bound (3.15). Suppose first that
$$\begin{eqnarray}Y>N^{3\unicode[STIX]{x1D70C}/2}S^{-1/2}.\end{eqnarray}$$
In view of (3.16), (3.18), we need to verify the four bounds
$$\begin{eqnarray}\displaystyle Z^{1/2}(L^{3}Y^{2})^{1/2}\biggl(\frac{SLY^{2}}{Z}\biggr)^{1/2} & \ll & \displaystyle YN^{1-24\unicode[STIX]{x1D702}},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle Z^{1/2}(L^{3}Y^{2})^{1/2}\biggl(\frac{SLY^{2}}{q}\biggr)^{1/2} & \ll & \displaystyle YN^{1-24\unicode[STIX]{x1D702}},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle Z^{1/2}(L^{3}Y^{2})^{1/2}\biggl(\frac{q}{Z}\biggr)^{1/2} & \ll & \displaystyle YN^{1-24\unicode[STIX]{x1D702}},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle Z^{1/2}(L^{3}Y^{2})^{1/2} & \ll & \displaystyle YN^{1-24\unicode[STIX]{x1D702}}.\end{eqnarray}$$
First of all, (3.21) holds since
$$\begin{eqnarray}L^{2}Y^{2}S^{1/2}(YN^{1-24\unicode[STIX]{x1D702}})^{-1}\ll YN^{4\unicode[STIX]{x1D70C}-1}\ll 1.\end{eqnarray}$$
Next, (3.22) holds since
$$\begin{eqnarray}\displaystyle Z^{1/2}L^{2}S^{1/2}Y^{2}q^{-1/2}(YN^{1-24\unicode[STIX]{x1D702}})^{-1} & \ll & \displaystyle S^{-1/2}NY^{-1}L^{2}YS^{1/2}N^{-3/2-\unicode[STIX]{x1D70C}/4+\unicode[STIX]{x1D700}/4}\nonumber\\ \displaystyle & \ll & \displaystyle N^{7\unicode[STIX]{x1D70C}/4-1/2+\unicode[STIX]{x1D700}}\ll 1.\nonumber\end{eqnarray}$$
Next, (3.23) holds since
$$\begin{eqnarray}L^{3/2}Y\,q^{1/2}(YN^{1-24\unicode[STIX]{x1D702}})^{-1}\ll N^{7\unicode[STIX]{x1D70C}/4-1/2}\ll 1.\end{eqnarray}$$
Finally, (3.24) holds since
$$\begin{eqnarray}Z^{1/2}L^{3/2}Y(YN^{1-24\unicode[STIX]{x1D702}})^{-1}\ll N^{3\unicode[STIX]{x1D70C}/2}S^{-1/2}Y^{-1}\ll 1\end{eqnarray}$$
from (3.20).
Now suppose that
$$\begin{eqnarray}Y\ll (N^{2-4\unicode[STIX]{x1D70C}-26\unicode[STIX]{x1D702}}S^{-1})^{1/3};\end{eqnarray}$$
we employ (3.16) and (3.19). We need to verify the bounds
$$\begin{eqnarray}\displaystyle Z^{1/2}LY(SLY^{2}/Z) & \ll & \displaystyle YN^{1-25\unicode[STIX]{x1D702}},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle Z^{1/2}LY(SLY^{2}/q) & \ll & \displaystyle YN^{1-25\unicode[STIX]{x1D702}},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle Z^{1/2}LY\,q/Z & \ll & \displaystyle YN^{1-25\unicode[STIX]{x1D702}},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle Z^{1/2}LY & \ll & \displaystyle YN^{1-25\unicode[STIX]{x1D702}}.\end{eqnarray}$$
First of all, (3.26) holds since
$$\begin{eqnarray}L^{2}SY^{3}Z^{-1/2}(YN^{1-25\unicode[STIX]{x1D702}})^{-1}\leqslant L^{4}SY^{3}N^{26\unicode[STIX]{x1D702}-2}\ll 1\end{eqnarray}$$
from (3.25). Next, (3.27) holds since
$$\begin{eqnarray}\displaystyle Z^{1/2}L^{2}SY^{3}q^{-1}(YN^{1-25\unicode[STIX]{x1D702}})^{-1} & \ll & \displaystyle \frac{N}{Y}\,S^{1/2}L^{2}Y^{2}N^{-2-\unicode[STIX]{x1D70C}/2+25\unicode[STIX]{x1D702}+\unicode[STIX]{x1D700}/4}\nonumber\\ \displaystyle & \ll & \displaystyle YN^{-1+7\unicode[STIX]{x1D70C}/2+\unicode[STIX]{x1D700}}\ll 1.\nonumber\end{eqnarray}$$
Next, (3.28) holds since
$$\begin{eqnarray}Z^{-1/2}LYq(YN^{1-25\unicode[STIX]{x1D702}})^{-1}\ll YN^{-1+7\unicode[STIX]{x1D70C}/2+26\unicode[STIX]{x1D702}}\ll 1.\end{eqnarray}$$
Finally, (3.29) holds since
$$\begin{eqnarray}Z^{1/2}LY(YN^{1-25\unicode[STIX]{x1D702}})^{-1}\ll \frac{N^{25\unicode[STIX]{x1D702}}L}{Y}\ll 1.\end{eqnarray}$$
In order to complete the proof, we show that the ranges of
$Y$
in (3.20) and (3.25) overlap. We have
$$\begin{eqnarray}N^{3\unicode[STIX]{x1D70C}/2}S^{-1/2}<(N^{2-4\unicode[STIX]{x1D70C}-26\unicode[STIX]{x1D702}}S^{-1})^{1/3},\end{eqnarray}$$
that is
$$\begin{eqnarray}S^{1/6}>N^{(17\unicode[STIX]{x1D70C}-4+52\unicode[STIX]{x1D702})/6},\end{eqnarray}$$
since
$\unicode[STIX]{x1D70C}<1/5$
. This completes the proof of (3.15), and Theorem 3 follows.◻
4 Asymptotic formulae via the Harman sieve and the generalized Vaughan identity
In the present section and the next, we suppose that
$\unicode[STIX]{x1D70E}=\unicode[STIX]{x1D70C}-1/6$
satisfies
$$\begin{eqnarray}{\textstyle \frac{1}{120}}<\unicode[STIX]{x1D70E}\leqslant {\textstyle \frac{1}{102}}.\end{eqnarray}$$
We write
$b=\frac{1}{6}-5\unicode[STIX]{x1D70E}$
,
$f=\frac{1}{3}-4\unicode[STIX]{x1D70E}$
,
$z=N^{b}$
and
$$\begin{eqnarray}P(s)=\mathop{\prod }_{p<s}p\quad (s>1).\end{eqnarray}$$
For a finite set
${\mathcal{E}}\subset \mathbb{N}$
, let
$$\begin{eqnarray}\displaystyle {\mathcal{E}}_{d} & = & \displaystyle \{m:dm\in {\mathcal{E}}\},\nonumber\\ \displaystyle S({\mathcal{E}},w) & = & \displaystyle |\{m\in {\mathcal{E}}:\langle m,P(w)\rangle =1\}|.\nonumber\end{eqnarray}$$
As in [Reference Baker1], our claim in Theorem 1 is a corollary of the lower bound
$$\begin{eqnarray}S({\mathcal{A}},(2N)^{1/2})>\frac{1}{7}\frac{\unicode[STIX]{x1D6FF}N}{\log N}\end{eqnarray}$$
for
$\unicode[STIX]{x1D70C}=\unicode[STIX]{x1D70C}_{2}$
.
We introduce some ‘comparison’ results for the pair
$S({\mathcal{A}},w)$
and
$2\unicode[STIX]{x1D6FF}\,S({\mathcal{B}},w)$
, and similar pairs, that will be needed in §6. First of all, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\substack{ p_{1}\sim P_{1} \\ N^{\unicode[STIX]{x1D70F}}\leqslant p_{\ell }\leqslant \cdots \leqslant p_{1} \\ p_{1}\leqslant p_{1}^{\prime }\leqslant \cdots \leqslant p_{t}^{\prime }}}\cdots \mathop{\sum }_{p_{\ell }\sim P_{\ell }}\mathop{\sum }_{p_{1}^{\prime }\sim Q_{1}}\cdots \mathop{\sum }_{p_{t}^{\prime }\sim Q_{t}}\biggl(\mathop{\sum }_{p_{1}\ldots p_{\ell }\,p_{1}^{\prime }\ldots p_{t}^{\prime }\in {\mathcal{A}}}1-2\unicode[STIX]{x1D6FF}\mathop{\sum }_{p_{1}\ldots p_{\ell }\,p_{1}^{\prime }\ldots p_{t}^{\prime }\in {\mathcal{B}}}1\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \ll \unicode[STIX]{x1D6FF}N^{-1-\unicode[STIX]{x1D702}}\end{eqnarray}$$
whenever
$\unicode[STIX]{x1D70F}$
is a positive constant and some subproduct
$R$
of
$P_{1}\ldots P_{\ell }$
$Q_{1}\ldots Q_{t}$
satisfies
$$\begin{eqnarray}N^{\unicode[STIX]{x1D70C}}\ll R\ll N^{f}.\end{eqnarray}$$
This is a consequence of Theorem 3; compare the discussion in [Reference Harman8, §§3.2 and 3.5]. Additional inequalities such as ‘
$p_{j}\leqslant K$
’ may be included in the summation in (4.3) without affecting its validity, as explained in [Reference Harman8, §3.2].
