${\delta}^{2{\gamma}}N^{\gamma}R^{2}\ll\sum_{r,r^{\prime}\leqslant
R}|\hat{\mu}({\theta}_{r}-{\theta}^{\prime}_{r})|^{\gamma}.$ (B.6)

Consider the value of ${\theta}={\theta}_{r}-{\theta}^{\prime}_{r}$ in the
right-hand side of (B.6). Define a set of ‘minor arcs’

${\mathfrak{n}}=\\{{\alpha}\in{\mathbb{T}}:|\hat{\mu}({\alpha})|\leqslant
N^{10^{-8}+3/4}\\}.$

In light of (B.4), the contribution from ${\theta}\in{\mathfrak{n}}$ to the
right-hand side of (B.6) is $o({\delta}^{2{\gamma}}N^{\gamma}R^{2})$, and so

${\delta}^{2{\gamma}}N^{\gamma}R^{2}\ll\sum_{\begin{subarray}{c}r,r^{\prime}\leqslant
R:\\\
{\theta}\notin{\mathfrak{n}}\end{subarray}}|\hat{\mu}({\theta}_{r}-{\theta}^{\prime}_{r})|^{\gamma}.$
(B.7)

Next, suppose ${\theta}\in{\mathbb{T}}\setminus{\mathfrak{n}}$, and fix a
small $\varepsilon>0$. By [Cho16, Lemma 2.3], there exist relatively prime
$q\in{\mathbb{N}}$ and $a\in{\mathbb{Z}}$ such that

$q\leqslant N^{3/4},\qquad|q{\theta}-a|\leqslant N^{-9/4}$

and

$\hat{\mu}({\theta})\ll
q^{\varepsilon-\frac{1}{3}}N(1+N^{3}|{\theta}-a/q|)^{-1/3}.$

With $C$ a large positive constant, put $Q=C+{\delta}^{-9}$. The contribution
to the right-hand side of (B.7) from denominators $q>Q$ is
$O(R^{2}N^{\gamma}Q^{{\gamma}(\varepsilon-\frac{1}{3})})$, which is negligible
compared to the left-hand side.

Hence

${\delta}^{2{\gamma}}R^{2}\ll\sum_{1\leqslant r,r^{\prime}\leqslant
R}G({\theta}_{r}-{\theta}^{\prime}_{r}),$ (B.8)

where

$G({\theta})=\sum_{q\leqslant
Q}\sum_{a=0}^{q-1}\frac{q^{{\gamma}(\varepsilon-\frac{1}{3})}}{(1+N^{3}|\sin({\theta}-a/q)|)^{{\gamma}/3}}.$

The inequality (B.8) is a cubic version of [Bou89, Eq. (4.16)]. As
${\gamma}(\varepsilon-\frac{1}{3})>1$, Bourgain’s argument carries through,
and yields (B.5). ∎

Finally, we need an upper bound on the number of trivial solutions.

######  Lemma B.4.

Let $k\geqslant 2$, and let $c_{1},\ldots,c_{s}$ be non-zero integers summing
to zero. Then there exists $s_{0}(k)$ such that if $s\geqslant s_{0}(k)$ then

$\\#\Bigl{\\{}{\mathbf{x}}\in
S(N;N^{\eta})^{s}:\sum_{i=1}^{s}c_{i}x_{i}^{k}=0\text{ and }x_{i}=x_{j}\text{
for some }i\neq j\Bigr{\\}}=o\left(N^{s-k}\right).$

Moreover, one can take $s_{0}(2)=5$, $s_{0}(3)=8$, and $s_{0}(k)$ satisfying
(1.3).

###### Proof.

Let $s_{0}(k)$ be as in Lemma B.3. By the union bound, it suffices to prove an
estimate of the required shape for the number of solutions with
$x_{s-1}=x_{s}$. In this case we are estimating

$\\#\Bigl{\\{}x\in
S(N;N^{\eta})^{s-1}:\sum_{i=1}^{s-2}c_{i}x_{i}^{k}+(c_{s-1}+c_{s})x_{s-1}^{k}=0\Bigr{\\}}.$

It may be that $c_{s-1}+c_{s}=0$, so we estimate the contribution from the
$x_{s-1}$ variable trivially. Using orthogonality and Hölder’s inequality, it
therefore suffices to prove that

$\int_{\mathbb{T}}\Bigl{|}\sum_{x\in S(N;N^{\eta})}e(\alpha
x^{k})\Bigr{|}^{s-2}\mathrm{d}\alpha=o(N^{s-1-k}).$ (B.9)

Let $p=s_{0}(k)-10^{-8}$. When $s-2\geqslant p$, the estimate (B.9) follows
from Lemma B.3 and the trivial estimate $\Bigl{|}\sum_{x\in
S(N;N^{\eta})}e(\alpha x^{k})\Bigr{|}\leqslant N$. When $s-2<p$, we apply
Hölder’s inequality and Lemma B.3 to obtain

$\displaystyle\int_{\mathbb{T}}\Bigl{|}\sum_{x\in S(N;N^{\eta})}e(\alpha
x^{k})\Bigr{|}^{s-2}\mathrm{d}\alpha$
$\displaystyle\leqslant\left(\int_{\mathbb{T}}\Bigl{|}\sum_{x\in
S(N;N^{\eta})}e(\alpha
x^{k})\Bigr{|}^{p}\mathrm{d}\alpha\right)^{\frac{s-2}{p}}$ $\displaystyle\ll
N^{s-2-\frac{k(s-2)}{p}}.$

It remains to check that $s-2-\frac{k(s-2)}{p}<s-1-k,$ or equivalently that
$2+p(1-\tfrac{1}{k})<s.$ Since $s>p$, this follows if $p/k\geqslant 2$, which
we can certainly ensure without affecting the bound (1.3). ∎

###  Appendix C A generalised von Neumann lemma

Recall the notion of $p$-restriction introduced in Definition 5.4.

######  Lemma C.1.

Let $\nu_{1},\nu_{2}:[N]\to[0,\infty)$. If both $\nu_{1}$ and $\nu_{2}$
satisfy a $p$-restriction estimate with constant $K$, then so does
$\nu_{1}+\nu_{2}$.

###### Proof.

Let $|\phi|\leqslant\nu_{1}+\nu_{2}$. Then $\phi=\psi\times\theta$, where
$\psi:[N]\to[0,\infty)$ satisfies $\psi\leqslant\nu_{1}+\nu_{2}$ and
$\theta:[N]\to\mathbb{C}$ satisfies $|\theta|\leqslant 1$. Put
$\psi_{1}:=\min\left\\{\psi,\nu_{1}\right\\}$ and $\psi_{2}:=\psi-\psi_{1}$.
On setting $\phi_{i}:=\psi_{i}\theta$, we have $\phi=\phi_{1}+\phi_{2}$ with
$|\phi_{i}|\leqslant\nu_{i}$. Applying the triangle inequality and restriction
estimates for each $\nu_{i}$ gives

$\displaystyle\big{\|}\hat{\phi}\big{\|}_{p}$
$\displaystyle\leqslant\big{\|}\hat{\phi}_{1}\big{\|}_{p}+\big{\|}\hat{\phi}_{2}\big{\|}_{p}$
$\displaystyle\leqslant(K/N)^{1/p}\left(\left\|\nu_{1}\right\|_{1}+\left\|\nu_{2}\right\|_{1}\right).$

Positivity gives that
$\left\|\nu_{1}\right\|_{1}+\left\|\nu_{2}\right\|_{1}=\left\|\nu_{1}+\nu_{2}\right\|_{1}$,
and the result then follows on taking $p$th powers. ∎

######  Lemma C.2.

Let $c_{1},\dots,c_{s}\in\mathbb{Z}\setminus\left\\{0\right\\}$,
$\delta\in(0,1)$ and suppose that $\nu_{1},\dots,\nu_{s}:[N]\to[0,\infty)$
each satisfy a $(s-\delta)$-restriction estimate with constant $K$. Then for
any $|f_{i}|\leqslant\nu_{i}$ we have

$\Biggl{|}\sum_{\mathbf{c}\cdot\mathbf{x}=0}\frac{f_{1}(x_{1})}{\left\|\nu_{1}\right\|_{1}}\dotsm\frac{f_{s}(x_{s})}{\left\|\nu_{s}\right\|_{1}}\Biggr{|}\leqslant\frac{K}{N}\min_{i}\bigg{\|}\frac{\hat{f}_{i}}{\left\|\nu_{i}\right\|_{1}}\biggr{\|}_{\infty}^{\delta}.$
(C.1)

###### Proof.

We prove the upper bound with $i=1$, the remaining cases following by re-
labelling indices. Let $p=s-\delta$. By orthogonality and Hölder’s inequality,
we have

$\displaystyle\Bigl{|}\sum_{\mathbf{c}\cdot\mathbf{x}=0}f_{1}(x_{1})\dotsm
f_{s}(x_{s})\Bigr{|}$
$\displaystyle=\Bigl{|}\int_{\mathbb{T}}\hat{f}_{1}(c_{1}\alpha)\dotsm\hat{f}_{s}(c_{s}\alpha)\mathrm{d}\alpha\Bigr{|}\leqslant\int_{\mathbb{T}}\Bigl{|}\hat{f}_{1}(c_{1}\alpha)\dotsm\hat{f}_{s}(c_{s}\alpha)\Bigr{|}\mathrm{d}\alpha$
$\displaystyle\leqslant\big{\|}\hat{f}_{1}\big{\|}_{\infty}^{\delta}\big{\|}\hat{f}_{1}\big{\|}_{p}^{1-\delta}\big{\|}\hat{f}_{2}\big{\|}_{p}\dotsm\big{\|}\hat{f}_{s}\big{\|}_{p}.$

