# Simultaneously Small Fractional Parts of Polynomials

Cheuk Fung (Joshua) Lau  University College, Oxford OX1 4BH, UK
<EMAIL_ADDRESS>
###### Abstract.

Let $f_{1},\dots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
with $f_{1}(0)=\dots=f_{k}(0)=0$. We show that there is an $n<x$ such that
$\|f_{i}(n)\|\ll x^{-1/10.5kd(d-1)+o(1)}$ for all $1\leq i\leq k$. This
improves on an earlier result of Maynard, who obtained the same exponent
dependency on $k$ but not on $d$.

##  1\. Introduction

The question of how small the fractional part $\lVert f(n)\rVert$ of a
polynomial $f(n)$ (with $f(0)=0$) can be made has been investigated since the
early 20th century, with the current record being

(1.1) $\min_{n\leq x}\ \lVert f(n)\rVert\ll_{d}x^{-1/2d(d-1)+o(1)},$

which is due to Baker (2016) (for $d\geq 8$). Here $\|\cdot\|$ denotes the
distance to the nearest integer. We impose the condition $f(0)=0$ so as to
avoid examples like $f(n)=n+1/2$, which clearly doesn’t attain arbitrarily
small fractional parts. Note that the bound depends only on $x$ and $d$, and
is otherwise completely uniform over all such polynomials $f$. The exponent
$1/2d(d-1)$ is based on proving Vinogradov’s Mean Value Theorem for $k\geq 4$
by Bourgain et al. (2016). This bound is not expected to be tight, and it is
conjectured (Baker, 1986) that this should be improvable to $1+o(1)$.

One of the first results of this kind is from Heilbronn (1948), who obtained
for any real $\alpha$, $\varepsilon>0$ and sufficiently large $x>0$,

$\min_{n\leq x}\lVert\alpha n^{2}\rVert<x^{-1/2+\varepsilon}.$

For the problem of a general polynomial $f\in\mathbb{R}[X]$ vanishing at the
origin, Selberg (1955) obtained

$\min_{n\leq x}\lVert f(n)\rVert\ll_{d}x^{-1/(8+o(1))d^{2}\log d}.$

Wooley (2012) improved the exponent to $-1/4d(d-1)$ for $d\geq 2$, and as
noted Baker (2016) improved it to $-1/2d(d-1)$ for $d\geq 8$.

We can ask the same question, but instead for $k$ polynomials
$f_{1},\dots,f_{k}\in\mathbb{R}[X]$ of degree at most $d$ with
$f_{1}(0)=\dots=f_{k}(0)=0$, the current record due to Maynard (2021) is for
some $c_{d}>0$ depending only on $d$,

(1.2) $\min_{n\leq x}\max_{i\leq
k}\|f_{i}(n)\|\ll_{k,d}\frac{1}{x^{c_{d}/k}}.$

This refines the work of Baker (1980), where he obtained a hybrid bound

(1.3) $\min_{n\leq x}\max_{i\leq
k}\|f_{i}(n)\|\ll_{k,d}\frac{1}{x^{1/(k^{2}+kc_{d})+o(1)}}.$

A heuristic based on choosing the coefficients of $f_{1},\dots,f_{k}$
uniformly at random shows that one could only expect

(1.4) $\min_{n\leq x}\max_{i\leq k}\|f_{i}(n)\|\ll\frac{1}{x^{1/k}}.$

Our main result is to establish a bound for (1.2) with an exponent of the form
$1/10.5kd(d-1)+o(1)$. By comparing this with (1.4) and (1.1), we see that this
bound is best in the $k$ and $d$ aspect, apart from hybrid bounds and the
constant 10.5. More precisely, our main result is the following.

######  Theorem 1.1.

Let $k,d\in\mathbb{Z}^{+},\,\varepsilon>0,$ and $M\in\mathbb{R}^{+}$ satisfy

$M\geq\max\left\\{4,\frac{1}{2}+\frac{\log\varepsilon^{-1}}{2\log
2}\right\\}.$

There is a constant $C_{d,k}>2$ depending only on $d$, $k$ (and
$\varepsilon,M$) such that the following holds.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
such that $f_{1}(0)=\cdots=$ $f_{k}(0)=0$. Let
$\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1/100]$, and put
$\Delta=\prod_{i=1}^{k}\varepsilon_{i}$. Define

$c_{2}=10.5+9k/(2k)^{M}+\varepsilon.$

If $\Delta^{-1}\leq x^{1/c_{2}d(d-1)}$ and $x>C_{d,k}$, then there is a
positive integer $n<x$ such that

$\left\|f_{i}(n)\right\|\leq\varepsilon_{i}\quad\text{ for all
}i\in\\{1,\ldots,k\\}\text{. }$

By taking $\varepsilon_{1}=\cdots=\varepsilon_{k}=x^{-1/c_{2}d(d-1)k}$, and
for any $\varepsilon>0$ letting $M$ large such that
$2^{M}\geq\varepsilon^{-1}$, we arrive at the following:

######  Corollary 1.2.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
such that $f_{1}(0)=\cdots=f_{k}(0)=0$. Then there is a positive integer $n<x$
such that for all $\varepsilon>0$,

$\lVert f_{i}(n)\rVert\ll_{d,k,\varepsilon}x^{-1/(10.5+\varepsilon)kd(d-1)}$

for all $i=1,2,\ldots,k$.

The proof of Theorem 1.1 is obtained by following the argument of Maynard
(2021), but making more careful choices of constants in the proofs. If we
follow strictly the argument based on Weyl differencing in Maynard (2021),
$c_{d}$ would be of the form $1/d^{d}$. To improve on the $d$-quantification,
one can use arguments based on the Vinogradov Mean Value Theorem such as
Wooley (2012), which gives $c_{d}$ to be of the form $1/2^{d}$. The main
improvement made is found in Lemma 6.2, which allows us to take $c_{d}$ to be
of the form $1/d^{2}$.

##  2\. Outline

In this section we outline the main improvement in this paper. The main method
follows closely the argument of Maynard (2021), where in Section 5 we prove a
version of Maynard (2021, Proposition 5.1), in Section 6 we prove a version of
Maynard (2021, Proposition 5.2), and in Section 7 we prove a version of
Maynard (2021, Proposition 5.3). Sections 5 and 7 are very similar to the
corresponding sections in Maynard (2021). To emphasize on the improvement made
in Section 6, we first review the arguments in the proof of (Maynard, 2021,
Lemma 7.3).

Given $f_{i}(X)=\sum_{j=1}^{d}f_{i,j}X^{j}\in\mathbb{R}[X]$, Fourier analysis
shows that, given any intervals $I_{1},\ldots,I_{k}$ of length $\varepsilon>0$
(for some small $\varepsilon>0$) and $x>\varepsilon^{-kC_{d}}$, either there
is an $n<x$ such that the vector of fractional parts $\mathbf{v}(n)=(\lVert
f_{1}(n)\rVert,\ldots,\lVert f_{k}(n)\rVert)$ lie in $I_{1}\times\cdots\times
I_{k}$, or there are at least $Q^{1/cd(d-1)}$ integer tuples
$(h_{1},\ldots,h_{k})$ with $h_{i}\leq\varepsilon^{-1-o(1)}$ such that

(2.1) $h_{1}f_{1,j}+\cdots+h_{k}f_{k,j}\approx\frac{a_{j}}{q_{j}}\text{ for
all }1\leq j\leq d,$

for some $a_{j}\in\mathbb{Z}$ and $Q\leq\varepsilon^{-ckd(d-1)}$ with
$q_{j}<Q^{1/cd}$. Maynard (2021) proved that, given any $\delta>0$ small,
$r\in\mathbb{Z}^{+}$ and
$\mathcal{S}\subseteq\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}^{k}$ set of
triples $(a,q,\mathbf{h})$ with $\gcd(a,q)=1$ and $q\leq Q^{1/cd}$ such that
$\\#\mathcal{S}\geq Q^{\delta}$, then one of the following holds:

  1. (1)

At least $\\#\mathcal{S}^{1/2}$ of triples $(a,q,\mathbf{h})$ have the same
$q=q_{0}$.

  2. (2)

There are at least $\\#\mathcal{S}^{r/5}$ distinct values of
$a_{1}/q_{1}+\cdots+a_{r}/q_{r}$, amongst all
$(a_{i},q_{i},\mathbf{h}^{(i)})\in\mathcal{S}$.

Indeed, if $(1)$ is false, then one can find an integer $m_{0}$ and a subset
of the $q$s such that they are all multiples of $m_{0}$, and not too many of
them are multiples of $m_{0}\ell$ for $\ell>1$. Letting $b=q/m_{0}$ for such
$m_{0}$, one can prove that the size of the set of all tuples
$(b_{1},\ldots,b_{r})$ with pairwise small gcd is not small. Focusing on these
tuples, one can prove that there are many distinct values of
$a_{1}/m_{0}b_{1}+\cdots+a_{r}/m_{0}b_{r}$.

Therefore, if we are in the case of (2.1), we can focus on the case $j=1$ to
prove there are many possible values of $\sum a_{1}/q_{1}$, hence there are at
least $Q^{1/2cd(d-1)}$ triples
$(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$
satisfying (2.1), with $q_{1}$ being identical amongst all triples. Doing this
procedure $d$ times, there are at least $Q^{1/2^{d}cd(d-1)}$ triples
$(\mathbf{a},\mathbf{q},\mathbf{h})$ satisfying (2.1) with $\mathbf{q}$
identical amongst all triples. This eventually leads to an exponential
dependence on $d$ in the final exponent.

To get around this issue, we consider a multi-dimensional argument, instead of
iterating a $1$-dimensional argument $d$ times. We prove that given
$\mathcal{S}\subseteq\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$
set of triples $(\mathbf{a},\mathbf{q},\mathbf{h})$ with $\gcd(a_{i},q_{i})=1$
and $q_{i}\leq Q^{1/cd}$ (for $i=1,2,\ldots,d$) such that $\\#\mathcal{S}\geq
Q^{\delta}$, then one of the following holds:

  1. (1)

At least $\\#\mathcal{S}^{1/2}$ of triples
$(\mathbf{a},\mathbf{q},\mathbf{h})$ have the same
$\mathbf{q}=\mathbf{q}^{(0)}$.

