# Quadratic Forms in Prime Variables with small off-diagonal ranks

Jakub Dobrowolski

###### Abstract.

The main goal of this note is to establish the limits of L. Zhao’s techniques
for counting solutions to quadratic forms in prime variables. Zhao considered
forms with rank at least 9, and showed that these equations have solutions in
primes provided there are no local obstructions. We consider in detail the
degenerate cases of off-diagonal rank 1 and 2, and improve the rank lower
bounds to at least 6 and at least 8 respectively. These results complement a
recent breakthrough of Green on the non-degenerate rank 8 case.

##  1\. Introduction

This entire paper concentrates on a problem of finding solutions to the
equation

(1) $\textbf{x}^{T}A\textbf{x}=t$

where $A\in M_{n}(\mathbb{Z})$ is a symmetric matrix, $t\in\mathbb{Z}$ and x
is a vector with all entries being prime. This problem was first looked at by
Hua [2]. He proved that in the case of $A$ an identity matrix, $n=5$ and
$t\equiv 5$ (mod 24) sufficiently large we can always find a solution to
equation (1) in primes. Not long ago Jianya Liu [3] was able to handle the
more general case. For $A$ satisfying $\operatorname{rank}(A)\geq 10$ and some
mild conditions, he was able to prove that an asymptotic number of solutions
to (1) in primes in a box $[0,X]^{n}$ is at least $C_{A}X^{n-2}(\log X)^{-n}$
where $C_{A}>0$ is some constant. Lilu Zhao [1] was able to prove that
$\operatorname{rank}(A)\geq 9$ suffices for all regular matrices $A$. We will
be mostly using his techniques throughout this paper. Most recently Ben Green
[4] was able to prove using techniques from representation theory that under
some mild non-degeneracy conditions if $\operatorname{rank}(A)=8$ and $A$ has
an off-diagonal submatrix of rank 4 then we can find $C_{A}X^{6}(\log X)^{-8}$
solutions to (1) in a box $[0,X]^{8}$ with $C_{A}>0$.
In this paper we will be working with

(2) $\mathcal{N}_{A,t}(X)=\sum_{\begin{subarray}{c}1\leq x_{1},...,x_{n}\leq
X\\\ \textbf{x}^{T}A\textbf{x}=t\end{subarray}}\prod_{i=1}^{n}\Lambda(x_{i})$

where $\Lambda(x)$ is the von Mangoldt function and $\mathcal{N}_{A,t}$ is
calculating number of solutions to (1) in prime powers in $[0,X]^{n}$ box.
This weighting is rather standard and makes the sum estimates much simpler.
Finally we will write $\operatorname{rank}_{\operatorname{off}}(A)=k$ if the
biggest rank of a submatrix of $A$ which does not contain diagonal entries is
exactly $k$. We can now state main theorem of this paper.

######  Theorem 1.1.

Let $\mathfrak{S}(A,t)$ be defined as in (11) and $A$ be an indefinite matrix
satisfying one of the following:
(i) $\operatorname{rank}(A)\geq 6$ and
$\operatorname{rank}_{\operatorname{off}}(A)=1$;
(ii) $\operatorname{rank}(A)\geq 8$ and
$\operatorname{rank}_{\operatorname{off}}(A)=2$.
Then:

(3) $\mathcal{N}_{A,t}(X)=\mathfrak{S}(A,t)X^{n-2}+O_{K}(X^{n-2}(\log
X)^{-K})$

where $K$ is an arbitrarily large constant.

This result was probably known to Zhao but he didn’t explicitly write down the
details. With this theorem we establish the limitations of his methods for
off-diagonal ranks 1 and 2.
It’s also important to point out how Theorem 1.1 relates to the result of Ben
Green[4]. For $A\in GL_{8}(\mathbb{Z})$, $0\leq
r=\operatorname{rank}_{\operatorname{off}}(A)\leq 4$. The case $r=0$ is a
result due to Hua[2] and for $r=1,2$ Theorem 1.1 applies. Under mild non-
degeneracy conditions the result of Ben Green deals with case $r=4$. Thus
there’s only case $r=3$ left to improve the restriction on $A$ to
$\operatorname{rank}(A)\geq 8$.

##  2\. Notation

Notation in this paper is mostly standard. We write $e(z)$ for $e^{2\pi iz}$
and we will let $X$ be a sufficiently large integer and $L=\log X$. When we
write $\lvert\textbf{x}\rvert\leq X$ or $\textbf{x}\leq X$ we mean that
coefficients $x_{i}$ of x satisfy $\lvert x_{i}\rvert\leq X$ or $x_{i}\leq X$
respectively.
In this paper matrix $A$ and matrices $A_{i}$ will always be symmetric. We
will write $M_{n,k}(S)$ for the set of $n$ by $k$ matrices with entries in $S$
and $M_{n}(S)$ for the set of $n$ by $n$ matrices with entries in $S$. Finally
as usual we denote by $GL_{n}(S)$ the set of $n$ by $n$ invertible matrices
with entries in $S$.
For $\textbf{x}^{T}=(x_{1},x_{2},...,x_{n})$ we will write
$\Lambda(\textbf{x})=\Lambda(x_{1})\Lambda(x_{2})...\Lambda(x_{n})$.
Many times in this paper we will be summing over the number of integer
solutions of certain linear and bilinear equations in some box. Instead of
writing that as

(4) $\sum_{\begin{subarray}{c}equation1\\\ equation2\\\ equation3\\\
\end{subarray}}1$

we will write it as

(5) $\Bigg{\\{}\begin{subarray}{c}equation1\\\ equation2\\\
equation3\end{subarray}\Bigg{\\}}$

to put a focus on the equations themselves. It will be clear from context
which unknowns are actually variables and all of them will be lying in the box
$[CX,C^{\prime}X]$ for $C,C^{\prime}$ some real constants.

##  3\. General approach to the problem

As usual in these kinds of problems we introduce the corresponding exponential
sum

(6) $S(\alpha)=\sum_{1\leq\textbf{x}\leq
X}\Lambda(\textbf{x})e(\alpha\textbf{x}^{T}A\textbf{x})$

and note that

(7) $\int_{0}^{1}S(\alpha)e(-\alpha
t)=\sum_{\begin{subarray}{c}1\leq\textbf{x}\leq X\\\
\textbf{x}^{T}A\textbf{x}=t\end{subarray}}\Lambda(\textbf{x})=\mathcal{N}_{A,t}(X).$

For $a,q$ coprime we will let:

(8)
$\mathcal{M}(a,q;Q)=\\{\alpha:\lvert\alpha-\tfrac{a}{q}\rvert\leq\tfrac{Q}{qX^{2}}\\}.$

For fixed $Q\leq X/2$ sets $\mathcal{M}(a,q;Q)$ are disjoint for different $a$
and $q$ provided that $q\leq Q$. We can then write

(9) $\mathcal{M}(Q)=\bigcup_{q=1}^{Q}\bigcup_{\begin{subarray}{c}a=1\\\
(a,q)=1\end{subarray}}^{q}\mathcal{M}(a,q;Q)$

We define major arcs to be $\mathfrak{M}=\mathcal{M}(P)$ where $P=L^{K}$. In
this paper $K$ and $K^{\prime}$ will denote arbitrarily large parameters
independent of $X$, satisfying $K>K^{\prime}$, which may change from line to
line.
Finally we will let $\mathfrak{m}=[X^{-1},1+X^{-1}]\setminus\mathfrak{M}$ be
the minor arcs.
Let’s define a few quantities:

(10) $C(q,a)=\sum_{\begin{subarray}{c}1\leq\mathbf{h}\leq q\\\
(\mathbf{h},q)=1\end{subarray}}e\Big{(}\mathbf{h}^{T}A\mathbf{h}\frac{a}{q}\Big{)},$
(11)
$\mathfrak{S}(A,t)=\sum_{q=1}^{\infty}\frac{1}{\varphi(q)}\sum_{\begin{subarray}{c}1\leq
a\leq q\\\ (a,q)=1\end{subarray}}C(q,a)e\Big{(}-\frac{at}{q}\Big{)}.$

In the literature $\mathfrak{S}(A,t)$ is referred to as the singular series.
In [1] (Lemma 3.6) Zhao proved that for invertible, indefinite $A$ with
$\operatorname{rank}(A)\geq 5$ major arcs bring the expected main contribution
in $\int_{0}^{1}S(\alpha)e(-\alpha t)\,d\alpha$. Namely that:

(12) $\int_{\mathfrak{M}}S(\alpha)e(-\alpha
t)\,d\alpha=\mathfrak{S}(A,t)X^{n-2}+O(X^{n-2}L^{-K})$

if equation (1) has positive real solutions and where $\mathfrak{S}(A,t)\gg 1$
provided that $\textbf{x}^{T}A\textbf{x}=t$ has local solutions in the set of
$p$-adic units for all primes $p$.
So all that’s left to do is to estimate the integral over the minor arcs. Our
goal is to show that for certain classes of matrices

(13) $\int_{\mathfrak{m}}S(\alpha)e(-\alpha t)\,d\alpha\ll X^{n-2}L^{-K}.$

This, combined with the result of Zhao about major arcs, will give us the
following:

(14) $\int_{[0,1]}S(\alpha)e(-\alpha
t)\,d\alpha=\mathfrak{S}(A,t)X^{n-2}+O(X^{n-2}L^{-K})\gg X^{n-2}$

under all the local conditions mentioned above and hence proving Theorem 1.1.
Let’s look at the trivial bound for the integral over the minor arcs first.

