# On The Partition Regularity of $ax+by=cw^{m}z^{n}$

Sohail Farhangi and Richard Magner

###### Abstract

Csikvári, Gyarmati, and Sárközy showed that the equation $x+y=z^{2}$ is not
partition regular (PR) over $\mathbb{N}$ and asked if the equation $x+y=wz$ is
PR over $\mathbb{N}$. Bergelson and Hindman independently answered this
question in the positive. We generalize this result by giving a partial
classification of the $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and
$m,n\in\mathbb{N}$ for which the equation $ax+by=cw^{m}z^{n}$ is PR over
$\mathbb{Z}\setminus\\{0\\}$. We show that if $m,n\geq 2$, then
$ax+by=cw^{m}z^{n}$ is PR over $\mathbb{Z}\setminus\\{0\\}$ if and only if
$a+b=0$. Next, we show that if $n$ is odd, then the equation $ax+by=cwz^{n}$
is PR over $\mathbb{Z}\setminus\\{0\\}$ if and only if one of
$\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is an $n$th power in
$\mathbb{Q}$. We come close to a similar characterization of the partition
regularity of $ax+by=cwz^{n}$ over $\mathbb{Z}\setminus\\{0\\}$ for even $n$,
and we examine some equations whose partition regularity remain unknown, such
as $16x+17y=wz^{8}$. Furthermore, we show that for an integral domain $R$,
$a,b,c\in R\setminus\\{0\\}$, and $n\in\mathbb{N}$, the equation
$ax+by=cwz^{n}$ is partition regular over $R\setminus\\{0\\}$ if one of
$\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is an $n$th power in the field
of fractions $K$ of $R$. Specializing to the case in which $R$ is a Dedekind
domain, we obtain results that suggest $ax+by=cwz^{n}$ is not partition
regular over $R\setminus\\{0\\}$ if none of $\frac{a}{b},\frac{b}{c},$ or
$\frac{a+b}{c}$ are $n$th powers in $K$ and we obtain some applications to
rings of integers of number fields. We also obtain some results supporting the
conjecture that a system of equations of the form
$a_{i}x_{i}+b_{i}y_{i}=c_{i}w_{i}z_{i}^{n},1\leq i\leq k$ is PR over
$\mathbb{Z}\setminus\\{0\\}$ if and only if
$\cap_{i=1}^{j}\\{\frac{a_{i}}{c_{i}},\frac{b_{i}}{c_{i}},\frac{a_{i}+b_{i}}{c_{i}}\\}$
contains an $n$th power in $\mathbb{Q}$.

In order to show that the equation $ax+by=cwz^{n}$ is not PR over
$\mathbb{Z}\setminus\\{0\\}$ for certain values of $a,b,c,$ and $n$, we prove
a partial generalization of the criteria of Grunwald and Wang for when
$\alpha\in\mathbb{Z}$ is an $n$th power modulo every prime $p$. In particular,
we show that for any odd $n$ and any $\alpha,\beta,\gamma\in\mathbb{Q}$ that
are not $n$th powers, there exist infinitely many primes $p\in\mathbb{N}$ for
which none of $\alpha,\beta,$ and $\gamma$ are $n$th powers modulo $p$.
Similarly, we show that for any even $n$ and any
$\alpha,\beta,\gamma\in\mathbb{Q}$ that are not $\frac{n}{2}$th powers, with
one not an $\frac{n}{4}$th power if $4|n$, there exist infinitely many primes
$p\in\mathbb{N}$ for which $\alpha,\beta,$ and $\gamma$ are not $n$th powers
modulo $p$. In order to show that for an integral domain $R$ the equation
$ax+by=cwz^{n}$ is PR over $R\setminus\\{0\\}$ for certain values of $a,b,c$
and $n$, we use ultrafilters and Algebra of the Stone-C̆ech Compactification.
This necessitates us to prove some new results of independent interest about
ultrafilters $v\in\beta R$ for which every $A\in v$ contains significant
additive and multiplicative structure.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 On the Partition Regularity of $ax+by=cwz^{n}$ over $\mathbb{N}$ and $\mathbb{Z}\setminus\\{0\\}$
  4. 4 Number Theoretic Results
  5. 5 Reduction to the case min(m,n) = 1
  6. 6 On the Partition Regularity of $ax+by=cwz^{n}$ over Integral Domains
  7. 7 Systems of equations
  8. 8 Conjectures and Concluding Remarks
  9. 9 The Existence of Special Ultrafilters

##  1 Introduction

We say that an equation is _partition regular over $S$_ if for any finite
partition $S=\bigcup_{i=1}^{r}C_{i}$, there exists some $C_{i}$ containing a
solution to the equation. One of the first results about partition regular
diophantine equations is the celebrated theorem of Schur ([18]), which
established the partition regularity of $x+y=z$ over $\mathbb{N}$. Schur’s
student Rado ([17]) classified which finite systems of linear homogeneous
equations are partition regular over $\mathbb{N}$. Since it is not known
whether or not the equation $x^{2}+y^{2}=z^{2}$ is partition regular over
$\mathbb{N}$, we are still far from achieving a classification for which
systems of polynomial equations are partition regular. One of the first
results in this direction is a theorem of Bergelson ([2], page 53), which
shows that the equation $x-y=p(z)$ is partition regular over $\mathbb{N}$ for
any polynomial $p(x)\in\mathbb{Z}[x]$ which satisfies $p(0)=0$. While the
result of Bergelson shows that the equation $x-y=z^{2}$ is partition regular
over $\mathbb{N}$, Csikvári, Gyarmati and Sárkozy ([7]) showed that the
equation $x+y=z^{2}$ is _not_ partition regular over $\mathbb{N}$, and asked
whether the equation $x+y=wz$ is partition regular over $\mathbb{N}$. Their
question was answered in the positive, independently by Bergelson ([3],
Section 6) and Hindman ([10]). Both proofs make use of ultrafilters and the
algebra of the Stone-Čech compactifation. More examples of partition regular
polynomial equations can be found in [8]. Results regarding necessary
conditions for a polynomial equation to be partition regular over $\mathbb{N}$
can be found in [1] and [8].

The polynomial equations mentioned so far have a simple form, but the proofs
of their partition regularity properties are quite specific to these cases.
Since there is currently no unified theory for the partition regularity of
polynomial equations, any class of equations whose partition regularity is
known may provide insight towards such a theory. In this paper, we give a
partial classification of the partition regularity of equations of the form
$ax+by=cw^{m}z^{n}$, where $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and
$m,n\in\mathbb{N}$ are parameters, and $x,y,z,w$ are variables. Theorem 1.1 is
the main result of this paper. Before stating Theorem 1.1, we note that we
remove $0$ when considering partition regularity over a ring $R$ in order to
avoid trivial solutions. We also recall that any equation which is partition
regular over a set $S_{1}$ (such as $\mathbb{N}$) is also partition regular
over any set $S_{2}$ that contains $S_{1}$ (such as
$\mathbb{Z}\setminus\\{0\\})$, but the converse is not true in general.