Inequality (C.1) then follows from our $p$-restriction assumption. ∎

######  Lemma C.3 (Generalised von Neumann).

Let $c_{1},\dots,c_{s}\in\mathbb{Z}\setminus\left\\{0\right\\}$,
$\delta\in(0,1)$ and suppose that $\nu_{i},\mu_{i}:[N]\to[0,\infty)$ each
satisfy a $(s-\delta)$-restriction estimate with constant $K$. Then for any
$|f_{i}|\leqslant\nu_{i}$ and $|g_{i}|\leqslant\mu_{i}$ we have

$\Biggl{|}\sum_{\mathbf{c}\cdot\mathbf{x}=0}\left(\frac{f_{1}(x_{1})}{\left\|\nu_{1}\right\|_{1}}\dotsm\frac{f_{s}(x_{s})}{\left\|\nu_{s}\right\|_{1}}\
-\
\frac{g_{1}(x_{1})}{\left\|\mu_{1}\right\|_{1}}\dotsm\frac{g_{s}(x_{s})}{\left\|\mu_{s}\right\|_{1}}\right)\Biggr{|}\\\
\leqslant\frac{sK}{N}\max_{i}\bigg{\|}\frac{\hat{f}_{i}}{\left\|\nu_{i}\right\|_{1}}-\frac{\hat{g}_{i}}{\left\|\mu_{i}\right\|_{1}}\biggr{\|}_{\infty}^{\delta}.$

###### Proof.

Let $p=s-\delta$. By Lemma C.1, the weight

$\frac{\nu_{i}}{\left\|\nu_{i}\right\|_{1}}+\frac{\mu_{i}}{\left\|\mu_{i}\right\|_{1}}$

satisfies a $p$-restriction estimate with constant $K$ and majorises the
difference

$\frac{f_{i}}{\left\|\nu_{i}\right\|_{1}}-\frac{g_{i}}{\left\|\mu_{i}\right\|_{1}}.$

Observing that this weight has $L^{1}$ norm equal to two, the lemma follows on
applying the telescoping identity

$a_{1}\dotsm a_{s}-b_{1}\dotsm
b_{s}=\sum_{i=1}^{s}(a_{i}-b_{i})\prod_{j<i}a_{j}\prod_{j>i}b_{j},$

together with Lemma C.2. ∎

###  Appendix D Pointwise exponential sum estimates

The primary objective of this section is to establish the Fourier decay
estimates in Lemmas 6.3, 12.4 and 16.1. Of these, Lemma 12.4 concerns an
exponential sum over smooth numbers. As before, put $R=P^{\eta}$, with
$\eta=1$ when $k=2$ and $\eta=\eta_{k}$ a small positive number when
$k\geqslant 3$, and define $P$ and $X$ by (12.3). Our weight function $\nu$ is
defined by (12.8), with $k=2$ when dealing with Lemmas 6.3 and 16.1 as well as
$\xi=1$ in the latter scenario. This is consistent with (6.5) and (16.2). We
assume throughout that $X$ is sufficiently large in terms of $w$.

Our goal is to prove the inequality (12.12), using the Hardy–Littlewood circle
method. More explicitly, we wish to show that if ${\alpha}\in{\mathbb{T}}$
then

$\frac{\hat{\nu}({\alpha})}{\left\|\nu\right\|_{1}}=\frac{\hat{1}_{[X]}({\alpha})}{X}+O_{\eta}(w^{-1/k}).$
(D.1)

We treat the $k\geqslant 3$ and $k=2$ cases separately, as smooth numbers are
used for the former.

####  D.1. Smooth Weyl sums

We first consider the case $k\geqslant 3$, recalling that here we choose
$\eta=\eta_{k}$ sufficiently small. The idea is to consider a rational
approximation $a/q$ to $\alpha$; there will ultimately be four regimes to
consider, according to the size of $q$. We begin with a variant of [Vau89,
Lemma 5.4], which is useful for low height major arcs. Let

$S_{q,a}=\sum_{r{\,\,\mathrm{mod}\,\,q}}e\Bigl{(}\frac{a}{q}\cdot\frac{(Wr+\xi)^{k}-\xi^{k}}{kW}\Bigr{)},\qquad
I({\beta})=\int_{0}^{X}e({\beta}z){\mathrm{d}}z.$

######  Lemma D.1 (First level).

Suppose $q\in{\mathbb{N}}$ and $a\in{\mathbb{Z}}$, with $q\leqslant R/W$ and
$\|q{\alpha}\|=|q{\alpha}-a|$. Then

$\hat{\nu}({\alpha})=\rho(1/\eta)q^{-1}S_{q,a}I\Bigl{(}{\alpha}-\frac{a}{q}\Bigr{)}+O_{\eta}\Bigl{(}\frac{P^{k}}{\log
P}(q+P^{k}\|q{\alpha}\|)\Bigr{)}.$

###### Proof.

The start of the proof of [Vau89, Lemma 5.4] yields

$\sum_{\begin{subarray}{c}x\in S(m;R)\\\ x\equiv
Wr+\xi{\,\,\mathrm{mod}\,\,W}q\end{subarray}}1=\frac{1}{Wq}\sum_{x\in
S(m;R)}1+O\Bigl{(}\frac{P}{\log P}\Bigr{)},$

valid for $r\in[q]$ and $m\leqslant P$. Therefore

$\sum_{\begin{subarray}{c}x\in S(m;R)\\\
x\equiv\xi{\,\,\mathrm{mod}\,\,W}\end{subarray}}e\Bigl{(}\frac{a}{q}\cdot\frac{x^{k}-\xi^{k}}{kW}\Bigr{)}=\frac{S_{q,a}}{Wq}\sum_{x\in
S(m;R)}1+O\Bigl{(}\frac{qP}{\log P}\Bigr{)}.$

In particular, if ${\alpha}(x)$ equals
$e\Bigl{(}\frac{a}{q}\cdot\frac{x^{k}-\xi^{k}}{kW}\Bigr{)}$ when
$x\equiv\xi{\,\,\mathrm{mod}\,\,W}$ is $R$-smooth and 0 otherwise, then

$\sum_{x\leqslant
m}\Bigl{(}{\alpha}(x)-\frac{S_{q,a}}{Wq}\rho\Bigl{(}\frac{\log m}{\log
R}\Bigr{)}\Bigr{)}\ll\frac{qP}{\log P}.$

By partial summation and the boundedness of $\rho^{\prime}$, we also have

$\sum_{x\leqslant m}\rho\Bigl{(}\frac{\log x}{\log
R}\Bigr{)}=m\rho\Bigl{(}\frac{\log m}{\log R}\Bigr{)}+O\Bigl{(}\frac{P}{\log
P}\Bigr{)}\qquad(1\leqslant m\leqslant P),$

and so

$\sum_{x\leqslant
m}\Bigl{(}{\alpha}(x)-\frac{S_{q,a}}{Wq}\rho\Bigl{(}\frac{\log x}{\log
R}\Bigr{)}\Bigr{)}\ll\frac{qP}{\log P}.$

Next, observe that with ${\beta}={\alpha}-a/q$ we have
$|{\beta}|=q^{-1}\|q{\alpha}\|$ and

$\displaystyle\hat{\nu}({\alpha})$
$\displaystyle=\sum_{\begin{subarray}{c}x\in S(P;R)\\\
x\equiv\xi{\,\,\mathrm{mod}\,\,W}\end{subarray}}e\Bigl{(}\frac{a}{q}\cdot\frac{x^{k}-\xi^{k}}{kW}\Bigr{)}\phi(x)$
$\displaystyle=\frac{S_{q,a}}{Wq}\sum_{x\leqslant P}\rho\Bigl{(}\frac{\log
x}{\log R}\Bigr{)}\phi(x)+E,$ (D.2)

where

$\phi(x)=x^{k-1}e\Bigl{(}{\beta}\frac{x^{k}-\xi^{k}}{kW}\Bigr{)}$

and

$E=\sum_{x\leqslant
P}\Bigl{(}{\alpha}(x)-\frac{S_{q,a}}{Wq}\rho\Bigl{(}\frac{\log x}{\log
R}\Bigr{)}\Bigr{)}\phi(x).$

Partial summation gives

$E\ll\frac{qP}{\log
P}(\|\phi\|_{L^{\infty}([1,P])}+P\|\phi^{\prime}\|_{L^{\infty}([1,P])})\ll\frac{qP^{k}}{\log
P}(1+P^{k}|{\beta}|),$

and with the boundedness of $\rho^{\prime}$ it also implies that

$\sum_{x\leqslant P}\rho\Bigl{(}\frac{\log x}{\log
R}\Bigr{)}\phi(x)=\rho(1/\eta)\sum_{x\leqslant
P}\phi(x)+O\Bigl{(}\frac{P^{k}}{\log P}\Bigr{)}.$

Meanwhile, Euler–Maclaurin summation [Vau97, Eq. (4.8)] yields

$\displaystyle\sum_{x\leqslant P}\phi(x)$
$\displaystyle=\int_{1}^{P}\phi(x){\mathrm{d}}x+O(P^{k-1}(1+P^{k}|{\beta}|))$
$\displaystyle=WI({\beta})+O(P^{k-1}(1+P^{k}|{\beta}|)).$

Substituting these estimates into (D.2) concludes the proof.