The following lemma is essentially the special case
$M=2X^{\unicode[STIX]{x1D6FC}}$
,
$S=1$
of [Reference Baker and Weingartner3, Lemma 14], and is a variant of [Reference Harman8, Theorem 3.1].
Lemma 5. Let
$w$
be a complex function with support in
$[1,N]$
,
$|w(n)|\leqslant N^{1/\unicode[STIX]{x1D702}}$
$(n\geqslant 1)$
. Let
$0<\unicode[STIX]{x1D703}<\unicode[STIX]{x1D703}+\unicode[STIX]{x1D713}<1/2$
. Let
$$\begin{eqnarray}S(r,v):=\mathop{\sum }_{\langle n,P(v)\rangle =1}w(rn).\end{eqnarray}$$
Suppose that, for some
$Y>1$
we have, (for any coefficients
$a_{m},|a_{m}|\leqslant 1$
,
$c_{n}$
,
$|c_{n}|\leqslant 1$
and
$b_{n}$
,
$|b_{n}|\leqslant \unicode[STIX]{x1D70F}(n)$
)
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{m\leqslant 2N^{\unicode[STIX]{x1D6FC}}}a_{m}\mathop{\sum }_{n}w(mn) & \ll & \displaystyle Y,\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{N^{\unicode[STIX]{x1D703}}\leqslant h\leqslant N^{\unicode[STIX]{x1D703}+\unicode[STIX]{x1D713}}}c_{h}\mathop{\sum }_{n}b_{n}w(hn) & \ll & \displaystyle Y.\end{eqnarray}$$
Let
$u_{r}$
$(r<N^{\unicode[STIX]{x1D703}})$
be complex numbers with
$|u_{r}|\!\leqslant \!1$
,
$u_{r}=0$
for
$\langle r,P(N^{\unicode[STIX]{x1D702}})\rangle >1$
. Then
$$\begin{eqnarray}\mathop{\sum }_{r<N^{\unicode[STIX]{x1D703}}}u_{r}S(r,N^{\unicode[STIX]{x1D713}})\ll Y(\log N)^{3}.\end{eqnarray}$$
We can deduce the following ‘bilinear’ lemma.
Lemma 6. Let
$w$
,
$\unicode[STIX]{x1D703}$
,
$\unicode[STIX]{x1D713}$
,
$S(r,v)$
be as in Lemma 5. Suppose that we have the hypothesis (4.6) and in addition, for some
$T\in [1,N)$
,
$$\begin{eqnarray}\mathop{\sum }_{m\leqslant 2N^{\unicode[STIX]{x1D703}}}a_{m}\mathop{\sum }_{t\leqslant T}c_{t}\mathop{\sum }_{n}w(mtn)\ll Y\end{eqnarray}$$
for any
$a_{m}$
,
$c_{t}$
with
$|a_{m}|\leqslant 1$
,
$|c_{t}|\leqslant 1$
. Then for any
$u_{r}$
$(r<N^{\unicode[STIX]{x1D703}})$
,
$v_{t}$
$(t\leqslant T)$
with
$|u_{r}|\leqslant 1$
,
$|v_{t}|\leqslant 1$
,
$u_{r}=0$
for
$\langle r,P(N^{\unicode[STIX]{x1D702}})\rangle >1$
, we have
$$\begin{eqnarray}\mathop{\sum }_{r\leqslant R}u_{r}\mathop{\sum }_{t\leqslant T}v_{t}S(rt,N^{\unicode[STIX]{x1D713}})\ll YN^{2\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Proof. We apply Lemma 5 with
$w$
replaced by
$w^{\ast }$
,
$$\begin{eqnarray}w^{\ast }(n)=\mathop{\sum }_{t\leqslant T}v_{t}w(nt),\end{eqnarray}$$
so that
$S(r,v)$
is replaced by
$$\begin{eqnarray}S^{\ast }(r,v)=\mathop{\sum }_{\langle n,P(v)\rangle =1}\mathop{\sum }_{t\leqslant T}v_{t}w(nt).\end{eqnarray}$$
From (4.8), (4.9) the hypotheses of Lemma 5 are satisfied with
$Y$
replaced by
$YN^{\unicode[STIX]{x1D702}}$
: for example,
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{N^{\unicode[STIX]{x1D703}}\leqslant m\leqslant N^{\unicode[STIX]{x1D703}+\unicode[STIX]{x1D713}}}a_{m}\mathop{\sum }_{n}b_{n}w^{\ast }(mn)\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{N^{\unicode[STIX]{x1D703}}\leqslant m\leqslant N^{\unicode[STIX]{x1D703}+\unicode[STIX]{x1D713}}}a_{m}\mathop{\sum }_{t\leqslant T}\mathop{\sum }_{n}\,b_{n}v_{t}w(mtn)\ll YN^{\unicode[STIX]{x1D702}}\nonumber\end{eqnarray}$$
(we may group the product
$nt$
as a single variable and apply (4.6)). The conclusion (4.7) with
$S$
replaced with
$S^{\ast }$
gives the desired bound (4.9).◻
We now apply Lemma 6 with
$$\begin{eqnarray}w(n)=\unicode[STIX]{x1D712}_{_{{\mathcal{A}}}}(n)-2\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D712}_{_{{\mathcal{B}}}}(n),\end{eqnarray}$$
$\unicode[STIX]{x1D703}=\unicode[STIX]{x1D70C}$
,
$\unicode[STIX]{x1D703}+\unicode[STIX]{x1D713}=f$
, where
$T=(2N)^{\unicode[STIX]{x1D708}}$
and the non-negative number
$\unicode[STIX]{x1D708}$
satisfies
$$\begin{eqnarray}\unicode[STIX]{x1D70C}+\unicode[STIX]{x1D708}\leqslant {\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}.\end{eqnarray}$$
It is easy to deduce the following inequalities from (4.10):
$$\begin{eqnarray}\displaystyle 3\unicode[STIX]{x1D70C}+2\unicode[STIX]{x1D708} & {\leqslant} & \displaystyle {\textstyle \frac{3}{2}}-3\unicode[STIX]{x1D70E},\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}+3\unicode[STIX]{x1D708} & {\leqslant} & \displaystyle \frac{7}{4}-\frac{15\unicode[STIX]{x1D70E}}{2}.\end{eqnarray}$$
Lemma 7. Suppose that (4.10) holds. Then
$$\begin{eqnarray}\mathop{\sum }_{r<N^{\unicode[STIX]{x1D70C}}}u_{r}~\mathop{\sum }_{t\leqslant (2N)^{\unicode[STIX]{x1D708}}}v_{t}(S({\mathcal{A}}_{rt},z)-2\unicode[STIX]{x1D6FF}\,S({\mathcal{B}}_{rt},z))\ll \unicode[STIX]{x1D6FF}N^{1-\unicode[STIX]{x1D702}},\end{eqnarray}$$
whenever
$|u_{r}|\leqslant 1$
,
$\langle r,P(N^{\unicode[STIX]{x1D702}})\rangle =1$
for
$u_{r}\neq 0$
, and
$|v_{t}|\leqslant 1$
.
Proof. We take
$Y=\unicode[STIX]{x1D6FF}N^{1-3\unicode[STIX]{x1D702}}$
. The hypothesis (4.8) is a consequence of Theorem 2 because of (4.10)–(4.12). The hypothesis (4.6) is a consequence of Theorem 3. Now the conclusion (4.9) may be written in the form (4.13). ◻
Lemma 8. Let
$0<g\leqslant \frac{1}{6}-2\unicode[STIX]{x1D70E}$
,
$\frac{1}{3}-4\unicode[STIX]{x1D70E}<\unicode[STIX]{x1D6FE}<\frac{2}{3}-8\unicode[STIX]{x1D70E}$
. Let
$\unicode[STIX]{x1D70C}_{1}\geqslant \cdots \geqslant \unicode[STIX]{x1D70C}_{t}\geqslant 0$
with
$\unicode[STIX]{x1D70C}_{1}+\cdots +\unicode[STIX]{x1D70C}_{t}=\unicode[STIX]{x1D6FE}$
,
$\unicode[STIX]{x1D70C}_{1}\leqslant \unicode[STIX]{x1D6FE}-g$
. There is a set
${\mathcal{C}}\subset \{1,\ldots ,t\}$
with
$\sum _{i\in {\mathcal{C}}}\unicode[STIX]{x1D70C}_{i}\in [g,\frac{1}{3}-4\unicode[STIX]{x1D70E}]$
.
Proof. Suppose that no such
${\mathcal{C}}$
exists. Now suppose first that
$\unicode[STIX]{x1D70C}_{1}\leqslant g$
. Since
$2g\leqslant \frac{1}{3}-4\unicode[STIX]{x1D70E}$
we can prove successively that
$\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2},\ldots ,\unicode[STIX]{x1D70C}_{1}+\cdots +\unicode[STIX]{x1D70C}_{t}$
are in
$[0,g]$
. This is absurd.