######  Theorem 1.1.

Fix $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $m,n\in\mathbb{N}$.

  * (i)

Suppose that $m,n\geq 2$.

    * (a)

If $a+b\neq 0$, then the equation

$ax+by=cw^{m}z^{n}$ (1)

is not partition regular over $\mathbb{Z}\setminus\\{0\\}$.

    * (b)

If $a+b=0$, then equation (1) is partition regular over $\mathbb{N}$.

  * (ii)

If one of $\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is an $n$th power in
$\mathbb{Q}$, then the equation

$ax+by=cwz^{n}$ (2)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$. If one of
$\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is an $n$th power in
$\mathbb{Q}_{\geq 0}$, then equation (2) is partition regular over
$\mathbb{N}$.

  * (iii)

Assume that equation (2) is partition regular over
$\mathbb{Q}\setminus\\{0\\}$.111To give a converse to part (ii) of Theorem 1.1
we would assume that equation (2) is partition regular over $\mathbb{N}$ or
$\mathbb{Z}\setminus\\{0\\}$. The assumption that equation (2) is partition
regular over $\mathbb{Q}\setminus\\{0\\}$ is weaker than either of the
previous assumptions since
$\mathbb{N}\subseteq\mathbb{Z}\setminus\\{0\\}\subseteq\mathbb{Q}\setminus\\{0\\}$.

    1. (a)

If $n$ is odd, then one of $\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is an
$n$th power in $\mathbb{Q}$.

    2. (b)

If $n\neq 4,8$ is even, then one of $\frac{a}{c},\frac{b}{c},$ or
$\frac{a+b}{c}$ is an $\frac{n}{2}$th power in $\mathbb{Q}$.

    3. (c)

If $n$ is even, then either one of $\frac{a}{c},\frac{b}{c},$ or
$\frac{a+b}{c}$ is a square in $\mathbb{Q}$, or
$(\frac{a}{c})(\frac{b}{c})(\frac{a+b}{c})$ is a square in $\mathbb{Q}$.

To prove item (i)(a) we use one of the Rado conditions for polynomial
equations that is proven in [1], and we will show that item (i)(b) is an easy
consequence of the partition regularity of the equation $x-y=z^{n}$. To prove
item (ii) we use ultrafilters similar to those that were used in [3] and [10].
To prove item (iii) we use the classic partitions that were used by Rado in
[17] when determining necessary conditions for a finite system of linear
homogeneous equations to be partition regular. In order to demonstrate that
these classic partitions yield our desired results, we are required to prove
Theorem 1.2, which is a partial generalization of the criteria of Grunwald and
Wang for when $\alpha\in\mathbb{Z}$ is an $n$th power modulo every prime
$p\in\mathbb{N}$ ([9],[20],[21]). In order to state Theorem 1.2 we define the
following notation. If $p\in\mathbb{N}$ is a prime and $r,s\in\mathbb{Z}$ are
such that $p\nmid s$, then $\frac{r}{s}\equiv rs^{-1}\pmod{p}$.

######  Theorem 1.2 (cf. Corollary 4.9).

Let $\alpha,\beta,\gamma\in\mathbb{Q}\setminus\\{0\\}$.

  1. (i)

Suppose $n$ is odd and $\alpha,\beta,\gamma$ are not $n$th powers; or

  2. (ii)

Suppose $n$ is even, $\alpha,\beta,\gamma$ are not $\frac{n}{2}$th powers, and
$\alpha$ is not an $\frac{n}{4}$th power if $4\mid n$.

Then there exists infinitely many primes $p\in\mathbb{N}$ modulo which none of
$\alpha,\beta,\gamma$ are $n$th powers.

We remark that we will apply Theorem 1.2 to
$\alpha=\frac{a}{c},\beta=\frac{b}{c},$ and $\gamma=\frac{a+b}{c}$. Since
$\alpha=\beta+\gamma$, the condition that at least one of $\alpha,\beta,$ and
$\gamma$ not be an $\frac{n}{4}$th power if $4\mid n$ is automatic by Fermat’s
Last Theorem when $n>8$. We would also like to apologize to the reader for
using $p$ to denote ultrafilters, generic polynomials, and primes in
$\mathbb{N}$. Thankfully, we do not have any proofs or statements that
simultaneously make use of an ultrafilter, a generic polynomial, and/or a
prime, so the meaning of $p$ will be clear from the context.

The structure of the paper is as follows. In Section 2 we provide a statement
of Rado’s Theorem and briefly review some facts about the usage of
ultrafilters in Ramsey Theory. A major goal of this section is to quickly
familiarize the reader who is inexperienced with Ramsey Theory with enough
basic knowledge of ultrafilters that they will be able to use as a blackbox
the special kinds of ultrafilters introduced in Theorems 2.8 and 6.1. In
Section 3 we prove items (i)(b), (ii), and (iii)(a-b) of Theorem 1.1 as
Corollary 3.2, Theorem 3.5 and Corollary 3.10 respectively.

The main result of Section 4 is Theorem 1.2. Since our proof of Theorem 1.2
already requires us to work in finite extensions of $\mathbb{Q}$, we also
prove a similar result as Lemma 4.8 in the more general setting of rings of
integers of number fields. We then prove item (iii)(c) of Theorem 1.1 as
Corollary 4.12, and we conclude the section with Lemma 4.13 which is an
analogue of Theorem 1.2 for $2$ variables.

In Section 5 we prove item (i)(a) of Theorem 1.1 as Theorem 5.1. We also
determine the partition regularity of some equations of the form
$ax+by=cwz^{n}$ which are not already addressed by Theorem 1.1 through the use
of Lemma 5.17. In Section 6 we investigate the partition regularity of
$ax+by=cwz^{n}$ over general integral domains $R$ and attain results analogous
to Theorem 1.1 while using methods very similar to those used in Section 3. In
Section 7 we investigate the partition regularity of systems of equations of
the form $a_{i}x_{i}+b_{i}y_{i}=c_{i}w_{i}z_{i}^{n},1\leq i\leq k$ over
integral domains $R$ and attain results that support Conjecture 1.3.

######  Conjecture 1.3 (cf. Conjecture 8.6).

Let
$a_{1},\cdots,a_{k},b_{1},\cdots,b_{k},c_{1},\cdots,c_{k}\in\mathbb{Z}\setminus\\{0\\}$
and $n\in\mathbb{N}$. The system of equations

$\begin{array}[]{ccccc}a_{1}x_{1}&+&b_{1}y_{1}&=&c_{1}w_{1}z_{1}^{n}\\\
a_{2}x_{2}&+&b_{2}y_{2}&=&c_{2}w_{2}z_{2}^{n}\\\ &&\vdots&&\\\
a_{k}x_{k}&+&b_{k}y_{k}&=&c_{k}w_{k}z_{k}^{n}\end{array}$ (3)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$ if and only if

$I:=\bigcap_{i=1}^{k}\left\\{\frac{a_{i}}{c_{i}},\frac{b_{i}}{c_{i}},\frac{a_{i}+b_{i}}{c_{i}}\right\\}$
(4)

contains an $n$th power in $\mathbb{Q}$.

