1 Introduction
The classical version of Waring’s problem asks whether every positive integer can be written as a sum of at most
$s$
positive integers, each of which is a
$k$
th power. In other words, is there an integer
$s$
(which depends on
$k$
) such that for each
$n\geqslant 1$
we have a solution to the equation
$$\begin{eqnarray}n=x_{1}^{k}+\cdots +x_{s}^{k}\end{eqnarray}$$
in non-negative integers
$x_{1},\ldots ,x_{s}$
? The least value of
$s$
which is admissible is usually referred to as
$g(k)$
, and Waring’s problem is thus the assertion that
$g(k)<\infty$
for any
$k\geqslant 1$
. Waring’s problem has a long history; for a nice exposition, see [Reference Vaughan and Wooley9].
The “easier” version of Waring’s problem, a name attributed to Wright [Reference Wright11], asks whether there is a solution to the equation
$$\begin{eqnarray}n=x_{1}^{k}\pm \cdots \pm x_{s}^{k}.\end{eqnarray}$$
The least
$s$
for which this equation is soluble for each
$n$
is usually referred to as
$v(k)$
, and establishing that
$v(k)<\infty$
is a fairly simple argument, which can be found in [Reference Nathanson6]. Clearly any upper bound for
$g(k)$
in the usual Waring problem extends to a bound for
$v(k)$
as well. However, the freedom to use negative summands may make
$v(k)$
considerably smaller.
One can verify that
$g(k)\geqslant 2^{k}-1$
. Indeed, in order for
$2^{k}-1$
to be written as a sum of
$k$
th powers, we only have
$1$
s at our disposal. For these reasons, one usually considers instead
$G(k)$
, which is the least
$s$
such that (1) is soluble for all
$n$
sufficiently large. Here, the bound
$G(k)\geqslant k$
is still quite simple. To represent each
$n$
in the range
$X\leqslant n\leqslant 2X$
, the variables
$x_{i}$
can be no larger than
$X^{1/k}$
. Thus, the vector
$(x_{1},\ldots ,x_{s})$
is a lattice point in the box
$[0,X^{1/k}]^{s}$
, and there are at most
$O(X^{s/k})$
such lattice points. To represent all integers in the desired range, we must therefore have
$s\geqslant k$
. The introduction of negative summands causes this argument to fail completely, because one is no longer counting lattice points in a bounded region. This motivates the following question, which was asked in [Reference Bukh2].
Problem.
For
$k$
sufficiently large, is it true that the set of integers of the form
$$\begin{eqnarray}n=x_{1}^{k}\pm x_{2}^{k}\pm \cdots \pm x_{s}^{k}\end{eqnarray}$$
has asymptotic density zero?
A result of Wooley (see [Reference Bukh2] and [Reference Wooley10]) asserts that, for
$k\gg s^{3}$
, the set of integers of the form
$$\begin{eqnarray}n=x_{1}^{k}\pm \cdots \pm x_{s}^{k}\end{eqnarray}$$
has density zero and in fact more is true—one can obtain fairly good decay rates in the proportion of integers up to
$X$
which can be represented. However, Wooley’s result is conditional on a generalized version of the
$abc$
-conjecture and, as far as the authors are aware, there seems to be little known unconditionally for large values of
$s$
, say
$s\geqslant 5$
. Moreover, proving such density results is expected to be very difficult in general since the variables, a priori, could be arbitrarily large, and since ruling out solubility for a specific vector of coefficients and a specific
$n$
is related to proving insolubility of generalized Fermat equations (see Wooley’s remarks, [Reference Bukh2]). Thus, we are content to prove a result in this direction which is much weaker, but unconditional. We will not be able to prove that the set of integers represented has zero density, but we will establish bounds on the asymptotic density of these integers. These bounds will, on average, decay quite rapidly with respect to
$k$
.
In fact, we will establish something a bit more general in that we will allow for arbitrary integer coefficients, not just
$1$
s and
$-1$
s. Let
$s\geqslant 2$
be fixed and let
$\text{}\text{a}=(a_{1},\ldots ,a_{s})\in \mathbb{Z}^{s}$
be arbitrary. We consider the form
$$\begin{eqnarray}F_{\text{}\text{a},k}()=a_{1}x_{1}^{k}+\cdots +a_{s}x_{s}^{k}\end{eqnarray}$$
and the set
$$\begin{eqnarray}{\mathcal{R}}(\text{}\text{a},k)=\{n:n=F_{\text{}\text{a},k}()\text{ for some }\in \mathbb{Z}^{s}\}\end{eqnarray}$$
of numbers which this form represents.