Thus we must have
$\unicode[STIX]{x1D70C}_{1}>\frac{1}{3}-4\unicode[STIX]{x1D70E}$
. But now
$\unicode[STIX]{x1D70C}_{2}+\cdots +\unicode[STIX]{x1D70C}_{t}=\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D70C}_{1}<\unicode[STIX]{x1D6FE}-(\frac{1}{3}-4\unicode[STIX]{x1D70E})<\frac{1}{3}-4\unicode[STIX]{x1D70E}$
, and
$\unicode[STIX]{x1D70C}_{2}+\cdots +\unicode[STIX]{x1D70C}_{t}\geqslant g$
since
$\unicode[STIX]{x1D70C}_{1}\leqslant \unicode[STIX]{x1D6FE}-g$
. This is absurd.◻
Lemma 9. Let
$F$
be a complex function on
$[1,N]$
. The sum
$\sum _{k\leqslant N}\unicode[STIX]{x1D6EC}(k)$
$F(k)$
may be decomposed into at most
$C(\log N)^{8}$
sums of the form
$$\begin{eqnarray}\mathop{\sum }_{n_{i}\in I_{i};~n_{1}\ldots n_{8}\leqslant N}(\log n_{1})\unicode[STIX]{x1D707}(n_{5})\ldots \unicode[STIX]{x1D707}(n_{8})\quad F(n_{1}\ldots n_{8})\end{eqnarray}$$
where
$I_{i}=(N_{i},2N_{i}]$
,
$\prod N_{i}<N$
and
$2N_{i}\leqslant N^{1/4}$
if
$i>4$
.
Proof. This is a case of Heath-Brown’s ‘generalized Vaughan identity’ [Reference Heath-Brown9]. ◻
Lemma 10. Let
$$\begin{eqnarray}(2N)^{3/8+33\unicode[STIX]{x1D70E}/4}\leqslant Q<Q^{\prime }\leqslant (2N)^{1/2},\qquad Q^{\prime }\leqslant 2Q.\end{eqnarray}$$
We have
$$\begin{eqnarray}\mathop{\sum }_{Q\leqslant p<Q^{\prime }}(S({\mathcal{A}}_{p},z)-2\unicode[STIX]{x1D6FF}\,S({\mathcal{B}}_{p},z))\ll \unicode[STIX]{x1D6FF}N^{1-\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Proof. Arguing as in the proof of (4.3), it will suffice to show that
$$\begin{eqnarray}\mathop{\sum }_{Q\leqslant p<Q^{\prime }}\mathop{\sum }_{\ell \leqslant L}c_{\ell }\mathop{\sum }_{\substack{ N/2<n\leqslant N \\ \langle n,P(z)\rangle =1}}e(\ell g(pn))\ll N^{1-2\unicode[STIX]{x1D702}}\end{eqnarray}$$
for
$|c_{\ell }|\leqslant 1$
. By a partial summation argument, it suffices to obtain
$$\begin{eqnarray}\mathop{\sum }_{Q\leqslant m<Q^{\prime }}\unicode[STIX]{x1D6EC}(m)\mathop{\sum }_{\ell \leqslant L}c_{\ell }\mathop{\sum }_{\substack{ N/2<n\leqslant N \\ \langle n,P(z)\rangle =1}}e(\ell g(mn))\ll N^{1-2\unicode[STIX]{x1D702}}.\end{eqnarray}$$
Applying Lemma 9, we need only show that
$$\begin{eqnarray}\mathop{\sum }_{\substack{ Q\leqslant n_{1}\ldots n_{8}<Q^{\prime } \\ n_{i}\sim N_{i}\text{ for all }i}}\!(\log n_{1})\unicode[STIX]{x1D707}(n_{5})\ldots \unicode[STIX]{x1D707}(n_{8})\mathop{\sum }_{\ell \leqslant L}c_{\ell }\!\mathop{\sum }_{\substack{ N/2<n\leqslant N \\ \langle n,P(z)\rangle =1}}e(\ell g(n_{1}\ldots n_{8}n))\ll N^{1-3\unicode[STIX]{x1D702}}\end{eqnarray}$$
whenever
$\prod _{i}N_{i}<N$
and
$2N_{i}\leqslant N^{1/4}$
for
$i>4$
. Thus
$Q\ll N_{1}\ldots N_{8}\ll Q$
. We write
$N_{1}\ldots N_{8}=N^{\unicode[STIX]{x1D6FE}}$
.
We may assume in view of Theorem 3 that no subproduct
$X$
of
$N_{1},\ldots ,N_{8}$
satisfies
$$\begin{eqnarray}N^{\unicode[STIX]{x1D70C}}\ll X\ll N^{f}.\end{eqnarray}$$
We now reorder
$N_{1},\ldots ,N_{8}$
as
$(N_{1}\ldots N_{8})^{\unicode[STIX]{x1D70C}_{j}}\;(1\leqslant j\!\leqslant \!8)$
with
$\unicode[STIX]{x1D70C}_{1}\geqslant \cdots \geqslant \!\unicode[STIX]{x1D70C}_{8}\!\geqslant 0$
. Let
$g=\min (3\unicode[STIX]{x1D6FE}-\frac{5}{4}-15\unicode[STIX]{x1D70E}/2,\unicode[STIX]{x1D6FE}-\frac{1}{3}-8\unicode[STIX]{x1D70E})$
, so that
$0<g\leqslant \unicode[STIX]{x1D6FE}-\frac{1}{3}-8\unicode[STIX]{x1D70E}$
and
$$\begin{eqnarray}g=\left\{\begin{array}{@{}ll@{}}3\unicode[STIX]{x1D6FE}-{\displaystyle \frac{5}{4}}-{\displaystyle \frac{15\unicode[STIX]{x1D70E}}{2}}\quad & \biggl(\unicode[STIX]{x1D6FE}\leqslant {\displaystyle \frac{11}{24}}-{\displaystyle \frac{\unicode[STIX]{x1D70E}}{4}}\biggr)\\ \unicode[STIX]{x1D6FE}-{\displaystyle \frac{1}{3}}-8\unicode[STIX]{x1D70E}\quad & \biggl(\unicode[STIX]{x1D6FE}>{\displaystyle \frac{11}{24}}-{\displaystyle \frac{\unicode[STIX]{x1D70E}}{4}}\biggr).\end{array}\right.\end{eqnarray}$$
We divide the argument into two cases.
Case 1. We have
$\unicode[STIX]{x1D70C}_{1}\geqslant \unicode[STIX]{x1D6FE}-g>\frac{1}{4}$
. Thus
$i\leqslant 4$
and (after a partial summation if necessary) we can apply Theorem 2 to the sum in (4.16) with
$$\begin{eqnarray}V\ll N^{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D70C}_{1}},\qquad W\ll N^{1-\unicode[STIX]{x1D6FE}}.\end{eqnarray}$$
We verify the hypotheses of Theorem 2. First,
$$\begin{eqnarray}3(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D70C}_{1})+2(1-\unicode[STIX]{x1D6FE})\leqslant 3g+2-2\unicode[STIX]{x1D6FE}\leqslant {\textstyle \frac{3}{2}}-3\unicode[STIX]{x1D70E}=2-3\unicode[STIX]{x1D70C},\end{eqnarray}$$
since
$$\begin{eqnarray}g\leqslant \unicode[STIX]{x1D6FE}-{\textstyle \frac{1}{3}}-8\unicode[STIX]{x1D70E}\leqslant {\textstyle \frac{1}{3}}(2\unicode[STIX]{x1D6FE}-3\unicode[STIX]{x1D70E}-{\textstyle \frac{1}{2}}).\end{eqnarray}$$
Next,
$$\begin{eqnarray}(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D70C}_{1})+3(1-\unicode[STIX]{x1D6FE})\leqslant g+3-3\unicode[STIX]{x1D6FE}\leqslant \frac{7}{4}-\frac{15\unicode[STIX]{x1D70E}}{2}=3-\frac{15\unicode[STIX]{x1D70C}}{2}\end{eqnarray}$$
and
$$\begin{eqnarray}(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D70C}_{1})+(1-\unicode[STIX]{x1D6FE})\leqslant g+1-\unicode[STIX]{x1D6FE}\leqslant {\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}=2-8\unicode[STIX]{x1D70C}\end{eqnarray}$$
from the definition of
$g$
. Now (4.16) follows from Theorem 2.