In Section 8 we state some conjectures and examine some equations and systems
of equations whose partition regularity remains unknown. We also elaborate on
the distinction between partition regularity of a polynomial equation over
$\mathbb{N}$ instead of $\mathbb{Z}\setminus\\{0\\}$ by considering some
illustrative examples of polynomial equations over $\mathbb{Z}[\sqrt{2}]$.

The main purposes of Section 9 is to give a thorough proof of the existence of
the ultrafilters in Theorems 2.8 and 6.1 as Theorem 9.12 and 9.18
respectively. To this end, we begin the section with a detailed introduction
to the theory of ultrafilters and its applications to semigroup theory. While
the ultrafilter that we use in Theorem 9.12 is the same as the ultrafilter
used in [3] and [10] (cf. Remark 9.13), an analogous ultrafilter need not
exist over a general integral domain $R$. Consequently, we prove Corollary
9.26 and Theorem 9.28 to obtain a characterization of the integral domains $R$
which possess an ultrafilters analogous to the one in Theorem 9.12.

Acknowledgements. We would like to thank to Vitaly Bergelson for proposing
this topic of research and for his helpful guidance.

##  2 Preliminaries

In this section we review some facts and useful tools in the study of
partition regular equations.

######  Definition 2.1.

Given a set $S$, a ring $R$, and functions
$f_{1},\cdots,f_{k}:S^{n}\rightarrow R$, the system of equations

$\begin{array}[]{ccc}f_{1}(x_{1},\cdots,x_{n})&=&0\\\
f_{2}(x_{1},\cdots,c_{n})&=&0\\\ \vdots&\vdots&\vdots\\\
f_{k}(x_{1},\cdots,x_{n})&=&0\end{array}$ (5)

is partition regular over S if for any finite partition
$S=\bigcup_{i=1}^{r}C_{i}$, there exists $1\leq i_{0}\leq r$ and
$x_{1},\cdots,x_{n}\in C_{i_{0}}$ which satisfy (5). If the set $S$ is
understood from context, then we simply say that the system of equations is
partition regular.

######  Definition 2.2.

Let $R$ be an integral domain with field of fractions $K$. A matrix ${\bf
M}\in\mathrm{M}_{m\times n}(R)$ satisfies the columns condition if there
exists a partition $C_{1},\cdots,C_{k}$ of the column indices such that for
$\vec{s}_{i}=\sum_{j\in C_{i}}\vec{c}_{j}$ we have

  * (i)

$\vec{s}_{1}=(0,\cdots,0)^{T}$.

  * (ii)

For all $i\geq 2$, we have

$\vec{s}_{i}\in\text{Span}_{K}\\{\vec{c}_{j}\ |\ j\in
C_{\ell},1\leq\ell<i\\}.$ (6)

The columns condition was used by Rado to classify when a finite system of
homogeneous linear equations is partition regular.

######  Theorem 2.3 (Rado, [17]).

Given ${\bf M}\in\mathrm{M}_{m\times n}(\mathbb{Z})$, the system of equations

${\bf M}(x_{1},\cdots,x_{n})^{T}=0$ (7)

is partition regular over $\mathbb{N}$ if and only if ${\bf M}$ satisfies the
columns condition.

######  Corollary 2.4.

For $a_{1},\cdots,a_{s}\in\mathbb{Z}$, the equation

$a_{1}x_{1}+\cdot+a_{s}x_{s}=0$ (8)

is partition regular over $\mathbb{N}$ if and only if there exists
$\emptyset\neq F\subseteq[1,S]$ for which $\sum_{i\in F}a_{i}=0$.

Rado also characterized which finite, not necessarily homogeneous, linear
systems of equations are partition regular.

######  Theorem 2.5 (Rado, [17]).

Given ${\bf M}\in M_{m\times n}(\mathbb{Z})$ and
$(b_{1},\cdots,b_{n})\in\mathbb{Z}^{n}$, the equation

${\bf M}(x_{1},\cdots,x_{n})^{T}=(b_{1},\cdots,b_{n})^{T}$ (9)

is partition regular over $\mathbb{Z}$ if and only if equation (9) admits an
integral solution in which $x_{1}=x_{2}=\cdots=x_{n}$.

We point out to the reader that in Theorem 2.5 it is possible to only obtain
partition regularity in a trivial sense. For example, since $(x,y)=(0,0)$ is
the only solution the system of equations

$\begin{array}[]{ccrcc}x&-&y&=&0\\\ x&-&2y&=&0\end{array},$ (10)

we see that the system is not partition regular over
$\mathbb{Z}\setminus\\{0\\}$ even though it is partition regular over
$\mathbb{Z}$. As we will see in Section 5, Corollary 2.4 and Theorem 2.5 can
be used in conjunction to determine whether or not a single linear equations
is partition regular over $\mathbb{N}$ or $\mathbb{Z}$ in a nontrivial
fashion.

######  Theorem 2.6 (Bergelson, [2], page 53).

If $p(x)\in\mathbb{Z}[x]$ satisfies $p(0)=0$, then the equation $x-y=p(z)$ is
partition regular over $\mathbb{N}$.

The theory of _ultrafilters_ has been very useful in the study of Ramsey
theory and partition regular equations. We briefly recall some basic facts
here and give a more detailed introduction in Section 9.

######  Definition 2.7.

Given a set $S$ let $\mathcal{P}(S)$ be the power set of the $S$.
$p\subseteq\mathcal{P}(S)$ is an ultrafilter over $S$ if it satisfies the
following properties:

  1. (i)

$\emptyset\notin p$.

  2. (ii)

If $A\in p$ and $A\subseteq B$ then $B\in p$.

  3. (iii)

If $A,B\in p$ then $A\cap B\in p$.

  4. (iv)

For any $A\subseteq\mathbb{N}$, either $A\in p$ or $A^{c}\in p$.

$\beta S$ denotes the space of all ultrafilters over $S$.

It is often useful to think about $\beta S$ as the set of finitely additive
$\\{0,1\\}$-valued measures on the set $S$. For now, we only require the
following facts about ultrafilters. First, we see that for any finite
partition of $S=\bigcup_{i=1}^{r}C_{i}$ and any ultrafilter $p$, there exists
$1\leq i_{0}\leq r$ for which $C_{i}\in p$ if and only if $i=i_{0}$. In fact,
for any $A\in p$ and any finite partition $A=\bigcup_{i=1}^{r}C_{i}$, there
exists $1\leq i_{0}\leq r$ for which $C_{i}\in p$ if and only if $i=i_{0}$.
Secondly, we note that if $p$ is an ultrafilter, $A\in p$ and $B\notin p$,
then $A\setminus B=A\cap B^{c}\in p$. Lastly, we require the existence of a
special kind of ultrafilter.