  2. (2)

There are at least $\\#\mathcal{S}^{r/5}$ distinct values of

$\left(\frac{a_{1}^{(1)}}{q_{1}^{(1)}}+\cdots+\frac{a_{1}^{(r)}}{q_{1}^{(r)}},\ldots,\frac{a_{d}^{(1)}}{q_{d}^{(1)}}+\cdots+\frac{a_{d}^{(r)}}{q_{d}^{(r)}}\right),$

amongst all
$(\mathbf{a}^{(i)},\mathbf{q}^{(i)},\mathbf{h}^{(i)})\in\mathcal{S}$.

Indeed, if $(1)$ is false, then one can find an integer tuple
$\mathbf{m}^{(0)}$ and a subset of the $\mathbf{q}$s such that all $q_{i}$ are
multiples of $m_{i}^{(0)}$, and not too many of them are multiples of
$m_{i}^{(0)}\ell_{i}$, where $\bm{\ell}\neq(1,\ldots,1)$. Letting
$b_{i}:=q_{i}/m_{i}^{(0)}$ for such $\mathbf{m}^{(0)}$, one can prove that the
size of the set of all tuples $(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})$
with pairwise small gcd (at all entries) is not small. Focusing on these
tuples, we can prove that there are many distinct vectors of the
aforementioned form.

Therefore, if we have (2.1), we can prove there are many possibilities of
$(\sum a_{1}/q_{1},\ldots,\sum a_{d}/q_{d})$, hence there are at least
$Q^{1/2cd(d-1)}$ triples
$(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$
satisfying (2.1), with $\mathbf{q}$ identical amongst all triples. This multi-
dimensional argument avoids the $2^{d}$ cost in the exponent, which eventually
allows a quadratic dependency on $d$ in the final exponent.

##  3\. Acknowledgements

We would like to thank James Maynard for suggesting this question, and also
the guidance we received throughout the writing of this paper.

##  4\. Notation

Throughout the paper we assume that we have polynomials
$f_{1},\dots,f_{k}\in\mathbb{R}[X]$ of degree at most $d$ with
$f_{1}(0)=\dots=f_{k}(0)=0$. We let these polynomials be given by
$f_{i}(X)=\sum_{j=1}^{d}f_{i,j}X^{j}$. Furthermore, we have reals
$\varepsilon_{1},\dots,\varepsilon_{k}\in(0,1/100]$, and we put
$\Delta:=\prod_{i=1}^{k}\varepsilon_{i}$.

We often use Vinogradov’s notation, where $X\ll Y$ denotes $|X|\leq CY$ for
some constant $C$. If the implied constant depends on $\varepsilon$ say, we
write $X\ll_{\varepsilon}Y$. We also use Landau’s Big O notation, where
$X=O(Y)$ if $X\ll Y$.

##  5\. Small Fractional Parts or Many Relations

Throughout this paper, we follow closely the arguments of Maynard (2021). For
the sake of clarity, we include the proofs for statements that are not
identical to the analogous ones in Maynard (2021), whereas for identical
results we just state without proof.

######  Lemma 5.1.

Let $f(X)=\sum_{i=1}^{d}f_{i}X^{i}\in\mathbb{R}[X]$ be a polynomial of degree
$d\geq 2$ with $f(0)=0$ and $\varepsilon>0$. If there is some
$Q\in[2,x^{1/2d(d-1)-\varepsilon}]$ such that

$\left|\sum_{n<x}e(f(n))\right|\geq\frac{x}{Q},$

then there is an integer $q<x^{\varepsilon}Q^{d}$ and integers
$a_{1},\ldots,a_{d}$ such that

$f_{j}=\frac{a_{j}}{q}+O\left(\frac{Q^{d}}{x^{j-\varepsilon}}\right)$

for $j=1,2,\ldots,d$. The implied constant only depends on $d$ and
$\varepsilon$.

###### Proof.

Follows immediately from Baker (2016, Theorem 4). ∎

######  Lemma 5.2.

Let $\varepsilon>0,M\geq 4$, $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be real
valued functions, and $\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1/2]$ be
real numbers, with $\Delta:=\prod_{i=1}^{k}\varepsilon_{i}$. Suppose $x$ is
sufficiently large in terms of $\varepsilon$. Then at least one of the
following holds:

  1. (1)

We have

$\\#\\{n\leq x:\lVert f_{i}(n)\rVert\leq\varepsilon_{i}\ \forall i\\}>0.$

  2. (2)

There is a quantity $Q\geq 2$ of the form $Q=2^{j}$ such that there are at
least $Q^{1/(1+\varepsilon)}$ distinct values of
$\mathbf{h}\in\mathbb{Z}^{k}\backslash\\{\mathbf{0}\\}$ with
$|h_{i}|<\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ such that

$\frac{x}{Q}\leq\left|\sum_{n\leq
x}e\left(\sum_{i=1}^{k}h_{i}f_{i}(n)\right)\right|\leq\frac{2x}{Q}.$

###### Proof.

The proof is nearly identical to that of Maynard (2021, Lemma 6.3). We fix a
smooth function $\phi:\mathbb{R}\to[0,1]$ with $\phi(t)$ supported on $|t|<1$,
which is $1$ on $|t|<1/2$. Let

$\Phi_{i}(t)=\sum_{m\in\mathbb{Z}}\phi\left(\frac{t+m}{\varepsilon_{i}}\right),$

which is clearly 1-periodic, smooth, and supported on $\lVert
t\rVert<\varepsilon_{i}$. By Poisson summation,

$\Phi_{i}(f_{i}(t))=\varepsilon_{i}\sum_{h\in\mathbb{Z}}\hat{\phi}(\varepsilon_{i}h)e(hf_{i}(t)).$

Since $\phi$ is fixed and smooth, $\phi^{(j)}(t)\ll_{j}1$, so
$|\hat{\phi}(u)|\ll_{j}u^{-j}$ for $j\geq 0$. Thus we see that the terms with
$|h|\geq\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ contribute
$O(\Delta^{100})$, and so

$\Phi_{i}(f(t))=\varepsilon_{i}\sum_{|h|\leq\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}}\hat{\phi}(\varepsilon_{i}h)e(hf_{i}(t))+O(\Delta^{100}).$

Thus we find that

$\displaystyle\\#\\{n\leq x:\lVert f_{i}(n)\rVert\leq\varepsilon_{i}\forall
i\\}$ $\displaystyle\geq\sum_{n\leq x}\prod_{i=1}^{k}\Phi_{i}(f_{i}(n))$
$\displaystyle=\Delta\sum_{\begin{subarray}{c}h_{1},\ldots,h_{k}\\\
|h_{i}|<\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}\end{subarray}}\left(\prod_{i=1}^{k}\hat{\phi}(\varepsilon_{i}h_{i})\right)\sum_{n\leq
x}e\left(\sum_{i=1}^{k}h_{i}f_{i}(n)\right)+O(x\Delta^{99})$
$\displaystyle=x\Delta\hat{\phi}(0)^{k}+O\left(\Delta\sum_{\begin{subarray}{c}\mathbf{h}\in\mathbb{Z}^{k}\setminus\\{0\\}\\\
|h_{i}|<\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}\end{subarray}}\left|\sum_{n\leq
x}e\left(\sum_{i=1}^{k}h_{i}f_{i}(n)\right)\right|\right)+O(x\Delta^{99}).$

For $\Delta$ sufficiently small, we see that
$\Delta\hat{\phi}(0)^{k}+O(\Delta^{99})\gg\Delta$, and so either

$\\#\\{n\leq x:\lVert f_{i}(n)\rVert\leq\varepsilon_{i}\forall i\\}>0,$

or

$\sum_{\begin{subarray}{c}\mathbf{h}\in\mathbb{Z}^{k}\setminus\\{0\\}\\\
|h_{i}|<\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}\end{subarray}}\left|\sum_{n\leq
x}e\left(\sum_{i=1}^{k}h_{i}f_{i}(n)\right)\right|\gg x.$

In the latter case, by pigeonhole principle (or more specifically Bloom and
Sisask (2020, Lemma 2.1)), there is some $Q=2^{j}$ such that there are at
least $Q^{1-\varepsilon}$ choices of $\mathbf{h}$ in the outer summation
satisfying

$\frac{x}{Q}\leq\left|\sum_{n\leq
x}e\left(\sum_{i=1}^{k}h_{i}f_{i}(n)\right)\right|\leq\frac{2x}{Q}.$

∎

Using this, we can prove the analogous version of (Maynard, 2021, Proposition
5.1).

######  Proposition 5.3.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
such that $f_{1}(0)=\cdots=f_{k}(0)=0$. Put
$f_{i}(X)=\sum_{j=1}^{d}f_{i,j}X^{j}$. Let
$\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1/100]$, and put
$\Delta=\prod_{i=1}^{k}\varepsilon_{i}$. Let $\varepsilon>0,M\geq 4$. Define
constants

$c_{0}=1+\varepsilon,\quad c_{1}=1+k/(2k)^{M}+10\varepsilon,\quad
c_{2}=2+1/(2k)^{M-1}+20\varepsilon.$

Then provided $\Delta^{-c_{2}d(d-1)}<x$, at least one of the following holds:

  1. (1)

We have

$\\#\\{n\leq x:\lVert f_{i}(n)\rVert<\varepsilon_{i}\ \forall i\\}>0.$

  2. (2)

There is some $Q\leq\Delta^{-c_{1}d(d-1)}$ such that there are at least
$Q^{1/(1+\varepsilon)c_{0}d(d-1)}$ triples
$(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$
satisfying:

    1. (a)

$q_{1}=\cdots=q_{d}$ and $1\leq q_{j}\leq x^{\varepsilon}Q^{1/c_{0}(d-1)}$ for
$1\leq j\leq d$.