Case 2. We have
$\unicode[STIX]{x1D70C}_{1}<\unicode[STIX]{x1D6FE}-g$
. By Lemma 8 and the absence of a product
$X$
satisfying (4.17), there is a subsum
$u=\sum _{i\in {\mathcal{C}}}\unicode[STIX]{x1D70C}_{i}$
such that
$$\begin{eqnarray}g\leqslant u<{\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E}.\end{eqnarray}$$
We are now in a position to apply Lemma 6 with
$\unicode[STIX]{x1D703}=\unicode[STIX]{x1D70C}$
,
$\unicode[STIX]{x1D703}+\unicode[STIX]{x1D713}=f$
,
$$\begin{eqnarray}w(n)=\left\{\begin{array}{@{}ll@{}}\displaystyle \mathop{\sum }_{\ell \leqslant L}c_{\ell }e(\ell g(n))\quad & \biggl({\displaystyle \frac{N}{2}}<n\leqslant N\biggr)\\ 0\quad & \text{(otherwise)},\end{array}\right.\end{eqnarray}$$
and
$$\begin{eqnarray}T\ll N^{\unicode[STIX]{x1D6FE}-u}.\end{eqnarray}$$
We need to verify (4.6), (4.8). Clearly (4.6) is a consequence of Theorem 3. As for (4.8), we need to verify the hypotheses of Theorem 2 with
$V\leqslant 2N^{1/6+\unicode[STIX]{x1D70E}}$
,
$W\ll N^{\unicode[STIX]{x1D6FE}-u}$
. Suppose first that
$\unicode[STIX]{x1D6FE}>\frac{11}{24}-\unicode[STIX]{x1D70E}/4$
. Then
$\unicode[STIX]{x1D6FE}-u\leqslant \frac{1}{2}-9\unicode[STIX]{x1D70E}$
. Since (4.10) implies (4.11)–(4.12), the hypotheses follow from
$$\begin{eqnarray}({\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E})+({\textstyle \frac{1}{2}}-9\unicode[STIX]{x1D70E})={\textstyle \frac{1}{2}}+9\unicode[STIX]{x1D70E}\leqslant {\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}.\end{eqnarray}$$
Now suppose that
$\unicode[STIX]{x1D6FE}\leqslant \frac{11}{24}-\unicode[STIX]{x1D70E}/4$
. Then
$\unicode[STIX]{x1D6FE}-u\leqslant \frac{5}{4}+15\unicode[STIX]{x1D70E}/2-2\unicode[STIX]{x1D6FE}$
, and as above the hypotheses follow from
$$\begin{eqnarray}\biggl(\frac{1}{6}+\unicode[STIX]{x1D70E}\biggr)+\biggl(\frac{5}{4}+\frac{15\unicode[STIX]{x1D70E}}{2}-2\unicode[STIX]{x1D6FE}\biggr)=\frac{17}{12}+\frac{17\unicode[STIX]{x1D70E}}{2}-2\unicode[STIX]{x1D6FE}\leqslant \frac{2}{3}-8\unicode[STIX]{x1D70E}.\end{eqnarray}$$
Thus Theorem 2 yields the desired estimate (4.8). Now the lemma follows from Lemma 6. ◻
5 Application of the linear sieve
In order to obtain an upper bound for
$$\begin{eqnarray}\mathop{\sum }_{P\leqslant p<P^{\prime }}S({\mathcal{A}}_{p},p)\end{eqnarray}$$
whenever
$$\begin{eqnarray}(2N)^{1/2-9\unicode[STIX]{x1D70E}}\leqslant P\leqslant P^{\prime }\leqslant (2N)^{3/8+33\unicode[STIX]{x1D70E}/4},\quad P^{\prime }\leqslant 2P,\end{eqnarray}$$
we apply Iwaniec [Reference Iwaniec10, Theorem 4], which we state in a form sufficient for our needs. The quantity estimated will actually exceed that in (5.1), which will be exploited in §6.
Let
${\mathcal{E}}$
be a set of integers in
$[1,N]$
. Fix an approximation
$X$
to
$|{\mathcal{E}}|$
and write
$$\begin{eqnarray}r({\mathcal{E}},d)=|{\mathcal{E}}_{d}|-\frac{X}{d}.\end{eqnarray}$$
Let
$F(s)$
be the upper bound function for the linear sieve [Reference Iwaniec10, p. 309]. In the following lemma, let
$D\geqslant Z\geqslant 2$
. Let
$$\begin{eqnarray}\displaystyle {\mathcal{G}} & = & \displaystyle \{D^{\unicode[STIX]{x1D700}^{2}(1+\unicode[STIX]{x1D702})^{n}}:n\geqslant 0\},\qquad V(Z)=\mathop{\prod }_{p<Z}\biggl(1-\frac{1}{p}\biggr),\nonumber\\ \displaystyle {\mathcal{H}} & = & \displaystyle \{\!\mathbf{D}=(D_{1},\ldots ,D_{r}):r\geqslant 1,D_{\ell }\in {\mathcal{G}}~\text{for }1\leqslant \ell \leqslant r,\nonumber\\ \displaystyle & & \displaystyle \,\,D_{r}\leqslant \cdots \leqslant D_{1}<D^{1/2}\!\},\nonumber\\ \displaystyle {\mathcal{D}}^{+} & = & \displaystyle \biggl\{\mathbf{D}\in {\mathcal{H}}:D_{1}\ldots D_{2\ell }D_{2\ell +1}^{3}<D~\text{for }0\leqslant \ell \leqslant \frac{r-1}{2}\biggr\}.\nonumber\end{eqnarray}$$
Lemma 11. With the above notation, we have
$$\begin{eqnarray}S({\mathcal{E}},Z)\leqslant V(Z)X\biggl\{F\biggl(\frac{\log D}{\log Z}\biggr)+E\biggr\}+R_{1}^{+}+R^{+},\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle E & {<} & \displaystyle C(\unicode[STIX]{x1D700}+\unicode[STIX]{x1D700}^{-8}(\log D)^{-1/3}),\nonumber\\ \displaystyle R_{1}^{+} & = & \displaystyle \mathop{\sum }_{\substack{ d<D^{\unicode[STIX]{x1D700}} \\ d\mid P(D^{\unicode[STIX]{x1D700}^{2}})}}\unicode[STIX]{x1D719}_{d}^{+}(D^{\unicode[STIX]{x1D700}})r({\mathcal{A}},d),\nonumber\\ \displaystyle R^{+} & = & \displaystyle \mathop{\sum }_{\mathbf{D}\in {\mathcal{D}}^{+}}~\mathop{\sum }_{\substack{ d<D^{\unicode[STIX]{x1D700}} \\ d\mid P(D^{\unicode[STIX]{x1D700}^{2}})}}\unicode[STIX]{x1D6EC}_{d}^{+}(\unicode[STIX]{x1D700},D)H_{d}({\mathcal{E}},Z,\unicode[STIX]{x1D700},\mathbf{D}),\nonumber\end{eqnarray}$$
with some coefficients
$\unicode[STIX]{x1D719}_{d}^{+}(D^{\unicode[STIX]{x1D700}})$
and
$\unicode[STIX]{x1D6EC}_{d}^{+}(\unicode[STIX]{x1D700},\mathbf{D})$
bounded by
$1$
in absolute value. Here
$$\begin{eqnarray}H_{d}({\mathcal{E}},Z,\unicode[STIX]{x1D700},\mathbf{D})=\mathop{\sum }_{D_{i}\leqslant p_{i}<D_{i}^{1+\unicode[STIX]{x1D702}},\,p_{i}<Z~(1\leqslant i\leqslant r)}r({\mathcal{E}},dp_{1}\ldots p_{r}).\end{eqnarray}$$
In our application, we shall take
$$\begin{eqnarray}{\mathcal{E}}={\mathcal{A}}_{p}\end{eqnarray}$$
for any
$p\in [P,P^{\prime })$
, and
$$\begin{eqnarray}X=2\unicode[STIX]{x1D6FF}|{\mathcal{B}}_{p}|.\end{eqnarray}$$
We write
$P=N^{\unicode[STIX]{x1D6FE}}$
. We shall take
$D=N^{2/3-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FE}-2\unicode[STIX]{x1D700}}$
and
$Z=D^{1/3}$
, so that
$$\begin{eqnarray}V(Z)F\biggl(\frac{\log D}{\log Z}\biggr)=\frac{2}{\log D}\,(1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700}).\end{eqnarray}$$
We apply Lemma 11 and sum over
$p$
, obtaining (on noting that
$P>D^{\unicode[STIX]{x1D6FF}}Z$
)
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{P\leqslant p<P^{\prime }}S({\mathcal{A}}_{p},Z)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant V(Z)\biggl\{F\biggl(\frac{\log D}{\log Z}\biggr)+C\unicode[STIX]{x1D700}\biggr\}\mathop{\sum }_{P\leqslant p<P^{\prime }}2\unicode[STIX]{x1D6FF}|{\mathcal{B}}_{p}|+E_{1}+E_{2}.\end{eqnarray}$$
Here
$$\begin{eqnarray}\displaystyle E_{1} & = & \displaystyle \mathop{\sum }_{\substack{ d<D^{\unicode[STIX]{x1D700}} \\ d\mid P(D^{\unicode[STIX]{x1D700}^{2}})}}\mathop{\sum }_{P\leqslant p<P^{\prime }}r({\mathcal{A}},pd),\nonumber\\ \displaystyle E_{2} & = & \displaystyle \mathop{\sum }_{\substack{ d<D^{\unicode[STIX]{x1D700}} \\ d\mid P(D^{\unicode[STIX]{x1D700}^{2}})}}\mathop{\sum }_{P\leqslant p<P^{\prime }}\mathop{\sum }_{\mathbf{D}\in {\mathcal{D}}^{+}}\unicode[STIX]{x1D6EC}_{d}^{+}(\unicode[STIX]{x1D700},\mathbf{D})\mathop{\sum }_{\substack{ D_{i}\leqslant p_{i}<D_{i}^{1+\unicode[STIX]{x1D702}}\,(1\leqslant i\leqslant r) \\ p_{i}<Z~(1\leqslant i\leqslant r)}}r({\mathcal{A}},pdp_{1}\ldots p_{r}).\nonumber\end{eqnarray}$$
We shall show that
$$\begin{eqnarray}E_{2}\ll \unicode[STIX]{x1D6FF}N^{1-\unicode[STIX]{x1D702}};\end{eqnarray}$$
the proof that
$E_{1}\ll \unicode[STIX]{x1D6FF}N^{1-\unicode[STIX]{x1D702}}$
is similar but simpler.