(15) $\displaystyle\left\lvert\int_{\mathfrak{m}}S(\alpha)e(-\alpha
t)\,d\alpha\right\rvert\leq\sup_{\alpha\in\mathfrak{m}}\displaystyle\left\lvert
S(\alpha)\right\rvert\leq\sum_{\textbf{x}\leq X}\Lambda(\textbf{x})\leq
L^{n}X^{n}.$

Let’s note that we’re still an arbitrarily large power of $\log$ and power of
$X^{2}$ off of our goal in (13). So the goal for this paper is to save this
much over the trivial bound above.
To get arbitrarily large logarithm savings we will use a standard result:

######  Lemma 3.1.

Let $\alpha\in\mathfrak{m}$, $\beta\in\mathbb{R}$ and
$d\in\mathbb{Q}\setminus\\{0\\}$, then we have

(16) $\Big{\lvert}\sum_{x\leq X}\Lambda(x)e(\alpha dx^{2}+\beta
x)\Big{\rvert}\ll_{d,K}XL^{-K}.$

The proof of it can be found in [5] as Theorem 10. We now state some lemmas
which are the current tool set for saving power of $X^{2}$ in the integral
over the minor arcs. They can be found with proofs in [1] as Lemmas 5.6, 5.7,
5.8 respectively.

######  Lemma 3.2.

Let $C\in M_{n}(\mathbb{Q})$ be a symmetric matrix and $H\in
M_{n,k}(\mathbb{Q})$. For $\alpha\in\mathbb{R}$ and
$\bm{\beta}\in\mathbb{R}^{k}$, we let:

(17)
$\mathcal{F}(\alpha,\bm{\beta})=\sum_{\textbf{x}\in\mathcal{X}}\Lambda(\textbf{x})e(\alpha\textbf{x}^{T}C\textbf{x}+\textbf{x}^{T}H\bm{\beta})$

for $\mathcal{X}\subset\mathbb{Z}^{n}$ a finite subset. Let

(18)
$\mathcal{N}(\mathcal{F})=\sum_{\begin{subarray}{c}\textbf{x}\in\mathcal{X}\textbf{y}\in\mathcal{X}\\\
\textbf{x}^{T}C\textbf{x}=\textbf{y}^{T}C\textbf{y}\\\
\textbf{x}^{T}H=\textbf{y}^{T}H\end{subarray}}\Lambda(\textbf{x})\Lambda(\textbf{y})$

Then

(19)
$\int_{[0,1]^{k+1}}\displaystyle\left\lvert\mathcal{F}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\ll\mathcal{N}(\mathcal{F})$

with the implied constant depending only on the matrices $C$ and $H$.

For the proof of Lemma 3.2 we simply expand
$\displaystyle\left\lvert\mathcal{F}(\alpha,\bm{\beta})\right\rvert^{2}$ into
a double sum and use orthogonal property of the exponential.

######  Lemma 3.3.

With $C$ and $H$ as above one has

(20)
$\Bigg{\\{}\begin{subarray}{c}\textbf{x}^{T}C\textbf{x}=\textbf{y}^{T}C\textbf{y}\\\
\textbf{x}^{T}H=\textbf{y}^{T}H\end{subarray}\Bigg{\\}}\ll\Bigg{\\{}\begin{subarray}{c}\textbf{x}^{T}C\textbf{y}=0\\\
\textbf{x}^{T}H=0\end{subarray}\Bigg{\\}}$

The proof of this Lemma is just a clever change of coordinates
$\mathbf{u}=\textbf{x}+\textbf{y}$ and $\mathbf{v}=\textbf{x}-\textbf{y}$.

######  Lemma 3.4.

If $C\in M_{n,k}(\mathbb{Q})$ and $\operatorname{rank}(C)\geq 2$ then

(21)
$\Big{\\{}\begin{subarray}{c}\textbf{x}^{T}C\textbf{y}=0\end{subarray}\Big{\\}}\ll
LX^{k+n-2}$

with the implied constant depending only on $C$.

This is the key step to Zhao’s estimates. Let’s note that we can trivially
bound
$\big{\\{}\begin{subarray}{c}\textbf{x}^{T}C\textbf{y}=0\end{subarray}\big{\\}}$
by $X^{k+n}$. So we’re getting $X^{2}$ saving over the trivial estimate with
this lemma.
Let’s briefly discuss the use of mentioned lemmas. In all of the problems in
section 4 we will be dealing with integrals of the form:

(22) $\int_{[0,1]^{k+1}}\sum_{\textbf{x}\leq
X}\Lambda(\textbf{x})e(\alpha\textbf{x}^{T}C\textbf{x}+\textbf{x}^{T}H\bm{\beta})\,d\alpha\,d\bm{\beta}$

where $C\in M_{n}(\mathbb{Q})$ satisfies $\operatorname{rank}(C)\geq 2$. We
can apply the Cauchy-Schwarz inequality to get it to the form in Lemma 3.2
with. So we will be able to trivially estimate

(23) $\mathcal{N}(\mathcal{F})\ll
L^{2n}\Bigg{\\{}\begin{subarray}{c}\textbf{x}^{T}C\textbf{x}=\textbf{y}^{T}C\textbf{y}\\\
\textbf{x}^{T}H=\textbf{y}^{T}H\end{subarray}\Bigg{\\}}$

and apply Lemma 3.3. From this, after some change of coordinates we will apply
Lemma 3.4. Notice what would happen if we tried to estimate the integral in
lemma 3.2 in a more crude way.

(24) $\displaystyle\left\lvert\int_{[0,1]^{k+1}}\sum_{\textbf{x}\leq
X}\Lambda(\textbf{x})e(\alpha\textbf{x}^{T}C\textbf{x}+\textbf{x}^{T}H\bm{\beta})\,d\alpha\,d\bm{\beta}\right\rvert\leq\sum_{\textbf{x}\leq
X}\Lambda(\textbf{x})\ll L^{n}X^{n}$

On the other hand after applying Lemma 3.2-3.4 we get a big improvement. The
saving in this case is of order $L^{O(1)}X^{(n-k-2)/2}$ over the trivial
estimate. Since in most of the cases we’ll be working with $k\leq n$ this will
bring us closer to the desired $X^{2}$ saving that we need in our problem.

##  4\. Optimising Zhao’s results

###  4.1. Off-diagonal rank 1

Firstly, as in Zhao’s paper, we need to work out the form of the matrix $A$.
As $\operatorname{rank_{\operatorname{off}}}(A)=1$, we know $A$ is not
diagonal. So suppose $a_{1,2}\neq 0$. Then we have:

######  Lemma 4.1.

If $A$ satisfies $\operatorname{rank_{\operatorname{off}}}(A)=1$ then $A$ can
be written as

(25) $A=\begin{pmatrix}a&\bm{\xi}^{T}\\\
\bm{\xi}&D+h\bm{\xi}\bm{\xi}^{T}\end{pmatrix}$

where $D=\operatorname{diag}(d_{1}$,…,$d_{n-1}$) with $d_{i}\in\mathbb{Q}$,
$h\in\mathbb{Q}$, $a\in\mathbb{Z}$ and
$\textbf{0}\neq\bm{\xi}\in\mathbb{Z}^{n-1}$.

###### Proof.

Let’s define $c_{i}=a_{1,i+1}$ for $i=1,...,n-1$. We begin by noting that as
$c_{1}\neq 0$, we have
$\bm{\xi}^{T}\vcentcolon=(c_{1},...,c_{n-1})\neq\textbf{0}$. Now let’s define
for $3\leq j\leq n$,
$\bm{\eta}_{j}=(a_{j,2},...,a_{j,j-1},a_{j,j+1},...,a_{j,n})^{T}$,
$\bm{\xi}_{j}=(a_{1,2},...a_{1,j-1},a_{1,j+1},...,a_{1,n})^{T}$ and
$\bm{\eta}_{2}=(a_{2,3},a_{2,4},...,a_{2,n})^{T}$,
$\bm{\xi}_{2}=(a_{1,3},a_{1,4},...,a_{1,n})^{T}$. We know that
$\operatorname{rank_{\operatorname{off}}}(A)=1$ so for each $j$,
$\bm{\eta}_{j}$ is a rational multiple of $\bm{\xi}_{j}$ as both of those are
vectors not on the diagonal. Therefore there is exactly one
$a_{j,j}^{\prime}\in\mathbb{Q}$ and $h_{j}\in\mathbb{Q}$ such that
$(a_{j,2},...,a_{j,j-1},a_{j,j}^{\prime},a_{j,j+1},...,a_{j,n})=h_{j}\bm{\xi}^{T}$.
Let’s write

(26) $A=\begin{pmatrix}a&\bm{\xi}^{T}\\\ \bm{\xi}&A_{1}\end{pmatrix}$

and let $\textbf{h}^{T}=(h_{2},...,h_{n})$. By comparing coefficients we note
that $\bm{h}\bm{\xi}^{T}-A_{1}$ must be a diagonal matrix, call it $D$.
Finally $\bm{h}\bm{\xi}^{T}$ is a symmetric matrix so for each $1\leq
k,\,l\leq n-1$ we have $c_{k}h_{l}=c_{l}h_{k}$. As $c_{1}\neq 0$ we can write

(27) $h_{k}=c_{k}\frac{h_{1}}{c_{1}}.$

So $\bm{h}=h\bm{\xi}$ for $h=\tfrac{h_{1}}{c_{1}}\in\mathbb{Q}$ and we get the
desired form. ∎

######  Lemma 4.2.

Let A be given by the form above with $\operatorname{rank}(A)\geq 6$ and let
$c_{i}=a_{1,i+1}$. We can find pairwise distinct $b_{i}$ with $1\leq
b_{1},b_{2},b_{3},b_{4},b_{5}\leq n-1$ such that
$c_{b_{1}}d_{b_{2}}d_{b_{3}}d_{b_{4}}d_{b_{5}}\neq 0$.

###### Proof.

By $\operatorname{rank}(A)\geq 6$ we have $\operatorname{rank}(D)\geq 4$. We
also assumed w.l.o.g. that $c_{1}\neq 0$ so let’s choose $b_{1}=1$. Suppose
that the above doesn’t hold. Namely, for any quadruple
$k_{1},k_{2},k_{3},k_{4}\geq 2$ we have
$d_{k_{1}}d_{k_{2}}d_{k_{3}}d_{k_{4}}=0$. This gives the upper bound
$\operatorname{rank}(D)\leq 4$, so we only have the case of
$\operatorname{rank}(D)=4$ left to consider.
We may suppose that $d_{1}d_{2}d_{3}d_{4}\neq 0$ and $d_{j}=0$ for $j>4$. But
then by looking at the rank of the whole matrix we conclude that
$\operatorname{rank}(c_{5},c_{6},...,c_{n-1})\neq 0$. That means $c_{j}\neq 0$
for some $j\geq 5$ giving us a quintuple $1,2,3,4,j$. ∎

With the desired form we can move on to the estimates of $S(\alpha)$ over the
minor arcs.