######  Theorem 2.8 (cf. Theorem 9.12).

There exists an ultrafilter $p\in\beta\mathbb{N}$ with the following
properties:

  1. (i)

For any $A\in p$ and $\ell\in\mathbb{N}$, there exists $b,g\in A$ with
$\\{bg^{j}\\}_{j=0}^{\ell}\subseteq A$.

  2. (ii)

For any $A\in p$ and $h,\ell\in\mathbb{N}$, there exists $a,d\in\mathbb{N}$
for which $\\{hd\\}\cup\\{ha+id\\}_{i=-\ell}^{\ell}\subseteq A$.

  3. (iii)

For any $s\in\mathbb{N}$, we have $s\mathbb{N}\in p$.

The proof of Theorem 2.8 requires more technical knowledge about ultrafilters.
Since this technical knowledge is not needed in Sections 2-8, we defer the
proof of Theorem 2.8 to Section 9.

##  3 On the Partition Regularity of $ax+by=cwz^{n}$ over $\mathbb{N}$ and
$\mathbb{Z}\setminus\\{0\\}$

The purpose of this section is to prove items (i)(b), (ii), (iii)(a), and
(iii)(b) of Theorem 1.1. We begin by proving item (i)(b) of Theorem 1.1 since
it is an easy consequence of the knowledge from the existing literature.

######  Lemma 3.1.

If $a,s\in\mathbb{N}$ and $p(x)\in\mathbb{Z}[x]$ satisfies $p(0)=0$, then the
equation

$ax-ay=p(z)$ (11)

is partition regular over $s\mathbb{N}$.

###### Proof.

Given a partition $s\mathbb{N}=\bigcup_{i=1}^{r}C_{i}$, we let
$\mathbb{N}=\bigcup_{i=1}^{r}(C_{i}\cap as\mathbb{N})/as$ be a partition of
$\mathbb{N}$. By Theorem 2.6, we see that there exists $1\leq i_{0}\leq r$ and
$x,y,z\in(C_{i_{0}}\cap as\mathbb{N})/as$ for which

$x-y=ca^{n-1}s^{n}z^{n+1}.$ (12)

The desired result in this case follows from the fact that $asx,asy,asz\in
C_{i_{0}}$ and

$a(asx)-a(asy)=c(asz)^{n+1}.$ (13)

∎

######  Corollary 3.2 (Item (i)(b) of Theorem 1.1).

For any $a,c\in\mathbb{Z}\setminus\\{0\\}$ and $m,n,s\in\mathbb{N}$ the
equation

$ax-ay=cw^{m}z^{n}$ (14)

is partition regular over $s\mathbb{N}$.

###### Proof.

A consequence of Lemma 3.1 is that the equation $ax-ay=cz^{m+n}$ is partition
regular over $s\mathbb{N}$, so the desired result follows from taking $w=z$. ∎

We now provide a simple lemma that will not be used later on in the paper, but
helps provide some context for items (ii) and (iii) of Theorem 1.1.

######  Lemma 3.3.

Given $a,c\in\mathbb{Z}\setminus\\{0\\}$ and $n\in\mathbb{N}$, the equation

$ax=cwz^{n}$ (15)

is partition regular over $\mathbb{Q}\setminus\\{0\\}$ if and only if
$\frac{a}{c}$ is an $n$th power in $\mathbb{Q}$.

A short proof of Lemma 3.3 can be obtained through the use of Theorem 3 of
[15] by viewing $\mathbb{Q}\setminus\\{0\\}$ as a $\mathbb{Z}$-module. We
choose to give a slightly longer proof since it familiarizes the reader with
techniques that will be used repeatedly throughout the rest of the paper. We
recall that for a prime $p\in\mathbb{N}$,
$v_{p}:\mathbb{Q}\setminus\\{0\\}\rightarrow\mathbb{Z}$ is the $p$-adic
valuation and is given by $v_{p}(\frac{t}{s})=k$ such that $p$ does not divide
the numerator or denominator of $\frac{t}{s}p^{-k}$ after simplification.

###### Proof of Lemma 3.3.

For the first direction, let us assume that $\frac{a}{c}$ is not an $n$th
power in $\mathbb{Q}$, and let $p$ be a prime for which $n\nmid
v_{p}(\frac{a}{c})$. Let $\mathbb{Q}\setminus\\{0\\}=\bigcup_{i=1}^{n}C_{i}$
be the partition given by

$C_{i}=\\{\frac{t}{s}\in\mathbb{Q}\setminus\\{0\\}\ |\
v_{p}(\frac{t}{s})\equiv i\pmod{n}\\}.$ (16)

We see that if $w,x,z\in C_{i_{0}}$ for some $1\leq i_{0}\leq n$, then

$v_{p}(ax)-v_{p}(cwz^{n})\equiv
v_{p}(\frac{a}{c})+v_{p}(x)-v_{p}(w)-nv_{p}(z)\equiv
v_{p}(\frac{a}{c})\not\equiv 0\pmod{n},$ (17)

so we cannot have $ax=cwz^{n}$.

For the next direction, let us assume that $\frac{a}{c}=(\frac{u}{v})^{n}$ for
some coprime $u,v\in\mathbb{Z}\setminus\\{0\\}$. Let $p$ be an ultrafilter
satisfying the conditions of Theorem 2.8 and let
$\mathbb{Z}\setminus\\{0\\}=\bigcup_{i=1}^{r}C_{i}$ be a partition. By
condition (i) of Theorem 2.8, we see that for every $A\in p$, there exists
$x,w,z\in A$ for which $x=wz^{n}$. We observe that

$v\mathbb{Z}\setminus\\{0\\}=\bigcup_{i=1}^{r}\frac{v}{u}(C_{i}\cap
u\mathbb{Z}\setminus\\{0\\})$ (18)

is a partition, so we may assume without loss of generality that
$\frac{v}{u}(C_{1}\cap u\mathbb{Z}\setminus\\{0\\})\in p$. It follows that
there exists $x,w,z\in C_{1}\cap u\mathbb{Z}\setminus\\{0\\}$ for which

$(\frac{v}{u}x)=(\frac{v}{u}w)(\frac{v}{u}z)^{n}\Rightarrow
x=(\frac{v}{u})^{n}wz^{n}=\frac{c}{a}wz^{n}\Rightarrow ax=cwz^{n}.$ (19)

The desired result folows after recalling that
$\mathbb{Z}\setminus\\{0\\}\subseteq\mathbb{Q}\setminus\\{0\\}$. ∎

Our next result, Lemma 3.4, is the basis for proving item (ii) of Theorem 1.1.
While Lemma 3.4 is an immediate corollary of Theorem 2.11 of [8], we decide to
give an independent proof for the sake of completeness and to further
familiarize the reader with methods that will be used later on in this paper.

######  Lemma 3.4.

Let $p\in\beta\mathbb{N}$ be an ultrafilter satisfying the conditions of
Theorem 2.8. For any $A\in p$, $a,b\in\mathbb{Z}\setminus\\{0\\}$ and
$n\in\mathbb{N}$, the equation

$ax+by=cwz^{n}$ (20)

has a solution in $A$ if $c\in\\{a,b,a+b\\}$.