Reducing the task of bounding
$E_{2}$
to estimating exponential sums as in previous sections, it suffices to prove that for
$|c_{\ell }|\leqslant 1$
, and a fixed
$\mathbf{D}\in {\mathcal{H}}$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\ell \leqslant L}c_{\ell }\mathop{\sum }_{\substack{ d<D^{\unicode[STIX]{x1D700}} \\ d\mid P(D^{\unicode[STIX]{x1D700}^{2}})}}\unicode[STIX]{x1D6EC}_{d}^{+}(\unicode[STIX]{x1D700},\mathbf{D})\mathop{\sum }_{P\leqslant p<P^{\prime }}\mathop{\sum }_{\substack{ D_{i}\leqslant p_{i}<\min (D_{i}^{1+\unicode[STIX]{x1D702}},Z) \\ (1\leqslant i\leqslant r)}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{N/2<dp_{1}\ldots p_{r}n\leqslant N}e(\ell g(dpp_{1}\ldots p_{r}n))\ll N^{1-3\unicode[STIX]{x1D702}}.\end{eqnarray}$$
This is a consequence of Theorem 2, grouping the variables as
$v=dp_{1}\ldots p_{r}$
and
$p$
,
$$\begin{eqnarray}v\ll N^{\unicode[STIX]{x1D700}}D^{1+\unicode[STIX]{x1D702}}\ll N^{2/3-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FE}},\qquad p\leqslant 2N^{\unicode[STIX]{x1D6FE}}.\end{eqnarray}$$
We make the usual verifications:
$$\begin{eqnarray}\displaystyle 3({\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FE})+2\unicode[STIX]{x1D6FE} & = & \displaystyle 2-24\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FE}<{\textstyle \frac{3}{2}}-3\unicode[STIX]{x1D70E},\nonumber\\ \displaystyle ({\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FE})+3\unicode[STIX]{x1D6FE} & = & \displaystyle {\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}+2\unicode[STIX]{x1D6FE}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{2}{3}-8\unicode[STIX]{x1D70E}+\frac{3}{4}+\frac{33\unicode[STIX]{x1D70E}}{2}<\frac{7}{4}-\frac{15\unicode[STIX]{x1D70E}}{2},\nonumber\\ \displaystyle ({\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FE})+\unicode[STIX]{x1D6FE} & = & \displaystyle {\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}.\nonumber\end{eqnarray}$$
The desired bound (5.5), and the same bound for
$E_{1}$
, now follows. Now (5.3), (5.4) yield
$$\begin{eqnarray}\mathop{\sum }_{P\leqslant p<P^{\prime }}S({\mathcal{A}}_{p},Z)\leqslant \frac{2(1+C\unicode[STIX]{x1D700})}{2/3-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FE}}\frac{\unicode[STIX]{x1D6FF}N}{\log N}\mathop{\sum }_{P\leqslant p<P^{\prime }}\frac{1}{p}.\end{eqnarray}$$
6 The sieve decomposition
For the final stage of our work we take
$\unicode[STIX]{x1D70C}=\frac{3}{17}$
so that
$\unicode[STIX]{x1D70E}=\frac{1}{102}$
. In this section, each
$S_{j}\;(j\geqslant 0)$
that occurs takes the form
where the asterisk indicates a restriction of
$p_{1}\ldots p_{r}$
and
$p_{1}^{\prime }\ldots p_{t}^{\prime }$
to certain subsets of
$[1,N]$
depending on
$j$
;
$S_{j}^{\prime }$
is obtained from
$S_{j}$
on replacing
${\mathcal{A}}$
by
${\mathcal{B}}$
. For some of these values of
$j$
we write
$$\begin{eqnarray}S_{j}=K_{j}+D_{j},\end{eqnarray}$$
where
$K_{j}$
is defined by the following additional condition of summation within
$S_{j}$
: a subproduct
$R$
of
$p_{1}\ldots p_{r}$
$p_{1}^{\prime }\ldots p_{t}^{\prime }$
satisfies
$$\begin{eqnarray}N^{\unicode[STIX]{x1D70C}}\ll R\ll N^{f}.\end{eqnarray}$$
We split up
$S_{j}^{\prime }$
as
$S_{j}^{\prime }=K_{j}^{\prime }+D_{j}^{\prime }$
in the same way. As noted in §5,
$$\begin{eqnarray}\displaystyle S_{j}\geqslant K_{j} & {\geqslant} & \displaystyle 2\unicode[STIX]{x1D6FF}K_{j}^{\prime }(1-C\unicode[STIX]{x1D700})\nonumber\\ \displaystyle & {\geqslant} & \displaystyle 2\unicode[STIX]{x1D6FF}K_{j}^{\prime }-\frac{C\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D700}N}{\log N}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle 2\unicode[STIX]{x1D6FF}S_{j}^{\prime }-2\unicode[STIX]{x1D6FF}D_{j}^{\prime }-\frac{C\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D700}N}{\log N}\end{eqnarray}$$
whenever (6.1) is used.
Conversion of sums into integrals, with an acceptable error, in the following is along familiar lines (see [Reference Harman8]). Concerning Buchstab’s function
$\unicode[STIX]{x1D714}(t)$
, we note that
$$\begin{eqnarray}\unicode[STIX]{x1D714}(t)\leqslant \unicode[STIX]{x1D705}\quad \text{for }t\geqslant \frac{1}{\unicode[STIX]{x1D705}}\end{eqnarray}$$
provided that
$\unicode[STIX]{x1D705}\geqslant 0.5672$
, and that
$$\begin{eqnarray}\unicode[STIX]{x1D714}(t)\geqslant \text{e}^{-\unicode[STIX]{x1D6FE}}-2.1\times 10^{-8}\quad (t\geqslant 6);\end{eqnarray}$$
see Cheer and Goldston [Reference Cheer and Goldston5]. We make the convention that whenever the lower limit of an integral exceeds the upper limit, the integrand is assumed to be zero.
Let
$S_{0}=S({\mathcal{A}},(2N)^{1/2})$
. Using Buchstab’s identity and writing
$p_{i}=(2N)^{\unicode[STIX]{x1D6FC}_{i}}$
, we have
$$\begin{eqnarray}S_{0}=S_{1}-\mathop{\sum }_{j=2}^{6}S_{j},\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle S_{1} & = & \displaystyle S({\mathcal{A}},z),\nonumber\\ \displaystyle S_{j} & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\in I_{j}}S({\mathcal{A}}_{p_{1}},p_{1})\quad (2\leqslant j\leqslant 6),\nonumber\end{eqnarray}$$
with
$I_{2}=[{\textstyle \frac{1}{6}}-5\unicode[STIX]{x1D70E},{\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E})$
,
$I_{3}=[\frac{1}{6}+\unicode[STIX]{x1D70E},\frac{1}{3}-4\unicode[STIX]{x1D70E})$
,
$I_{4}=[\frac{1}{3}-4\unicode[STIX]{x1D70E},\frac{1}{2}-9\unicode[STIX]{x1D70E})$
,
$I_{5}=[\frac{1}{2}-9\unicode[STIX]{x1D70E},\frac{3}{8}+33\unicode[STIX]{x1D70E}/4)$
and
$I_{6}=[\frac{3}{8}+33\unicode[STIX]{x1D70E}/4,\frac{1}{2}]$
.
Recalling (4.3) and Lemma 7 (with
$\unicode[STIX]{x1D708}=0$
), we have
$$\begin{eqnarray}S_{j}=2\unicode[STIX]{x1D6FF}S_{j}^{\prime }(1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700})\end{eqnarray}$$
for
$j=1,3$
. For
$j=2$
, we apply Buchstab’s identity three further times to obtain
$$\begin{eqnarray}S_{2}=S_{7}-S_{8}+S_{9}-S_{10},\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle & \displaystyle S_{7}=\mathop{\sum }_{b\leqslant \unicode[STIX]{x1D6FC}_{1}<\unicode[STIX]{x1D70C}}S({\mathcal{A}}_{p_{1}},z),\qquad S_{8}=\mathop{\sum }_{b\leqslant \unicode[STIX]{x1D6FC}_{2}\leqslant \unicode[STIX]{x1D6FC}_{1}<\unicode[STIX]{x1D70C}}S({\mathcal{A}}_{p_{1}p_{2}},z), & \displaystyle \nonumber\\ \displaystyle & \displaystyle S_{9}=\mathop{\sum }_{b\leqslant \unicode[STIX]{x1D6FC}_{3}\leqslant \unicode[STIX]{x1D6FC}_{2}\leqslant \unicode[STIX]{x1D6FC}_{1}<\unicode[STIX]{x1D70C}}S({\mathcal{A}}_{p_{1}p_{2}p_{3}},z) & \displaystyle \nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}S_{10}=\mathop{\sum }_{b\leqslant \unicode[STIX]{x1D6FC}_{4}\leqslant \unicode[STIX]{x1D6FC}_{3}\leqslant \unicode[STIX]{x1D6FC}_{2}\leqslant \unicode[STIX]{x1D6FC}_{1}<\unicode[STIX]{x1D70C}}S({\mathcal{A}}_{p_{1}p_{2}p_{3}p_{4}},p_{4}).\end{eqnarray}$$
We can apply Lemma 7 to
$S_{7}$
,
$S_{8}$
,
$S_{9}$
to obtain (6.7): for example, in
$S_{9}$
,
$$\begin{eqnarray}\unicode[STIX]{x1D70C}+(\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2})\leqslant 3\unicode[STIX]{x1D70C}\leqslant {\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}.\end{eqnarray}$$
We treat
$S_{10}$
via (6.3).