We shall estimate the average asymptotic (upper)-density
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{k}=\limsup _{N\rightarrow \infty }\frac{|{\mathcal{R}}(\text{}\text{a},k)\cap [1,N]|}{N}\end{eqnarray}$$
as a function of
$k$
. This number implicitly depends on
$\text{}\text{a}$
, but the results we shall prove about
$\unicode[STIX]{x1D6FF}_{k}$
are uniform over
$\text{}\text{a}$
. For
$s\geqslant 3$
, the following theorem establishes that the value of
$\log (1/\unicode[STIX]{x1D6FF}_{k})$
is large on average.
Theorem 1. Let
$s\geqslant 3$
be fixed and
$\text{}\text{a}\in \mathbb{Z}^{s}$
be arbitrary. Let
$X\geqslant 3$
be sufficiently large depending at most on
$s$
and define
$\unicode[STIX]{x1D6FF}_{k}$
as in (4). Then there exists a constant
$c_{s}>0$
depending only on
$s$
such that
$$\begin{eqnarray}\biggl(\mathop{\prod }_{1\leqslant k\leqslant X}\unicode[STIX]{x1D6FF}_{k}\biggr)^{1/X}\leqslant \exp \biggl(-\frac{c_{s}X^{1/(s-1)}}{\log X}\biggr).\end{eqnarray}$$
Remark.
The constant
$c_{s}$
in Theorem 1 (and
$C_{s}$
in Corollary 2) could be made explicit with an effective variant of the Bombieri–Vinogradov theorem though we did not apply such a version here.
We expect that the quantity on the left-hand side in (5) is zero for all
$X$
sufficiently large depending only on
$s$
and
$\text{}\text{a}$
. To see that this theorem does make some progress on the problem, we observe the following immediate consequence.
Corollary 2. Let
$s\geqslant 3$
be fixed and
$\text{}\text{a}\in \mathbb{Z}^{s}$
be arbitrary. Then for any
$\unicode[STIX]{x1D700}>0$
, there exists a constant
$C_{s}>0$
depending only on
$s$
and a positive integer
$k\leqslant C_{s}(\log (1/\unicode[STIX]{x1D700}))^{s-1}\log \log (1/\unicode[STIX]{x1D700})$
, such that
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{k}=\limsup _{N\rightarrow \infty }\frac{|{\mathcal{R}}(\text{}\text{a},k)\cap [1,N]|}{N}<\unicode[STIX]{x1D700}.\end{eqnarray}$$
In particular, the density of integers in
${\mathcal{R}}(\text{}\text{a},k)$
can be made arbitrarily small by choosing an appropriate
$k$
of modest size.
Perhaps it is instructive to compare Theorem 1 to a conditional result. Let
$\unicode[STIX]{x1D70B}(X;k,1)$
denote the number of primes
$p<X$
satisfying
$p\equiv 1(\text{mod}\,k)$
and let
$\unicode[STIX]{x1D711}(k)$
denote Euler’s totient function.
Proposition 3. Let
$s\geqslant 3$
be fixed and
$\text{}\text{a}\in \mathbb{Z}^{s}$
be arbitrary. Let
$k\geqslant 3$
be sufficiently large depending at most on
$s$
and define
$\unicode[STIX]{x1D6FF}_{k}$
as in (4). If
$$\begin{eqnarray}\unicode[STIX]{x1D70B}(X;k,1)=\frac{1}{\unicode[STIX]{x1D711}(k)}\text{Li}(X)+O_{\unicode[STIX]{x1D700}}\biggl(\frac{X^{1/2+\unicode[STIX]{x1D700}}}{k^{1/2}}\biggr)\end{eqnarray}$$
for any
$\unicode[STIX]{x1D700}>0$
and
$X\geqslant k^{1+\unicode[STIX]{x1D700}}$
then there is a constant
$c_{s}>0$
depending only on
$s$
such that
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{k}\leqslant \exp \biggl(-\frac{c_{s}k^{1/(s-1)}}{\log k}\biggr).\end{eqnarray}$$
Assumption (6) is one of the strongest widely believed conjectures regarding the distribution of primes in arithmetic progressions. It is essentially a conjecture due to Montgomery; see [Reference Soundararajan7, p. 18] for example. Theorem 1 unconditionally obtains the corresponding geometric average result for
$\unicode[STIX]{x1D6FF}_{k}$
by appealing to the Bombieri–Vinogradov theorem. Note that the special case
$s=2$
is addressed by classical work of Mahler [Reference Mahler5], which implies
$\unicode[STIX]{x1D6FF}_{k}=0$
for
$k\geqslant 3$
. For further details on the case
$s=2$
, see for example [Reference Bennett, Dummigan and Wooley1] and [Reference Stewart and Xiao8].