###### Proof.

Let

$A^{\prime}=\\{v\in A\ |\ v=wz^{n}\text{ for some }z,w\in A\\}.$ (21)

Since $A\in p$, to see that $A^{\prime}=A\setminus(A\setminus A^{\prime})\in
p$ it suffices to observe that $A\setminus A^{\prime}\notin p$ because
$A\setminus A^{\prime}$ does not satisfy condition $(i)$ of Theorem 2.8. Our
first case is when $c=a+b$, and in this case we let $x\in A^{\prime}$ be
arbitrary and let $w,z\in A$ be such that $x=wz^{n}$. Since

$ax+bx=cx=cwz^{n},$ (22)

we see that $x,x,w,z$ is a solution to equation (20) coming from $A$. For our
second case it suffices to consider $c=a$ since the case of $c=b$ is handled
similarly. By replacing $a,b,c$ with $-a,-b,-c$ if necessary, we may assume
without loss of generality that $a>0$. Observe that
$A^{\prime}_{a}:=A^{\prime}\cap a\mathbb{N}\in p$ since
$A^{\prime},a\mathbb{N}\in p$ and consider

$A^{\prime\prime}=\\{x_{1}\in A^{\prime}_{a}\ |\ \text{there exists }x_{2}\in
A^{\prime}_{a}\text{ satisfying }x_{1}+\lambda\frac{x_{2}}{a}\in
A^{\prime}_{a}\ \forall\ \lambda\in[-|b|,|b|]\\}.$ (23)

Since $A^{\prime}_{a}\in p$, to see that
$A^{\prime\prime}=A^{\prime}_{a}\setminus(A^{\prime}_{a}\setminus
A^{\prime\prime})\in p$ it suffices to observe that $A^{\prime}_{a}\setminus
A^{\prime\prime}\notin p$ because $A^{\prime}_{a}\setminus A^{\prime\prime}$
does not satisfy condition $(ii)$ of Theorem 2.8 with $(h,\ell)=(c,b)$. Now
let $x_{1}\in A^{\prime\prime}$ be arbitrary and let $x_{2}\in A^{\prime}_{a}$
be as in equation (23). then we observe that

$ax_{1}+bx_{2}=a(x_{1}+b\frac{x_{2}}{a}).$ (24)

Since $x_{1}+b\frac{x_{2}}{a}\in A^{\prime}$, so we may pick $w,z\in A$ for
which $x_{1}+b\frac{x_{2}}{a}=wz^{n}$. In this case we observe that

$ax_{1}+bx_{2}=c(x_{1}+b\frac{x_{2}}{a})=cwz^{n},$ (25)

so $x_{1},x_{2},w,z$ is a solution to equation (20) coming from $A$. ∎

######  Theorem 3.5 (Item (ii) of Theorem 1.1).

If $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $n\in\mathbb{N}$ are such that one
of $\frac{a}{c},\frac{b}{c}$, or $\frac{a+b}{c}$ is an $n$th power in
$\mathbb{Q}_{\geq 0}$, then the equation

$ax+by=cwz^{n}$ (26)

is partition regular over $s\mathbb{N}$ for any $s\in\mathbb{N}$. If one of
$\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is an $n$th power in
$\mathbb{Q}$, then equation (26) is partition regular over
$s\mathbb{Z}\setminus\\{0\\}$ for any $s\in\mathbb{N}$.

###### Proof.

Let $d\in\\{a,b,a+b\\}$ be such that $\frac{d}{c}=(\frac{u}{v})^{n}$ with
$u,v\in\mathbb{Z}$. We see that if $d=0$, then $a=-b$ and the desired result
follows from Lemma 3.1 after taking $w=z$. Let us now assume that $d\neq 0$,
so we also have that $u\neq 0$. Using Lemma 3.4 we see that if
$p\in\beta\mathbb{N}$ is an ultrafilter satisfying the properties of Theorem
2.8, then for any $A\in p$ there exists $w,x,y,z\in A$ for which

$ax+by=dwz^{n}.$ (27)

We now consider the cases of $\frac{u}{v}\in\mathbb{Q}^{+}$ and
$\frac{u}{v}\in\mathbb{Q}$ separately. If $\frac{u}{v}\in\mathbb{Q}^{+}$ and
$s\mathbb{N}=\bigcup_{i=1}^{r}C_{i}$ is a partition, then

$vs\mathbb{N}=\bigcup_{i=1}^{r}\frac{v}{u}(C_{i}\cap us\mathbb{N})$ (28)

is also a partition. Similarly, if $\frac{u}{v}\in\mathbb{Q}$ and
$s\mathbb{Z}\setminus\\{0\\}=\bigcup_{i=1}^{r}C_{i}$ is a partition, then

$vs\mathbb{N}=\bigcup_{i=1}^{r}\left(\frac{v}{u}(C_{i}\cap us\mathbb{N})\cap
vs\mathbb{N}\right)$ (29)

is also a partition. In either case, since $vs\mathbb{N}\in p$, there exists
$1\leq i_{0}\leq r$ for which $\frac{v}{u}(C_{i_{0}}\cap us\mathbb{N})\cap
vs\mathbb{N}\in p$, so there exist $w,x,y,z\in C_{i_{0}}\cap us\mathbb{N}$ for
which

$a(\frac{v}{u}x)+b(\frac{v}{u}y)=d(\frac{v}{u}w)(\frac{v}{u}z)^{n}\Rightarrow
ax+by=d(\frac{v}{u})^{n}wz^{n}=cwz^{n}.$ (30)

∎

A particularly aesthetic result arises when we set $n=1$ in Theorem 3.5.

######  Corollary 3.6.

For any $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ the equation

$ax+by=cwz$ (31)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$.

######  Remark 3.7.

It is interesting to note that the equation $x+y=-wz$ is partition regular
over $\mathbb{Z}\setminus\\{0\\}$ as a consequence of Theorem 3.5, but not
over $\mathbb{N}$ due to sign obstructions. We are currently unable to
determine whether equations such as $2x-8y=wz^{3}$ are partition regular over
$\mathbb{N}$ since there are no sign obstructions preventing the partition
regularity.

Now that we have proven (ii) of Theorem 1.1, we are ready to state Theorem
3.8, which will be a crucial tool in our efforts to prove item (iii) of
Theorem 1.1. The techniques that we use to prove Theorem 3.8 are similar to
techniques used in [1], [8], and [17], to show that certain equations are not
partition regular over $\mathbb{N}$. We note that if $p\in\mathbb{N}$ is a
prime and $r,s\in\mathbb{Z}$ are such that $p\nmid s$, then we define
$\frac{r}{s}\equiv rs^{-1}\pmod{p}$.

######  Theorem 3.8.

Given $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $n\in\mathbb{N}$, the equation

$ax+by=cwz^{n}$ (32)

is not partition regular over $\mathbb{Q}\setminus\\{0\\}$ if there exists a
prime $p>\text{max}(|a|+|b|,|c|)$ for which $\frac{a}{c},\frac{b}{c}$, and
$\frac{a+b}{c}$ are not $n$th powers mod $p$.