For
$j=4$
we apply Buchstab’s identity once. Iterating once more for part of the sum over
$p_{1}$
,
$p_{2}$
, this gives
$$\begin{eqnarray}S_{4}=S_{11}-S_{12}-S_{13}+S_{14},\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle S_{11} & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\in I_{4}}S({\mathcal{A}}_{p_{1}},z),\qquad S_{12}=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FC}_{1}\in I_{4} \\ \unicode[STIX]{x1D6FC}_{2}\in [f/2,\unicode[STIX]{x1D6FC}_{1})}}S({\mathcal{A}}_{p_{1}p_{2}},p_{2}),\nonumber\\ \displaystyle S_{13} & = & \displaystyle \mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FC}_{1}\in I_{4} \\ \unicode[STIX]{x1D6FC}_{2}\in [b,f/2)}}S({\mathcal{A}}_{p_{1}p_{2}},z),\qquad S_{14}=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FC}_{1}\in I_{4} \\ b\leqslant \unicode[STIX]{x1D6FC}_{3}\leqslant \unicode[STIX]{x1D6FC}_{2}<f/2}}S({\mathcal{A}}_{p_{1}p_{2}p_{3}},p_{3}).\nonumber\end{eqnarray}$$
We have (6.7) for
$S_{11}$
,
$S_{13}$
since Lemma 7 is applicable. For example, for
$S_{13}$
we have
$$\begin{eqnarray}\unicode[STIX]{x1D70C}+\unicode[STIX]{x1D6FC}_{1}\leqslant {\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E}+{\textstyle \frac{1}{2}}-9\unicode[STIX]{x1D70E}={\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}.\end{eqnarray}$$
We have (6.7) also for
$S_{14}$
, this time using (4.3), since
$\unicode[STIX]{x1D70C}\leqslant 2b\leqslant \unicode[STIX]{x1D6FC}_{2}+\unicode[STIX]{x1D6FC}_{3}<f$
in
$S_{14}$
. For
$S_{12}$
, we use the lower bound (6.3).
We also apply Buchstab once more to
$S_{6}$
,
$$\begin{eqnarray}S_{6}=S_{15}-S_{16}\end{eqnarray}$$
where
$$\begin{eqnarray}S_{15}=\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\in I_{6}}S({\mathcal{A}}_{p_{1}},z)\end{eqnarray}$$
satisfies (6.7) by Lemma 10, and
$$\begin{eqnarray}S_{16}=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FC}_{1}\in I_{6} \\ b\leqslant \unicode[STIX]{x1D6FC}_{2}<\unicode[STIX]{x1D6FC}_{1}}}S({\mathcal{A}}_{p_{1}p_{2}},p_{2})\end{eqnarray}$$
is bounded below as in (6.3).
For
$S_{5}$
, we proceed differently. We have
$$\begin{eqnarray}S_{5}+S_{17}=\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\in I_{5}}S({\mathcal{A}}_{p_{1}},Z)\end{eqnarray}$$
where
$$\begin{eqnarray}S_{17}=\biggl|\biggl\{p_{1}p_{1}^{\prime }\ldots p_{r}^{\prime }\in \biggl(\frac{N}{2},N\biggr]:\unicode[STIX]{x1D6FC}_{1}\in I_{5},r\geqslant 2,Z\leqslant p_{1}^{\prime }\leqslant \cdots \leqslant p_{r}^{\prime }\biggr\}\biggr|\end{eqnarray}$$
is treated as in (6.3).
By (5.7),
$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\in I_{5}}S({\mathcal{A}}_{p_{1}},Z)\leqslant \frac{\unicode[STIX]{x1D6FF}N(1+C\unicode[STIX]{x1D700})}{\log N}\int _{I_{5}}\,\frac{d\unicode[STIX]{x1D6FC}_{1}}{\unicode[STIX]{x1D6FC}_{1}}\frac{2}{\frac{2}{3}-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FC}_{1}}.\end{eqnarray}$$
It is convenient to introduce the notation
$$\begin{eqnarray}h(x)=\frac{2}{9}-\frac{8\unicode[STIX]{x1D70E}}{3}-\frac{x}{3}.\end{eqnarray}$$
Recalling that
$Z=N^{h(\unicode[STIX]{x1D6FC}_{1})-\unicode[STIX]{x1D700}/3}$
, we have
$$\begin{eqnarray}S_{5}^{\prime }+S_{17}^{\prime }=\frac{\unicode[STIX]{x1D6FF}N}{\log N}\,(1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700})\int _{I_{5}}\,\frac{d\unicode[STIX]{x1D6FC}_{1}}{\unicode[STIX]{x1D6FC}_{1}}\frac{1}{h(\unicode[STIX]{x1D6FC}_{1})}\,\unicode[STIX]{x1D714}\biggl(\frac{1-\unicode[STIX]{x1D6FC}_{1}}{\unicode[STIX]{x1D6FC}_{1}}\biggr).\end{eqnarray}$$
Combining (6.11)–(6.13) and (6.3) with
$j=17$
,
$$\begin{eqnarray}\displaystyle S_{5} & {\leqslant} & \displaystyle 2\unicode[STIX]{x1D6FF}S_{5}^{\prime }+\frac{\unicode[STIX]{x1D6FF}N}{\log N}\int _{I_{5}}\,\frac{d\unicode[STIX]{x1D6FC}_{1}}{\unicode[STIX]{x1D6FC}_{1}}\,\biggl(\frac{2}{{\textstyle \frac{2}{3}}-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FC}_{1}}-\frac{1}{h(\unicode[STIX]{x1D6FC}_{1})}\unicode[STIX]{x1D714}\biggl(\frac{1-\unicode[STIX]{x1D6FC}_{1}}{h(\unicode[STIX]{x1D6FC}_{1})}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,2\unicode[STIX]{x1D6FF}D_{17}^{\prime }+\frac{C\unicode[STIX]{x1D700}\unicode[STIX]{x1D6FF}N}{\log N}.\end{eqnarray}$$
Our sieve decomposition, obtained by combining (6.6), (6.8), (6.9) and (6.10), is
$$\begin{eqnarray}\displaystyle S_{0} & = & \displaystyle S_{1}-S_{3}-(S_{7}-S_{8}+S_{9}-S_{10})\nonumber\\ \displaystyle & & \displaystyle -\,(S_{11}-S_{12}-S_{13}+S_{14})-(S_{15}-S_{16})-S_{5}\nonumber\end{eqnarray}$$
and also holds if
$S_{j}$
is replaced by
$S_{j}^{\prime }$
. Combining all applications of (6.7) and (6.3) with (6.14), we end up with
$$\begin{eqnarray}\displaystyle S_{0} & {\geqslant} & \displaystyle 2\unicode[STIX]{x1D6FF}S_{0}^{\prime }-2\unicode[STIX]{x1D6FF}(D_{10}^{\prime }+D_{12}^{\prime }+D_{16}^{\prime }+D_{17}^{\prime })\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\unicode[STIX]{x1D6FF}N}{\log N}\int _{I_{5}}\,\frac{d\unicode[STIX]{x1D6FC}_{1}}{\unicode[STIX]{x1D6FC}_{1}}\biggl(\frac{2}{\frac{2}{3}-8\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FC}_{1}}-\frac{1}{h(\unicode[STIX]{x1D6FC}_{1})}\unicode[STIX]{x1D714}\biggl(\frac{1-\unicode[STIX]{x1D6FC}_{1}}{h(\unicode[STIX]{x1D6FC}_{1})}\biggr)\biggr)-\frac{C\unicode[STIX]{x1D700}\unicode[STIX]{x1D6FF}N}{\log N}\nonumber\\ \displaystyle & = & \displaystyle 2\unicode[STIX]{x1D6FF}S_{0}^{\prime }-2\unicode[STIX]{x1D6FF}(D_{10}^{\prime }+D_{12}^{\prime }+D_{16}^{\prime }+D_{17}^{\prime })\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\unicode[STIX]{x1D6FF}N}{\log N}\,(L_{1}-L_{2})-\frac{C\unicode[STIX]{x1D700}\unicode[STIX]{x1D6FF}N}{\log N},\end{eqnarray}$$
say.