∎

We supplement this by bounding $S_{q,a}$ and $I({\beta})$.

######  Lemma D.2.

If $(q,a)=1$ then $S_{1,0}=1$,

$S_{q,a}=0\qquad(2\leqslant q\leqslant w)$ (D.3)

and

$S_{q,a}\ll q^{1-1/k}.$ (D.4)

###### Proof.

Plainly $S_{1,0}=1$, so let $q\geqslant 2$, and let $a\in{\mathbb{Z}}$ with
$(q,a)=1$. The binomial expansion gives

$S_{q,a}=\sum_{r{\,\,\mathrm{mod}\,\,q}}e_{q}\Biggl{(}a\sum_{\ell=1}^{k}\frac{{k\choose\ell}W^{\ell-1}}{k}\xi^{k-\ell}r^{\ell}\Biggr{)},$

and we note that $\frac{{k\choose\ell}W^{\ell-1}}{k}\in{\mathbb{Z}}$
($1\leqslant\ell\leqslant k$). Write $q=uv$, where $u$ is $w$-smooth and
$(v,W)=1$. Since $(u,v)=1$, a standard calculation reveals that

$S_{q,a}=S_{u,a_{1}}S_{v,a_{2}},$ (D.5)

where $a_{1}=av^{-1}\in({\mathbb{Z}}/u{\mathbb{Z}})^{\times}$ and
$a_{2}=au^{-1}\in({\mathbb{Z}}/v{\mathbb{Z}})^{\times}$ (see [Vau97, Lemma
2.10]).

Put $u=hu^{\prime}$, where $h=(u,W/k)$. Representing
$r{\,\,\mathrm{mod}\,\,q}$ as $r=r_{1}+u^{\prime}r_{2}$, where $0\leqslant
r_{1}<u^{\prime}$ and $0\leqslant r_{2}<h$, gives

$\displaystyle S(u,a_{1})$ $\displaystyle=\sum_{\begin{subarray}{c}0\leqslant
r_{1}<u^{\prime}\\\ 0\leqslant
r_{2}<h\end{subarray}}e_{hu^{\prime}}\Biggl{(}a_{1}\sum_{\ell=1}^{k}{k\choose\ell}\frac{{k\choose\ell}W^{\ell-1}}{k}\xi^{k-\ell}(r_{1}+u^{\prime}r_{2})^{\ell}\Biggr{)}$
$\displaystyle=\sum_{r_{1}=0}^{u^{\prime}-1}e_{hu^{\prime}}\Biggl{(}a_{1}\sum_{\ell=1}^{k}\frac{{k\choose\ell}W^{\ell-1}}{k}\xi^{k-\ell}r_{1}^{\ell}\Biggr{)}$
$\displaystyle\qquad\cdot\sum_{r_{2}=0}^{h-1}e_{h}\Biggl{(}a_{1}\sum_{\ell=1}^{k}\frac{{k\choose\ell}W^{\ell-1}}{k}\xi^{k-\ell}(u^{\prime})^{\ell-1}r_{2}^{\ell}\Biggr{)}.$

As $h$ divides $W/k$, the inner sum is

$\sum_{r_{2}{\,\,\mathrm{mod}\,\,h}}e_{h}(a_{1}\xi^{k-1}r_{2}),$

which vanishes unless $h\mid a_{1}\xi^{k-1}$. As $(h,a_{1})=(h,\xi)=1$, and as

$(u,W/k)=1\Leftrightarrow(u,W)=1\Leftrightarrow u=1,$

we conclude that

$S_{u,a_{1}}=\begin{cases}0&\text{if }u\neq 1\\\ 1&\text{if }u=1.\end{cases}$
(D.6)

Moreover, note that if $2\leqslant q\leqslant w$ then $u=q$ and $v=1$. Now
(D.5) and (D.6) complete the proof of (D.3).

Next we prove (D.4). By (D.5) and (D.6), we may assume $u=1$. Consider

$e_{kWv}(a_{2}\xi^{k})S_{v,a_{2}}=\sum_{r{\,\,\mathrm{mod}\,\,v}}e_{v}\Bigl{(}a_{2}\frac{(Wr+\xi)^{k}}{kW}\Bigr{)}.$

As $(v,W)=1$, we can change variables by $t=\xi
W^{-1}+r\in{\mathbb{Z}}/v{\mathbb{Z}}$, which gives

$e_{kWv}(a_{2}\xi^{k})S_{v,a_{2}}=\sum_{t{\,\,\mathrm{mod}\,\,v}}e_{v}\Bigl{(}a_{2}\frac{W^{k-1}}{k}t^{k}\Bigr{)}.$

Since $\Bigl{(}v,a_{2}\frac{W^{k-1}}{k}\Bigr{)}=1$, we may apply [Vau97,
Theorem 4.2], which gives

$S_{v,a_{2}}\ll v^{1-1/k}=q^{1-1/k}.$

By (D.5) and (D.6), we now have $S_{q,a}\ll q^{1-1/k}$. ∎

A standard calculation provides the following bound.

######  Lemma D.3.

We have

$I({\beta})\ll\min\\{X,\|{\beta}\|^{-1}\\}.$

Before continuing in earnest, we briefly describe the plan. We can modify
[Vau89, Theorem 1.8] to handle a set of minor arcs. At that stage, our major
and minor arcs fail to cover the entire torus ${\mathbb{T}}$, but we can
bridge the gap using a classical circle method contraption known as _pruning_
(also used in Appendix B). Adapting [VW91, Lemma 7.2], we can prune down to
$q\leqslant(\log P)^{A}$. Finally, by adapting [VW91, Lemma 8.5], we prune
down to $q\leqslant(\log P)^{1/4}$.

In order to tailor the classical theory to suit our needs, we begin with the
observation that

$\displaystyle\sum_{\begin{subarray}{c}x\in S(m;R)\\\
x\equiv\xi{\,\,\mathrm{mod}\,\,W}\end{subarray}}e\Bigl{(}{\alpha}\frac{x^{k}-\xi^{k}}{kW}\Bigr{)}=$
$\displaystyle\qquad\frac{1}{W}\sum_{t{\,\,\mathrm{mod}\,\,W}}e\Bigl{(}-\frac{{\alpha}}{kW}\xi^{k}-\frac{t}{W}\xi\Bigr{)}\sum_{x\in
S(m;R)}e\Bigl{(}\frac{{\alpha}}{kW}x^{k}+\frac{t}{W}x\Bigr{)}.$ (D.7)

The inner summation is a classical quantity with a linear twist.

######  Lemma D.4 (Minor arcs).

Suppose $0<{\delta}<(2k)^{-1}$, and let ${\mathfrak{m}}_{1}$ denote the set of
real numbers ${\gamma}$ with the property that if $a\in{\mathbb{Z}}$,
$q\in{\mathbb{N}}$, $(a,q)=1$ and $|q{\gamma}-a|\leqslant
P^{\frac{1}{2}-k+{\delta}k}$ then $q>P^{\frac{1}{2}+{\delta}k}$. Put

$\iota(k)=\max_{{\lambda}\in{\mathbb{Z}}_{\geqslant
2}}\frac{1}{4{\lambda}}(1-(k-2)(1-1/k)^{{\lambda}-2}).$

Then, assuming $\eta\leqslant\eta_{0}(\varepsilon,k)$, we have

$\hat{\nu}({\alpha})\ll
P^{k+\varepsilon}(P^{-{\delta}}+P^{-\iota(k)})\qquad({\alpha}\in
kW{\mathfrak{m}}_{1}).$

######  Remark D.5.

We will later apply this with $\varepsilon=\varepsilon_{k}$, so that the
condition $\eta\leqslant\eta_{0}(\varepsilon,k)$ will be met.

###### Proof.

Following the proof of [Vau89, Theorem 1.8], we find that if
${\alpha}\in{\mathfrak{m}}_{1}$ and $1\leqslant m\leqslant P$ then

$\sum_{x\in S(m;R)}e\Bigl{(}\frac{{\alpha}}{kW}x^{k}+\frac{t}{W}x\Bigr{)}\ll
P^{1+\varepsilon}(P^{-{\delta}}+P^{-\iota(k)}).$

Indeed, already built into that proof are bounds uniform over linear twists;
see [Vau89, Eq. (10.9)]. The sum above is over $x\in S(m;R)$, where
$1\leqslant m\leqslant P$, rather than over $x\in S(P;R)$, however we can
assume that $\sqrt{P}\leqslant m\leqslant P$ and then run Vaughan’s argument.