    2. (b)

$h_{i}\ll\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ for $1\leq i\leq k$.

    3. (c)

For each $j=1,2,\ldots,d$, we have

$\sum_{i=1}^{k}h_{i}f_{i,j}=\frac{a_{j}}{q_{j}}+O\left(\frac{Q^{1/d}}{x^{j-\varepsilon}}\right).$

All implied constants depend only on $d,k,\varepsilon$ and $M$..

###### Proof.

The proof is very similar to that of Maynard (2021, Proposition 5.1). Suppose
condition (1) does not hold. Then, by Lemma 5.2, there is $Q_{1}\geq 2$ such
that there are at least $Q_{1}^{1/(1+\varepsilon)}$ distinct values of
$\mathbf{h}\in\mathbb{Z}^{k}\backslash\\{\mathbf{0}\\}$ with
$|h_{i}|<\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ such that

$\left|\sum_{n\leq
x}e\left(\sum_{i=1}^{k}h_{i}f_{i}(n)\right)\right|\geq\frac{x}{Q_{1}}.$

Letting $Q=Q_{1}^{c_{0}d(d-1)}$, there are at least
$Q^{1/(1+\varepsilon)c_{0}d(d-1)}$ distinct values of
$\mathbf{h}\in\mathbb{Z}^{k}\backslash\\{\mathbf{0}\\}$ with
$|h_{i}|<\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ such that

$\left|\sum_{n\leq
x}e\left(\sum_{i=1}^{k}h_{i}f_{i}(n)\right)\right|\geq\frac{x}{Q^{1/c_{0}d(d-1)}}$

Note that $Q_{1}\leq\Delta^{(-1-k/(2k)^{M})(1+\varepsilon)}$, so we have
$Q_{1}^{c_{0}d(d-1)}=Q\leq\Delta^{-c_{1}d(d-1)}$ since $c_{1}\geq
c_{0}(1+k/(2k)^{M})(1+\varepsilon)$, and

$Q_{1}=Q^{1/c_{0}d(d-1)}\leq\Delta^{-(1+k/(2k)^{M})(1+\varepsilon)}\leq\Delta^{-(c_{2}+2\varepsilon)d(d-1)/2d(d-1)}<x^{1/2d(d-1)-\varepsilon},$

since $c_{2}\geq 2+1/(2k)^{M-1}+20\varepsilon$. Therefore applying Lemma 5.1
to $Q_{1}$, there are integers
$q_{1},\ldots,q_{d}<x^{\varepsilon}Q_{1}^{d}=x^{\varepsilon}Q^{1/c_{0}(d-1)}$
and integers $a_{1},\ldots,a_{d}$ such that $\gcd(a_{j},q_{j})=1$ for all
$j=1,2,\ldots,d$ and

$f_{j}=\frac{a_{j}}{q_{j}}+O\left(\frac{Q^{1/d}}{x^{j-\varepsilon}}\right)$

for $j=1,2,\ldots,d$. ∎

##  6\. Many Relations Have Same Denominator

The proof of the following lemma can be found in the proof of (Maynard, 2021,
Lemma 7.1). We state it below for sake of clarity.

######  Lemma 6.1.

Let $B>1,\delta>0,$ and $r\in\mathbb{Z}^{+}$. For
$b_{1},\ldots,b_{r}\in\mathbb{Z}$ satisfying

$B\leq b_{i}<2B,\quad\gcd(b_{i},b_{j})<B^{\delta^{2}/r^{2}}\text{ for all
}1\leq i<j\leq r,$

define the set $\mathcal{R}(b_{1},\ldots,b_{r})$ consisting of
$b_{1}^{\prime},\ldots,b_{r}^{\prime}\in\mathbb{Z}$ satisfying

  1. (1)

$B\leq b_{i}^{\prime}<2B,$
$\gcd(b_{i}^{\prime},b_{j}^{\prime})<B^{\delta^{2}/r^{2}}\text{ for all }1\leq
i<j\leq r,$

  2. (2)

there exists
$a_{1},\ldots,a_{r},a_{1}^{\prime},\ldots,a_{r}^{\prime}\in\mathbb{Z}$ such
that

    1. (a)

$\gcd(a_{i},b_{i}),\gcd(a_{i}^{\prime},b_{i}^{\prime})=1$ for all $1\leq i\leq
r$,

    2. (b)

$\frac{a_{1}}{b_{1}}+\cdots+\frac{a_{r}}{b_{r}}=\frac{a_{1}^{\prime}}{b_{1}^{\prime}}+\cdots+\frac{a_{r}^{\prime}}{b_{r}^{\prime}}$.

Then $\\#\mathcal{R}(b_{1},\ldots,b_{r})\leq B^{5\delta^{2}}$.

###### Proof.

For any choice of $a_{1},\ldots,a_{r}$ with $\gcd(a_{i},b_{i})=1$, the
denominator of $a_{1}/b_{1}+\cdots+a_{r}/b_{r}$ is a multiple of $p^{\ell}$ if
$p^{\ell}$ divides exactly one of $b_{1},\ldots,b_{r}$, and $p^{\ell+1}$
divides none of them. Let $\gcd(b,p^{\infty})$ denote the largest power of $p$
dividing $b>1$, and $\gcd(b_{i},b_{j},p^{\infty})$ the largest power of $p$
dividing both $b_{i}$ and $b_{j}$. Define

$g_{p}:=\frac{\prod_{i=1}^{r}\gcd(b_{i},p^{\infty})}{\prod_{1\leq i<j\leq
r}\gcd(b_{i},b_{j},p^{\infty})^{2}}.$

We see that if $p^{\ell}$ divides exactly one of $b_{1},\ldots,b_{r}$, say
$b_{1}$, and $p^{\ell+1}$ divides none of them, then $g_{p}\leq p^{\ell}$
since $\gcd(b_{1},b_{j},p^{\infty})^{2}\geq\gcd(b_{j},p^{\infty})$ for all
$1<j\leq r$. Similarly, if $p^{\ell}$ divides at least 2 of the $b_{i}$ but
$p^{\ell+1}$ divides none of them, then $g_{p}\leq 1$.

Therefore, for any $a_{1},\ldots,a_{r}$ with $\gcd(a_{i},b_{i})=1$, the
denominator of $a_{1}/b_{1}+\cdots+a_{r}/b_{r}$ is of size at least
$\prod_{p}g_{p}$ by considering the prime factorisation of $b_{i}$. However,

$\prod_{p}g_{p}=\frac{\prod_{i=1}^{r}b_{i}}{\prod_{1\leq i<j\leq
r}\gcd(b_{i},b_{j})^{2}}\geq B^{r-2\delta^{2}}.$

Clearly any such denominator is of size $O(B^{r})$, so there are
$O(B^{2\delta^{2}})$ possible denominators of
$a_{1}/b_{1}+\cdots+a_{r}/b_{r}$. Given such a denominator
$q>B^{r-2\delta^{2}}$, if it is also the denominator of
$a_{1}^{\prime}/b_{1}^{\prime}+\cdots+a_{r}^{\prime}/b_{r}^{\prime}$, then $q$
divides $\prod_{i=1}^{r}b_{i}^{\prime}$.

As $\prod_{i=1}^{r}b_{i}^{\prime}\ll B^{r}$, this implies there are
$O(B^{2\delta^{2}})$ choices of $\prod_{i=1}^{r}b_{i}^{\prime}$, so
$O(B^{2\delta^{2}+o(1)})$ choices of $b_{1}^{\prime},\ldots,b_{r}^{\prime}$
using the divisor bound. Therefore given $(b_{1},\ldots,b_{r})$ satisfying the
above requirements, there are at most $B^{5\delta^{2}}$ choices of
$(b_{1}^{\prime},\ldots,b_{r}^{\prime})$ in total, so
$\\#\mathcal{R}(b_{1},\ldots,b_{r})\leq B^{5\delta^{2}}$. ∎

We prove a generalisation of (Maynard, 2021, Lemma 7.1). In Maynard (2021),
Lemma 7.3 was proven with induction, which introduced extra factors of $d$ on
the exponent. We directly prove a generalisation which avoids this issue.

######  Lemma 6.2.

(Expansion or Same Denominators) Let $\delta,\varepsilon\in(0,1)$, $r$ a
positive integer, $d\geq 2$, and $X\in\mathbb{R}_{\geq 1}$. Let $Q>0$ be
sufficiently large in terms of $r,\ d$, $\delta$ and $\varepsilon$, and let
$\mathcal{S}\subseteq\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$
be a set of triples $(\mathbf{a},\mathbf{q},\mathbf{h})$ with
$\gcd(a_{i},q_{i})=1$ and $q_{i}\leq XQ^{1/(d-1)}$ (for $i=1,2,\ldots,d$) such
that $\\#\mathcal{S}\geq Q^{\delta}$.

Then one of the following holds:

  1. (1)

There is a $\mathbf{q}^{(0)}$ such that at least $\\#\mathcal{S}^{1/2}$ of the
triples $(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathcal{S}$ have
$\mathbf{q}=\mathbf{q}^{(0)}$.

  2. (2)

The set

$\mathcal{A}:=\left\\{\left(\frac{a_{1}^{(1)}}{q_{1}^{(1)}}+\cdots+\frac{a_{1}^{(r)}}{q_{1}^{(r)}},\ldots,\frac{a_{d}^{(1)}}{q_{d}^{(1)}}+\cdots+\frac{a_{d}^{(r)}}{q_{d}^{(r)}}\right):\exists\mathbf{h}^{(1)},\ldots,\mathbf{h}^{(r)}\in\mathbb{Z}^{k}\text{
s.t. }(\mathbf{a}^{(i)},\mathbf{q}^{(i)},\mathbf{h}^{(i)})\in\mathcal{S}\text{
for }1\leq i\leq r\right\\},$

(where $\mathbf{a}^{(i)}=(a_{1}^{(i)},\ldots,a_{d}^{(i)})$ and
$\mathbf{q}^{(i)}=(q_{1}^{(i)},\ldots,q_{d}^{(i)}))$ has cardinality at least
$X^{-\varepsilon\delta
dr/40-5\varepsilon\delta^{2}d/r}\\#\mathcal{S}^{(\frac{1}{2}-\varepsilon)r}$.