Our next task is to evaluate
$D_{10}^{\prime }$
,
$D_{12}^{\prime }$
,
$D_{16}^{\prime }$
and
$D_{17}^{\prime }$
with sufficient accuracy. Using
$2b>\unicode[STIX]{x1D70C}$
we find that in
$D_{10}$
,
$\unicode[STIX]{x1D6FC}_{3}+\unicode[STIX]{x1D6FC}_{4}\notin [\unicode[STIX]{x1D70C},f]$
implies
$\unicode[STIX]{x1D6FC}_{3}+\unicode[STIX]{x1D6FC}_{4}>f$
. With a little thought we find that
$$\begin{eqnarray}\displaystyle D_{10}^{\prime } & {\leqslant} & \displaystyle \mathop{\sum }_{\substack{ b\leqslant \unicode[STIX]{x1D6FC}_{4}\leqslant \unicode[STIX]{x1D6FC}_{3}\leqslant \unicode[STIX]{x1D6FC}_{2}\leqslant \unicode[STIX]{x1D6FC}_{1}<\unicode[STIX]{x1D70C},\,\unicode[STIX]{x1D6FC}_{3}+\unicode[STIX]{x1D6FC}_{4}>f \\ p_{1}p_{2}p_{3}p_{4}n_{4}\in (N/2,N],\,p\mid n_{4}\rightarrow p\geqslant p_{4}}}1\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (1+\unicode[STIX]{x1D700})\,\frac{N}{2\log N}\,J_{1},\end{eqnarray}$$
where
$$\begin{eqnarray}J_{1}=\underset{\substack{ b\leqslant w\leqslant z\leqslant y\leqslant x<\unicode[STIX]{x1D70C} \\ z+w>f}}{\iiiint }\,\frac{dx}{x}\,\frac{dy}{y}\,\frac{dz}{z}\,\frac{dw}{w^{2}}\unicode[STIX]{x1D714}\biggl(\frac{1-x-y-z-w}{w}\biggr).\end{eqnarray}$$
It is simplest to replace the
$\unicode[STIX]{x1D714}$
factor by
$0.596775$
using (6.4). Carrying out the
$x$
and
$w$
integrations (and using
$z>f/2$
) leads to
$$\begin{eqnarray}J_{1}\leqslant 0.596775\int _{f/2<z\leqslant y\leqslant \unicode[STIX]{x1D70C}}\,\frac{dy}{y}\,\frac{dz}{z}\biggl(\frac{1}{f-z}-\frac{1}{z}\biggr)\log \frac{\unicode[STIX]{x1D70C}}{y}<0.000391.\end{eqnarray}$$
For
$D_{12}^{\prime }$
,
$D_{16}^{\prime }$
we write the numbers
$p_{1},p_{2},p_{1}^{\prime },\ldots ,p_{t}^{\prime }$
that appear in
$D_{j}^{\prime }$
as
$$\begin{eqnarray}(2N)^{\unicode[STIX]{x1D6FC}_{1}},(2N)^{\unicode[STIX]{x1D6FC}_{2}},(2N)^{\unicode[STIX]{x1D6FD}_{1}},\ldots ,(2N)^{\unicode[STIX]{x1D6FD}_{k}},(2N)^{\unicode[STIX]{x1D6FE}_{1}},\ldots ,(2N)^{\unicode[STIX]{x1D6FE}_{\ell }},\end{eqnarray}$$
where
$k\geqslant 0$
,
$\ell \geqslant 0$
,
$k+\ell \geqslant 1$
,
$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FC}_{2}\leqslant \unicode[STIX]{x1D6FD}_{1}\leqslant \cdots \leqslant \unicode[STIX]{x1D6FD}_{k}<\unicode[STIX]{x1D70C}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle f<\unicode[STIX]{x1D6FE}_{1}\leqslant \cdots \leqslant \unicode[STIX]{x1D6FE}_{\ell } & \displaystyle \nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2}+\unicode[STIX]{x1D6FD}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{k}+\unicode[STIX]{x1D6FE}_{1}+\cdots +\unicode[STIX]{x1D6FE}_{\ell }=1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700}.\end{eqnarray}$$
We now consider
$D_{12}^{\prime }$
. First we treat the contribution
$D^{(1)}$
from
$\unicode[STIX]{x1D6FC}_{2}>f$
. Thus
$k=0$
. We cannot have
$\ell \geqslant 2$
, since
$$\begin{eqnarray}4({\textstyle \frac{1}{3}}-4\unicode[STIX]{x1D70E})>1.\end{eqnarray}$$
If we consider
$\ell =1$
, we have
$\unicode[STIX]{x1D6FC}_{1}+2\unicode[STIX]{x1D6FC}_{2}<1$
and obtain
$$\begin{eqnarray}D^{(1)}=(1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700})\frac{\unicode[STIX]{x1D6FF}N}{\log N}J_{2},\end{eqnarray}$$
where
$$\begin{eqnarray}J_{2}=\int _{1/3-4\unicode[STIX]{x1D70E}}^{1/2-9\unicode[STIX]{x1D70E}}\int _{1/3-4\unicode[STIX]{x1D70E}}^{\min (x,(1-x)/2)}\,\frac{dy\,dx}{xy(1-x-y)}<0.062921.\end{eqnarray}$$
Let
$D^{(2)}$
be the contribution to
$D_{12}^{\prime }$
from
$\unicode[STIX]{x1D6FC}_{2}<\unicode[STIX]{x1D70C}$
. We have
$\unicode[STIX]{x1D6FC}_{2}+\unicode[STIX]{x1D6FD}_{1}\geqslant f$
,
$\unicode[STIX]{x1D6FD}_{1}\geqslant f/2$
, by an argument used above. We cannot have
$\ell \geqslant 2$
since
$$\begin{eqnarray}3({\textstyle \frac{1}{3}}-4\unicode[STIX]{x1D70E})+{\textstyle \frac{1}{6}}-2\unicode[STIX]{x1D70E}={\textstyle \frac{7}{6}}-14\unicode[STIX]{x1D70E}>1.\end{eqnarray}$$
If
$\ell =1$
, we must have
$k\leqslant 1$
(use (6.19) again). If
$\ell =0$
, we must have
$k\leqslant 3$
similarly; however,
$k>2$
, since
$$\begin{eqnarray}{\textstyle \frac{1}{2}}-9\unicode[STIX]{x1D70E}+3({\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E})<1.\end{eqnarray}$$
The three remaining cases lead to
$$\begin{eqnarray}D^{(2)}=(1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700})(J_{3}+J_{4}+J_{5}),\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle J_{3} & = & \displaystyle \int _{1/3-4\unicode[STIX]{x1D70E}}^{1/2-9\unicode[STIX]{x1D70E}}\int _{1/6-2\unicode[STIX]{x1D70E}}^{1/6+\unicode[STIX]{x1D70E}}\,\frac{dy\,dx}{xy(1-x-y)}<0.126097\quad (\ell =1,k=0),\nonumber\\ \displaystyle J_{4} & = & \displaystyle \int _{1/3-4\unicode[STIX]{x1D70E}}^{1/2-9\unicode[STIX]{x1D70E}}\int _{1/6-2\unicode[STIX]{x1D70E}}^{1/6+\unicode[STIX]{x1D70E}}\int _{y}^{\min (1/6+\unicode[STIX]{x1D70E},2/3+4\unicode[STIX]{x1D70E}-x-y)}\,\frac{dz\,dy\,dx}{xyz(1-x-y-z)}\nonumber\\ \displaystyle & {<} & \displaystyle 0.012943\quad (\ell =1,k=1),\nonumber\\ \displaystyle J_{5} & = & \displaystyle \int _{1/3-4\unicode[STIX]{x1D70E}}^{1/2-9\unicode[STIX]{x1D70E}}\!\int _{1/6-2\unicode[STIX]{x1D70E}}^{1/6+\unicode[STIX]{x1D70E}}\!\int _{y}^{1/6+\unicode[STIX]{x1D70E}}\!\int _{\max (z,5/6-\unicode[STIX]{x1D70E}-x-y-z)}^{(1-x-y-z)/2}\,\frac{dw\,dz\,dy\,dx}{xyzw(1-x-y-z-w)}\nonumber\\ \displaystyle & {<} & \displaystyle 0.000131\quad (\ell =0,k=3).\nonumber\end{eqnarray}$$
A few comments are in order for the domains of integration of
$J_{4}$
,
$J_{5}$
. When
$\ell =k=1$
, we have
$\unicode[STIX]{x1D6FE}_{1}>\frac{1}{3}-4\unicode[STIX]{x1D70E}$
, and hence
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2}+\unicode[STIX]{x1D6FD}_{1}<{\textstyle \frac{2}{3}}+4\unicode[STIX]{x1D70E}+C\unicode[STIX]{x1D700},\qquad \unicode[STIX]{x1D6FD}_{1}<{\textstyle \frac{2}{3}}+4\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D6FC}_{1}-\unicode[STIX]{x1D6FC}_{2}+C\unicode[STIX]{x1D700}.\end{eqnarray}$$
When
$\ell =0$
,
$k=3$
, we have
$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FD}_{3}=1-(\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2}+\unicode[STIX]{x1D6FD}_{1}+\unicode[STIX]{x1D6FD}_{2}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700})<{\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D6FD}_{2}>{\textstyle \frac{5}{6}}-\unicode[STIX]{x1D70E}-(\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2}+\unicode[STIX]{x1D6FD}_{1})-C\unicode[STIX]{x1D700}. & \displaystyle \nonumber\end{eqnarray}$$
We remark that
$J_{5}$
was bounded above by an integral in
$x$
,
$y$
,
$z$
, using
$1-x-y-z-w\geqslant w$
, before carrying out the
$w$
integration.