Now, by (D.7), we have

$\sum_{\begin{subarray}{c}x\in S(m;R)\\\
x\equiv\xi{\,\,\mathrm{mod}\,\,W}\end{subarray}}e\Bigl{(}{\alpha}\frac{x^{k}-\xi^{k}}{kW}\Bigr{)}\ll
P^{1+\varepsilon}(P^{-{\delta}}+P^{-\iota(k)}).$

From here, partial summation gives

$\hat{\nu}({\alpha})=\sum_{\begin{subarray}{c}x\in S(P;R)\\\
x\equiv\xi{\,\,\mathrm{mod}\,\,W}\end{subarray}}x^{k-1}e\Bigl{(}{\alpha}\frac{x^{k}-\xi^{k}}{kW}\Bigr{)}\ll
P^{k+\varepsilon}(P^{-{\delta}}+P^{-\iota(k)}).$

∎

######  Lemma D.6 (First pruning step).

Suppose $R\leqslant M\leqslant P$, where $R=P^{\eta}$ as before. Suppose
$a\in{\mathbb{Z}}$, $q\in{\mathbb{N}}$ with $(a,q)=1$ and
$|q{\alpha}-a|\leqslant M/(P^{k}R)$. Then for any $\varepsilon>0$ we have

$\displaystyle\hat{\nu}({\alpha})\ll_{\varepsilon,k,W,\eta}$ $\displaystyle
P^{k}(\log P)^{3}q^{\varepsilon}$
$\displaystyle\quad\cdot((q+P^{k}|q{\alpha}-a|)^{-1/(2k)}+(MR/P)^{1/2}+q^{\frac{1}{2}-\frac{1}{2k}}(R/M)^{1/2}).$

###### Proof.

By partial summation and (D.7), it suffices to show that if $\sqrt{P}\leqslant
m\leqslant P$ then

$\displaystyle\sum_{x\in
S(m;R)}e\Bigl{(}\frac{{\alpha}}{kW}x^{k}+\frac{t}{W}x\Bigr{)}$
$\displaystyle\quad\ll P(\log
P)^{3}q^{\varepsilon}((q+P^{k}|q{\alpha}-a|)^{-1/(2k)}+(MR/P)^{1/2}+q^{\frac{1}{2}-\frac{1}{2k}}(R/M)^{1/2}).$

To show this, we work through the proof of [VW91, Lemma 7.2]; the inner sum of
Eq. (7.4) therein becomes

$S=\sum_{y\in
I\cap{\mathbb{Z}}}e\Bigl{(}\frac{{\alpha}}{kW}p^{k}(u_{2}^{k}-u_{1}^{k})y^{k}+\frac{t}{W}p(u_{2}-u_{1})y\Bigr{)},$

where

$I=(V/p,\min\\{2V/p,m/(u_{1}p),m/(u_{2}p)\\}]$

is an interval of length at most $V/p$ and $u_{1},u_{2},p$ are the outer
summation variables in [VW91, Eq. (7.4)]. With reference to that proof, we
have

$D=\gcd(kWq,ap^{k}(u_{2}^{k}-u_{1}^{k}),tqkp(u_{2}-u_{1}))\ll_{k,W}(q,p^{k}(u_{2}^{k}-u_{1}^{k})),$

and

$S=\sum_{y\in
I\cap{\mathbb{Z}}}e_{q^{\prime}}(a^{\prime}y^{k}+b^{\prime}y)e({\beta}y^{k}),$

where

$Dq^{\prime}=kWq,\quad Da^{\prime}=ap^{k}(u_{1}^{k}-u_{2}^{k}),\quad
Db^{\prime}=tkqp(u_{2}-u_{1}),$

$\gcd(a^{\prime},b^{\prime},q^{\prime})=1$ and

${\beta}=\frac{p^{k}(u_{2}^{k}-u_{1}^{k})}{kW}({\alpha}-a/q).$

Continuing to follow the proof of [VW91, Lemma 7.2], we now apply [Bak86,
Lemma 4.4] and [Vau97, Theorems 7.1 and 7.3] in lieu of the more specific
[Vau97, Lemma 2.8 and Theorems 4.1 and 4.2]. One can check that

$|q{\beta}|\leqslant(2k^{2})^{-1}(V/p)^{1-k}.$

As $|I|\leqslant V/p$, this condition enables us to apply [Bak86, Lemma 4.4],
giving

$S=(q^{\prime})^{-1}\sum_{x\leqslant
q^{\prime}}e_{q^{\prime}}(a^{\prime}x^{k}+b^{\prime}x)\int_{I}e({\beta}z^{k}){\mathrm{d}}z+O(q^{1-\frac{1}{k}+\varepsilon}).$

The error term $RU^{2}q^{\frac{1}{2}+\varepsilon}$ in [VW91, Eq. (7.5)] is
enlarged to $RU^{2}q^{1-\frac{1}{k}+\varepsilon}$, and the effect of applying
[Vau97, Theorems 7.1 and 7.3] is to increase the quantity $S_{3}$ appearing
therein by a multiplicative factor of
$O_{\varepsilon,k,W}(q^{\varepsilon/8})$.

The remainder of the proof of [VW91, Lemma 7.2] carries through in the present
context, _mutatis mutandis_. The eventual outcome of the changes above is to
increase the term $q^{1/4}P(R/M)^{1/2}$ to
$q^{\frac{1}{2}-\frac{1}{2k}}P(R/M)^{1/2}$, and we obtain the asserted bound.
∎

######  Lemma D.7 (Second pruning step).

Suppose $R=P^{\eta}$ with $0<\eta<1/2$, and that $a,q\in{\mathbb{Z}}$ with
$(a,q)=1$ and $1\leqslant q\leqslant(\log P)^{A}$. Then for some $c=c(\eta,A)$
we have

$\hat{\nu}({\alpha})\ll_{\varepsilon,k,w,\eta,A,c}P^{k}(q+P^{k}|q{\alpha}-a|)^{\varepsilon-\frac{1}{k}}+P^{k}\cdot\exp(-c\sqrt{\log
P})(1+P^{k}|{\alpha}-a/q|).$

###### Proof.

Again we apply partial summation and (D.7), leaving us to show that if
$P^{0.99}\leqslant m\leqslant P$ then

$\displaystyle g({\alpha})$ $\displaystyle:=\sum_{x\in
S(m;R)}e\Bigl{(}\frac{{\alpha}}{kW}x^{k}+\frac{t}{W}x\Bigr{)}$
$\displaystyle\ll
P(q+P^{k}|q{\alpha}-a|)^{\varepsilon-\frac{1}{k}}+P\cdot\exp(-c\sqrt{\log
P})(1+P^{k}|{\alpha}-a/q|).$

This time we follow the proof of [VW91, Lemma 8.5]. Writing
${\beta}={\alpha}-a/q$, this initially formats our smooth Weyl sum as

$\displaystyle g({\alpha})=\sum_{\begin{subarray}{c}d\mid kWq\\\ q/d\in
S(m;R)\end{subarray}}\sum_{\begin{subarray}{c}y=1\\\
(y,d)=1\end{subarray}}^{d}$ $\displaystyle e((kWq/d)^{k-1}y^{k}a/d+tkqy/d)$
$\displaystyle\quad\cdot\Psi\Bigl{(}\frac{md}{kWq},R;d,y,\frac{{\beta}(kWq/d)^{k}}{kW}\Bigr{)},$

where

$\Psi(Q,R;d,y,{\gamma})=\sum_{\begin{subarray}{c}z\in S(Q;R)\\\ z\equiv
y{\,\,\mathrm{mod}\,\,d}\end{subarray}}e({\gamma}z^{k}).$

The calculation by Vaughan and Wooley in the proof of [VW91, Lemma 8.5]
ensures that $\Psi(Q,R;d,y,{\gamma})$ is, up to a small additive error,
independent of $y$. As $m\leqslant P$ and $kW\ll_{k,W}1$, the outcome of this
calculation is unaffected, and we obtain

$\displaystyle g({\alpha})$ $\displaystyle\ll P(1+P^{k}|{\beta}|)^{-1/k}$
$\displaystyle\qquad\cdot\Bigl{(}\exp(-c\sqrt{\log P})+\sum_{d\mid
kWq}\frac{d}{q\varphi(d)}|{\mathcal{W}}(d,a(kWq/d)^{k-1},tkq)|\Bigr{)},$

where

${\mathcal{W}}(Q,A,B)=\sum_{\begin{subarray}{c}y{\,\,\mathrm{mod}\,\,Q}\\\
(y,Q)=1\end{subarray}}e_{Q}(Ay^{k}+By).$

Our final task is to show that if $d\mid kWq$ then

${\mathcal{W}}(d,a(kWq/d)^{k-1},tkq)\ll_{k,w,\varepsilon}q^{1-\frac{1}{k}+\varepsilon}.$

One may readily verify the usual multiplicativity property: if
$(Q_{1},Q_{2})=1$ then

${\mathcal{W}}(Q_{1}Q_{2},A,B)={\mathcal{W}}(Q_{1},AQ_{2}^{k-1},B)\cdot{\mathcal{W}}(Q_{2},AQ_{1}^{k-1},B);$
(D.8)

see [Hua65, Lemma 8.1]. Next we analyse

${\mathcal{W}}(p^{i},A,tkq),$

when $p$ is prime and $p^{i}\|d$. If $p>w$ then $p^{i}\mid tkq$, so

${\mathcal{W}}(p^{i},A,tkq)={\mathcal{W}}(p^{i},A,0)\ll
p^{i/2}(p^{i},A)^{1/2},$

using [VW91, Lemma 8.4]. Meanwhile, if $p\leqslant w$ then we use the identity

${\mathcal{W}}(p^{i},A,tkq)=S(p^{i},A,tkq)-S(p^{i-1},Ap^{k-1},tkq),$

where

$S(Q,A,B)=\sum_{y\leqslant Q}e_{Q}(Ay^{k}+By).$

Since $p\leqslant w$, we have

$\gcd(p^{i},A,tkq),\>\gcd(p^{i-1},Ap^{k-1},tkq)\ll_{k,w}(p^{i},A),$

so we may use [Vau97, Eq. (7.9)] to infer that

${\mathcal{W}}(p^{i},A,tkq)\ll_{k,w}(p^{i},A)^{1/k}(p^{i})^{1-\frac{1}{k}}.$

In both cases we have

$|{\mathcal{W}}(p^{i},A,tkq)|\leqslant
c_{k,w}(p^{i},A)^{1/k}(p^{i})^{1-\frac{1}{k}},$

and inputting this into (D.8) reveals that

${\mathcal{W}}(d,A,tkq)\ll q^{\varepsilon}(d,A)^{1/k}d^{1-\frac{1}{k}}.$

Apply this with $A=a(kWq/d)^{k-1}$. With this choice of $A$, we have

$\displaystyle(d,A)\leqslant(d,a)(d,(kWq/d)^{k-1})$
$\displaystyle\leqslant(kWq,a)(d,(kWq/d)^{k-1})$ $\displaystyle\leqslant
kW(d,(kWq/d)^{k-1}).$