###### Proof.

The proof is a generalisation of Lemma 7.1 found in Maynard (2021). Throughout
the lemma we will assume that $Q$ is large enough in terms of
$\varepsilon,\delta$ and $r$. We first restrict our attention to a suitable
subset of the $\mathbf{q}^{\prime}s$ appearing in $\mathcal{S}$. For
$j_{1},j_{2},\ldots,j_{d}\in\mathbb{N}_{0}$, let

$\mathcal{B}_{j_{1},\ldots,j_{d}}=\left\\{\mathbf{q}\in\prod_{i=1}^{d}[2^{j_{i}},2^{j_{i}+1}):\exists(\mathbf{a},\mathbf{h})\in\mathbb{Z}^{d}\times\mathbb{Z}^{k}\text{
with }\gcd(a_{i},q_{i})=1\ \forall i\text{ and
}(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathcal{S}\right\\}.$

Clearly $\mathcal{B}_{j_{1},\ldots,j_{d}}=\emptyset$ if
$j_{i}>2\log(XQ^{2/d})$ for some $i$. Note if

$\\#\left\\{\mathbf{q}:\exists(\mathbf{a},\mathbf{h})\in\mathbb{Z}^{d}\times\mathbb{Z}^{k}\text{
with
}(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathcal{S}\right\\}=\sum_{2^{j_{i}}\leq
Q^{1/d}\,\forall
i}\\#\mathcal{B}_{j_{1},\ldots,j_{d}}\leq\\#\mathcal{S}^{1/2}$

then by pigeonhole principle, there is a
$\mathbf{q}\in\mathbb{Z}^{d}\cap[1,XQ^{2/d}]^{d}$ such that there are at least
$\\#\mathcal{S}^{1/2}$ choices of $(\mathbf{a},\mathbf{h})$ with
$(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathcal{S}$. Thus condition (1) is
satisfied in this case.

Therefore we may assume that

$\sum_{j_{i}}\\#\mathcal{B}_{j_{1},\ldots,j_{d}}>\\#\mathcal{S}^{1/2}\geq
Q^{\delta/2},$

and so there is some $j_{1},\ldots,j_{d}\leq 2\log(XQ^{2/d})$ such that
$\\#\mathcal{B}_{j_{1},\ldots,j_{d}}>\\#\mathcal{S}^{\frac{1}{2}-\varepsilon_{1}}$,
where $\varepsilon_{1}=\varepsilon/20$. Note we must have
$2^{j_{1}+\cdots+j_{d}}>Q^{\delta/3}$ from the trivial bound
$\\#\mathcal{B}_{j}\leq 2^{j_{1}+\cdots+j_{d}}$.

For convenience, we define the following terminology. We say
$\mathbf{m}\in\mathbb{Z}^{d}$ divides $\mathbf{q}\in\mathbb{Z}^{d}$ if
$m_{i}\mid q_{i}$ for all $1\leq i\leq d$, and define
$\mathbf{m}\mathbf{q}:=(m_{1}q_{1},\ldots,m_{d}q_{d})$.

If there is a $d$-tuple $\mathbf{m}$ which divides at least
$\\#\mathcal{B}_{j_{1},\ldots,j_{d}}/m_{1}^{\varepsilon_{1}\delta/2}\cdots
m_{d}^{\varepsilon_{1}\delta/2}$ elements of
$\mathcal{B}_{j_{1},\ldots,j_{d}}$, we restrict our attention to this subset.
By performing this repeatedly, we may assume there is a fixed $d$-tuple
$\mathbf{m}$ and a set
$\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}\subseteq\mathcal{B}_{j_{1},\ldots,j_{d}}$
such that
$\\#\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}\geq\\#\mathcal{B}_{j_{1},\ldots,j_{d}}/m_{1}^{\varepsilon_{1}\delta/2}\cdots
m_{d}^{\varepsilon_{1}\delta/2}$, $\mathbf{m}$ divides all elements of
$\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}$, and there is no $d$-tuple
$\mathbf{m}^{\prime}\neq(1,\ldots,1)$ such that
$\mathbf{m}^{\prime}\mathbf{m}$ divides at least
$\\#\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}/(m_{1}^{\prime}\cdots
m_{d}^{\prime})^{\varepsilon_{1}\delta/2}$ elements of
$\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}$. Since $m_{i}\leq 2^{j_{i}+1}\leq
2XQ^{2/d}$, we have

$\\#\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}\geq\frac{\\#\mathcal{S}^{\frac{1}{2}-\varepsilon_{1}}}{(2X)^{\varepsilon_{1}\delta
d/2}Q^{\varepsilon_{1}\delta}}\geq X^{-\varepsilon_{1}\delta
d/2}\\#\mathcal{S}^{\frac{1}{2}-3\varepsilon_{1}}.$

Since
$\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}\subseteq\\{\mathbf{b}\in\prod_{i=1}^{d}[2^{j_{i}},2^{j_{i}+1}):m_{i}\mid
b_{i}\\}$ is a set of size $O(2^{j_{1}+\cdots+j_{d}}/m_{1}\cdots m_{d})$, we
see this also implies $m_{1}\cdots m_{d}<2^{j_{1}+\cdots+j_{d}}/Q^{\delta/4}$.
We let

$\mathcal{B}:=\left\\{\mathbf{b}:(m_{1}b_{1},\ldots,m_{d}b_{d})\in\mathcal{B}_{j_{1},\ldots,j_{d}}^{\prime}\right\\},$

and note $\mathcal{B}\subseteq\prod_{i=1}^{d}[B_{i},2B_{i})$, where we set
$B_{i}:=2^{j_{i}}/m_{i}$. Let $B=\max_{i}B_{i}$, and note that
$\\#\mathcal{B}\geq(2X)^{-\varepsilon_{1}\delta
d/2}\\#\mathcal{S}^{\frac{1}{2}-2\varepsilon_{1}}$ and $B\leq XQ^{2/d}$.

Consider the graph $G=(\mathcal{V},\mathcal{E})$, where the vertex set
$\mathcal{V}$ is taken to be $\mathcal{B}$, and the edge set $\mathcal{E}$ is
defined by

$\mathcal{E}:=\left\\{(\mathbf{b}^{(1)},\mathbf{b}^{(2)})\in\mathcal{B}^{2}:\text{exists
}1\leq i\leq d\ \text{ s.t. }\gcd(b_{i}^{(1)},b_{i}^{(2)})\geq
B^{\varepsilon\delta^{2}/6r^{2}}\right\\}.$

We consider separately two cases.

Case 1: $\\#\mathcal{E}\geq\varepsilon_{1}\\#\mathcal{V}^{2}/2r^{2}$.
In this case, there are many pairs with a gcd of some size. If we pick a
vertex $v$ in $G$ at random, then the expected number of vertices connected to
$v$ is at least ${\varepsilon_{1}\\#\mathcal{V}}/{2r^{2}}$, and so (by the
pigeonhole principle) there is some $\mathbf{b}^{(0)}\in\mathcal{B}$ and
$1\leq i\leq d$ such that there are at least
${\varepsilon_{1}\\#\mathcal{B}}/{2r^{2}d}$ elements
$\mathbf{b}\in\mathcal{B}$ with
$\gcd(b_{i},b_{i}^{(0)})>B^{\varepsilon\delta^{2}/6r^{2}}$. Since there are at
most $B^{o(1)}$ divisors of $b_{i}^{(0)}$, there must be a divisor
$m>B^{\varepsilon\delta^{2}/6r^{2}}$ of $b_{i}^{(0)}$ such that $m|b_{i}$ for
at least
${\varepsilon_{1}\\#\mathcal{B}}/{2r^{2}dB^{o(1)}}>{\\#\mathcal{B}}/{m^{\varepsilon_{1}\delta/2}}$
elements $\mathbf{b}\in\mathcal{B}$. But this contradicts the fact that
$\mathcal{B}$ is constructed to have no such integers. Thus we must instead
have $\\#\mathcal{E}<{\varepsilon_{1}\\#\mathcal{V}^{2}}/{2r^{2}}$.