######  Lemma 4.3.

Let $A\in M_{n}(\mathbb{Z})$ satisfy
$\operatorname{rank_{\operatorname{off}}}(A)=1$ with $n\geq 6$, then
$\int_{\mathfrak{m}}\displaystyle\left\lvert S(\alpha)\right\rvert\,d\alpha\ll
X^{n-2}L^{-K}$

###### Proof.

We start by noting that we may use lemmas above to read information on the
form of $A$. Then we can write $S(\alpha)$ as below:

(28) $S(\alpha)=\sum_{x\leq
X}\sum_{\begin{subarray}{c}\textbf{y}\in\mathbb{N}^{n-1}\\\ \textbf{y}\leq
X\end{subarray}}\Lambda(x)\Lambda(\textbf{y})e(\alpha(ax^{2}+2x\bm{\xi}^{T}\textbf{y}+\textbf{y}^{T}D\textbf{y}+h\textbf{y}^{T}\bm{\xi}\bm{\xi}^{T}\textbf{y})).$

For each $\textbf{y}\leq X$ we will let $z=\bm{\xi}^{T}\textbf{y}$. We can use
orthogonality property of the exponential function to write

(29) $\displaystyle S(\alpha)=$ $\displaystyle\int_{0}^{1}\sum_{\lvert
z\rvert\ll X}\sum_{x\leq X}\sum_{\textbf{y}\leq
X}\Lambda(\textbf{y})\Lambda(x)e(\alpha(ax^{2}+2xz+\textbf{y}^{T}D\textbf{y}+hz^{2}))$
$\displaystyle\times e((\bm{\xi}\textbf{y}^{T}-z)\beta)\,d\beta$

Let’s introduce new sums using Zhao’s notation.

(30) $\displaystyle\mathcal{F}(\alpha,\bm{\beta})=\sum_{\lvert z\rvert\ll
X}\sum_{x\leq X}\Lambda(x)e(\alpha(ax^{2}+2xz+hz^{2})-z\beta)$ $\displaystyle
f_{j}(\alpha,\bm{\beta})=\sum_{y_{j}\leq X}\Lambda(y_{j})e(\alpha
d_{j}y_{j}^{2}+y_{j}c_{j}\beta)$

for $j=1,...,n-1$. By the above lemma we can find a quintuple
$b_{1},b_{2},b_{3},b_{4},b_{5}$ and suppose for notational simplicity that
it’s $1,2,3,4,5$ i.e. $c_{1}d_{2}d_{3}d_{4}d_{5}\neq 0$. We bound the integral
using the Cauchy-Schwarz inequality.

(31) $\int_{\mathfrak{m}}\displaystyle\left\lvert
S(\alpha)\right\rvert\,d\alpha\leq(\mathcal{I}_{1}\mathcal{I}_{2})^{\tfrac{1}{2}}\prod_{i=5}^{n-1}\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\beta\in[0,1]\end{subarray}}\Big{\lvert}\sum_{y_{i}\leq
X}\Lambda(y_{i})e(\alpha d_{i}y_{i}^{2}+c_{i}y_{i}\beta)\Big{\rvert}$

where

(32)
$\mathcal{I}_{1}=\int_{[0,1]^{2}}\Big{\lvert}\mathcal{F}(\alpha,\beta)f_{4}(\alpha,\beta)\Big{\rvert}^{2}\,d\beta\,d\alpha$

and

(33)
$\mathcal{I}_{2}=\int_{[0,1]^{2}}\Big{\lvert}f_{1}(\alpha,\beta)f_{2}(\alpha,\beta)f_{3}(\alpha,\beta)\Big{\rvert}^{2}\,d\beta\,d\alpha.$

As $d_{5}\neq 0$ we can bound the product in (31) using Lemma 3.1 by

(34) $\prod_{i=5}^{n-1}\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\beta\in[0,1]\end{subarray}}\Big{\lvert}\sum_{y_{i}\leq
X}\Lambda(y_{i})e(\alpha d_{i}y_{i}^{2}+c_{i}y_{i}\beta)\Big{\rvert}\ll
X^{n-5}L^{-K}$

where for $i\geq 6$ we used a trivial bound on the sum: $LX$.
Now we turn to the two integrals. Note that by Lemma 3.2

(35) $\mathcal{I}_{1}\ll
L^{4}\Bigg{\\{}\begin{subarray}{c}ax^{2}+2xz+hz^{2}+d_{4}y_{4}^{2}=ax^{\prime
2}+2x^{\prime}z^{\prime}+hz^{\prime 2}+d_{4}y_{4}^{\prime 2}\\\
c_{4}y_{4}-z=c_{4}y_{4}^{\prime}-z^{\prime}\end{subarray}\Bigg{\\}}.$

We can therefore apply Lemma 3.3 to get

(36) $\mathcal{I}_{1}\ll
L^{4}\Bigg{\\{}\begin{subarray}{c}axx^{\prime}+xz^{\prime}+x^{\prime}z+hzz^{\prime}+d_{4}y_{4}y_{4}^{\prime}=0\\\
c_{4}y_{4}-z=0\end{subarray}\Bigg{\\}}\\\ $

and if we plug in $z=c_{4}y_{4}$ note that we get a bilinear equation in 5
variables.

(37) $\mathcal{I}_{1}\ll
L^{4}\Bigg{\\{}axx^{\prime}+xz^{\prime}+c_{4}y_{4}x^{\prime}+hc_{4}y_{4}z^{\prime}+d_{4}y_{4}y_{4}^{\prime}=0\Bigg{\\}}$

The corresponding matrix is of rank at least 2 as coefficients by
$xz^{\prime}$ and $y_{4}y_{4}^{\prime}$ are nonzero. So by Lemma 3.4 we can
bound the last sum by $L^{5}X^{3}$. We do the same procedure for the second
integral.

(38)
$\mathcal{I}_{2}=\int_{[0,1]^{2}}\Big{\lvert}f_{1}(\alpha,\beta)f_{2}(\alpha,\beta)f_{3}(\alpha,\beta)\Big{\rvert}^{2}\,d\beta\,d\alpha\ll
L^{6}\Bigg{\\{}\begin{subarray}{c}d_{1}y_{1}y_{1}^{\prime}+d_{2}y_{2}y_{2}^{\prime}+d_{3}y_{3}y_{3}^{\prime}=0\\\
c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}=0\end{subarray}\Bigg{\\}}$

where we already used Lemma 3.2 and 3.3. Now since $c_{1}\neq 0$ we can divide
by it and write $y_{1}$ in terms of $y_{2}$ and $y_{3}$. So we get:

(39) $\mathcal{I}_{2}\ll
L^{6}\Bigg{\\{}-d_{1}y_{1}^{\prime}(c_{2}y_{2}+c_{3}y_{3})/c_{1}+d_{2}y_{2}y_{2}^{\prime}+d_{3}y_{3}y_{3}^{\prime}=0\Bigg{\\}}$

a bilinear equation in 5 variables with rank of a corresponding matrix at
least two as this time $d_{2}d_{3}\neq 0$. So we can apply Lemma 3.4 to get

(40) $\mathcal{I}_{2}\ll L^{7}X^{3}.$

We combine the results to get that gives us

(41) $\int_{\mathfrak{m}}\displaystyle\left\lvert
S(\alpha)\right\rvert\,d\alpha\ll L^{-K}X^{n-5}L^{5/2}X^{3/2}L^{5/2}X^{3/2}\ll
L^{-K}X^{n-2}$

as we wanted. ∎

###  4.2. Off-diagonal rank 2

Suppose now without loss of generality that
$\operatorname{rank}_{\operatorname{off}}(A)=\operatorname{rank}(B)=2$ where
$B=\begin{pmatrix}a_{1,3}&a_{1,4}\\\ a_{2,3}&a_{2,4}\end{pmatrix}$. We
consider different cases depending on ranks of following matrices:

(42) $\displaystyle
B_{1}=\begin{pmatrix}a_{1,3}&a_{1,5}&a_{1,6}&.&.&.&a_{1,n}\\\
a_{2,3}&a_{2,5}&a_{2,6}&.&.&.&a_{2,n}\end{pmatrix}$ (43) $\displaystyle
B_{2}=\begin{pmatrix}a_{1,4}&a_{1,5}&a_{1,6}&.&.&.&a_{1,n}\\\
a_{2,4}&a_{2,5}&a_{2,6}&.&.&.&a_{2,n}\end{pmatrix}.$

There are three cases to consider (up to reordering of the matrix $A$) and so
three lemmas which we will need to prove:

######  Lemma 4.4.

If $\operatorname{rank}(B_{1})=\operatorname{rank}(B_{2})=1$ and
$\operatorname{rank}(A)\geq 8$ then we have
$\int_{\mathfrak{m}}\displaystyle\left\lvert S(\alpha)\right\rvert\,d\alpha\ll
X^{n-2}L^{-K}$

######  Lemma 4.5.

If $\operatorname{rank}(B_{1})=2,\operatorname{rank}(B_{2})=1$ and
$\operatorname{rank}(A)\geq 8$ then we have
$\int_{\mathfrak{m}}\displaystyle\left\lvert S(\alpha)\right\rvert\,d\alpha\ll
X^{n-2}L^{-K}$

######  Lemma 4.6.

If $\operatorname{rank}(B_{1})=\operatorname{rank}(B_{2})=2$ and
$\operatorname{rank}(A)\geq 8$ then we have
$\int_{\mathfrak{m}}\displaystyle\left\lvert S(\alpha)\right\rvert\,d\alpha\ll
X^{n-2}L^{-K}$ .