Letting $p^{i}\|d$ and $p^{j}\|kWq$ gives

$(d,A)\ll_{k,w}\prod_{p}p^{\min\\{i,(k-1)(j-i)\\}},$

and so

$\displaystyle{\mathcal{W}}(d,a(kWq/d)^{k-1},tkq)$ $\displaystyle\ll
q^{\varepsilon}\prod_{\begin{subarray}{c}p^{i}\|d\\\
p^{j}\|kWq\end{subarray}}p^{i(1-\frac{1}{k})+k^{-1}\min\\{i,(k-1)(j-i)\\}}$
$\displaystyle\ll q^{\varepsilon}\prod_{p^{j}\|kWq}(p^{j})^{1-\frac{1}{k}}\ll
q^{1-\frac{1}{k}+\varepsilon}.$

∎

To tie together what we have gleaned, we make a Hardy–Littlewood dissection.
For $q\in{\mathbb{N}}$ and $a\in{\mathbb{Z}}$, let ${\mathfrak{M}}(q,a)$ be
the set of ${\alpha}\in{\mathbb{T}}$ such that $|{\alpha}-a/q|\leqslant(\log
P)^{1/4}/P^{k}$. Let ${\mathfrak{M}}(q)$ be the union of the sets
${\mathfrak{M}}(q,a)$ over integers $a$ such that $(a,q)=1$, and let
${\mathfrak{M}}$ be the union of the sets ${\mathfrak{M}}(q)$ over
$q\leqslant(\log P)^{1/4}$. By identifying ${\mathbb{T}}$ with a unit
interval, we may write ${\mathfrak{M}}(q)$ as a disjoint union

${\mathfrak{M}}(q)=\bigcup_{\begin{subarray}{c}a=0\\\
(a,q)=1\end{subarray}}^{q-1}{\mathfrak{M}}(q,a).$

First we consider the minor arcs
${\mathfrak{m}}:={\mathbb{T}}\setminus{\mathfrak{M}}$.

######  Lemma D.8.

If $\varepsilon>0$ and ${\alpha}\in{\mathfrak{m}}$ then
$\hat{\nu}({\alpha})\ll_{\varepsilon,W,\eta}X(\log
X)^{\varepsilon-\frac{1}{4k}}$.

###### Proof.

Let ${\alpha}\in{\mathfrak{m}}$. If
$\frac{{\alpha}}{kW}\in{\mathfrak{m}}_{1}$, where ${\mathfrak{m}}_{1}$ is as
in Lemma D.4 with ${\delta}=(4k)^{-1}$, then Lemma D.4 applies and is more
than sufficient (recall (12.3)). We may therefore assume that
$\frac{{\alpha}}{kW}\notin{\mathfrak{m}}_{1}$, and then deduce the existence
of relatively prime integers $q>0$ and $a$ for which $q+P^{k}|q{\alpha}-a|\ll
P^{3/4}$. If the ‘natural height’ $q+P^{k}|q{\alpha}-a|$ exceeds $(\log
P)^{9k}$, then an application of Lemma D.6 with $M\asymp RP^{3/4}$ suffices.
So we may suppose instead that $q+P^{k}|q{\alpha}-a|\leqslant(\log P)^{9k}$.
As ${\alpha}\notin{\mathfrak{M}}$, we must also have

$q+P^{k}|q{\alpha}-a|\geqslant\max\Bigl{\\{}q,P^{k}\Bigl{|}{\alpha}-\frac{a}{q}\Bigr{|}\Bigr{\\}}>(\log
P)^{1/4},$

and now Lemma D.7 delivers the sought inequality. ∎

We are ready to prove Lemma 12.4, in the case $k\geqslant 3$. As discussed at
the beginning of this appendix, our task is to establish the estimate (D.1).
It will be useful to have (12.3) and (12.9) in mind. By a geometric series
calculation, we have

$\widehat{1_{[X]}}({\alpha})=\sum_{x\leqslant
X}e({\alpha}x)\ll\|{\alpha}\|^{-1}.$ (D.9)

First suppose ${\alpha}\in{\mathfrak{m}}$. By Dirichlet’s approximation
theorem, we obtain relatively prime integers $q$ and $a$ such that

$1\leqslant q\leqslant(\log P)^{1/4},\qquad|q{\alpha}-a|\leqslant(\log
P)^{-1/4}.$

As ${\alpha}\notin{\mathfrak{M}}$, we must have

$\|q{\alpha}\|=|q{\alpha}-a|>\frac{q(\log P)^{1/4}}{kWX},$

so

$\widehat{1_{[X]}}({\alpha})\ll\|{\alpha}\|^{-1}\ll\frac{q}{\|q{\alpha}\|}\ll\frac{WX}{(\log
P)^{1/4}}.$

By Lemma D.8, we now have (D.1).

Next we consider the case in which $q=1$ and ${\alpha}\in{\mathfrak{M}}(q)$,
in other words $|{\alpha}|\leqslant(\log P)^{1/4}/P^{k}$. By Lemma D.1, we
have

$\hat{\nu}({\alpha})-\rho(1/\eta)I({\alpha})\ll\frac{P^{k}}{\log
P}(1+P^{k}\|{\alpha}\|)\ll\frac{P^{k}}{\sqrt{\log P}}.$ (D.10)

By Euler–Maclaurin summation [Vau97, Eq. (4.8)], we have

$\widehat{1_{[X]}}({\alpha})-I({\alpha})\ll 1+X\|{\alpha}\|\ll\sqrt{\log P}.$
(D.11)

Coupling (D.10) with (D.11) yields

$\hat{\nu}({\alpha})-\rho(1/\eta)\widehat{1_{[X]}}({\alpha})\ll\frac{P^{k}}{\sqrt{\log
P}}\ll Xw^{-1/k},$

and now (12.9) confirms (D.1).

Finally, let ${\alpha}\in{\mathfrak{M}}(q,a)$ with $2\leqslant q\leqslant(\log
P)^{1/4}$ and $(a,q)=1$, and put

${\beta}={\alpha}-\frac{a}{q}\in\Bigl{[}-\frac{(\log
P)^{1/4}}{P^{k}},\frac{(\log P)^{1/4}}{P^{k}}\Bigr{]}.$

Substituting

$\|{\alpha}\|\geqslant q^{-1}-|{\beta}|\geqslant q^{-1}-\frac{(\log
P)^{1/4}}{P^{k}}\geqslant\frac{1}{2q}$

into (D.9) gives

$\widehat{1_{[X]}}({\alpha})\ll q\ll(\log P)^{1/4}.$

By Lemma D.1, we also have

$\hat{\nu}({\alpha})\ll\frac{P^{k}}{\sqrt{\log P}}+X|q^{-1}S_{q,a}|,$

and now Lemma D.2 yields (D.1).

We have established (D.1) for all ${\alpha}\in{\mathbb{T}}$, assuming
$k\geqslant 3$.

####  D.2. Quadratic Weyl sums

The purpose of this subsection will be a proof of Lemmas 6.3 and 16.1,
together with the $k=2$ case of Lemma 12.4. In all of these cases $k=2$, so
$\eta=1$, and the weight function is simpler, namely

$\nu(n)=\begin{cases}x,&\text{if }n=\frac{x^{2}-\xi^{2}}{2W}\text{ for some
}x\in[P]\text{ with }x\equiv\xi\bmod W\\\ 0,&\text{otherwise.}\end{cases}$

For the Fourier transform of this weight function, we can obtain a power
saving on the minor arcs, as in [BP17]. This will be used in the next
appendix, in the proof of the restriction estimate. We keep this brief, as the
analysis is essentially the same as that of [BP17].

As discussed at the beginning of this appendix, we seek to establish (D.1).
The Fourier transform is given by

$\hat{\nu}({\alpha})=\sum_{\begin{subarray}{c}x\leqslant P\\\
x\equiv\xi{\,\,\mathrm{mod}\,\,W}\end{subarray}}xe\Bigl{(}{\alpha}\frac{x^{2}-\xi^{2}}{2W}\Bigr{)}.$

The following is a straightforward adaptation of [BP17, Lemma 5.1].

######  Lemma D.9 (Major arc asymptotic).

Suppose that $\|q{\alpha}\|=|q{\alpha}-a|$ for some $q,a\in{\mathbb{Z}}$ with
$q>0$. Then

$\hat{\nu}({\alpha})=q^{-1}S_{q,a}I\Bigl{(}{\alpha}-\frac{a}{q}\Bigr{)}+O_{w}(\sqrt{X}(q+X\|q{\alpha}\|)).$

Lemmas D.2 and D.3 still hold when $k=2$, with the same proof.