###### Proof.

Let $p>\text{max}(|a|+|b|,|c|)$ be a prime for which $a,b$, and $a+b$ are not
$n$th powers modulo $p$. Let
$\chi:\mathbb{Q}\setminus\\{0\\}\rightarrow[1,p-1]$ be given by

$\frac{x}{p^{v_{p}(x)}}\equiv\chi(x)\pmod{p}.$ (33)

Note that for all $r,s\in\mathbb{Z}$ we have
$\chi(rs)\equiv\chi(r)\chi(s)\pmod{p}$ and for all nonzero $-p<r<p$ we have
$r\equiv\chi(r)\pmod{p}$. We also see that for all $r,s\in\mathbb{Z}$ we have

$\chi(r+s)\equiv\begin{cases}\chi(r)+\chi(s)\hfill\pmod{p}&\text{ if
}v_{p}(r)=v_{p}(s)\text{ and }r+s\not\equiv 0\pmod{p}\\\
\chi(s)\hfill\pmod{p}&\text{ if }v_{p}(r)>v_{p}(s)\\\
\chi(r)\hfill\pmod{p}&\text{ if }v_{p}(s)>v_{p}(r)\end{cases}.$ (34)

Let $\mathbb{Q}\setminus\\{0\\}=\bigcup_{i=1}^{p-1}C_{i}$ be the partition
given by $C_{i}=\chi^{-1}(\\{i\\})$. Let us assume for the sake of
contradiction that there exists $d\in[1,p-1]$ and $w,x,y,z\in C_{d}$
satisfying equation (32). We now have 3 cases to consider. If
$v_{p}(x)=v_{p}(y)$, then we see that

$0\not\equiv(a+b)d\equiv\chi(a)\chi(x)+\chi(b)\chi(y)\equiv\chi(ax+by)\equiv\chi(cwz^{n})\equiv
cd^{n+1}\pmod{p}$ (35)

$\implies(a+b)c^{-1}\equiv d^{n}\pmod{p},$ (36)

which yields the desired contradiction in this case. For our next case we
assume that $v_{p}(x)<v_{p}(y)$ and note that

$0\not\equiv ad\equiv\chi(a)\chi(x)\equiv\chi(ax+by)\equiv\chi(cwz^{n})\equiv
cd^{n+1}\pmod{p}$ (37)

$\implies ac^{-1}\equiv d^{n}\pmod{p},$ (38)

which once again yields a contradiction. Similarly, in our final case when
$v_{p}(x)>v_{p}(y)$ we have

$0\not\equiv bd\equiv\chi(b)\chi(y)\equiv\chi(ax+by)\equiv\chi(cwz^{n})\equiv
cd^{n+1}\pmod{p}$ (39)

$\implies bc^{-1}\equiv d^{n}\pmod{p},$ (40)

which once more yields a contradiction. ∎

######  Lemma 3.9 (cf. Corollary 4.9).

Let $n$ be odd and suppose that $\alpha,\beta,\gamma\in\mathbb{Q}$ are not
$n$th powers, or let $n$ be even and suppose that
$\alpha,\beta,\gamma\in\mathbb{Q}$ are not $\frac{n}{2}$th powers with at
least one not $\frac{n}{4}$th power if $4\mid n$. There exist infinitely many
primes $p\in\mathbb{N}$ for which $\alpha,\beta,$ and $\gamma$ are
simultaneously not $n$th powers modulo $p$.

######  Corollary 3.10 (cf. Theorem 1.1(iii)(a-b)).

Let $n\in\mathbb{N}$ and $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ be such that
either $n$ is odd and none of $\frac{a}{c},\frac{b}{c},\frac{a+b}{c}$ are
$n$th powers in $\mathbb{Q}$, or $n\neq 4,8$ is even and none of
$\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ are $\frac{n}{2}$th powers in
$\mathbb{Q}$. Then the equation

$ax+by=wz^{n}$ (41)

is not partition regular over $\mathbb{Q}\setminus\\{0\\}$.

###### Proof.

By Theorem 3.8 it suffices to construct a prime $p>\text{max}(|a|+|b|,|c|)$
for which none of $\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ are perfect
$n$th powers modulo $p$. Firstly, we see that if $n$ is odd, then we may use
Lemma 3.9 to show that the desired prime $p$ exists. Next, we see that if
$n=2m$ with $m$ odd, then none of $\frac{a}{c},\frac{b}{c},\frac{a+b}{c}$ are
$m$th powers by assumption, so we may once again use Lemma 3.9 to find the a
prime $p>\text{max}(|a|+|b|,|c|)$ for which none of $\frac{a}{c},\frac{b}{c},$
or $\frac{a+b}{c}$ are $m$th powers mod $p$. We note that none of
$\frac{a}{c},\frac{b}{c},\frac{a+b}{c}$ are $n$th powers mod $p$ since $m|n$,
so $p$ is the desired prime in this case. Lastly, we see that if $4|n$ and
$n\geq 12$, then $\frac{n}{4}\geq 3$, so by Fermat’s Last Theorem, at least
one of $\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is not an $\frac{n}{4}$th
power in $\mathbb{Q}$, so we may once again use Lemma 3.9 to show that the
desired prime $p$ exists. ∎

##  4 Number Theoretic Results

In this section we assume that the reader has had an introduction to algebraic
number theory. Specifically, we assume familiarity with the content appearing
in chapters i@, ii@, and iv@ of [13] and the Chebotarev Density Theorem. The
main goal of this section is to prove Lemma 3.9 as Corollary 4.9. The reader
willing to take the existence of such primes on faith can safely skip this
section and the algebraic number theory content appearing here. We first
handle the odd exponent case as some aspects of the argument are simplified
and very general. Afterwards, we add a few details to handle the even exponent
case.

We briefly recall some of the concepts we will need. We call $K$ a number
field if it is a finite field extension of $\mathbb{Q}$. We write
$\mathcal{O}_{K}$ for the ring of integers of $K$, which is the integral
closure of $\mathbb{Z}$ in $K$. This is a Dedekind domain, so nonzero ideals
factor uniquely into a product of prime ideals.

Given an extension of number fields $L/K$, one can ask how a prime ideal
$\mathfrak{p}\subset\mathcal{O}_{K}$ factors in $\mathcal{O}_{L}$. We have
$\mathfrak{p}\mathcal{O}_{L}=\mathfrak{q}_{1}^{e_{1}}\cdots\mathfrak{q}_{g}^{e_{g}}$
for some prime ideals $\mathfrak{q}_{i}\subset\mathcal{O}_{L}$. Recall that
$e(\mathfrak{q}_{i}/\mathfrak{p}):=e_{i}$ is the ramification degree of
$\mathfrak{q}_{i}$ over $\mathfrak{p}$ and
$f_{i}=f(\mathfrak{q}_{i}/\mathfrak{p}):=[\mathcal{O}_{L}/\mathfrak{q}_{i}:\mathcal{O}_{K}/\mathfrak{p}]$
is the inertia degree of $\mathfrak{q}_{i}$ over $\mathfrak{p}$. We say
$\mathfrak{p}$ is unramified in $\mathcal{O}_{L}$, or just in $L$, if
$e_{i}=1$ for all $i$. It is a fact that only finitely many prime ideals of
$\mathcal{O}_{K}$ are ramified in $L$.