Strategy for all three lemmas is similar as for proof of off-diagonal rank 1
case. In each case we will firstly set up a structure lemma for the form of
matrix $A$. With that we will introduce a new variable which will allow us to
bound the integral using the Cauchy-Schwarz inequality. In the end we will use
Lemmas 3.1-3.4 to get the desired result.

####  4.2.1. Proof of Lemma 4.4

######  Lemma 4.7.

If $A$ is a matrix with
$\operatorname{rank}(B_{1})=\operatorname{rank}(B_{2})=1$ then $A$ is of the
form

$A=\begin{pmatrix}A_{1}&B&0\\\ B^{T}&A_{2}&C\\\ 0&C^{T}&D\end{pmatrix}$

where $B\in GL_{2}(\mathbb{Z})$, $C\in M_{2,n-4}(\mathbb{Z})$ and
$D=\operatorname{diag}(d_{1},...,d_{n-4})$

###### Proof.

Let’s introduce vectors $\gamma_{j}=(a_{1,2+j},a_{2,2+j})^{T}$ for
$j=1,2,...,n-2$. By looking at ranks of $B_{1}$ and $B_{2}$ we note that for
any $j\geq 3$ $\operatorname{rank}(\gamma_{1},\gamma_{j})=$
$\operatorname{rank}(\gamma_{2},\gamma_{j})=1$, but as $\gamma_{1}$ and
$\gamma_{2}$ are linearly independent we conclude that $\gamma_{j}=0$ for
$j\geq 3$. Finally let’s consider matrices

$M_{i,j}=\begin{pmatrix}a_{1,3}&a_{1,4}&a_{1,j}\\\ a_{2,3}&a_{2,4}&a_{2,j}\\\
a_{i,3}&a_{i,4}&a_{i,j}\end{pmatrix}$

for $i\neq j$ and $i,j>4$.
Each such $M_{i,j}$ is an off-diagonal matrix so it must have rank 2 (as $B$
is its submatrix). Therefore as $a_{1,j}=a_{2,j}=0$ we must have $a_{i,j}=0$,
and we get the desired form. ∎

######  Proof of Lemma 4.4.

We now turn to the estimate of $S(\alpha)$. By the structure lemma above we
get that:

(44) $S(\alpha)=\sum_{\begin{subarray}{c}\textbf{x},\textbf{y}\leq X\\\
\textbf{x},\textbf{y}\in\mathbb{N}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{z}\leq
X\\\
\textbf{z}\in\mathbb{N}^{n-4}\end{subarray}}\Lambda(\textbf{x})\Lambda(\textbf{y})\Lambda(\textbf{z})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+2\textbf{x}^{T}B\textbf{y}+\textbf{y}^{T}A_{2}\textbf{y}+2\textbf{y}^{T}C\textbf{z}+\textbf{z}^{T}D\textbf{z}))$

As in the proof of Lemma 4.3 for each $\textbf{x},\textbf{y}\in\mathbb{N}^{2}$
and $\textbf{z}\in\mathbb{N}^{n-4}$ satisfying
$\textbf{x},\textbf{y},\textbf{z}\leq X$ we let
$\textbf{w}^{T}=2\textbf{x}^{T}B+\textbf{y}^{T}A_{2}+2\textbf{z}^{T}C^{T}$. By
orthogonality property we have then

(45) $\displaystyle S(\alpha)={}$
$\displaystyle\int_{[0,1]^{2}}\sum_{\begin{subarray}{c}\lvert\textbf{w}\rvert\ll
X\\\
\textbf{w}\in\mathbb{Z}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{x}\leq
X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{y}\leq
X\\\
\textbf{y}\in\mathbb{N}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{z}\leq
X\\\
\textbf{z}\in\mathbb{N}^{n-4}\end{subarray}}\Lambda(\textbf{x})\Lambda(\textbf{y})\Lambda(\textbf{z})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+\textbf{w}^{T}\textbf{y}+\textbf{z}^{T}D\textbf{z}))$
$\displaystyle\times
e((2\textbf{x}^{T}B+\textbf{y}^{T}A_{2}+2\textbf{z}^{T}C^{T}-\textbf{w}^{T})\bm{\beta})\,d\bm{\beta}.$

where $\bm{\beta}$ is a two dimensional vector with
$d\bm{\beta}=d\beta_{1}\,d\beta_{2}$ We introduce exponential sums.

(46)
$\displaystyle\mathcal{F}(\alpha,\bm{\beta})=\sum_{\begin{subarray}{c}\textbf{x}\leq
X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\Lambda(\textbf{x})e(\alpha\textbf{x}^{T}A_{1}\textbf{x}+2\textbf{x}^{T}B\bm{\beta}))$
$\displaystyle\mathcal{H}_{j}(\alpha,\bm{\beta})=\sum_{\begin{subarray}{c}\lvert
w_{j}\rvert\ll X\\\
w_{j}\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}y_{j}\leq X\\\
y_{j}\in\mathbb{N}\end{subarray}}e(\alpha
w_{j}y_{j}+y_{j}\bm{\gamma_{j}}^{T}\bm{\beta}-w_{j}\bm{e}_{j}^{T}\bm{\beta})\Lambda(y)$

where $j=1,2$, $\bm{\gamma}_{j}^{T}=(a_{2+j,3},a_{2+j,4})$, $\bm{e}_{j}$
standard basis vectors of size 2 and $y_{j},w_{j}$ are entries of vectors y
and w respectively. Finally we let

(47) $f_{j}(\alpha,\bm{\beta})=\sum_{z_{j}\leq X}\Lambda(z_{j})e(\alpha
d_{j}z_{j}^{2}+2z_{j}\bm{\xi}_{j}^{T}\bm{\beta})$

with $\bm{\xi}_{j}=(a_{3,4+j},a_{4,4+j})^{T}$ and $z_{j}$ entries of vector z
for $j=1,2,...,n-4$.
We will consider 2 different cases depending on $\operatorname{rank}(D)$.
Let’s note that as $\operatorname{rank}(A)\geq 8$ we have that
$\operatorname{rank}(D)\geq 2$. Let’s suppose first that
$\operatorname{rank}(D)\geq 3$ with $d_{1}d_{2}d_{3}\neq 0$. By applying CS
inequality we get:

(48) $\displaystyle\int_{\mathfrak{m}}\left\lvert
S(\alpha)\right\rvert\,d\alpha$
$\displaystyle\leq\Bigg{(}\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\Big{\lvert}\prod_{i=3}^{n-4}f_{i}(\alpha,\bm{\beta})\Big{\rvert}\Bigg{)}\left(\int_{[0,1]^{3}}\left\lvert\mathcal{F}(\alpha,\bm{\beta})f_{1}(\alpha,\bm{\beta})f_{2}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$
$\displaystyle\times\left(\int_{[0,1]^{3}}\left\lvert\mathcal{H}_{1}(\alpha,\bm{\beta})\mathcal{H}_{2}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$

As $d_{3}\neq 0$ in the supremum we can save an arbitrarily large power of log
over the trivial bound. Namely:

(49) $\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\displaystyle\left\lvert\prod_{i=3}^{n-4}f_{i}(\alpha,\bm{\beta})\right\rvert\ll
X^{n-6}L^{-K}$

by Lemma 3.1. After applying Lemmas 3.2 and 3.3 the first integral becomes

(50)
$\int_{[0,1]^{3}}\displaystyle\left\lvert\mathcal{F}(\alpha,\bm{\beta})f_{1}(\alpha,\bm{\beta})f_{2}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\ll
L^{8}\Bigg{\\{}\begin{subarray}{c}\textbf{x}^{T}A_{1}\textbf{x}^{\prime}+d_{1}z_{1}z_{1}^{\prime}+d_{2}z_{2}z_{2}^{\prime}=0\\\
\textbf{x}^{T}B+z_{1}\bm{\xi}_{1}^{T}+z_{2}\bm{\xi}_{2}^{T}=0\end{subarray}\Bigg{\\}}.$

Let’s observe that as $B$ is invertible we can write x in terms of $z_{1}$ and
$z_{2}$. So we can get a bilinear equation in 6 variables and as we chose
$d_{1},d_{2}\neq 0$, the matrix corresponding to it is of rank at least 2. So
first integral can be bounded, using Lemma 3.4, by

(51) $\ll
L^{8}\Bigg{\\{}-(z_{1}\bm{\xi}_{1}^{T}+z_{2}\bm{\xi}_{2}^{T})B^{-1}A_{1}\textbf{x}^{\prime}+d_{1}z_{1}z_{1}^{\prime}+d_{2}z_{2}z_{2}^{\prime}=0\Bigg{\\}}\ll
L^{9}X^{4}.$

So after taking square root we are left with $L^{9/2}X^{2}$. Note that we’ve
saved a power of $X^{2}$ over the trivial bound. One power of $X$ came from
$B$ being invertible and the other from applying Lemma 3.4. So we hope that
the second integral can be bounded from above by $L^{O(1)}X^{4}$ and indeed it
can. Second integral is bounded by:

(52) $\ll
L^{4}\Bigg{\\{}\begin{subarray}{c}w_{1}y_{1}^{\prime}+w_{1}^{\prime}y_{1}+w_{2}y_{2}^{\prime}+w_{2}^{\prime}y_{2}=0\\\
y_{1}\bm{\gamma_{1}^{T}}+y_{2}\bm{\gamma_{2}^{T}}-w_{1}\bm{e}_{1}^{T}-w_{2}\bm{e}_{2}^{T}=0\end{subarray}\Bigg{\\}}.$

Because matrix $(\bm{e}_{1},\bm{e}_{2})$ is invertible we can write $w_{1}$
and $w_{2}$ as a linear combination of $y_{1}$ and $y_{2}$. So we get a bound:

(53) $\ll
L^{4}\Bigg{\\{}\textbf{y}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})^{T}\textbf{y}^{\prime}+\textbf{y}^{T}\textbf{w}^{\prime}=0\Bigg{\\}}\ll
L^{5}X^{4}$

where for the last inequality we applied Lemma 3.4 again. Therefore after
combining those results we get $\int_{\mathfrak{m}}\displaystyle\left\lvert
S(\alpha)\right\rvert\,d\alpha\ll X^{n-2}L^{-K}$
Now suppose $\operatorname{rank}(D)=2$ and after permuting suppose
$d_{1}d_{2}\neq 0$ and $d_{k}=0$ for $k=3,...,n-4$. Because
$\operatorname{rank}(A)\geq 8$ and $\operatorname{rank}(D)=2$ we may find
$i,j$ with $1\leq i\leq 2$ and $j\geq 3$ such that
$\operatorname{rank}(\bm{e}_{i},\bm{\xi}_{j})=2$. Without loss of generality
let’s suppose $i=1,j=3$. In this case we split the starting integral as
follows:

(54) $\displaystyle\int_{\mathfrak{m}}\left\lvert
S(\alpha)\right\rvert\,d\alpha$
$\displaystyle\leq\Bigg{(}\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\Big{\lvert}\prod_{i\neq
2,3}f_{i}(\alpha,\bm{\beta})\Big{\rvert}\Bigg{)}\left(\int_{[0,1]^{3}}\left\lvert\mathcal{F}(\alpha,\bm{\beta})\mathcal{H}_{2}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$
$\displaystyle\times\left(\int_{[0,1]^{3}}\left\lvert\mathcal{H}_{1}(\alpha,\bm{\beta})f_{2}(\alpha,\bm{\beta})f_{3}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$

Because $d_{1}\neq 0$ we again can save an arbitrarily large power of log over
the trivial bound. Namely:

(55) $\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\displaystyle\left\lvert\prod_{i\neq
2,3}f_{i}(\alpha,\bm{\beta})\right\rvert\ll X^{n-6}L^{-K}$

by Lemma 3.1. First integral can be bounded by:

(56) $\ll
L^{6}\Bigg{\\{}\begin{subarray}{c}2\textbf{x}^{T}A_{1}\textbf{x}^{\prime}+w_{2}y_{2}^{\prime}+w_{2}^{\prime}y_{2}=0\\\
2\textbf{x}^{T}B+y_{2}\bm{\gamma}_{2}^{T}-w_{2}\bm{e}_{2}^{T}=0\end{subarray}\Bigg{\\}}\ll
L^{6}\Bigg{\\{}-(y_{2}\bm{\gamma}_{2}^{T}-w_{2}\bm{e}_{2}^{T})B^{-1}A_{1}\textbf{x}^{\prime}+w_{2}y_{2}^{\prime}+w_{2}^{\prime}y_{2}=0\Bigg{\\}}$

where we again used the fact that $B$ was invertible to write x in terms of
$y_{2}$ and $w_{2}$. So we have a bilinear equation in 6 variables with
corresponding matrix of rank at least 2. So we can apply Lemma 3.4 to say that
the first integral is bounded by:

(57)
$\left(\int_{[0,1]^{3}}\displaystyle\left\lvert\mathcal{F}(\alpha,\bm{\beta})\mathcal{H}_{2}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}\ll(L^{7}X^{4})^{1/2}=L^{7/2}X^{2}.$

We again apply Lemmas 3.2 and 3.3 to the second integral in (53) to get that
it’s bounded by

(58) $\ll
L^{6}\Bigg{\\{}\begin{subarray}{c}w_{1}y_{1}^{\prime}+w_{1}^{\prime}y_{1}+2d_{2}z_{2}z_{2}^{\prime}+2d_{3}z_{3}z_{3}^{\prime}=0\\\
y_{1}\bm{\gamma_{1}^{T}}-w_{1}\bm{e}_{1}^{T}+2z_{2}\bm{\xi}_{2}^{T}+2z_{3}\bm{\xi}_{3}^{T}=0\end{subarray}\Bigg{\\}}.$

Because $d_{3}=0$ we note that variable $z_{3}^{\prime}$ doesn’t have any
restrictions on it, so can take $O(X)$ possible values. Therefore

(59)
$\int_{[0,1]^{3}}\displaystyle\left\lvert\mathcal{H}_{1}(\alpha,\bm{\beta})f_{2}(\alpha,\bm{\beta})f_{3}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\ll
L^{6}X\Bigg{\\{}\begin{subarray}{c}w_{1}y_{1}^{\prime}+w_{1}^{\prime}y_{1}+2d_{2}z_{2}z_{2}^{\prime}=0\\\
y_{1}\bm{\gamma_{1}^{T}}-w_{1}\bm{e}_{1}^{T}+2z_{2}\bm{\xi}_{2}^{T}+2z_{3}\bm{\xi}_{3}^{T}=0\end{subarray}\Bigg{\\}}.$

By what we said above $(\bm{e}_{1},\bm{\xi}_{3})$ is an ivertible matrix. So
we can write $w_{1}$ and $z_{3}$ in terms of $y_{1}$ and $z_{2}$. This leaves
us with a bilinear equation in 5 variables of rank at least 2 as $d_{2}\neq 0$
and coefficient by $w_{1}^{\prime}y_{1}$ is non zero. So using Lemma 3.4 again
we get:

(60)
$\displaystyle\left(\int_{[0,1]^{3}}\displaystyle\left\lvert\mathcal{H}_{1}(\alpha,\bm{\beta})f_{2}(\alpha,\bm{\beta})f_{3}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}\ll(XL^{7}X^{3})^{1/2}=L^{7/2}X^{2}.$

After combining the result we get the exact expression we need.
∎

####  4.2.2. Proof of Lemma 4.5

######  Lemma 4.8.

If $A$ is a matrix with
$\operatorname{rank}(B_{1})=2,\operatorname{rank}(B_{2})=1$ then $A$ is of the
form

$A=\begin{pmatrix}A_{1}&\bm{\gamma}_{1}&\bm{\gamma}_{2}\bm{\xi}^{T}\\\
\bm{\gamma}_{1}^{T}&a&\bm{v}^{T}\\\
\bm{\xi}\bm{\gamma}_{2}^{T}&\bm{v}&D+h\bm{\xi}\bm{\xi}^{T}\end{pmatrix}$

where $\bm{\gamma}_{1}\in\mathbb{Z}^{2}$, $\bm{\gamma}_{2}\in\mathbb{Q}^{2}$,
$\bm{\xi},\bm{v}\in\mathbb{Z}^{n-3}$, $h\in\mathbb{Q}$ and
$D=\operatorname{diag}(d_{1},...,d_{n-3})$. Finally one also has
$(\bm{\gamma}_{1},\bm{\gamma}_{2})\in GL_{2}(\mathbb{Q})$.

###### Proof.

We begin by noting that as $\operatorname{rank}(B_{2}$)=1 and
$(a_{1,4},a_{2,4})^{T}\neq 0$ we can say that
$B_{2}=(a_{1,4},a_{2,4})^{T}\bm{\xi}^{\prime T}$ with
$\bm{\xi}^{\prime}\in\mathbb{Q}^{n-3}$. So after rescaling appropriately we
have $B_{2}=\bm{\gamma}_{2}\bm{\xi}^{T}$ for some
$\bm{\xi}\in\mathbb{Z}^{n-3}$ and $\bm{\gamma}_{2}$ satisfying
$\operatorname{rank}(B)=\operatorname{rank}(\bm{\gamma}_{1},\bm{\gamma}_{2})=2$.
Let’s define for $4\leq j\leq n$, $1\leq i\leq 2$
$\bm{\eta}_{j}=(a_{j,3},...,a_{j,j-1},a_{j,j+1},...,a_{j,n})^{T}$,
$\bm{\theta}_{i,j}=(a_{i,3},...a_{i,j-1},a_{i,j+1},...,a_{i,n})^{T}$. We know
that $\operatorname{rank}(B)=\operatorname{rank}(B_{1})=2$ so for each $j$,
$\bm{\eta}_{j}$ is a linear combination of
$\bm{\theta}_{1,j},\bm{\theta}_{2,j}$. Because of that if we let
$\bm{\theta}_{i}=(a_{i,4},...,a_{i,n})^{T}$ for $i=1,2$ then for each $4\leq
j\leq n$ there is exactly one $a_{j,j}^{\prime}\in\mathbb{Q}$ such that
$\bm{\varphi}_{j}=(a_{j,4},...,a_{j,j-1},a_{j,j}^{\prime},a_{j,j+1},...,a_{j,n})^{T}$
is a linear combination of $\bm{\theta}_{1}$ and $\bm{\theta}_{2}$. Because
$\operatorname{rank}(B_{2}$)=1 we have that $\bm{\theta}_{1}$ and
$\bm{\theta}_{2}$ are both a rational multiple of $\bm{\xi}$ i.e.
$\bm{\varphi}_{j}=h_{j}\bm{\xi}$ for some $h_{j}\in\mathbb{Q}$. If we now let
$\mathbf{h}^{T}=(h_{2},...,h_{n})$ then

$A=\begin{pmatrix}A_{1}&\bm{\gamma}_{1}&\bm{\gamma}_{2}\bm{\xi}^{T}\\\
\bm{\gamma}_{1}^{T}&a&\bm{v}^{T}\\\
\bm{\xi}\bm{\gamma}_{2}^{T}&\bm{v}&D+\bm{h}\bm{\xi}^{T}\end{pmatrix}$

for $D$ a diagonal matrix. But as $A$ is symmetric we get
$\mathbf{h}=h\bm{\xi}$ for some $h\in\mathbb{Q}$. ∎

######  Proof of Lemma 4.5.

Note that
$\operatorname{rank}(D)+\operatorname{rank}(\bm{v})+1+3\geq\operatorname{rank}(A)\geq
8$ which gives us $\operatorname{rank}(D)\geq 3$. If we let
$\bm{\xi}^{T}=(\epsilon_{1},...,\epsilon_{n-3})$, then as
$\operatorname{rank}(B)=\operatorname{rank}(B_{1}$)=2 we can find $2\leq l\leq
n-3$ such that $\epsilon_{1}\epsilon_{l}\neq 0$. As
$\operatorname{rank}(D)\geq 3$ we know that $d_{k}\neq 0$ for some $2\leq
k\neq l$. So suppose w.l.o.g. that $l=2$ and $k=3$. Let’s now proceed with the
estimates.