2 Local densities and a conditional result
The method of proof for Theorem 1 and Proposition 3 is to bound the density
$\unicode[STIX]{x1D6FF}_{k}$
by considering local constraints. For instance, a very simple first observation is that for
$p$
prime
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{p-1}\leqslant \frac{2^{s}}{p}.\end{eqnarray}$$
This is just the trivial observation that the set of
$(p-1)\text{th}$
powers modulo
$p$
consist of the residue classes
$0$
and
$1$
modulo
$p$
. Thus, there are at most
$2^{s}$
admissible values of the diagonal form
$F_{\text{}\text{a},k}$
modulo
$p$
. Our aim is then to improve this estimate for a given
$k$
, and subsequently obtain good density estimates for the average exponent
$k$
. To this end, define
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{k}(p)=\frac{1}{p}|\{F_{\text{}\text{a},k}(\text{}\text{z})~\text{mod}\;p:\text{}\text{z}\in \mathbb{F}_{p}^{s}\}|.\end{eqnarray}$$
The Chinese remainder theorem then gives the following.
Lemma 4. For any integer
$k\geqslant 1$
,
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{k}\leqslant \mathop{\prod }_{p}\unicode[STIX]{x1D6FF}_{k}(p).\end{eqnarray}$$
We will combine this with a simple development of the idea we used to bound
$\unicode[STIX]{x1D6FF}_{p-1}$
. Let
$(m,n)$
denote the greatest common divisor of two integers
$m$
and
$n$
.
Lemma 5. Let
$k\geqslant 2$
and let
$p$
be a prime. Then
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{k}(p)\leqslant \unicode[STIX]{x1D6FC}_{k,p},\end{eqnarray}$$
where
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{k,p}=\frac{1}{p}\biggl(\frac{p-1}{(k,p-1)}+1\biggr)^{s}.\end{eqnarray}$$
Proof. By the structure theorem for cyclic groups, the set non-zero
$k\text{th}$
powers modulo
$p$
forms a subgroup of the unit group of size
$(p-1)/(k,p-1)$
. Adding in the
$0$
class modulo
$p$
, there are
$(p-1)/(k,p-1)+1$
values of
$z^{k}$
modulo
$p$
. Thus, the proportion of admissible residue classes modulo
$p$
is at most
$\unicode[STIX]{x1D6FC}_{k,p}$
.
Using these two lemmas, we can establish Proposition 3.
Proof of Proposition 3.
Suppose
$p$
is a prime satisfying
$p\equiv 1(\text{mod}~k)$
. With
$\unicode[STIX]{x1D6FC}_{k,p}$
defined as in Lemma 5, observe that
$\unicode[STIX]{x1D6FC}_{k,p}<1$
if and only if
$$\begin{eqnarray}\frac{p-1}{k}+1<p^{1/s}.\end{eqnarray}$$
Since
$(p-1)/k+1<p/k+p/k=2p/k$
, we have that
$\unicode[STIX]{x1D6FC}_{k,p}<1$
whenever
$$\begin{eqnarray}p<{\textstyle \frac{1}{4}}k^{1/(1-1/s)}={\textstyle \frac{1}{4}}k^{1+1/(s-1)}.\end{eqnarray}$$
Set
$Z=(1/e^{3})k^{1+1/(s-1)}$
. Thus, by Lemmas 4 and 5 and noting
$\log (a+b)\leqslant \log a+\log b$
for
$a,b\geqslant 2$
, it follows that
$$\begin{eqnarray}\displaystyle \log (1/\unicode[STIX]{x1D6FF}_{k})\geqslant \mathop{\sum }_{p}\log (1/\unicode[STIX]{x1D6FF}_{k}(p)) & {\geqslant} & \displaystyle \mathop{\sum }_{\substack{ p\equiv 1(\text{mod}~k) \\ p<Z}}\biggl(\log p-s\log \biggl(\frac{p-1}{k}+1\biggr)\biggr)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \mathop{\sum }_{\substack{ p\equiv 1(\text{mod}~k) \\ p<Z}}(s\log k-(s-1)\log p-s).\nonumber\end{eqnarray}$$