These invariants are bounded via the following classic formula:

$[L:K]=\sum_{i=1}^{g}e_{i}f_{i}.$ (42)

In the case $L/K$ is Galois, we have all of the $e_{i}$’s and $f_{i}$’s are
equal, so in fact $e_{i}$ and $f_{i}$ both divide $[L:K]$.

To compute these numbers in practice, one uses modular arithmetic and
factoring polynomials. The process can be summarized as follows. Suppose
$L=K(\alpha)$ with $\alpha\in\mathcal{O}_{L}$. Then
$\mathcal{O}_{K}[\alpha]\subset\mathcal{O}_{L}$ and both are finite free
$\mathcal{O}_{K}$-modules of rank $[L:K]$, so
$[\mathcal{O}_{L}:\mathcal{O}_{K}[\alpha]]$ is finite. If the residue
characteristic of $\mathfrak{p}\subset\mathcal{O}_{K}$ does not divide
$[\mathcal{O}_{L}:\mathcal{O}_{K}[\alpha]]$, then the factorization behavior
of $\mathfrak{p}$ in $\mathcal{O}_{L}$ can be detected by factoring the
minimal polynomial of $\alpha\bmod\mathfrak{p}$.

More precisely, let $f(x)\in\mathcal{O}_{K}[x]$ be the minimal polynomial for
$\alpha$. Then, under the divisibility assumption above, we have

$\mathfrak{p}\mathcal{O}_{L}=\mathfrak{q}_{1}^{e_{1}}\cdots\mathfrak{q}_{g}^{e_{g}}\iff
f(x)\equiv q_{1}(x)^{e_{1}}\cdots q_{g}(x)^{e_{g}}\bmod\mathfrak{p}$ (43)

and $\deg(q_{i}(x))=f(\mathfrak{q}_{i}/\mathfrak{p})$. Hence for all but
finitely many $\mathfrak{p}\subset\mathcal{O}_{K}$, its factoring behavior in
$\mathcal{O}_{L}$ is detected by factoring $f(x)\bmod\mathfrak{p}$.

When $L/K$ is Galois with group $G$, there is an important relationship
between the arithmetic and algebra of the fields expressed via Frobenius
elements. Suppose $\mathfrak{q}\subset\mathcal{O}_{L}$ divides
$\mathfrak{p}\subset\mathcal{O}_{K}$. Then there exists a unique
$\operatorname{Frob}_{\mathfrak{q}/\mathfrak{p}}\in G$ defined by the property

$\operatorname{Frob}_{\mathfrak{q}/\mathfrak{p}}(x)\equiv
x^{q}\bmod\mathfrak{q}$ (44)

where $q=\\#\mathcal{O}_{L}/\mathfrak{p}$. If $\mathfrak{q}$ and
$\mathfrak{q}^{\prime}$ are primes dividing $\mathfrak{p}$ in
$\mathcal{O}_{L}$, then $\operatorname{Frob}_{\mathfrak{q}/\mathfrak{p}}$ and
$\operatorname{Frob}_{\mathfrak{q}^{\prime}/\mathfrak{p}}$ are conjugate.
Conversely, for every $\sigma\in G$ in the conjugacy class of
$\operatorname{Frob}_{\mathfrak{q}/\mathfrak{p}}$, there exists
$\mathfrak{q}^{\prime}$ dividing $\mathfrak{p}$ so that
$\sigma=\operatorname{Frob}_{\mathfrak{q}^{\prime}/\mathfrak{p}}$. Thus we can
speak of a well-defined Frobenius conjugacy class
$\operatorname{Frob}_{\mathfrak{p}}\subset G$. When $G$ is abelian, Frobenius
elements associated to primes in $K$ are therefore well-defined.

In order for this to be useful to us, we need a way to construct primes with
given Frobenius elements. The major tool for achieving this is the Chebotarev
density theorem.

######  Theorem 4.1 (Chebotarev Density Theorem).

Let $L/K$ be a Galois extension of number fields. Let $C\subset
G=\operatorname{Gal}(L/K)$ be a fixed conjugacy class. Then the natural
density of primes $\mathfrak{p}\subset K$ with
$\operatorname{Frob}_{\mathfrak{p}}\in C$ is given by $\\#C/\\#G$, i.e.

$\lim_{x\to\infty}\frac{\\#\\{\mathfrak{p}\in\mathcal{O}_{K}:\\#\mathcal{O}_{K}/\mathfrak{p}\leq
x\mathrm{,and}\operatorname{Frob}_{\mathfrak{p}}\in
C\\}}{\\#\\{\mathfrak{p}\in\mathcal{O}_{K}:\\#\mathcal{O}_{K}/\mathfrak{p}\leq
x\\}}=\frac{\\#C}{\\#G}.$ (45)

Fix $n>1$. We turn our attention to studying the polynomial $x^{n}-a$ for
$a\in\mathcal{O}_{K}$ and its factoring behavior modulo various primes in
$\mathcal{O}_{K}$. Suppose $a\in\mathcal{O}_{K}$ is not an $n$th power.
Suppose and $a=\alpha^{d}$ for $\alpha\in\mathcal{O}_{K(\zeta_{n})}$ with
$d\mid n$ maximal. Set $m=n/d$. Consider the diagram of fields:

${L=C(\alpha^{1/m})}$${C=K(\zeta_{n})}$${K}$

where $\zeta_{n}$ is a primitive $n$th root of unity, and $\alpha^{1/m}$ is an
arbitrary root of the polynomial $x^{m}-\alpha$. We will show the field $L$ is
a well-defined radical extension of $C$ using the following lemma.

######  Lemma 4.2 ([14] Theorem VI.9.1).

Let $k$ be a field and $a\in k$ nonzero. Assume for all primes $p\mid n$, we
have $a\not\in k^{p}$ and if $4\mid n$, then $a\not\in-4k^{4}$. Then $x^{n}-a$
is irreducible in $k[x]$.

From here we deduce the following.

######  Lemma 4.3.

The polynomial $x^{m}-\alpha$ is irreducible over $C$. In particular, the
field $L$ above is a well-defined radical extension of $C$, and $[L:C]=m$.

###### Proof.

By maximality of $d$, we see for all primes $p\mid n$ we have $\alpha\not\in
C^{p}$, so the polynomial $x^{m}-\alpha$ satisfies the first criteria of Lemma
4.2, and the polynomial is irreducible as long as $4\nmid n$. In the case
$4\mid n$, we check $\alpha\not\in-4C^{4}$. Suppose otherwise. Then
$\alpha=-4\beta^{4}$ for some $\beta\in C$. But since $4\mid n$ and
$\zeta_{n}\in C$, we have $\sqrt{-1}=\zeta_{4}\in C$, so
$\alpha=(\zeta_{4}\cdot 2\cdot\beta^{2})^{2}$, and $a=(\zeta_{4}\cdot
2\cdot\beta^{2})^{2d}$ contradicting maximality of $d$. In either case, we
find $x^{m}-\alpha$ is irreducible.