(61) $\displaystyle S(\alpha)=$
$\displaystyle\sum_{\begin{subarray}{c}\textbf{x}\leq X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\sum_{\begin{subarray}{c}y\leq X\\\
y\in\mathbb{N}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{z}\leq X\\\
\textbf{z}\in\mathbb{N}^{n-3}\end{subarray}}\Lambda(\textbf{x})\Lambda(y)\Lambda(\textbf{z})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+2\textbf{x}^{T}\bm{\gamma_{1}}y+ay^{2}+2\textbf{x}^{T}\bm{\gamma_{2}}\bm{\xi}^{T}z))$
$\displaystyle\times
e(\alpha(2y\bm{v}^{T}\textbf{z}+\textbf{z}^{T}D\textbf{z}+\textbf{z}^{T}h\bm{\xi}\bm{\xi}^{T}\textbf{z}))$

For each $\textbf{x}\in\mathbb{N}^{2}$, $y\in\mathbb{N}$ and
$\textbf{z}\in\mathbb{N}^{n-3}$ satisfying $\textbf{x},y,\textbf{z}\leq X$ we
introduce new variables $s=\bm{\xi}^{T}\textbf{z}$,
$w=2\textbf{x}^{T}\bm{\gamma_{1}}+ay+2\textbf{z}^{T}\bm{v}$ and use
orthogonality property of exponential function to write $S(\alpha)$ as

(62) $\displaystyle S(\alpha)={}$
$\displaystyle\int_{[0,1]^{2}}\sum_{\begin{subarray}{c}\lvert s\rvert\ll X\\\
s\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}\lvert w\rvert\ll X\\\
w\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{x}\leq X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\sum_{\begin{subarray}{c}y\leq X\\\
y\in\mathbb{N}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{z}\leq X\\\
\textbf{z}\in\mathbb{N}^{n-3}\end{subarray}}\Lambda(\textbf{x})\Lambda(y)\Lambda(\textbf{z})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+wy+\textbf{z}^{T}D\textbf{z}))$
$\displaystyle\times
e(\alpha(hs^{2}+2\textbf{x}^{T}\bm{\gamma}_{2}s)+(2\textbf{x}^{T}\bm{\gamma}_{1}+ay+2\textbf{z}^{T}\bm{v}-w)\beta_{1}+(\bm{\xi}^{T}\textbf{z}-s)\beta_{2})d\bm{\beta}$

where $d\bm{\beta}=d\beta_{1}\,d\beta_{2}$ As before we will split the
integral into different parts. First we introduce different exponential sums.

(63)
$\displaystyle\mathcal{F}(\alpha,\bm{\beta})=\sum_{\begin{subarray}{c}\lvert
s\rvert\ll X\\\
s\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{x}\leq X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\Lambda(\textbf{x})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+hs^{2}+2\textbf{x}^{T}\bm{\gamma}_{2}s)+2\textbf{x}^{T}\bm{\gamma}_{1}\beta_{1}-s\beta_{2})$
$\displaystyle\mathcal{H}(\alpha,\bm{\beta})=\sum_{\begin{subarray}{c}\lvert
w\rvert\ll X\\\ w\in\mathbb{Z}\end{subarray}}\sum_{\begin{subarray}{c}y\leq
X\\\ y\in\mathbb{N}\end{subarray}}e(\alpha
wy+ay\beta_{1}-w\beta_{1})\Lambda(y)$

and finally for $j=1,2,...,n-3$ we let

(64) $f_{j}(\alpha,\bm{\beta})=\sum_{\begin{subarray}{c}z_{j}\leq X\\\
z_{j}\in\mathbb{N}\end{subarray}}\Lambda(\textbf{z})e(\alpha
d_{j}z_{j}^{2}+2z_{j}v_{j}\beta_{1}+\epsilon_{j}z_{j}\beta_{2})$

where $\bm{v}^{T}=(v_{1},...,v_{n-3})$. Let’s bound the integral as below.

(65) $\displaystyle\int_{\mathfrak{m}}\left\lvert
S(\alpha)\right\rvert\,d\alpha\leq$
$\displaystyle\Bigg{(}\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\big{\lvert}\prod_{k\neq
2,3}f_{k}(\alpha,\bm{\beta})\big{\rvert}\Bigg{)}\left(\int_{[0,1]^{3}}\left\lvert\mathcal{F}(\alpha,\bm{\beta})f_{3}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$
$\displaystyle\times\left(\int_{[0,1]^{3}}\left\lvert\mathcal{H}(\alpha,\bm{\beta})f_{2}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$

From our discussion above we know that $d_{3}\neq 0$. So we can bound the
first integral by:

(66) $\displaystyle\ll
L^{6}\Bigg{\\{}\begin{subarray}{c}2\textbf{x}^{T}A_{1}\textbf{x}^{\prime}+2hss^{\prime}+2d_{3}z_{3}z_{3}^{\prime}+\textbf{x}^{T}\bm{\gamma}_{2}s^{\prime}+s\bm{\gamma}_{2}^{T}\textbf{x}^{\prime}=0\\\
\textbf{x}^{T}\bm{\gamma}_{1}+z_{3}v_{3}=0\\\
s=\epsilon_{3}z_{3}\end{subarray}\Bigg{\\}}$ (67) $\displaystyle\ll
L^{6}\Bigg{\\{}\begin{subarray}{c}2\textbf{x}^{T}A_{1}\textbf{x}^{\prime}+2h\epsilon_{3}z_{3}s^{\prime}+2d_{3}z_{3}z_{3}^{\prime}+\textbf{x}^{T}\bm{\gamma}_{2}s^{\prime}+\epsilon_{3}z_{3}\bm{\gamma}_{2}^{T}\textbf{x}^{\prime}=0\\\
\textbf{x}^{T}\bm{\gamma}_{1}+z_{3}v_{3}=0\end{subarray}\Bigg{\\}}.$

Let’s introduce a new variable $q=\textbf{x}^{T}\bm{\gamma}_{2}$. We can
rewrite the above set of equations as

(68)
$\Bigg{\\{}\begin{subarray}{c}2\textbf{x}^{T}A_{1}\textbf{x}^{\prime}+2h\epsilon_{3}z_{3}s^{\prime}+2d_{3}z_{3}z_{3}^{\prime}+qs^{\prime}+\epsilon_{3}z_{3}\bm{\gamma}_{2}^{T}\textbf{x}^{\prime}=0\\\
\textbf{x}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})+(z_{3}v_{3},-q)=0\end{subarray}\Bigg{\\}}.$

Note that as $(\bm{\gamma}_{1},\bm{\gamma}_{2})$ is invertible we can again
write x as a combination of $z_{3}$ and $q$. After that we are left with
bilinear equation in 6 variables with rank at least 2

(69)
$\Bigg{\\{}-2(z_{3}v_{3},-q)(\bm{\gamma}_{1},\bm{\gamma}_{2})^{-1}A_{1}\textbf{x}^{\prime}+2h\epsilon_{3}z_{3}s^{\prime}+2d_{3}z_{3}z_{3}^{\prime}+qs^{\prime}+\epsilon_{3}z_{3}\bm{\gamma}_{2}^{T}\textbf{x}^{\prime}=0\Bigg{\\}}.$

So we again apply Lemma 3.4 to get:

(70)
$\left(\int_{[0,1]^{3}}\displaystyle\left\lvert\mathcal{F}(\alpha,\bm{\beta})f_{3}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}\ll(L^{7}X^{4})^{1/2}=L^{7/2}X^{2}.$

For the second integral we use Lemmas 3.2 and 3.3 to arrive at:

(71)
$\int_{[0,1]^{3}}\displaystyle\left\lvert\mathcal{H}(\alpha,\bm{\beta})f_{2}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\ll
L^{4}\Bigg{\\{}\begin{subarray}{c}2d_{2}z_{2}z_{2}^{\prime}+wy^{\prime}+w^{\prime}y=0\\\
2z_{2}v_{2}+ya-w=0\\\ \epsilon_{2}z_{2}=0\end{subarray}\Bigg{\\}}.$

As $\epsilon_{2}\neq 0$ we must have $z_{2}=0$. Leaving us with:

(72) $L^{4}\Bigg{\\{}\begin{subarray}{c}wy^{\prime}+w^{\prime}y=0\\\
ya=w\end{subarray}\Bigg{\\}}\ll
L^{4}\Bigg{\\{}y(ay^{\prime}+w^{\prime})=0\Bigg{\\}}\ll L^{4}X^{2}.$

Finally as $\operatorname{rank}(D)\geq 3$ we can find $k\neq 2,3$ such that
$d_{k}\neq 0$. Therefore by Lemma 3.1 we get

(73) $\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\displaystyle\left\lvert\prod_{k\neq
2,3}f_{k}(\alpha,\bm{\beta})\right\rvert\ll L^{-K}X^{n-5}$

After combining all three estimates we get the desired result. ∎

####  4.2.3. Proof of Lemma 4.6

######  Lemma 4.9.

Let $\operatorname{rank}(B_{1})=\operatorname{rank}(B_{2})=2$. Then $A$ is of
the form:

$A=\begin{pmatrix}A_{1}&(\bm{\gamma}_{1},\bm{\gamma}_{2})C\\\
C^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})^{T}&D+C^{T}HC\end{pmatrix}$

where $(\bm{\gamma}_{1},\bm{\gamma}_{2})\in GL_{2}(\mathbb{Q})$, $C\in
M_{2,n-2}(\mathbb{Z})$, $H\in M_{2}(\mathbb{Q})$ and
$D=\operatorname{diag}(d_{1},...,d_{n-2})$