Since
$p<Z$
and
$s\geqslant 3$
, the above is
$$\begin{eqnarray}{\geqslant}\mathop{\sum }_{\substack{ p\equiv 1(\text{mod}~k) \\ p<Z}}(3(s-1)-s)\geqslant \mathop{\sum }_{\substack{ p\equiv 1(\text{mod}~k) \\ p<Z}}s.\end{eqnarray}$$
Using assumption (6) to bound the sum on the right in the above inequality, it follows that
$$\begin{eqnarray}\log (1/\unicode[STIX]{x1D6FF}_{k})\gg _{s}\frac{k^{1+1/(s-1)}}{\unicode[STIX]{x1D711}(k)\log k}\gg _{s}\frac{k^{1/(s-1)}}{\log k},\end{eqnarray}$$
after bounding
$\unicode[STIX]{x1D711}(k)$
trivially by
$k$
. This proves the proposition.◻
3 Global density is small on average
This section is dedicated to proving Theorem 1. Roughly speaking, we will modify the proof of Proposition 3 by replacing Montgomery’s conjecture with the Bombieri–Vinogradov theorem. Thus, we establish a lower bound on average for
$\log (1/\unicode[STIX]{x1D6FF}_{k})$
, rather than on
$\log (1/\unicode[STIX]{x1D6FF}_{k})$
individually. First, we will require one additional lemma. For integers
$a$
and
$q$
, let
$$\begin{eqnarray}\unicode[STIX]{x1D713}(X;q,a)=\mathop{\sum }_{\substack{ n<X \\ n\equiv a(\text{mod}\,q)}}\unicode[STIX]{x1D6EC}(n),\end{eqnarray}$$
where
$\unicode[STIX]{x1D6EC}(n)$
equals
$\log p$
if
$n$
is a power of a prime
$p$
and equals
$0$
otherwise.
Lemma 6. Let
$X\geqslant 1$
be arbitrary. For
$Y\leqslant X^{1/2}$
,
$$\begin{eqnarray}\mathop{\sum }_{1\leqslant m\leqslant Y}\unicode[STIX]{x1D713}(mX;m,1)\geqslant \frac{\unicode[STIX]{x1D701}(2)\unicode[STIX]{x1D701}(3)\log 2}{\unicode[STIX]{x1D701}(6)}XY+O(X\log ^{2}X+XY(\log X)^{-2}).\end{eqnarray}$$
Additionally, if
$s\geqslant 2$
is fixed then
$$\begin{eqnarray}\mathop{\sum }_{1\leqslant m<Y}\int _{(m+1)^{s}}^{mX}{\displaystyle \frac{\unicode[STIX]{x1D713}(t;m,1)}{t(\log t)^{2}}}\,dt\ll \frac{XY}{\log ^{2}X}.\end{eqnarray}$$
Proof. We divide the sum in (9) dyadically. For
$1\leqslant M\leqslant 2M\leqslant X^{1/2}$
, note
$M\leqslant (MX)^{1/3}$
. Hence, by the Bombieri–Vinogradov theorem [Reference Iwaniec and Kowalski3, Theorem 17.1], we have that
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{M\leqslant m<2M}\unicode[STIX]{x1D713}(mX;m,1) & {\geqslant} & \displaystyle \mathop{\sum }_{M\leqslant m<2M}\unicode[STIX]{x1D713}(MX;m,1)\nonumber\\ \displaystyle & = & \displaystyle \biggl(\mathop{\sum }_{M\leqslant m<2M}\frac{MX}{\unicode[STIX]{x1D711}(m)}\biggr)+O(MX(\log MX)^{-2})\nonumber\\ \displaystyle & = & \displaystyle \frac{\unicode[STIX]{x1D701}(2)\unicode[STIX]{x1D701}(3)\log 2}{\unicode[STIX]{x1D701}(6)}MX+O(X\log M+MX(\log MX)^{-2}).\nonumber\end{eqnarray}$$
In the last step, we applied the classical fact [Reference Landau4] that, for
$x\geqslant 2$
,
$$\begin{eqnarray}\mathop{\sum }_{n\leqslant x}\frac{1}{\unicode[STIX]{x1D711}(n)}=\frac{\unicode[STIX]{x1D701}(2)\unicode[STIX]{x1D701}(3)}{\unicode[STIX]{x1D701}(6)}\biggl(\log x+\unicode[STIX]{x1D6FE}-\mathop{\sum }_{p}\frac{\log p}{p^{2}-p+1}\biggr)+O(x^{-1}\log x).\end{eqnarray}$$