Case 2: $\\#\mathcal{E}<\varepsilon_{1}\\#\mathcal{V}^{2}/2r^{2}$.
Picking $r$ distinct vertices in $G$ uniformly at random, then the expected
number of edges between these vertices is less than $\varepsilon_{1}$. In
particular, the probability that there are no edges between any of the $r$
chosen vertices is at least $1-\varepsilon_{1}$. Thus, if we define

$\mathcal{C}=\left\\{(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})\in\mathcal{B}^{r}:\gcd(b^{(i)}_{s},b^{(j)}_{s})<B^{\varepsilon\delta^{2}/6r^{2}}\text{
for }1\leq i<j\leq r,\ 1\leq s\leq d\right\\},$

then $\\#\mathcal{C}\gg_{r,\varepsilon}\\#\mathcal{B}^{r}$. Given
$(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})\in\mathcal{C}$, let

$\displaystyle\mathcal{R}(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})=$
$\displaystyle\Big{\\{}(\mathbf{b}^{(1)^{\prime}},\ldots,\mathbf{b}^{(r)^{\prime}})\in\mathcal{C}:\exists\mathbf{a}^{(1)},\ldots,\mathbf{a}^{(r)},\mathbf{a}^{(1)^{\prime}},\ldots,\mathbf{a}^{(r)^{\prime}}\text{
s.t. }$
$\displaystyle\gcd(a_{i}^{(j)},m_{i}b_{i}^{(j)})=\gcd(a_{i}^{(j)^{\prime}},m_{i}b_{i}^{(j)^{\prime}})=1,\frac{a_{i}^{(1)}}{b_{i}^{(1)}}+\cdots+\frac{a_{i}^{(r)}}{b_{i}^{(r)}}=\frac{a_{i}^{(1)^{\prime}}}{b_{i}^{(1)^{\prime}}}+\cdots+\frac{a_{i}^{(r)^{\prime}}}{b_{i}^{(r)^{\prime}}},\forall
i,j\Big{\\}}.$

Abusing notation and defining $\mathcal{R}$ the same as in Lemma 6.1, we have

$\\#\mathcal{R}(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})\leq\prod_{i=1}^{d}\\#\mathcal{R}(b_{i}^{(1)},\ldots,b_{i}^{(r)})\leq
B^{5d\varepsilon\delta^{2}/6}.$

For each $\mathbf{b}\in\mathcal{B}$, let $\mathbf{a}(\mathbf{b})$ be an
integer tuple where each entry is coprime to that of $\mathbf{m}\mathbf{b}$,
and such that
$(\mathbf{a}(\mathbf{b}),\mathbf{m}\mathbf{b},\mathbf{h})\in\mathcal{S}$ for
some $\mathbf{b}$. Thus,

$\displaystyle\\#\mathcal{A}$
$\displaystyle\geq\\#\left\\{\frac{{a}(\mathbf{b}^{(1)})_{i}}{m_{i}b_{i}^{(1)}}+\cdots+\frac{{a}(\mathbf{b}^{(r)})_{i}}{m_{i}b_{i}^{(r)}}:(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})\in\mathcal{C}\right\\}$
$\displaystyle\geq\sum_{(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})\in\mathcal{C}}\frac{1}{\\#\mathcal{R}(\mathbf{b}^{(1)},\ldots,\mathbf{b}^{(r)})}$
$\displaystyle\geq\frac{\\#\mathcal{C}}{B^{5d\varepsilon\delta^{2}/6}}\geq\frac{\\#\mathcal{B}^{r}}{X^{5d\varepsilon\delta^{2}/6}Q^{(5+\frac{\varepsilon}{20})\varepsilon\delta^{2}/6}}.$

where we recall $r\geq 1>\delta$ and $\\#\mathcal{B}>X^{-\varepsilon_{1}\delta
d/2}\\#\mathcal{S}^{\frac{1}{2}-3\varepsilon_{1}}\geq
X^{-\varepsilon_{1}\delta d/2}Q^{(\frac{1}{2}-3\varepsilon_{1})\delta}$. This
gives condition (2), and we are done. ∎

Using this, we obtain an improvement to (Maynard, 2021, Lemma 7.3).

######  Lemma 6.3.

Let $\varepsilon>0$, $\delta\in(0,1)$ and $x\in\mathbb{R}_{\geq 1}$. Let
$M\in\mathbb{R}^{+}$ satisfy

$M\geq\frac{1}{2}+\frac{\log\varepsilon^{-1}}{2\log 2}.$

Let $k,d\geq 2$ be positive integers and let $\alpha_{i,j}\in[0,1)$ for $1\leq
i\leq k$, $1\leq j\leq d$ be reals. Let $Q$ be large enough in terms of
$\delta,d,k,\varepsilon$, and let
$\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1]$ be such that
$\Delta=\prod_{i=1}^{k}\varepsilon_{i}$ satisfies
$Q^{(1+\varepsilon)\delta}\leq\Delta^{-1}\leq Q^{(2k)^{M}}$.

Let
$\mathcal{S}\subseteq\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$
be a set of triples $(\mathbf{a},\mathbf{q},\mathbf{h})$ satisfying:

  1. (1)

$\gcd(a_{j},q_{j})=1$ and $q_{j}\leq x^{\varepsilon}Q^{1/(d-1)}$ for
$j\in\\{1,2,\ldots,d\\}$.

  2. (2)

$h_{i}\leq\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ for
$i\in\\{1,2,\ldots,k\\}$.

  3. (3)

$\\#\mathcal{S}\geq Q^{\delta}$.

  4. (4)

For each $j\in\\{1,2,\ldots,d\\}$, we have

$\left|h_{1}\alpha_{1,j}+\cdots+h_{k}\alpha_{k,j}-\frac{a_{j}}{q_{j}}\right|\leq
x^{-\varepsilon}\left(\prod_{i=1}^{k}\varepsilon_{i}\right)^{(2+k/(2k)^{M}+200\varepsilon)/\delta}.$

Then there is a $\mathbf{q}_{0}\in\mathbb{Z}^{d}$ such that at least
$\\#\mathcal{S}^{1/2}$ of the triples
$(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathcal{S}$ have
$\mathbf{q}=\mathbf{q}_{0}$.

###### Proof.

The proof is similar to that of (Maynard, 2021, Lemma 7.2). Choose an integer
$r$ such that $Q^{\delta
r/(2+k/(2k)^{M}+10\varepsilon)}>x^{\varepsilon}\prod_{i=1}^{k}\varepsilon_{i}^{-1}\geq
Q^{\delta r/(2+k/(2k)^{M}+100\varepsilon)}$. A computation shows $r$ must
exist and satisfy

$r\in[(2+k/(2k)^{M}+10\varepsilon)(1+\varepsilon\log_{Q}x+\varepsilon),(2+k/(2k)^{M}+100\varepsilon)(\varepsilon\log_{Q}x+(2k)^{M})/\delta].$

Therefore we may assume that $Q$ is sufficiently large in terms of $r$.
If
$(\mathbf{a}^{(1)},\mathbf{q}^{(1)},\mathbf{h}^{(1)}),\ldots(\mathbf{a}^{(r)},\mathbf{q}^{(r)},\mathbf{h}^{(r)})\in\mathcal{S}$,
then we have for $1\leq s\leq r$ such that for all $j=1,2,\ldots,d$,

$\sum_{i=1}^{k}h_{i}^{(s)}\alpha_{i,j}=\frac{a_{j}}{q_{j}}+O\left(x^{-\varepsilon}\left(\prod_{i=1}^{k}\varepsilon_{i}\right)^{(2+k/(2k)^{M}+200\varepsilon)/\delta}\right)=\frac{a_{j}^{(s)}}{q_{j}^{(s)}}+O(Q^{-(1+\varepsilon/1000)r}).$

Adding these together and observing $rQ^{-\varepsilon r/1000}\leq 1$ for $Q$
sufficiently large gives

$\frac{a_{j}^{(1)}}{q_{j}^{(1)}}+\cdots+\frac{a_{j}^{(r)}}{q_{j}^{(r)}}+O(Q^{-r})=\sum_{i=1}^{k}\alpha_{i,j}\tilde{h}_{i},$

where $\tilde{h}_{i}=\sum_{s=1}^{r}h_{i}^{(s)}$. Since the denominator of
$\frac{a_{j}^{(1)}}{q_{j}^{(s)}}+\cdots+\frac{a_{j}^{(r)}}{q_{j}^{(r)}}$ (when
written as a single fraction) is at most $Q^{r}$, we see that this fraction is
uniquely determined by the integers $\tilde{h}_{1},\ldots,\tilde{h}_{k}$. Note
$|\tilde{h}_{j}|\ll r\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$, so we have

$\displaystyle\\#\left\\{\left(\frac{a_{1}^{(1)}}{q_{1}^{(1)}}+\cdots+\frac{a_{1}^{(r)}}{q_{1}^{(r)}},\ldots,\frac{a_{d}^{(1)}}{q_{d}^{(1)}}+\cdots+\frac{a_{d}^{(r)}}{q_{d}^{(r)}}\right):\exists\mathbf{h}^{(1)},\ldots,\mathbf{h}^{(r)}\text{
s.t.
}(\mathbf{a}^{(1)},\mathbf{q}^{(1)},\mathbf{h}^{(1)}),\ldots,(\mathbf{a}^{(r)},\mathbf{q}^{(r)},\mathbf{h}^{(r)})\in\mathcal{S}\right\\}$
$\displaystyle\leq\\#\left\\{(\tilde{h}_{1},\ldots,\tilde{h}_{k})\in\mathbb{Z}^{k}:|\tilde{h}_{i}|\ll
r\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}\text{ for }i=1,2,\ldots,k\right\\}$
$\displaystyle\leq\left(\prod_{i=1}^{k}\varepsilon_{i}^{-1}\right)^{1+k/(2k)^{M}+\varepsilon/1000}$
$\displaystyle\leq
x^{-(1+k/(2k)^{M}+\varepsilon/1000)\varepsilon}Q^{\frac{1+k/(2k)^{M}+\varepsilon/1000}{2+k/(2k)^{M}+100\varepsilon}\delta
r}\leq
x^{-(1+k/(2k)^{M}+\varepsilon/1000)\varepsilon}Q^{\frac{1}{2}-\varepsilon}$

Here we used the fact that
$r\leq\delta^{-1}\log\prod_{i=1}^{k}\varepsilon_{i}^{-1}$ and
$x^{-\varepsilon}Q^{\delta
r/(2+k/(2k)^{M}+100\varepsilon)}\leq\prod_{i=1}^{k}\varepsilon^{-1}$, followed
by $\prod_{i=1}^{k}\varepsilon_{i}^{-1}<x^{-\varepsilon}Q^{\delta
r/(2+k/(2k)^{M}+10\varepsilon)}$.

Note our assumptions satisfy all hypotheses of Lemma 6.2, so by the lemma and
also the above bound, (1) must hold, so there must be a
$\mathbf{q}_{0}\in\mathbb{Z}^{d}$ such that at least $\\#\mathcal{S}^{1/2}$ of
the triples $(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathcal{S}$ have
$\mathbf{q}=\mathbf{q}_{0}$. ∎

Therefore, we can prove a proposition akin to (Maynard, 2021, Proposition
5.2).