###### Proof.

The proof goes along the same lines as before. Note that each vector
$(a_{1,2+j},a_{2,2+j})^{T}$ for $j=1,2,...,n-2$ is a linear combination of
$(a_{1,3},a_{2,3})^{T}$ and $(a_{1,4},a_{2,4})^{T}$. Therefore we have

$A=\begin{pmatrix}A_{1}&(\bm{\gamma}_{1},\bm{\gamma}_{2})C\\\
C^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})^{T}&A_{2}\end{pmatrix}$

after rescaling we get $C\in M_{2,n-2}(\mathbb{Z})$ and
$\operatorname{rank}((\bm{\gamma}_{1},\bm{\gamma}_{2}))=2$. For the final part
let’s define for $3\leq j\leq n$ and $1\leq i\leq 2$,
$\bm{\eta}_{j}=(a_{j,3},...,a_{j,j-1},a_{j,j+1},...,a_{j,n})^{T}$ and
$\bm{\theta}_{i,j}=(a_{i,3},...,a_{i,j-1},a_{i,j+1},...,a_{i,n})^{T}$.
Similarly as before we can say that for each $j$, $\bm{\eta}_{j}$ is a linear
combination of $\bm{\theta}_{1,j},\bm{\theta}_{2,j}$. For $j=3$ it’s because
$\operatorname{rank}(B_{2})=2$ for $j=4$ because
$\operatorname{rank}(B_{1})=2$ and for $j\geq 5$ because
$\operatorname{rank}(B)=2$. Finally for $i=1,2$ we define
$\bm{\theta}_{i}=(a_{i,3},...,a_{i,n})^{T}$ and for $3\leq j\leq n$ we observe
that there is a unique $a_{j,j}^{\prime}\in\mathbb{Q}$ such that
$\bm{\varphi}_{j}=(a_{j,3},...,a_{j,j-1},a_{j,j}^{\prime},a_{j,j+1},...,a_{j,n})^{T}$
is a linear combination of $\bm{\theta}_{1}$ and $\bm{\theta}_{2}$. This means
that $A$ has the form

$A=\begin{pmatrix}A_{1}&(\bm{\gamma}_{1},\bm{\gamma}_{2})C\\\
C^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})^{T}&D+MC\end{pmatrix}$

and as $MC$ is symmetric it can be written in the form $C^{T}HC$ for $H\in
M_{2}(\mathbb{Q})$. ∎

######  Lemma 4.10.

Let $A$ be given by the form above. Let
$C=(\bm{\xi}_{1},\bm{\xi}_{2},...,\bm{\xi}_{n-2})$. Then we can find pairwise
distinct $b_{i}$ with $1\leq b_{1},b_{2},b_{3},b_{4},b_{5}\leq n-2$ such that
$\operatorname{rank}((\bm{\xi}_{b_{1}},\bm{\xi}_{b_{2}}))=2$ and
$d_{b_{3}}d_{b_{4}}d_{b_{5}}\neq 0$

###### Proof.

We know that as $\operatorname{rank}(B)=2$ we have
$\operatorname{rank}(\bm{\xi}_{1},\bm{\xi}_{2})=2$. So suppose the above
statement doesn’t happen. Namely $\forall k_{1},k_{2},k_{3}\geq 3$ we have
$d_{k_{1}}d_{k_{2}}d_{k_{3}}=0$. This gives the upper bound
$\operatorname{rank}(D)\leq 4$. But we know from looking at form of $A$ and
the fact that $\operatorname{rank}(A)\geq 8$ we have
$\operatorname{rank}(D)\geq 4$.
So we only have case $\operatorname{rank}(D)=4$ left. We may suppose that
$d_{1}d_{2}d_{3}d_{4}\neq 0$ and $d_{j}=0$ for $j\geq 4$. But then by looking
at the rank of whole matrix we conclude that
$\operatorname{rank}(\bm{\xi}_{5},\bm{\xi}_{6},...,\bm{\xi}_{n-2})=2$. That
then obviously gives us a desired quintuple. ∎

######  Proof of Lemma 4.6.

Using Lemma 4.9 we know the structure of our exponential sum.

(74) $S(\alpha)=\sum_{\begin{subarray}{c}\textbf{x}\leq X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{y}\leq
X\\\
\textbf{y}\in\mathbb{N}^{n-2}\end{subarray}}\Lambda(\textbf{x})\Lambda(\textbf{y})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+2\textbf{x}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})C\textbf{y}+\textbf{y}^{T}D\textbf{y}+\textbf{y}^{T}C^{T}HC^{T}\textbf{y}))$

For each $\textbf{y}\in\mathbb{N}^{n-2}$ satisfying $\textbf{y}\leq X$ we let
$\textbf{z}=C\textbf{y}$ and use orthogonality of exponential function to get

(75) $\displaystyle S(\alpha)={}$
$\displaystyle\int_{[0,1]^{2}}\sum_{\begin{subarray}{c}\lvert\textbf{z}\rvert\ll
X\\\
\textbf{z}\in\mathbb{Z}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{x}\leq
X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{y}\leq
X\\\
\textbf{y}\in\mathbb{N}^{n-2}\end{subarray}}\Lambda(\textbf{x})\Lambda(\textbf{y})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+2\textbf{x}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})\textbf{z}+\textbf{y}^{T}D\textbf{y})$
$\displaystyle\times
e(\alpha\textbf{z}^{T}H\textbf{z}+(\textbf{y}^{T}C^{T}-\textbf{z}^{T})\bm{\beta})\,d\bm{\beta}.$

We introduce new sums

(76)
$\mathcal{F}(\alpha,\bm{\beta})=\sum_{\begin{subarray}{c}\lvert\textbf{z}\rvert\ll
X\\\
\textbf{z}\in\mathbb{Z}^{2}\end{subarray}}\sum_{\begin{subarray}{c}\textbf{x}\leq
X\\\
\textbf{x}\in\mathbb{N}^{2}\end{subarray}}\Lambda(\textbf{x})e(\alpha(\textbf{x}^{T}A_{1}\textbf{x}+2\textbf{x}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})\textbf{z}+\textbf{z}^{T}H\textbf{z})-\textbf{z}^{T}\bm{\beta})$

and for $i=1,...,n-2$

(77) $f_{i}(\alpha,\bm{\beta})=\sum_{\begin{subarray}{c}y_{i}\leq X\\\
y_{i}\in\mathbb{N}\end{subarray}}\Lambda(y_{i})e(\alpha
d_{i}y_{i}^{2}+y_{i}\bm{\xi}_{i}^{T}\bm{\beta})$

where $\bm{\xi}_{i}$ is the $i$-th column vector of $C$. We again use CS
inequality to get that:

(78) $\displaystyle\int_{\mathfrak{m}}\left\lvert
S(\alpha)\right\rvert\,d\alpha\leq$
$\displaystyle\Bigg{(}\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\big{|}\prod_{k\neq
b_{1},b_{2},b_{3},b_{4}}f_{k}(\alpha,\bm{\beta})\big{|}\Bigg{)}\left(\int_{[0,1]^{3}}\left\lvert\mathcal{F}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$
$\displaystyle\times\left(\int_{[0,1]^{3}}\left\lvert
f_{b_{1}}(\alpha,\bm{\beta})f_{b_{2}}(\alpha,\bm{\beta})f_{b_{3}}(\alpha,\bm{\beta})f_{b_{4}}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}$

where $b_{i}$ are chosen according to the rules from the previous lemma. The
supremum can be bounded using Lemma 3.1 (as $d_{b_{5}}\neq 0$) by

(79) $\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}\\\
\bm{\beta}\in[0,1]^{2}\end{subarray}}\displaystyle\left\lvert\prod_{k\neq
b_{1},b_{2},b_{3},b_{4}}f_{k}(\alpha,\bm{\beta})\right\rvert\ll
X^{n-6}L^{-K}.$

First integral can be bounded using Lemma 3.2 and 3.3 by

(80) $\ll
L^{4}\Bigg{\\{}\begin{subarray}{c}2\textbf{x}^{T}A_{1}\textbf{x}^{\prime}+2\textbf{x}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})\textbf{z}^{\prime}+2\textbf{z}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})^{T}\textbf{x}^{\prime}+2\textbf{z}^{T}H\textbf{z}^{\prime}=0\\\
z_{1}=0\\\ z_{2}=0\end{subarray}\Bigg{\\}}.$

After plugging in $z_{1}=z_{2}=0$ we are left with

(81) $\ll
L^{4}\Bigg{\\{}2\textbf{x}^{T}A_{1}\textbf{x}^{\prime}+2\textbf{x}^{T}(\bm{\gamma}_{1},\bm{\gamma}_{2})\textbf{z}^{\prime}=0\Bigg{\\}}\ll
L^{5}X^{4}$

where we again used Lemma 3.4 as we’re working with 6 variables and
$(\bm{\gamma}_{1},\bm{\gamma}_{2})$ is invertible. If we look at the second
integral and use Lemma 3.2 and 3.3 we obtain the bound:

(82) $\ll
L^{8}\Bigg{\\{}\begin{subarray}{c}d_{b_{1}}y_{b_{1}}y_{b_{1}}^{\prime}+d_{b_{2}}y_{b_{2}}y_{b_{2}}^{\prime}+d_{b_{3}}y_{b_{3}}y_{b_{3}}^{\prime}+d_{b_{4}}y_{b_{4}}y_{b_{4}}^{\prime}=0\\\
y_{b_{1}}\bm{\xi}_{b_{1}}+y_{b_{2}}\bm{\xi}_{b_{2}}+y_{b_{3}}\bm{\xi}_{b_{3}}+y_{b_{4}}\bm{\xi}_{b_{4}}=0\end{subarray}\Bigg{\\}}.$

Because $\operatorname{rank}((\bm{\xi}_{b_{1}},\bm{\xi}_{b_{2}}))=2$ we can
write $y_{b_{1}}$ and $y_{b_{2}}$ in terms of $y_{b_{3}}$ and $y_{b_{4}}$ and
plug in into the first equation. We are left with bilinear equation in 6
variables and as $d_{b_{3}}d_{b_{4}}\neq 0$ it has rank at least 2. So after
applying Lemma 3.4 we get

(83) $\left(\int_{[0,1]^{3}}\displaystyle\left\lvert
f_{b_{1}}(\alpha,\bm{\beta})f_{b_{2}}(\alpha,\bm{\beta})f_{b_{3}}(\alpha,\bm{\beta})f_{b_{4}}(\alpha,\bm{\beta})\right\rvert^{2}\,d\alpha\,d\bm{\beta}\right)^{\tfrac{1}{2}}\ll(L^{9}X^{4})^{1/2}=L^{9/2}X^{2}$

Once we combine all three estimated we get the desired result and finishes the
whole discussion on off diagonal rank 2 matrices. ∎

##  5\. Some Thanks

I would like to thank my supervisor Aled Walker for an indescribable support
and Department of Pure Mathematics and Mathematical Statistics for funding my
work on this project.