Summing the prior estimate over
$M=Y/2^{j+1}$
with
$0\leqslant j\leqslant \lfloor \log Y\rfloor$
and recalling
$Y\leqslant X^{1/2}$
yields the desired result. To prove (10), we proceed similarly. For
$t\geqslant (m+1)^{s}$
and
$s\geqslant 2$
, we may apply the Brun–Titchmarsh inequality [Reference Iwaniec and Kowalski3, Theorem 6.6] to
$\unicode[STIX]{x1D713}(t;m,1)$
and deduce that
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{M\leqslant m<2M}\int _{(m+1)^{s}}^{mX}{\displaystyle \frac{\unicode[STIX]{x1D713}(t;m,1)}{t(\log t)^{2}}}\,dt & \ll & \displaystyle \mathop{\sum }_{M\leqslant m<2M}\int _{(m+1)^{s}}^{mX}{\displaystyle \frac{1}{\unicode[STIX]{x1D711}(m)(\log t)^{2}}}\,dt\nonumber\\ \displaystyle & \ll & \displaystyle \mathop{\sum }_{M\leqslant m<2M}\frac{1}{\unicode[STIX]{x1D711}(m)}\int _{(M+1)^{s}}^{2MX}{\displaystyle \frac{1}{(\log t)^{2}}}\,dt\nonumber\\ \displaystyle & \ll & \displaystyle \mathop{\sum }_{M\leqslant m<2M}\frac{MX}{\unicode[STIX]{x1D711}(m)\log ^{2}(MX)}\nonumber\\ \displaystyle & \ll & \displaystyle \frac{MX}{\log ^{2}X}.\nonumber\end{eqnarray}$$
The desired bound follows by dyadically summing this estimate.◻
Proof of Theorem 1.
By Lemma 4, observe that
$$\begin{eqnarray}\mathop{\sum }_{1\leqslant k<X}\log (1/\unicode[STIX]{x1D6FF}_{k})\geqslant S_{1},\end{eqnarray}$$
where
$$\begin{eqnarray}S_{1}=\mathop{\sum }_{1\leqslant k<X}\mathop{\sum }_{p}\log (1/\unicode[STIX]{x1D6FF}_{k}(p)).\end{eqnarray}$$
It suffices to show
$S_{1}\gg _{s}X^{1+1/(s-1)}/\text{log}\,X$
. By Lemma 5,
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{k}(p)\leqslant \min \{\unicode[STIX]{x1D6FC}_{k,p},1\}\end{eqnarray}$$
for each prime
$p$
. Let
$m=(p-1)/(k,p-1)$
, so
$k=(p-1)/m\cdot d$
for some integer
$d\geqslant 1$
with
$(d,p-1)=1$
. Thus, each pair
$(k,p)$
in the above sum corresponds to a unique triple
$(m,d,p)$
of positive integers
$m$
and
$d$
such that
$m\mid p-1$
and
$(d,p-1)=1$
. Moreover,
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{k,p}<1\;\Longleftrightarrow \;p>(m+1)^{s}.\end{eqnarray}$$
Collecting these observations, we deduce that
$$\begin{eqnarray}S_{1}\geqslant \mathop{\sum }_{m\geqslant 1}\mathop{\sum }_{\substack{ m\mid p-1 \\ p>(m+1)^{s}}}\mathop{\sum }_{\substack{ 1\leqslant d\leqslant mX/(p-1) \\ (d,p-1)=1}}(\log p-s\log (m+1)).\end{eqnarray}$$
Whenever
$p<mX$
, the inner sum over
$d$
contains
$d=1$
so may trivially lower bound it by
$1$
. This simple estimate will only affect the implied constant in our final result, since the main contribution comes from large primes
$p\approx mX$
in the second sum. Thus, the above is
$$\begin{eqnarray}{\geqslant}\mathop{\sum }_{m\geqslant 1}\mathop{\sum }_{\substack{ p\equiv 1(\text{mod}\,m) \\ (m+1)^{s}<p<mX}}(\log p-s\log (m+1)).\end{eqnarray}$$
By positivity, we may restrict the outer sum to
$1\leqslant m<Y$
for some parameter
$Y\geqslant 1$
satisfying
$$\begin{eqnarray}(Y+1)^{s}<XY.\end{eqnarray}$$