In
$D_{16}^{\prime }$
, we cannot have
$\unicode[STIX]{x1D6FC}_{2}\geqslant \unicode[STIX]{x1D70C}$
, since then
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{2}<\frac{1}{2}\,(1-\unicode[STIX]{x1D6FC}_{1})<\frac{1}{2}\biggl(\frac{5}{8}-\frac{33\unicode[STIX]{x1D70E}}{4}\biggr)<\frac{1}{3}-4\unicode[STIX]{x1D70E},\end{eqnarray}$$
which is absurd. Thus
$\unicode[STIX]{x1D6FC}_{2}<\unicode[STIX]{x1D70C}$
; we cannot have
$\ell \geqslant 2$
or
$\ell =1$
,
$k\geqslant 1$
or
$\ell =0$
,
$k\geqslant 3$
since
$$\begin{eqnarray}\frac{3}{8}+\frac{33\unicode[STIX]{x1D70E}}{4}+2\biggl(\frac{1}{3}-4\unicode[STIX]{x1D70E}\biggr)>1.\end{eqnarray}$$
For
$\ell =0$
, we have
$k>1$
since
$$\begin{eqnarray}{\textstyle \frac{1}{2}}+2({\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E})<1.\end{eqnarray}$$
Note that for
$\ell =0$
,
$k=2$
, we have
$$\begin{eqnarray}1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700}=\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2}+\unicode[STIX]{x1D6FD}_{1}+\unicode[STIX]{x1D6FD}_{2}<{\textstyle \frac{5}{6}}+2\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D6FC}_{2},\qquad \unicode[STIX]{x1D6FC}_{2}>{\textstyle \frac{1}{6}}-2\unicode[STIX]{x1D70E}-C\unicode[STIX]{x1D700}.\end{eqnarray}$$
Thus
$$\begin{eqnarray}D_{16}^{\prime }=(1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700})\frac{N}{2\log N}(J_{6}+J_{7}),\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle J_{6} & = & \displaystyle \int _{3/8+33\unicode[STIX]{x1D70E}/4}^{1/2}\int _{1/6-5\unicode[STIX]{x1D70E}}^{1/6+\unicode[STIX]{x1D70E}}\,\frac{dy\,dx}{xy(1-x-y)}<0.099581\quad (\ell =1,k=0);\nonumber\\ \displaystyle J_{7} & = & \displaystyle \underset{1/6+\unicode[STIX]{x1D70E}>1-x-y-z>z}{\int _{3/8+33\unicode[STIX]{x1D70E}/4}^{1/2}\int _{1/6-2\unicode[STIX]{x1D70E}}^{1/6+\unicode[STIX]{x1D70E}}\int _{y}^{1/6+\unicode[STIX]{x1D70E}}}\,\frac{dz\,dy\,dx}{xyz(1-x-y-z)}<0.000298\nonumber\\ \displaystyle & & \displaystyle \quad (\ell =0,k=2).\nonumber\end{eqnarray}$$
In
$D_{17}^{\prime }$
, we write the numbers that appear as
$p_{1},p_{1}^{\prime },\ldots ,p_{r}^{\prime }$
as
$$\begin{eqnarray}(2N)^{\unicode[STIX]{x1D6FC}_{1}},(2N)^{\unicode[STIX]{x1D6FD}_{1}}\leqslant \cdots \leqslant (2N)^{\unicode[STIX]{x1D6FD}_{k}},(2N)^{\unicode[STIX]{x1D6FE}_{1}}\leqslant \cdots \leqslant (2N)^{\unicode[STIX]{x1D6FE}_{k}},\end{eqnarray}$$
where
$k+\ell =r\geqslant 2$
,
$k\geqslant 0$
,
$\ell \geqslant 0$
,
$\unicode[STIX]{x1D6FD}_{k}<\unicode[STIX]{x1D70C}$
,
$\unicode[STIX]{x1D6FE}_{1}>f$
,
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FD}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{k}+\unicode[STIX]{x1D6FE}_{1}+\cdots +\unicode[STIX]{x1D6FE}_{k}=1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700}.\end{eqnarray}$$
We claim that no subsum
$\unicode[STIX]{x1D70F}$
of
$\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{k}$
,
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{\ell }$
can be in the interval
$[\unicode[STIX]{x1D70C},1-\unicode[STIX]{x1D6FC}_{1}-\unicode[STIX]{x1D70C}]$
. For this we use a ‘reflection argument’:
$\unicode[STIX]{x1D70F}$
cannot satisfy
$$\begin{eqnarray}1-\unicode[STIX]{x1D6FC}_{1}-({\textstyle \frac{1}{3}}-4\unicode[STIX]{x1D70E})\leqslant \unicode[STIX]{x1D70F}\leqslant 1-\unicode[STIX]{x1D6FC}_{1}-\unicode[STIX]{x1D70C}.\end{eqnarray}$$
Moreover,
$$\begin{eqnarray}1-\unicode[STIX]{x1D6FC}_{1}-({\textstyle \frac{1}{3}}-4\unicode[STIX]{x1D70E})\leqslant {\textstyle \frac{1}{3}}-4\unicode[STIX]{x1D70E}\end{eqnarray}$$
since
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{1}>{\textstyle \frac{1}{2}}-9\unicode[STIX]{x1D70E}={\textstyle \frac{1}{3}}+8\unicode[STIX]{x1D70E}.\end{eqnarray}$$
This leads to our claim.
The length of
$[\unicode[STIX]{x1D70C},1-\unicode[STIX]{x1D6FC}_{1}-\unicode[STIX]{x1D70C}]$
is greater than
$\unicode[STIX]{x1D70C}$
, since
$$\begin{eqnarray}3\unicode[STIX]{x1D70C}=3\biggl(\frac{1}{6}+\unicode[STIX]{x1D70E}\biggr)<\frac{5}{8}-\frac{33\unicode[STIX]{x1D70E}}{4}\leqslant 1-\unicode[STIX]{x1D6FC}_{1}.\end{eqnarray}$$
It follows at once that
$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{k}\leqslant {\textstyle \frac{1}{6}}+\unicode[STIX]{x1D70E},\end{eqnarray}$$
and
$\ell \geqslant 1$
. We cannot have
$\ell \geqslant 2$
, since
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{1}+2(1-\unicode[STIX]{x1D6FC}_{1}-\unicode[STIX]{x1D70C})=2-2\unicode[STIX]{x1D70C}-\unicode[STIX]{x1D6FC}_{1}>{\textstyle \frac{3}{2}}-2\unicode[STIX]{x1D70C}>1.\end{eqnarray}$$
Hence
$\ell =1$
, and
$k\geqslant 1$
since
$\ell +k\geqslant 2$
.
Now either
$k\leqslant 3$
, or
$k=4$
and
$$\begin{eqnarray}\frac{3}{8}+\frac{33\unicode[STIX]{x1D70E}}{4}-\unicode[STIX]{x1D700}<\unicode[STIX]{x1D6FC}_{1}\leqslant \frac{3}{8}+\frac{33\unicode[STIX]{x1D70E}}{4}.\end{eqnarray}$$
To see this,
$$\begin{eqnarray}4\biggl(\frac{2}{9}-\frac{8\unicode[STIX]{x1D70E}}{3}-\frac{\unicode[STIX]{x1D6FC}_{1}}{3}-\frac{1}{3}\,\unicode[STIX]{x1D700}\biggr)\geqslant \frac{1}{6}+\unicode[STIX]{x1D70E}\end{eqnarray}$$
unless
$$\begin{eqnarray}\frac{4\unicode[STIX]{x1D6FC}_{1}}{3}>\frac{13}{18}-\frac{35\unicode[STIX]{x1D70E}}{3}-\frac{4\unicode[STIX]{x1D700}}{3},\end{eqnarray}$$
that is,
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{1}>\frac{13}{24}-\frac{35\unicode[STIX]{x1D70E}}{4}-\unicode[STIX]{x1D700}=\frac{3}{8}+\frac{33\unicode[STIX]{x1D70E}}{4}-\unicode[STIX]{x1D700}.\end{eqnarray}$$
Hence
$$\begin{eqnarray}D_{17}^{\prime }=\frac{N}{2\log N}(1+\unicode[STIX]{x1D706}\unicode[STIX]{x1D700})(L_{3}+L_{4}+L_{5}),\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle L_{3} & = & \displaystyle \int _{1/2-9\unicode[STIX]{x1D70E}}^{3/8+33\unicode[STIX]{x1D70E}/4}\int _{h(x)}^{1/6+\unicode[STIX]{x1D70E}}\,\frac{dy\,dx}{xy(1-x-y)}<0.272041,\nonumber\\ \displaystyle L_{4} & = & \displaystyle \int _{1/2-9\unicode[STIX]{x1D70E}}^{3/8+33\unicode[STIX]{x1D70E}/4}\int _{h(x)}^{1/12+\unicode[STIX]{x1D70E}/2}\int _{y}^{1/6+\unicode[STIX]{x1D70E}-y}\,\frac{dz\,dy\,dx}{xyz(1-x-y-z)}<0.063399,\nonumber\\ \displaystyle L_{5} & = & \displaystyle \int _{1/2-9\unicode[STIX]{x1D70E}}^{3/8+33\unicode[STIX]{x1D70E}/4}\int _{h(x)}^{1/18+\unicode[STIX]{x1D70E}/3}\int _{y}^{1/12+\unicode[STIX]{x1D70E}/2-y/2}\int _{z}^{1/6+\unicode[STIX]{x1D70E}-y-z}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{dw\,dz\,dy\,dx}{xyzw(1-x-y-z-w)}\nonumber\\ \displaystyle & {<} & \displaystyle 0.000902.\nonumber\end{eqnarray}$$
We note here that
$$\begin{eqnarray}L_{1}<1.324181\quad \text{and}\quad L_{2}>1.115210.\end{eqnarray}$$
Since
$S_{0}^{\prime }=(1-\unicode[STIX]{x1D706}\unicode[STIX]{x1D700})(N/2\log N)$
, we need only combine all our numerical estimates to obtain Theorem 1 from (6.15):
$$\begin{eqnarray}J_{1}+J_{2}+\cdots +J_{7}+L_{1}-L_{2}+L_{3}+L_{4}+L_{5}<{\textstyle \frac{6}{7}}.\end{eqnarray}$$
Acknowledgements
Integrals that appear here and in earlier drafts were calculated by Andreas Weingartner; thanks, Andreas, for your generosity. I would like to thank the referee for a thorough scrutiny of the previous version of the paper.