Following [BP17], put $\tau=\frac{1}{100}$, and to each reduced fraction $a/q$
with $0\leqslant a<q\leqslant X^{\tau}$ associate a major arc

${\mathfrak{M}}_{2}(q,a)=\\{{\alpha}\in{\mathbb{T}}:\Bigl{|}{\alpha}-\frac{a}{q}\Bigr{|}\leqslant
X^{\tau-1}\\}.$

Let ${\mathfrak{M}}_{2}$ denote the union of all major arcs, and define the
minor arcs by ${\mathfrak{m}}_{2}={\mathbb{T}}\setminus{\mathfrak{M}}_{2}$.
The following is a straightforward adaptation of [BP17, Eq. (5.3)].

######  Lemma D.10.

If $\varepsilon>0$ and ${\alpha}\in{\mathfrak{m}}_{2}$ then

$\hat{\nu}({\alpha})\ll_{\varepsilon}X^{1-\frac{\tau}{2}+\varepsilon}.$

We proceed towards (D.1). Let ${\alpha}\in{\mathbb{T}}$. By (6.6), it suffices
to prove that

$\hat{\nu}({\alpha})-\widehat{1_{[X]}}({\alpha})\ll Xw^{-1/2}.$ (D.12)

First suppose ${\alpha}\in{\mathfrak{m}}_{2}$. As in the proof of [BP17, Lemma
5.5], we have

$\widehat{1_{[X]}}({\alpha})\ll X^{1-\tau}.$

Pairing this with Lemma D.10 yields (D.12).

Next, suppose ${\alpha}\in{\mathfrak{M}}_{2}(q,a)$ for some coprime
$q,a\in{\mathbb{Z}}$ with $0\leqslant a<q\leqslant N^{\tau}$, where
$q\geqslant 2$. Lemmas D.2, D.3 and D.9 give

$\hat{\nu}({\alpha})\ll Xw^{-1/2}.$

Meanwhile $\|{\alpha}\|\geqslant(2q)^{-1}$, so

$\widehat{1_{[X]}}({\alpha})\ll\|{\alpha}\|^{-1}\ll q\ll X^{\tau},$

and now the triangle inequality yields (D.12).

Finally, when $q=1$ and ${\alpha}\in{\mathfrak{M}}_{2}(1,0)$, Lemma D.9 gives

$\hat{\nu}({\alpha})-I({\alpha})\ll_{w}X^{\frac{1}{2}+2\tau},$

and Euler–Maclaurin summation gives

$\widehat{1_{[X]}}({\alpha})-I({\alpha})\ll X^{2\tau}.$

The triangle inequality now furnishes (D.12).

We have examined all cases, thereby completing the proofs of Lemmas 6.3, 12.4
and 16.1.

###  Appendix E Restriction estimates

In this section we prove the restriction estimates claimed in Lemmas 6.4, 12.5
and 16.2. The core elements of our setup are the same as in Appendix D, but we
repeat all of this for clarity. Put $R=P^{\eta}$, and define $P$ and $X$ by
(12.3). In the cases of Lemmas 6.3 and 16.1 let $\eta=1$ and $k=2$, and
$\xi=1$ in the latter scenario. Our weight function $\nu$ is defined by
(12.8). When $k\geqslant 3$, we choose $\eta=\eta_{k}$ sufficiently small. We
assume throughout that $X$ is sufficiently large in terms of $w$.

Let $\phi:{\mathbb{Z}}\to{\mathbb{C}}$ with $|\phi|\leqslant\nu$ pointwise.
For an appropriate restriction exponent $p$, our task is to establish the
restriction inequality

$\int_{\mathbb{T}}\left|\hat{\phi}(\alpha)\right|^{p}\mathrm{d}\alpha\ll
X^{p-1}.$ (E.1)

The implied constant, in particular, will not depend on $w$. As

$\|\hat{\phi}\|_{\infty}\leqslant\|\phi\|_{1}\leqslant\|\nu\|_{1}\ll X,$ (E.2)

it suffices to show this when

$p=\begin{cases}5-\frac{1}{200},&\text{if }k=2\\\
s_{0}(k)+2-\frac{1}{200},&\text{if }k\geqslant 4\\\ 8-10^{-8},&\text{if
}k=3,\\\ \end{cases}$ (E.3)

where $s_{0}(k)\in{\mathbb{N}}$ is as in Theorem 1.3. Fix this choice of $p$.

To summarise what is written above, we seek to establish the restriction
inequality (E.1) when the exponent $p$ is given by (E.3). This will prove
Lemmas 6.4, 12.5 and 16.2 at one fell swoop.

Even moments play a key role, owing to the presence of an underlying
Diophantine equation. In particular, they allow bounded weights to be freely
removed. Let $2m$ be the greatest even integer strictly less than $p$.

######  Lemma E.1.

It holds that

$\int_{\mathbb{T}}|\hat{\phi}({\alpha})|^{2m}{\mathrm{d}}{\alpha}\ll_{k,\varepsilon}\begin{cases}(WX)^{2m-1},&\text{if
}k\geqslant 4\\\ X^{2m-1+\varepsilon},&\text{if }k=2\\\
P^{15.25-10^{-4}},&\text{if }k=3.\end{cases}$

######  Remark E.2.

The sixth moment estimate, for the case $k=3$, has a slightly different
flavour; it is a consequence of Wooley’s ‘subconvex’ mean value estimate
[Woo95]. It is this that ultimately enables us to procure a $p$-restriction
estimate with $p<8$.

###### Proof.

By orthogonality and the triangle inequality

$\int_{\mathbb{T}}|\hat{\phi}({\alpha})|^{2m}{\mathrm{d}}{\alpha}\ll
P^{2m(k-1)}{\mathcal{N}},$

where ${\mathcal{N}}$ is the number of solutions
$({\mathbf{x}},{\mathbf{y}})\in S(P;P^{\eta})^{m}\times S(P;P^{\eta})^{m}$ to
the Diophantine equation

$x_{1}^{k}+\cdots+x_{m}^{k}=y_{1}^{k}+\cdots+y_{m}^{k}.$

Note that adding a constant to $s_{0}(k)$ in the case $k\geqslant 4$ does not
cause it to violate (1.3), and so we may assume that $2m\geqslant s_{0}(k)$
for the quantity $s_{0}(k)$ appearing in Theorem B.1. For $k\geqslant 4$ we
therefore have, by Theorem B.1, that

$\int_{\mathbb{T}}|\hat{\phi}({\alpha})|^{2m}{\mathrm{d}}{\alpha}\ll
P^{2m(k-1)}P^{2m-k}=P^{k(2m-1)}\ll(WX)^{2m-1}.$

The case $k=2$ is similar, as the crude bound
${\mathcal{N}}\ll_{\varepsilon}P^{2+\frac{\varepsilon}{2}}$ is standard. When
$k=3$ the proof may be concluded using [Woo95, Theorem 1.2], which implies
that ${\mathcal{N}}\ll P^{3.25-10^{-4}}$. ∎

These estimates fall short of being sharp. By increasing the exponent, we are
able to make them sharp, using Bourgain’s epsilon-removal procedure [Bou89].
In the case $k=3$, an additional intermediate exponent is required.

####  E.1. Epsilon-removal

In this subsection we assume that $k\neq 3$. The case $k=3$ is treated in the
next subsection by incorporating a small finesse. Denote by ${\delta}$ a
parameter in the range

$0<{\delta}\ll 1.$

Define the large spectra

$\mathcal{R}_{\delta}=\\{{\alpha}\in{\mathbb{T}}:|\hat{\phi}({\alpha})|>{\delta}X\\},$

and note from (E.2) that $\mathcal{R}_{\delta}$ is empty unless ${\delta}\ll
1$. By the dyadic pigeonholing argument in [BP17, §6], it suffices to prove
that

$\mathrm{meas}(\mathcal{R}_{\delta})\ll\frac{1}{{\delta}^{p-10^{-8}}X}.$ (E.4)

Moreover, Lemma E.1 ensures that

$({\delta}X)^{2m}\mathrm{meas}(\mathcal{R}_{\delta})\leqslant\int_{\mathbb{T}}|\hat{\phi}({\alpha})|^{2m}{\mathrm{d}}{\alpha}\ll_{k,\varepsilon}\begin{cases}(WX)^{2m-1},&\text{if
}k\geqslant 4\\\ X^{3+\frac{\varepsilon}{2}},&\text{if }k=2,\end{cases}$

so we may assume without loss that

${\delta}>\begin{cases}W^{2(1-2m)},&\text{if }k\geqslant 4\\\
X^{-\varepsilon},&\text{if }k=2,\end{cases}$ (E.5)

for any $\varepsilon>0$. Let ${\theta}_{1},\ldots,{\theta}_{R}$ be
$X^{-1}$-spaced points in $\mathcal{R}_{\delta}$. As

$p-10^{-8}\geqslant 2k+0.3,$

it suffices to show that

$R\ll{\delta}^{-2k-0.3}.$ (E.6)

Put ${\gamma}=k+0.1$. By the calculation in [BP17, §6], we have

${\delta}^{2{\gamma}}X^{\gamma}R^{2}\ll\sum_{1\leqslant r,r^{\prime}\leqslant
R}|\hat{\nu}({\theta}_{r}-{\theta}_{r}^{\prime})|^{\gamma}.$ (E.7)

First suppose $k\geqslant 4$. Consider
${\theta}={\theta}_{r}-{\theta}^{\prime}_{r}$ in the summand on the right-hand
side of (E.7). By Lemma D.8, the contribution from ${\theta}\in{\mathfrak{m}}$
is