######  Proposition 6.4.

Let $\varepsilon>0$, and $f_{1},\ldots,f_{k}\in$ $\mathbb{R}[X]$ be
polynomials of degree at most $d$ such that $f_{1}(0)=\cdots=f_{k}(0)=0$. Put
$f_{i}(X)=\sum_{j=1}^{d}f_{i,j}X^{j}$. Let
$\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1/100]$, and put
$\Delta=\prod_{i=1}^{k}\varepsilon_{i}$. Define constants

$c_{0}=1+\varepsilon,\quad c_{1}=1+k/(2k)^{M}+10\varepsilon,\quad
c_{2}=2.5+1/(2k)^{M-1}+500\varepsilon,$

and let $M\in\mathbb{R}^{+}$ satisfy

$M\geq\frac{1}{2}+\frac{\log\varepsilon^{-1}}{2\log 2}.$

Let $Q\leq\Delta^{-c_{1}d(d-1)}$ be such that there are at least
$Q^{1/(1+\varepsilon)c_{0}d(d-1)}$ triples
$(\mathbf{a},\mathbf{q},\mathbf{h})\in$
$\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$ satisfying:

  1. (1)

$q_{1}=\cdots=q_{d}$ and $1\leq q_{j}\leq x^{\varepsilon}Q^{1/c_{0}(d-1)}$ for
$1\leq j\leq d$.

  2. (2)

$h_{i}\ll\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ for $1\leq i\leq k$.

  3. (3)

For each $j\in\\{1,\ldots,d\\}$ we have

$\sum_{i=1}^{k}h_{i}f_{i,j}=\frac{a_{j}}{q_{j}}+O\left(\frac{Q^{1/d}}{x^{j-\varepsilon}}\right).$

Then provided $\Delta^{-c_{2}d(d-1)}<x$ there is some positive integer $q\leq
x^{\varepsilon}Q^{1/c_{0}(d-1)}$ and at least
$Q^{1/2(1+\varepsilon)c_{1}d(d-1)}$ pairs $(\mathbf{a},\mathbf{h})\in$
$\mathbb{Z}^{d}\times\mathbb{Z}^{k}$ such that:

  1. (1)

$h_{i}\ll\varepsilon_{i}^{-1}\Delta^{-2/(2k)^{M}}$ for $i\in\\{1,\ldots,k\\}$.

  2. (2)

For each $j\in\\{1,\ldots,d\\}$ we have

$\sum_{i=1}^{k}h_{i}f_{i,j}=\frac{a_{j}}{q}+O\left(\frac{Q^{1/d}}{x^{j-\varepsilon}}\right).$

All implied constants depend only on $d,k,\varepsilon$ and $M$.

###### Proof.

The proof is nearly identical to (Maynard, 2021, Proposition 5.2). By
assumption, there is some
$Q\leq(\prod_{i=1}^{k}\varepsilon_{i}^{-1})^{c_{1}d(d-1)}$ such that there are
at least $Q^{1/2c_{1}d(d-1)}$ triples
$(\mathbf{a},\mathbf{q},\mathbf{h})\in\mathbb{Z}^{d}\times\mathbb{Z}^{d}\times\mathbb{Z}^{k}$
with $\gcd(a_{j},q_{j})=1$ and $q_{j}\leq x^{\varepsilon}Q^{1/c_{0}(d-1)}$ for
$j=1,2,\ldots,d$, and with
$|h_{i}|\ll\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}$ for $i=1,2,\ldots,k$ and
with

$\sum_{i=1}^{k}h_{i}f_{i,j}=\frac{a_{j}}{q_{j}}+O\left(\frac{Q^{1/d}}{x^{j-\varepsilon}}\right).$

If $Q\leq\Delta^{-1/(2k)^{M}}$, then we just take one such triple
$(\mathbf{a},\mathbf{q},\mathbf{h})$. In this case the triple
$(j\mathbf{a},j\mathbf{q},j\mathbf{h})$ for $j=1,2,\ldots,Q$ then give $Q$
relations of the desired type $C_{d}^{\prime}>2d(d-1)+1$, since $jh_{i}\ll
Q\varepsilon_{i}^{-1}\Delta^{-1/(2k)^{M}}\ll\varepsilon_{i}^{-1}\Delta^{-2/(2k)^{M}}$.
Thus we may assume $Q^{(2k)^{M}}>\Delta^{-1}$.

We now apply Lemma 6.3 with $\delta=((1+\varepsilon)c_{1}d(d-1))^{-1}$.
Provided $c_{2}=2.5+1/(2k)^{M-1}+500\varepsilon$ and
$c_{1}=1+k/(2k)^{M}+10\varepsilon$, we see that the bounds
$Q\leq(\prod_{i=1}^{k}\varepsilon_{i}^{-1})^{c_{1}d(d-1)}$ and
$(\prod_{i=1}^{k}\varepsilon_{i}^{-1})^{c_{2}d(d-1)}\leq x$ imply

$\frac{Q^{1/d}}{x^{j-\varepsilon}}<x^{-\varepsilon}\left(\prod_{i=1}^{k}\varepsilon_{i}\right)^{c_{2}(1-2\varepsilon)d(d-1)-c_{1}(d-1)}<x^{-\varepsilon}\left(\prod_{i=1}^{k}\varepsilon_{i}\right)^{(2+k/(2k)^{M}+200\varepsilon)/\delta}$

for $1\leq i\leq d$, since

$c_{2}(1-2\varepsilon)d(d-1)-c_{1}(d-1)\geq(2+k/(2k)^{M}+200\varepsilon)(1+\varepsilon)c_{1}d(d-1)$

is implied by the fact that

$d\geq\frac{1+k/(2k)^{M}+\varepsilon}{(2.5+1/(2k)^{M-1}+500\varepsilon)(1-2\varepsilon)-(2+k/(2k)^{M}+200\varepsilon)(1+k/(2k)^{M}+10\varepsilon)(1+\varepsilon)},$

and so all hypotheses of Lemma 6.3 are satisfied. Therefore, there is some
$\mathbf{q}_{0}\in\mathbb{Z}^{d}$ such that at least $Q^{\delta/2}$ of the
triples $(\mathbf{a},\mathbf{q},\mathbf{h})$ have $\mathbf{q}=\mathbf{q}_{0}$.
These all give rise to a rational $a_{j}/q$ due to the condition
$q_{1}=\cdots=q_{d}$, and so we are done. ∎

##  7\. Density Increment

We also need to alter (Maynard, 2021, Lemma 8.3) slightly.

######  Lemma 7.1.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
with $f_{1}(0)=\cdots=$ $f_{k}(0)=0$. Put
$f_{i}(X)=\sum_{j=1}^{d}f_{i,j}X^{j}$.

Let $c>0$, $B_{1},\ldots,B_{k}\geq 1$, $Q\in\mathbb{Z}_{>0}$, and $1\leq
q_{0}<x^{\varepsilon}Q^{2/d}$. Let $\eta\in[0,1/100]$ be such that

$\eta<\frac{Q^{1/d}}{x^{1-\varepsilon}}.$

Let $r\in\\{1,\ldots,k\\}$ and
$\mathbf{h}^{(1)},\ldots,\mathbf{h}^{(r)}\in\mathbb{Z}^{k}$ and
$\mathbf{a}^{(1)},\ldots,\mathbf{a}^{(r)}\in\mathbb{Z}^{d}$ satisfy:

  1. (1)

$h_{i}^{(\ell)}\leq B_{i}$ for $1\leq i\leq k$ and $1\leq\ell\leq r$.

  2. (2)

For $1\leq\ell\leq r$ and $1\leq j\leq d$ we have

$\left|\sum_{i=1}^{k}h_{i}^{(\ell)}f_{i,j}-\frac{a_{j}^{(\ell)}}{q_{0}}\right|\leq\eta^{j}.$

  3. (3)

Put $\tilde{h}_{i}^{(\ell)}=h_{i}^{(\ell)}/B_{i}$ for $1\leq\ell\leq r$ and
$1\leq i\leq k$. We have

$\left\|\tilde{\mathbf{h}}^{(1)}\wedge\cdots\wedge\tilde{\mathbf{h}}^{(r)}\right\|\asymp\left\|\tilde{\mathbf{h}}^{(1)}\right\|_{\infty}\cdots\left\|\tilde{\mathbf{h}}^{(r)}\right\|_{\infty}$

and

$\left\|\tilde{\mathbf{h}}^{(1)}\right\|_{\infty}\cdots\left\|\tilde{\mathbf{h}}^{(r)}\right\|_{\infty}\ll\frac{1}{Q^{1/cd(d^{2}-1)}}.$

Then there is an integer $k^{\prime}<k$, real polynomials
$g_{1},\ldots,g_{k^{\prime}}\in\mathbb{R}[X]$ of degree at most $d$ with
$g_{1}(0)=\cdots=g_{k}(0)=0$ and quantities
$B_{1}^{\prime},\ldots,B_{k^{\prime}}^{\prime}\geq 2$ and $y<x$ such that:

  1. (1)

If there is an integer $n^{\prime}<y$ such that

$\left\|g_{i}\left(n^{\prime}\right)\right\|<\frac{1}{B_{i}^{\prime}}\quad\text{
for all }1\leq i\leq k^{\prime}$

then there is an integer $n<x$ such that

$\left\|f_{i}(n)\right\|<\frac{1}{B_{i}}\quad\text{ for all }1\leq i\leq k.$

  2. (2)

We have

$\frac{y}{\left(B_{1}^{\prime}\cdots
B_{k^{\prime}}^{\prime}\right)^{d(d-1)(4.5c+1.5-1/k^{\prime
3})}}\gg\frac{x^{1-\varepsilon}}{\left(B_{1}\cdots
B_{k}\right)^{d(d-1)(4.5c+1.5-1/k^{4}-1/k^{4})}}.$

###### Proof.

The majority of the proof is identical to the corresponding proof in Maynard
(2021). Due to the proof being quite long, we only include the parts where it
differs. The following is a modified version of the proof, bottom half of Page
24 of Maynard (2021) onwards.