## References

  * [1] L. Zhao, The Quadratic Form In Nine Prime Variables, Nagoya Math. J., 223(1) (2016), 21–65 DOI 10.1017/nmj.2016.23
  * [2] L.K Hua, Some results in the additive prime number theory, Quart. J. Math., 9 (1938), 68 - 80.
  * [3] J. Liu Integral points on quadrics with prime coordinates, Monatsh Math (2011) 164:439–465 DOI 10.1007/s00605-010-0253-5
  * [4] B. J. Green, Quadratic forms in 8 prime variables arXiv:2108.10401 [math.NT]
  * [5] L.K Hua, Additive Theory of Prime Numbers, Providence, American Mathematical Society, 1965 (OCoLC)624471062


* [55] M. Feiginov, C. Sydlo, O. Cojocari, and P. Meissner, EPL 97, 58006 (2012).
  * [56] A. M. Dirac, Proceedings of the Royal Society of London A. 114, 24 (1927).
  * [57] S. Tomonaga, Progress of Theoretical Physics 1, 27 (1946).
  * [58] J. Schwinger, Physical Review 73, 416 (1946).
  * [59] R. P. Feynman, Physical Review 76, 769 (1949).
  * [60] J. George, T. Chervy, A. Shalabney, E. Devaux, H. Hiura, C. Genet, and T. W. Ebbesen, Phys. Rev. Lett. 117, 153601 (2016).
  * [61] J. Galego, F. J. Garcia-Vidal, and J. Feist, Nat. Commun. 7, 13841 (2016).
  * [62] X. Wang, E. Ronca, and M. A. Sentef, Phys. Rev. B 99, 235156 (2019).
  * [63] J. Li, D. Golez, G. Mazza, A. J. Millis, A. Georges, and M. Eckstein, Phys. Rev. B 101, 205140 (2020).
  * [64] Y. Ashida, A. Imamoglu, and E. Demler, Phys. Rev. Lett. 126, 153603 (2021).
  * [65] V. Rokaj, M. Penz, M. A. Sentef, M. Ruggenthaler, and A. Rubio, Phys. Rev. Lett. 123, 047202 (2019).
  * [66] J. Flick, D. M. Welakuh, M. Ruggenthaler, H. Appel, and A. Rubio, ACS Photonics 6, 2757 (2019).
  * [67] V. Rokaj, M. Ruggenthaler, F. G. Eich, and A. Rubio, Phys. Rev. Research 4, 013012 (2022).
  * [68] C. Cohen-Tannoudji, J. Dupont-Roc, and G. Grynberg, Photons and Atoms-Introduction to Quantum Electrodynamics (Wiley, 2004).
  * [69] G. Grynberg, A. Aspect, and C. Fabre, Introduction to Quantum Optics: From the Semi-Classical Approach to Quantized Light (Cambridge University Press, 2011).
  * [70] M. O. Scully and M. S. Zubairy, Quantum optics (Cambridge University Press, 2008).
  * [71] X. Oriols and J. Mompart, Applied Bohmian Mechanics From Nanoscale Systems to Cosmology (Jenny Stanford Publishing, 2019).
  * [72] D. Bohm, Phys. Rev. 85, 166–179 (1952).
  * [73] D. Bohm, Phys. Rev. 85, 180–193 (1952).
  * [74] The Coulomb gauge (also known as the transverse gauge) is the typical gauge used in quantum chemistry and condensed matter physics. It is particularly useful for the present calculations because it will allow us to quantize the vector potential, while still treating the Coulomb interaction in a classical way (as it is routinely done in electronics). Notice that the fact that the longitudinal electric field $\mathbf{E}_{\parallel}(\mathbf{r},t)$ instantaneously follows the evolution of the charge distribution does not imply the existence of faster than light phenomena. It can be shown that the transverse electrical field also has an instantaneous component which exactly balances that of the longitudinal electric field so that the total (measured) electric field remains purely retarded [68].
  * [75] C. G. Darwin, The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science 39, 537 (1920).
  * [76] X. Oriols, Phys. Rev. Lett. 98, 066803 (2007).
  * [77] We have neglected in our calculations the “self-energy” of an electron interacting with itself. This “self-energy” term would produce an infinite contribution of the longitudinal electric field $\mathbf{E}_{\parallel}(\mathbf{r},t)$ to the total electromagnetic energy. But, since such a term does not vary with time it can be considered as a simple shift in the zero energy and it is routinely ignored in the literature [68]. See also 28-4 in [95].
  * [78] The fact that $\mathbf{E}_{\perp}(\mathbf{r},t)$ and $\mathbf{B}(\mathbf{r},t)$ are the genuine degrees of freedom is not a consequence of our single-electron assumption. From Maxwell’s equations and the Coulomb gauge, we know that $\mathbf{E}_{\parallel}(\mathbf{r},t)$ depends only on the positions of the electrons. Thus $\mathbf{E}_{\parallel}(\mathbf{r},t)$ is not a genuine degree of freedom [68].
  * [79] The pertinent Maxwell’s equations in the reciprocal space are $\frac{\partial\tilde{\mathbf{B}}(\mathbf{k},t)}{\partial t}=-i\mathbf{k}\times\tilde{\mathbf{E}}_{\perp}(\mathbf{k},t),$ and $\frac{\partial\tilde{\mathbf{E}}_{\perp}(\mathbf{k},t)}{\partial t}=ic^{2}\mathbf{k}\times\tilde{\mathbf{B}}(\mathbf{k},t)-\frac{1}{\epsilon_{0}}\tilde{\mathbf{J}}_{c\perp}(\mathbf{k},t)$.
  * [80] Here, $\alpha$ depends on the type of electromagnetic field considered through $\Delta k$ while, in the literature, such dependence is typically not present and it is substituted by an arbitrary dependence on the size of a mathematical cavity that contains the electromagnetic field inside. Our paper uses Fourier integrals, while the use of a mathematical cavity implies Fourier series. Such different approaches do not have any implications in the results of this paper.
  * [81] An intuitive explanation of the non-local character of quantum phenomena in the Bohmian language is that an electron corresponding to some $a$-experiment guided by a Hamiltonian is affected by what happens to another electron in a different $b$-experiment also governed by the same Hamiltonian. In other words, the trajectory $x_{a}(t)$ is affected by the trajectory $x_{b}(t)$, even if they are different experiments and have (non-local) spatial separation. The appropriate element that can capture such non-classical (non-local) behavior is a wave function whose extension (coherence length) includes $x_{a}(t)$ and $x_{b}(t)$ in the relevant times.
  * [82] According to the orthodox theory, the properties of the studied system such as the position of the electron or the amplitude of the electromagnetic field have no well-defined values while the system follows the unitary evolution. Such properties are only well-defined when a (strong) measurement is done and the system collapse into a state with well-defined properties. The orthodox theory states that the measurement process is not governed by the unitary evolution, but by a new law named wave function collapse (or state reduction), that provides a definite value of the position of the electron, if measured, or of the amplitude of the electromagnetic field, if measured. This relation between the measurement and the reality of the properties of the quantum system is known as the eigenstate-eigenvalue-link. When one gets information of a quantum system at THz frequencies, the system is measured at time steps lower than $1$ ps. Within the orthodox theory, such multi-time measurement brings computational difficulties when changing from unitary to non-unitary evolutions of a quantum state that can be greatly simplified when using a Bohmian explanation of such measurement [71].
  * [83] X. Oriols and D. K. Ferry, Proc. IEEE 109, 955 (2021).
  * [84] S. Ramo, Proc. IRE 27, 584 (1939).
  * [85] W. Shockley, J. Appl. Phys. 9, 635 (1938).
  * [86] B. Pellegrini, Phys. Rev. B 34, 5921 (1986).
  * [87] Kevin L. Jensen, Joel L. Lebowitz, Jeanne M. Riga, Donald A. Shiffler, and Rebecca Seviour, Phys. Rev. B 103, 155427 (2021).
  * [88] Damien Querlioz, Jérôme Saint-Martin, Arnaud Bournel, and Philippe Dollfus, Phys. Rev. B 78, 165306 (2008).
  * [89] F. Mazzamuto, V. Hung Nguyen, Y. Apertet, C. Caer, C. Chassat, J. Saint-Martin, and P. Dollfus, Phys. Rev. B 83, 235426 (2011).
  * [90] E. D. Herbschleb, R. K. Puddy, P. Marconcini, J. P. Griffiths, G. A. C. Jones, M. Macucci, C. G. Smith, and M. R. Connolly, Phys. Rev. B 92, 119905(E) (2015).
  * [91] P. Marconcini, M. Macucci, G. Iannaccone, and B. Pellegrini, Phys. Rev. B 79, 241307(R) (2009).
  * [92] H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems (Oxford University Press, 2007).
  * [93] G. Lindblad, Communications in Mathematical Physics 48, 119 (1976).
  * [94] T. Norsen, Found. Phys. 40, 1858 (2010).
  * [95] R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Vol. II (Addison-Wesley Publishing Company,1965).