$O(R^{2}(X(\log X)^{-1/(8k)})^{\gamma}),$

and by (E.5) this is $o({\delta}^{2{\gamma}}X^{\gamma}R^{2})$. Hence

${\delta}^{2{\gamma}}X^{\gamma}R^{2}\ll\sum_{\begin{subarray}{c}1\leqslant
r,r^{\prime}\leqslant R:\\\
{\theta}={\theta}_{r}-{\theta}^{\prime}_{r}\in{\mathfrak{M}}\end{subarray}}|\hat{\nu}({\theta})|^{\gamma}.$
(E.8)

If ${\theta}\in{\mathfrak{M}}(q,a)$ with $(a,q)=1$ and $q\leqslant(\log
P)^{1/4}$ then, by Lemmas D.1, D.2 and D.3 we have

$\displaystyle\hat{\nu}({\theta})$ $\displaystyle\ll
q^{-1/k}\min\Bigl{\\{}X,\Bigl{\|}{\theta}-\frac{a}{q}\Bigr{\|}^{-1}\Bigr{\\}}+\frac{P^{k}}{\log
P}(q+P^{k}\|q{\alpha}\|)$
$\displaystyle\ll_{k}q^{-1/k}\frac{X}{1+X|{\theta}-\frac{a}{q}|}+\frac{WX}{\sqrt{\log
X}}.$

With $C$ a large positive constant, the contribution to the right-hand side of
(E.8) from denominators $q>Q_{1}:=C+{\delta}^{-3k}$ is therefore bounded, up
to a constant, by

$R^{2}X^{\gamma}(Q_{1}^{-{\gamma}/k}+W^{\gamma}(\log X)^{-{\gamma}/2})$

which, by (E.5), is negligible compared to the left-hand side of (E.8).
Therefore

${\delta}^{2{\gamma}}R^{2}\ll\sum_{1\leqslant r,r^{\prime}\leqslant
R}G({\theta}_{r}-{\theta}^{\prime}_{r}),$ (E.9)

where

$G({\alpha})=\sum_{q\leqslant
Q_{1}}\sum_{a=0}^{q-1}\frac{q^{-{\gamma}/k}}{(1+X|\sin({\alpha}-\frac{a}{q})|)^{\gamma}}.$

The inequality (E.9) is very similar to [Bou89, Eq. (4.16)], but with $N^{2}$
replaced by $X$, and with $Q_{1}\sim{\delta}^{-3k}$ rather than
$Q_{1}\sim{\delta}^{-5}$. The exponents differ but, since ${\gamma}>k$,
Bourgain’s argument carries through, and we obtain (E.6) in the case
$k\geqslant 4$.

Now suppose $k=2$. Consider ${\theta}={\theta}_{r}-{\theta}^{\prime}_{r}$ in
the summand on the right-hand side of (E.7). By Lemma D.10, the contribution
from ${\theta}\in{\mathfrak{m}}_{2}$ is

$O(R^{2}(X^{1-\frac{\tau}{2}+\tau^{2}})^{\gamma}),$

and by (E.5) this is $o({\delta}^{2{\gamma}}X^{\gamma}R^{2})$. Hence

${\delta}^{2{\gamma}}X^{\gamma}R^{2}\ll\sum_{\begin{subarray}{c}1\leqslant
r,r^{\prime}\leqslant R:\\\
{\theta}={\theta}_{r}-{\theta}^{\prime}_{r}\in{\mathfrak{M}}\end{subarray}}|\hat{\nu}({\theta})|^{\gamma}.$
(E.10)

If ${\theta}\in{\mathfrak{M}}_{2}(q,a)$ with $(a,q)=1$ and $q\leqslant
X^{\tau}$ then, by Lemmas D.2, D.3 and D.9, we have

$\hat{\nu}({\theta})\ll
q^{-1/2}\min\Bigl{\\{}X,\Bigl{\|}{\theta}-\frac{a}{q}\Bigr{\|}^{-1}\Bigr{\\}}+O_{w}(X^{\frac{1}{2}+2\tau})\ll\frac{Xq^{-1/2}}{1+X|{\theta}-\frac{a}{q}|}.$

We obtain (E.9), but with $k=2$ in the definition of $G(\cdot)$, and
Bourgain’s argument again completes the proof.

####  E.2. An intermediate exponent

In this subsection let $k=3$, and let $\eta$ be a small positive constant as
before. We proceed in two steps, effectively ‘pruning’ the large spectrum. In
the first step, we use a power-saving minor arc estimate for an auxiliary
majorant to come close to a sharp restriction estimate. In the second step, we
no longer require a power saving on the minor arcs, so we are able to obtain a
sharp restriction estimate by reverting to the majorant $\nu$.

#####  E.2.1. A close estimate

Here we concede a small loss. By slightly increasing the exponent, we will
recover it in the next subsection. Our goal for the time being is to establish
the following.

######  Lemma E.3.

We have

$\sup_{|\phi|\leqslant\nu}\int_{\mathbb{T}}|\hat{\phi}({\alpha})|^{8-10^{-6}}{\mathrm{d}}{\alpha}\ll
X^{-1}(WX)^{8-10^{-6}}.$

Similarly to the $k\neq 3$ case, it suffices to prove that

$\mathrm{meas}(\mathcal{R}_{\delta})\ll\frac{1}{{\delta}^{8-10^{-5}}X},$

where it is now convenient to redefine

$\mathcal{R}_{\delta}=\\{{\alpha}\in{\mathbb{T}}:|\hat{\phi}({\alpha})|>{\delta}WX\\}.$

Note that in this setting ${\delta}\ll W^{-1}$. Since Lemma E.1 implies that

$({\delta}WX)^{6}\mathrm{meas}(\mathcal{R}_{\delta})\leqslant\int_{\mathbb{T}}|\hat{\phi}({\alpha})|^{6}{\mathrm{d}}{\alpha}\ll
P^{15.25-10^{-4}},$

we may assume without loss that

${\delta}>P^{10^{-5}-\frac{1}{8}}.$ (E.11)

Let ${\theta}_{1},{\theta}_{2},\ldots,{\theta}_{R}$ be $X^{-1}$-spaced points
in $\mathcal{R}_{\delta}$. It suffices to show that

$R\ll{\delta}^{-6.3}.$ (E.12)

For some $a_{n}\in{\mathbb{C}}$ with $|a_{n}|\leqslant 1$, we have
$\phi(n)=a_{n}\mu(n)$, wherein we employ the majorant

$\mu(n)=\begin{cases}x^{2},&\text{if }n=\frac{x^{3}-\xi^{3}}{3W}\text{ for
some }x\leqslant P\text{ with }x\equiv\xi{\,\,\mathrm{mod}\,\,W}\\\
0,&\text{otherwise}.\end{cases}$

With ${\gamma}=3.1$, the calculation in [BP17, §6] implies

${\delta}^{2{\gamma}}(WX)^{\gamma}R^{2}\ll\sum_{r,r^{\prime}\leqslant
R}|\hat{\mu}({\theta}_{r}-{\theta}^{\prime}_{r})|^{\gamma}.$ (E.13)

Consider ${\theta}={\theta}_{r}-{\theta}^{\prime}_{r}$ in the summand. We
require a circle method analysis. The majorant $\mu$ is very similar to the
‘auxiliary majorant’ from [Cho17, §5]. Therein, the calculations are based on
partial summation and Roger Baker’s estimates, as packaged in [Cho16, §2]. The
same approach yields the following major arc estimate, where the corresponding
set of minor arcs is

${\mathfrak{n}}:=\\{{\alpha}\in{\mathbb{T}}:|\hat{\mu}({\alpha})|\leqslant
P^{2.75+10^{-6}}\\}.$

######  Lemma E.4.

If ${\alpha}\in{\mathbb{T}}\setminus{\mathfrak{n}}$ then there exist
$q,a\in{\mathbb{Z}}$ such that $0\leqslant a\leqslant q-1$ and

$\hat{\mu}({\alpha})\ll
WXq^{\varepsilon-\frac{1}{3}}\Bigl{(}1+X\Bigl{|}{\alpha}-\frac{a}{q}\Bigr{|}\Bigr{)}^{-1/3}.$

We continue the proof of Lemma E.3. In light of (E.11), the contribution from
${\theta}\in{\mathfrak{n}}$ to the right-hand side of (E.13) is
$o({\delta}^{2{\gamma}}(WX)^{\gamma}R^{2})$, and so

${\delta}^{2{\gamma}}(WX)^{\gamma}R^{2}\ll\sum_{\begin{subarray}{c}r,r^{\prime}\leqslant
R:\\\
{\theta}\notin{\mathfrak{n}}\end{subarray}}|\hat{\mu}({\theta}_{r}-{\theta}^{\prime}_{r})|^{\gamma}.$

With $C$ a large positive constant and $Q_{2}=C+{\delta}^{-9}$, we can use
Lemma E.4 to obtain

${\delta}^{2{\gamma}}R^{2}\ll\sum_{1\leqslant r,r^{\prime}\leqslant
R}G_{1}({\theta}_{r}-{\theta}^{\prime}_{r}),$

where

$G_{1}({\theta})=\sum_{q\leqslant
Q_{2}}\sum_{a=0}^{q-1}\frac{q^{{\gamma}(\varepsilon-\frac{1}{3})}}{(1+X|\sin({\theta}-\frac{a}{q})|)^{{\gamma}/3}}.$

As ${\gamma}(\varepsilon-\frac{1}{3})>1$, the proof may be completed by
Bourgain’s argument, as in the previous subsection.