We see that $g_{i}$ are polynomials of degree at most $d$ with $g_{i}(0)=0$
since $\tilde{f}_{i}$ are. Finally, we put $y=\delta
x^{1-\varepsilon}\min_{i}\lVert\tilde{\mathbf{h}}_{i}\rVert_{\infty}/q_{0}Q^{1/d}D_{2}$,
and note that since $\eta<Q^{1/d}/x^{1-\varepsilon}$, we have
$y<\delta\min_{i}\lVert\tilde{\mathbf{h}}_{i}\rVert_{\infty}/\eta q_{0}D_{2}$.
Putting everything together, we see that there is an $n^{\prime}<y$ such that

$\lVert g_{i}(n^{\prime})\rVert\leq\frac{1}{B_{i}^{\prime}}$

for $1\leq i\leq k^{\prime}=k-r$, then there is an $n=n^{\prime}q_{0}D_{2}$
with $n<x$ and
$n<\delta\min_{i}\lVert\tilde{\mathbf{h}}_{i}\rVert_{\infty}/\eta$ such that

$\lVert f_{i}(n)\rVert\leq\frac{1}{B_{i}}$

for $1\leq i\leq k$. Thus we are left to verify the size estimates with this
choice of $B_{1}^{\prime},\ldots,B_{k-r}^{\prime}$ and $y$. We have

$\prod_{i=1}^{k-r}B_{i}^{\prime}=\frac{\lVert\mathbf{z}_{1}\rVert_{\infty}\cdots\lVert\mathbf{z}_{k-r}\rVert_{\infty}\prod_{i=r+1}^{k}B_{i}}{\delta^{2(k-r)}}\ll\frac{D_{1}\prod_{i=r+1}^{k}B_{i}}{D_{2}}.$

This implies that (with $C_{2}$ a constant depending only on $d$ chosen later)

$\frac{y}{(\prod_{i=1}^{k-r}B_{i}^{\prime})^{C_{2}}}=\frac{\delta
x^{1-\varepsilon}\min_{i}\lVert\tilde{\mathbf{h}}_{i}\rVert_{\infty}}{q_{0}Q^{1/d}D_{2}(\prod_{i=1}^{k-r}B_{i}^{\prime})^{C_{2}}}\gg\frac{x^{1-2\varepsilon}D_{2}^{C_{2}-1}\min_{i}\lVert\tilde{\mathbf{h}}_{i}\rVert_{\infty}}{Q^{3/d}D_{1}^{C_{2}}(\prod_{i=r+1}^{k}B_{i})^{C_{2}}}.$

Recall that
$D_{1}=\det(H_{1})\asymp\lVert\mathbf{h}_{1}\rVert_{\infty}\cdots\lVert\mathbf{h}_{r}\rVert_{\infty}\ll
B_{1}\cdots B_{r}/Q^{1/cd(d-1)^{2}}$, and that
$\min_{i}\lVert\tilde{\mathbf{h}}_{i}\rVert_{\infty}\gg\prod_{i=1}^{r}\lVert\tilde{\mathbf{h}}_{i}\rVert_{\infty}\gg
D_{1}/(B_{1}\cdots B_{r})$. This gives

$\displaystyle\frac{y}{(\prod_{i=1}^{k-r}B_{i}^{\prime})^{C_{2}}}$
$\displaystyle\gg\frac{x^{1-2\varepsilon}}{(\prod_{i=r+1}^{k}B_{i})^{C_{2}}}\frac{D_{2}^{C_{2}-1}}{D_{1}^{C_{2}-1}Q^{3/d}B_{1}\cdots
B_{r}}$
$\displaystyle\gg\frac{x^{1-2\varepsilon}}{(\prod_{i=1}^{k}B_{i})^{C_{2}}}\cdot
Q^{(C_{2}-1)/cd(d-1)^{2}-3/d}D_{2}^{C_{2}-1}.$

Finally, we choose $C_{2}=d(d-1)(4.5c+1.5-1/(k-r)^{3})$. Since $D_{2},Q\geq 1$
and

$(C_{2}-1)/cd(d-1)^{2}-3/d\geq 0,$

which gives

$\frac{y}{(\prod_{i=1}^{k-r}B_{i}^{\prime})^{d(d-1)(4.5c+1.5-1/(k-r)^{3})}}\gg\frac{x^{1-\varepsilon}}{(\prod_{i=1}^{r}B_{i})^{d(d-1)(4.5c+1.5-1/(k-r)^{3})}}.$

Since $k>r\geq 1$, we have $1/(k-r)^{3}\geq 1/k^{3}+1/k^{4}$, and so we are
done. ∎

There are minimal changes to the proof of Maynard (2021) from here onwards.
For completeness, we include the remaining statements and proofs.

######  Lemma 7.2.

Let $\eta>0$ be sufficiently small in terms of $k$ and $d$. Let
$B_{1},\ldots,B_{k}>1$ satisfy $\prod_{i=1}^{k}B_{i}\leq\eta^{-1/2}$ and
$\beta_{i,j}\in\mathbb{R}$ for $1\leq i\leq k$, $1\leq j\leq d$. Let
$\mathcal{R}$ be the region in $\mathbb{R}^{k+d}$ defined by

$\displaystyle\mathcal{R}=\\{$
$\displaystyle(h_{1},\ldots,h_{k},a_{1},\ldots,a_{d})\in\mathbb{R}^{k+d}:$
$\displaystyle\left|\sum_{i=1}^{k}h_{i}\beta_{i,j}-a_{j}\right|\leq\eta^{j}\enspace\forall
1\leq j\leq d,\,|h_{i}|\leq B_{i}\enspace\forall 1\leq i\leq k\\},$

and assume that $\\#(\mathcal{R}\cap\mathbb{Z}^{k+d})=N$ is sufficiently large
in terms of $k$ and $d$. Then $\exists 1\leq r\leq k$ and
$\mathbf{h}^{(1)},\ldots,\mathbf{h}^{(r)}\in\mathbb{Z}^{k}$ and
$\mathbf{a}^{(1)},\ldots,\mathbf{a}^{(r)}\in\mathbb{Z}^{d}$ such that:

  1. (1)

$\forall j\in\\{1,\ldots,r\\}$,
$(h_{1}^{(j)},\ldots,h_{k}^{(j)},a_{1}^{(j)},\ldots,a_{d}^{(j)})\in\mathcal{R}\cap\mathbb{Z}^{k+d}.$

  2. (2)

$\lVert\mathbf{h}^{(1)}\wedge\cdots\wedge\mathbf{h}^{(r)}\rVert\asymp_{k,d}\lVert\mathbf{h}^{(1)}\rVert_{\infty}\cdots\lVert\mathbf{h}^{(r)}\rVert_{\infty}$

  3. (3)

$\lVert\mathbf{h}^{(1)}\rVert_{\infty}\cdots\lVert\mathbf{h}^{(r)}\rVert_{\infty}\ll_{k,d}{B_{1}\cdots
B_{k}}/{N^{1/(d+1)}}.$

All implied constants depend at most on $k$ and $d$.

###### Proof.

This is (Maynard, 2021, Lemma 8.1). ∎

######  Proposition 7.3.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most
$d\geq 2$ such that $f_{1}(0)=\cdots=f_{k}(0)=0$. Put
$f_{i}(X)=\sum_{j=1}^{d}f_{i,j}X^{j}$. Let
$\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1/100]$, and put
$\Delta=\prod_{i=1}^{k}\varepsilon_{i}$. For any $\varepsilon>0,M\geq 4$,
define constants

$c_{0}=1+\varepsilon,\quad c_{1}=1+k/(2k)^{M}+10\varepsilon,\quad
c_{2}=2.5+1/(2k)^{M-1}+500\varepsilon.$

Suppose $\Delta^{-c_{2}d(d-1)}\leq x$, and let $Q\leq\Delta^{-c_{1}d(d-1)}$.
Let $q$ be a positive integer with $q<x^{\varepsilon}Q^{1/c_{0}(d-1)}$. Let
$\mathcal{S}$ be the set of pairs
$(\mathbf{a},\mathbf{h})\in\mathbb{Z}^{d}\times\mathbb{Z}^{k}$ such that for
$j\in\\{1,\ldots,d\\}$, we have

$\left|\sum_{i=1}^{k}h_{i}f_{i,j}-\frac{a_{j}}{q}\right|\ll\frac{Q^{1/d}}{x^{j-\varepsilon}},$

and such that $|h_{i}|\ll\varepsilon_{i}^{-1}\Delta^{-2/(2k)^{M}}$. Assume
that $\\#\mathcal{S}>Q^{1/2(1+\varepsilon)c_{1}d(d-1)}$.

Then there is an integer $k^{\prime}<k$, polynomials
$g_{1},\ldots,g_{k^{\prime}}\in\mathbb{R}[X]$ of degree at most $d\geq 2$ with
$g_{1}(0)=\cdots=g_{k^{\prime}}(0)=0$ and quantities
$\varepsilon_{1}^{\prime},\ldots,\varepsilon_{k^{\prime}}^{\prime}\in(0,1/100]$
and $y<x$ such that for $c=9(1+\varepsilon)c_{1}+1.5$:

  1. (1)

If there is an integer $n^{\prime}<y$ such that for all $1\leq i\leq
k^{\prime}$

$\lVert g_{i}(n^{\prime})\rVert<\varepsilon_{i}^{\prime}$

then there is an integer $n<x$ such that for all $1\leq i\leq k$,

$\lVert f_{i}(n)\rVert<\varepsilon_{i}.$

  2. (2)

We have

$y(\varepsilon_{1}^{\prime}\cdots\varepsilon_{k^{\prime}}^{\prime})^{d(d-1)(c-1/k^{\prime
3})}\gg
x^{1-\varepsilon}(\varepsilon_{1}\cdots\varepsilon_{k})^{d(d-1)(c-1/k^{3})}.$

All implied constants depend only on $k$, $d$ and $\varepsilon$.