Recalling (8), it follows by partial summation that
$$\begin{eqnarray}\displaystyle & & \displaystyle S_{1}\geqslant \mathop{\sum }_{1\leqslant m<Y}\int _{(m+1)^{s}}^{mX}\biggl(1-\frac{s\log (m+1)}{\log t}\biggr)\,d\unicode[STIX]{x1D713}(t;m,1)\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \mathop{\sum }_{1\leqslant m<Y}\biggl(\unicode[STIX]{x1D713}(mX;m,1)\biggl(1-\frac{s\log (m+1)}{\log (mX)}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \quad -\,s\log (m+1)\int _{(m+1)^{s}}^{mX}{\displaystyle \frac{\unicode[STIX]{x1D713}(t;m,1)}{t(\log t)^{2}}}\,dt\biggr).\end{eqnarray}$$
Set
$Y=X^{1/(s-1+\unicode[STIX]{x1D702})}$
, where
$\unicode[STIX]{x1D702}=\unicode[STIX]{x1D702}(X)<1/2$
is a parameter which will be specified. For
$1\leqslant m<Y$
, we have that
$$\begin{eqnarray}1-\frac{s\log (m+1)}{\log (mX)}=1-\frac{s}{1+(\log X/\!\log m)}+O\biggl(\frac{1}{m\log X}\biggr).\end{eqnarray}$$
If
$X^{1/(2s-1)}\leqslant m\leqslant X^{1/(s-1+\unicode[STIX]{x1D702})}$
, then from (13) we see that
$$\begin{eqnarray}1-\frac{s\log (m+1)}{\log (mX)}\geqslant \frac{\unicode[STIX]{x1D702}}{s}+O(X^{-1/(2s-1)})\geqslant \frac{\unicode[STIX]{x1D702}}{s}+O\biggl(\frac{1}{\log X}\biggr).\end{eqnarray}$$
Otherwise, for
$m\leqslant X^{1/(2s-1)}$
, we similarly have that
$$\begin{eqnarray}1-\frac{s\log (m+1)}{\log (mX)}\geqslant \frac{1}{2}+O\biggl(\frac{1}{\log X}\biggr)\geqslant \frac{\unicode[STIX]{x1D702}}{s}+O\biggl(\frac{1}{\log X}\biggr).\end{eqnarray}$$
Substituting these bounds into (12) and noting
$s\log (m+1)<s\log (Y+1)\ll \log X$
, we deduce that
$$\begin{eqnarray}\displaystyle S_{1} & {\geqslant} & \displaystyle \biggl(\mathop{\sum }_{1\leqslant m<Y}\unicode[STIX]{x1D713}(mX;m,1)\biggr)\biggl(\frac{\unicode[STIX]{x1D702}}{s}+O\biggl(\frac{1}{\log X}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,O\biggl(\log X\mathop{\sum }_{1\leqslant m<Y}\int _{(m+1)^{s}}^{mX}{\displaystyle \frac{\unicode[STIX]{x1D713}(t;m,1)}{t(\log t)^{2}}}\,dt\biggr).\nonumber\end{eqnarray}$$
Since
$Y\leqslant X^{1/2}$
for
$s\geqslant 3$
, Lemma 6 therefore implies that
$$\begin{eqnarray}S_{1}\geqslant \frac{\unicode[STIX]{x1D701}(2)\unicode[STIX]{x1D701}(3)\log 2}{\unicode[STIX]{x1D701}(6)}\cdot \frac{\unicode[STIX]{x1D702}}{s}\cdot XY+O\biggl(\frac{XY}{\log X}\biggr).\end{eqnarray}$$
Note the implied constant is independent of
$\unicode[STIX]{x1D702}$
and depends only on
$s$
. Choose
$\unicode[STIX]{x1D702}=A_{s}/\text{log}\,X$
, where
$A_{s}$
is a fixed sufficiently large constant depending only on
$s$
. Thus,
$Y=X^{1/(s-1+\unicode[STIX]{x1D702})}\gg _{s}X^{1/(s-1)}$
satisfies (11) and it follows that
$S_{1}\gg _{s}XY/\text{log}\,X\gg _{s}X^{1+1/(s-1)}/\text{log}\,X$
.◻
Acknowledgements
This research was done while the first author was visiting the Fields Institute for the Thematic Program on Unlikely Intersections, Heights, and Efficient Congruencing. He thanks the Fields Institute for their support and superb working conditions. The authors would also like to thank Trevor Wooley for fruitful conversations.