#####  E.2.2. A sharp estimate

We are ready to prove (E.1); recall that $p=8-10^{-8}$. This time let

$\mathcal{\mathcal{R}}_{\delta}=\\{{\alpha}\in{\mathbb{T}}:|\hat{\phi}({\alpha})|>{\delta}X\\},$

where ${\delta}\ll 1$. Following the same strategy, it suffices to prove that

$\mathrm{meas}(\mathcal{R}_{\delta})\ll\frac{1}{{\delta}^{8-10^{-7}}X}.$

From Lemma E.3 we have

$({\delta}X)^{8-10^{-6}}\mathrm{meas}(\mathcal{R}_{\delta})\ll
W^{8}X^{7-10^{-6}},$

so we may assume without loss that

$W^{-10^{7}}<{\delta}\ll 1.$

With $1,2,\ldots,R$ being $X^{-1}$-spaced points in $\mathcal{R}_{\delta}$, it
again suffices to prove (E.12). This time put $\phi(n)=a_{n}\nu(n)$, where
$a_{n}\in{\mathbb{C}}$ with $|a_{n}|\leqslant 1$. The calculation in [BP17,
§6] then gives (E.7), and by the method of subsection E.1 (in the $k\geqslant
4$ case) we obtain (E.9); once again Bourgain’s argument carries through.

We have considered all cases, thereby completing the proof of Lemmas 6.4, 12.5
and 16.2.

###  Appendix F Lefmann’s criterion

In this section we prove Theorem 1.8, which is a consequence of Lefmann’s
lemma [Lef91, Fact 2.8]. The theorem is a special case of Theorem 1.3, but can
be established more simply, and we presently provide a proof. By rearranging
the variables, we may suppose that for some $t\in\\{6,7,\ldots,s\\}$ we have

$c_{1}+\cdots+c_{t}=0.$ (F.1)

Let

$a:=c_{t+1}+\cdots+c_{s}.$

The case $a=0$ was treated by Browning and Prendiville [BP17] so, for
simplicity, we assume henceforth that $a\neq 0$.

The following obscure fact was shown by Lefmann [Lef91, Fact 2.8].

######  Lemma F.1 (Lefmann).

Let $c_{1},\ldots,c_{s}$ be non-zero integers. Assume that there exists
$t\in[s]$ for which we have (F.1). Assume further that there exist
$y\in{\mathbb{Z}}\setminus\\{0\\}$ and $y_{1},\ldots,y_{t}\in{\mathbb{Z}}$
such that

$c_{1}y_{1}+\cdots+c_{t}y_{t}=0$ (F.2)

and

$ay^{2}+c_{1}y_{1}^{2}+\cdots+c_{t}y_{t}^{2}=0.$ (F.3)

Then (1.8) is partition regular over ${\mathbb{N}}$.

To complete the proof of Theorem 1.8, it remains to prove that the system has
a solution
$(y,{\mathbf{y}})\in({\mathbb{Z}}\setminus\\{0\\})\times{\mathbb{Z}}^{t}$. The
number of such solutions in $[-P,P]^{t+1}$ is
${\mathcal{N}}_{1}-{\mathcal{N}}_{2}$, where ${\mathcal{N}}_{1}$ is the total
number of integer solutions $(y,{\mathbf{y}})\in[-P,P]^{t+1}$ and
${\mathcal{N}}_{2}$ is the number of integer solutions
${\mathbf{y}}\in[-P,P]^{t}$ to

$c_{1}y_{1}+\cdots+c_{t}y_{t}=c_{1}y_{1}^{2}+\cdots+c_{t}y_{t}^{2}=0.$

Here $P$ is a large positive real number.

######  Lemma F.2.

We have

${\mathcal{N}}_{2}\ll P^{t-3}\log P.$

###### Proof.

Rogovskaya [Rog86] showed that the system

$\displaystyle x_{1}+x_{2}+x_{3}$ $\displaystyle=y_{1}+y_{2}+y_{3}$
$\displaystyle x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$
$\displaystyle=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}$

has $\frac{18}{\pi^{2}}P^{3}\log P+O(P^{3})$ solutions
$({\mathbf{x}},{\mathbf{y}})\in[P]^{6}$. By orthogonality, one can deduce from
this that

$\int_{{\mathbb{T}}^{2}}\Bigl{|}\sum_{|x|\leqslant
P}e({\alpha}_{1}x+{\alpha}_{2}x^{2})\Bigr{|}^{6}{\mathrm{d}}{\alpha}_{1}{\mathrm{d}}{\alpha}_{2}\ll
P^{3}\log P.$

As $t\geqslant 6$, the lemma now follows from orthogonality, Hölder’s
inequality, and the trivial bound $\sum_{|x|\leqslant
P}e({\alpha}_{1}x+{\alpha}_{2}x^{2})\ll P$. ∎

######  Lemma F.3.

We have

${\mathcal{N}}_{1}\gg P^{t-2}.$

###### Proof.

Let

$\displaystyle Q(y_{1},\ldots,y_{t-1})$
$\displaystyle=c_{t}^{-1}(c_{1}y_{1}+\cdots+c_{t-1}y_{t-1})^{2}+\sum_{i\leqslant
t-1}c_{i}y_{i}^{2}$ $\displaystyle=\sum_{i\leqslant
t-1}(c_{i}+c_{i}^{2}/c_{t})y_{i}^{2}+2\sum_{1\leqslant i<j\leqslant
t-1}(c_{i}c_{j}/c_{t})y_{i}y_{j},$

and put

$C=|c_{1}|+\cdots+|c_{t}|.$

Now ${\mathcal{N}}_{1}$ is greater than or equal to the number of integer
solutions

$(y,y_{1},\ldots,y_{t-1})\in[-P/C,P/C]^{t}$

to

$ay^{2}+Q(y_{1},\ldots,y_{t-1})=0$

with $c_{1}y_{1}+\cdots+c_{t-1}y_{t-1}\equiv 0\mod c_{t}$. By considering only
multiples of $c_{t}$, we find that ${\mathcal{N}}_{1}$ is greater than or
equal to the number of integer solutions
${\mathbf{x}}\in[-P/C^{2},P/C^{2}]^{t}$ to

$Q_{1}({\mathbf{x}}):=Q(x_{1},\ldots,x_{t-1})+ax_{t}^{2}=0.$

For the sake of brevity, we appeal to Birch’s very general theorem [Bir61,
Theorem 1]. The _Birch singular locus_ is the set ${\mathcal{S}}$ of
${\mathbf{x}}\in{\mathbb{C}}^{t}$ at which the gradient of $Q_{1}$ vanishes
identically. (In this instance, the Birch singular locus coincides with the
usual singular locus.) We compute that

$\frac{1}{2}\partial_{i}Q(y_{1},\ldots,y_{t-1})=(c_{i}+c_{i}^{2}/c_{t})y_{i}+\sum_{\begin{subarray}{c}j\leqslant
t-1\\\ j\neq i\end{subarray}}c_{i}c_{j}y_{j}/c_{t},$

and so

$\frac{c_{t}}{2c_{i}}\partial_{i}Q=(c_{t}+c_{i})y_{i}+\sum_{\begin{subarray}{c}j\leqslant
t-1\\\ j\neq i\end{subarray}}c_{j}y_{j}=c_{t}(y_{i}-y_{t})+\sum_{j\leqslant
t}c_{j}y_{j}=c_{t}(y_{i}-y_{t}),$

where $y_{t}:=-c_{t}^{-1}(c_{1}y_{1}+\cdots+c_{t-1}y_{t-1})$. Therefore

${\mathcal{S}}=\\{(x,x,\ldots,x,0)\in{\mathbb{C}}^{t}\\},$

and in particular $\dim{\mathcal{S}}=1$.

As $t-\dim{\mathcal{S}}>4$, Birch’s theorem [Bir61, Theorem 1] gives

${\mathcal{N}}_{1}={\mathfrak{S}}{\mathfrak{J}}P^{t-2}+O(P^{t-2-{\delta}}),$
(F.4)

for some constant ${\delta}>0$, where ${\mathfrak{S}}$ and ${\mathfrak{J}}$
are respectively the _singular series_ and _singular integral_ arising from
the circle method analysis. Birch notes in [Bir61, §7] that ${\mathfrak{S}}$
is positive as long as $Q_{1}$ has a non-singular $p$-adic zero for each prime
$p$, and that ${\mathfrak{J}}$ is positive as long as $Q_{1}$ has a real zero
outside of ${\mathcal{S}}_{1}$. Note that $Q$ has a zero
${\mathbf{x}}^{*}\in{\mathbb{Z}}^{t-1}$ with pairwise distinct coordinates;
this follows from [Kei14, Theorem 1.1], or from a circle method analysis. Now
$({\mathbf{x}}^{*},0)$ is a real zero of $Q_{1}$ outside of
${\mathcal{S}}_{1}$, and is also a non-singular $p$-adic zero for each $p$.
Hence ${\mathfrak{S}}{\mathfrak{J}}>0$, and by (F.4) the proof is now
complete. ∎

The previous two lemmas yield ${\mathcal{N}}_{1}>{\mathcal{N}}_{2}$, and this
completes the proof of Theorem 1.8.

######  Remark F.4.

Lefmann’s lemma generalises straightforwardly to higher degrees. We do not
explore this avenue further, as any results thus obtained are likely subsumed
by Theorem 1.3.

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