###### Proof.

Taking $\beta_{i,j}=f_{i,j}$, $\eta=Q^{1/d}/x$ and
$B_{i}\ll\varepsilon_{i}^{-1}\Delta^{-2/(2k)^{M}}$, Lemma 7.2 shows that we
can find a subset of essentially orthogonal generators, and we can apply Lemma
7.1. It suffices to verify the density estimates. Indeed,

$\displaystyle
y(\varepsilon^{\prime}_{1}\cdots\varepsilon^{\prime}_{k^{\prime}})^{d(d-1)(c-1/k^{\prime
3})}$ $\displaystyle=\frac{y}{(B_{1}^{\prime}\cdots
B_{k^{\prime}}^{\prime})^{d(d-1)(c-1/k^{\prime 3})}}$
$\displaystyle\gg\frac{x}{(B_{1}\cdots B_{k})^{d(d-1)(c-1/k^{4}-1/k^{4})}}$
$\displaystyle=x(\varepsilon_{1}\cdots\varepsilon_{k}\Delta^{2/(2k)^{M}})^{d(d-1)(c-1/k^{4}-1/k^{4})}$
$\displaystyle\gg
x(\varepsilon_{1}\cdots\varepsilon_{k})^{d(d-1)(c-1/k^{3})}.$

Note here $M\geq 4$ is required for the last step. ∎

##  8\. Putting Everything Together

As in Maynard (2021), combining Propositions 5.3, 6.4, 7.3, we obtain the
following version of (Maynard, 2021, Proposition 5.4).

######  Proposition 8.1.

Let $\varepsilon>0$, and $d,k$ be positive integers. Define constants

$c_{0}=1+\varepsilon,\quad c_{2}=2.5+1/(2k)^{M-1}+100\varepsilon,$

and let $M\in\mathbb{R}^{+}$ satisfy

$M\geq\max\left\\{4,\frac{1}{2}+\frac{\log\varepsilon^{-1}}{2\log
2}\right\\}.$

Let $c_{2}=2.5+1/(2k)^{M-1}+100\varepsilon$, and $C_{d,k}$ be a constant
depending only on $d$, $k$ (and implicitly on $\varepsilon,M$) such that the
following holds.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
such that $f_{1}(0)=\cdots=f_{k}(0)=0$. Put
$f_{i}(X)=\sum_{j=1}^{d}f_{i,j}X^{j}$. Let
$\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1/100]$, and put
$\Delta=\prod_{i=1}^{k}\varepsilon_{i}$. Let $\Delta^{-1}\leq
x^{1/c_{2}d(d-1)}$.

If there is no positive integer $n<x$ such that for all $i=1,2,\ldots,k$,

$\lVert f_{i}(n)\rVert<\varepsilon_{i},$

then there is a positive integer $k^{\prime}<k$ and polynomials
$g_{1},\ldots,g_{k^{\prime}}\in\mathbb{R}[X]$ of degree at most $d$, with
$g_{1}(0)=\cdots=g_{k^{\prime}}(0)=0$ and reals
$\varepsilon_{1}^{\prime},\ldots,\varepsilon_{k^{\prime}}^{\prime}\in(0,1/100]$
and $y\in\mathbb{R}$ with $y<x$ such that both of the following hold:

  1. (1)

There is no positive integer $n^{\prime}<y$ such that for all
$i=1,2,\ldots,k^{\prime}$,

$\lVert g_{i}(n^{\prime})\rVert<\varepsilon_{i}^{\prime}.$

  2. (2)

We have

$y(\varepsilon_{1}^{\prime}\cdots\varepsilon_{k^{\prime}}^{\prime})^{d(d-1)(c-1/k^{\prime
3})}\geq\frac{1}{C_{d,k}}x^{1-\varepsilon}(\varepsilon_{1}\cdots\varepsilon_{k})^{d(d-1)(c-1/k^{3})},$

where $c=9(1+\varepsilon)(1+k/(2k)^{M}+10\varepsilon)+1.5$.

All implied constants depend only on $k$, $d$ and $\varepsilon$.

######  Theorem 8.2.

Let $k,d\in\mathbb{Z}^{+},\,\varepsilon>0,$ and $M\in\mathbb{R}^{+}$ satisfy

$M\geq\max\left\\{4,\frac{1}{2}+\frac{\log\varepsilon^{-1}}{2\log
2}\right\\}.$

There is a constant $C_{d,k}>2$ depending only on $d$, $k$ (and
$\varepsilon,M$) such that the following holds.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
such that $f_{1}(0)=\cdots=$ $f_{k}(0)=0$. Let
$\varepsilon_{1},\ldots,\varepsilon_{k}\in(0,1/100]$, and put
$\Delta=\prod_{i=1}^{k}\varepsilon_{i}$. Define

$c_{2}=10.5+9k/(2k)^{M}+\varepsilon.$

If $\Delta^{-1}\leq x^{1/c_{2}d(d-1)}$ and $x>C_{d,k}$, then there is a
positive integer $n<x$ such that

$\left\|f_{i}(n)\right\|\leq\varepsilon_{i}\quad\text{ for all
}i\in\\{1,\ldots,k\\}\text{. }$

###### Proof.

The proof is essentially identical to (Maynard, 2021, Theorem 1.1). Assume for
a contradiction that there is no positive $n<x$ such that $\lVert
f_{i}(n)\rVert\leq\varepsilon_{i}$ for all $i=1,2,\ldots,k$. Let $C_{d,k}$ be
the constant in Proposition 8.1. Let $C_{0}=\sup_{j\leq k}C_{d,j}$.

Define a _System_ to be a tuple $(k,\mathbf{g},\bm{\delta},y)$ consisting of

  1. (1)

A positive integer $k$.

  2. (2)

A $k$-tuple $\mathbf{g}$ of real polynomials $(g_{1},\ldots,g_{k})$ of degree
at most $d$ satisfying $g_{1}(0)=\cdots=g_{k}(0)=0$.

  3. (3)

A $k$-tuple $\bm{\delta}$ of reals $(\delta_{1},\ldots,\delta_{k})$ with
$\delta_{i}\in(0,1/100]$ for all $i\in\\{1,2,\ldots,k\\}$.

  4. (4)

A real $y$ such that there is no positive integer $n<y$ satisfying

$\lVert g_{i}(n)\rVert\leq\delta_{i}$

for all $i=1,2,\ldots,k$.

Let $c=9(1+\varepsilon)(1+k/(2k)^{M}+\varepsilon)+1.5$. Given a system
$(k,\mathbf{g},\bm{\delta},y)$, let
$\Delta(\bm{\delta})=\prod_{i=1}^{k}\delta_{i}$. By Proposition 8.1, if a
system $(k_{j},\mathbf{g}_{j},\bm{\delta}_{j},y_{j})$ satisfies
$\Delta(\bm{\delta}_{j})^{-c_{2}d(d-1)}<y_{j}$, then there is a system
$(k_{j+1},\mathbf{g}_{j+1},\bm{\delta}_{j+1},y_{j+1})$ such that
$k_{j+1}<k_{j}$, $y_{j+1}\leq y_{j}$,

$y_{j+1}\Delta(\bm{\delta}_{j+1})^{d(d-1)(c-1/k_{j+1}^{2})}\geq\frac{y_{j}^{1-\varepsilon}\Delta(\bm{\delta}_{j})^{d(d-1)(c-1/k_{j}^{2})}}{C_{0}},$

Note that if

(8.1)
$y_{j}^{1-k_{j}\varepsilon}\Delta(\bm{\delta}_{j})^{d(d-1)(c-1/k_{j}^{2})}\geq
C_{0}^{k_{j}},$

then we have

$y_{j+1}^{1-k_{j+1}\varepsilon}\Delta(\bm{\delta}_{j+1})^{d(d-1)(c-1/k_{j+1}^{2})}\geq
y_{j+1}^{-k_{j+1}\varepsilon}\frac{y_{j}^{1-\varepsilon}\Delta(\bm{\delta}_{j})^{d(d-1)(c-1/k_{j}^{2})}}{C_{0}}\geq
C_{0}^{k_{j}-1}\geq C_{0}^{k_{j+1}}.$

Therefore, if (8.1) holds, we can find infinite systems with $k_{j}$ a
strictly descending list of positive integers, which is impossible. Therefore,
there does not exist a system satisfying (8.1).

Therefore, given an integer $k$, polynomials
$\mathbf{f}=(f_{1},\ldots,f_{k})\in\mathbb{R}[X]^{k}$ of degree at most $d$
satisfying $f_{1}(0)=\cdots=f_{k}(0)=0$, quantities
$\bm{\varepsilon}=(\varepsilon_{1},\ldots,\varepsilon_{k})\in(0,1/100]^{k}$
and a real $x$ satisfying $x^{1-k\varepsilon}\Delta^{d(d-1)(c-1/k^{2})}\geq
C_{0}^{k}$, $(k,\mathbf{f},\bm{\varepsilon},x)$ cannot form a system, so there
must exist some $n<x$ such that $\lVert f_{i}(n)\rVert\leq\varepsilon_{i}$ for
all $i=1,2,\ldots,k$. ∎

By taking $\varepsilon_{1}=\cdots=\varepsilon_{k}=x^{-1/c_{2}d(d-1)k}$, and
for any $\varepsilon>0$ letting $M$ large such that
$2^{M}\geq\varepsilon^{-1}$, we arrive at the following:

######  Corollary 8.3.

Let $f_{1},\ldots,f_{k}\in\mathbb{R}[X]$ be polynomials of degree at most $d$
such that $f_{1}(0)=\cdots=f_{k}(0)=0$. Then there is a positive integer $n<x$
such that for all $\varepsilon>0$,

$\lVert f_{i}(n)\rVert\ll_{d,k,\varepsilon}x^{-1/(10.5+\varepsilon)kd(d-1)}$

for all $i=1,2,\ldots,k$.

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