For the statement on degrees, we use Kummer theory. Recall this tells us that
since $C$ contains all $m$th roots of unity, extensions of the form
$C(\alpha^{1/m})/C$ are cyclic of degree equal to the order of $\alpha$ in
$C^{\times}/C^{\times,m}$. But we have just showed that $\alpha$ is not a
$d$th power for any $d\mid m$, so its order in this group is $m$. ∎

######  Remark 4.4.

The key idea of our argument is as follows. We will use density arguments to
produce a prime ideal $\mathfrak{p}$ in the ring of integers of
$C=K(\zeta_{n})$ modulo which $x^{m}-\alpha$ has no root. Given such a
$\mathfrak{p}$, the going down theorem provides a prime ideal of
$\mathcal{O}_{K}$ with the same property. In fact, if we can bound the density
of such $\mathfrak{p}$ well enough, then repeating with $b$ and $c$ can yield
a density bound on the set of prime ideals $\mathfrak{p}$ where at least one
of $a,b,c$ are $n$th powers. If this density is less than 1, then the lemma
will be established in that case.

The setup is as follows. Let $a=\alpha^{d_{a}};b=\beta^{d_{b}}$; and
$c=\gamma^{d_{c}}$ with the $d$’s maximal. Set $m_{a}=n/d_{a}$, etc. Let
$\mathfrak{p}\subset\mathcal{O}_{K}$. Then we have $x^{n}-a$ has a root mod
$\mathfrak{p}$ if and only if $x^{m}-\alpha$ has a root mod $\mathfrak{p}$,
and (with finitely many exceptions for $\mathfrak{p}$) the density of such
$\mathfrak{p}$ correspond to the density of $\mathfrak{p}$ splitting in
$L_{a}=C(\alpha^{1/m_{a}})$. By the Chebotarev density theorem, the latter is
given by $1/[L_{a}:C]$.

Let $\delta_{a}:=1/[L_{a}:C]=1/m_{a}$ and similarly define $\delta_{b}$ and
$\delta_{c}$. As mentioned, if $\delta_{a}+\delta_{b}+\delta_{c}<1$, then
there must exist a prime (infinitely many, in fact) in $C$ where none of
$a,b,c$ are $n$th powers. Unfortunately this sum can very well be at least 1
or more, so we will devote most of the rest of this section handling those
cases.

First, we gather some results to rule out the case that $\delta_{a}=1$, at
least when $K=\mathbb{Q}$ and many other cases. We recall a fact about the
interplay between roots of unity and radical extensions, due to Schinzel.
Since Lemmas 4.5 and 4.6 are generic, we will omit the subscripts and just
write $m,d,$ and $L$ in their statements and proofs.

######  Lemma 4.5.

Let $\omega_{m}$ be the number of roots of unity in a field $F$ of
characteristic 0. Suppose $x^{m}-\alpha$ is irreducible over $F$. Then
$F(\alpha^{1/m})/F$ is an abelian Galois extension if and only if
$\alpha^{\omega_{m}}=\beta^{m}$ for some $\beta\in F$.

###### Proof.

See [19]. ∎

This provides us with our first serious condition on $\delta_{a}$, by
restricting the degree $[L:C]$.

######  Lemma 4.6.

Let $\omega_{n}$ be the number of $n$th roots of unity in $K$. Suppose
$m=[L:C]=1$ with the notation as above. Then

$a^{\omega_{n}}=k^{n}$ (46)

for some $k\in K$.

###### Proof.

Because $C/K$ is a cyclotomic extension, it is abelian. But since $m=1$, we
have $a^{1/n}=\alpha\in C$. Hence $K(a^{1/n})/K$ is abelian. By Lemma 4.5, we
then have $a^{\omega_{n}}=k^{n}$ for some $k\in K$. ∎

In particular, we can control the size of $m$ by the assumptions we make on
$\alpha$, and hence $a$, in $K$. We will carry the details out towards the end
of the proof of the main lemma. For now, we investigate the density of primes
where both $a$ and $b$ are $n$th powers.

######  Lemma 4.7.

Let $m_{a}$ and $m_{b}$ be as in Remark 4.4. The density of primes in
$K(\zeta_{n})$ where both $a$ and $b$ are $n$th powers modulo is given by at
least $1/(m_{a}\cdot m_{b})$.

###### Proof.

Since both $C(\alpha^{1/m_{a}})/C$ and $C(\beta^{1/m_{b}})/C$ are Galois
extensions, their composite field $C(\alpha^{1/m_{a}},\beta^{1/m_{b}})$ is
Galois over $C$ and has degree at most $m_{a}\cdot m_{b}$. The primes where
both $x^{m_{a}}-\alpha$ and $x^{m_{b}}-\beta$ have roots modulo correspond to
those with trivial Frobenius element in
$G=\operatorname{Gal}(C(\alpha^{1/m_{a}},\beta^{1/m_{b}})/C)$. By Chebotarev,
these have density $1/\\#G\geq 1/(m_{a}\cdot m_{b})$. ∎

We can now prove the main existence result.

######  Lemma 4.8.

Let $K$ be a number field and $\omega_{n}$ the number of $n$th roots of unity
in $K$. Let $a,b,c\in\mathcal{O}_{K}$.

  1. (i)

Suppose $n$ is odd, and that $a^{\omega_{n}}$ is not an $n$th power in
$\mathcal{O}_{K}$, and similarly for $b,c$; or

  2. (ii)

Suppose $n$ is even, and $a,b,c$ satisfy the same conditions as in (i), but
$a^{2\omega_{n}}$ is also not an $n$th power.

Then there exists infinitely many primes of $K$ modulo which none of $a,b,c$
are $n$th powers.

###### Proof.

Recall that $C=K(\zeta_{n})$ and $L_{a}=C(\alpha^{1/m_{a}})$. Let
$\delta_{a,b}$ be the density of primes of $C$ modulo which both $a$ and $b$
are $n$th powers. By Lemma 4.7, we have $\delta_{a,b}\geq 1/(m_{a}\cdot
m_{b})$. Similarly we write $\delta_{a,b,c}$ for the density of primes modulo
which all three are $n$th powers. Also let $\Delta$ be the density of primes
modulo which at least one of $a,b,c$ is an $n$th power. We want to use
inclusion-exclusion to bound $\Delta$. We have

$\Delta=\delta_{a}+\delta_{b}+\delta_{c}-\delta_{a,b}-\delta_{b,c}-\delta_{a,c}+\delta_{a,b,c}.$

Suppose without loss of generality that $\delta_{a,c}$ is minimal among the
densities for the possible pairs. Then $\delta_{a,b,c}\leq\delta_{a,c}$, so we
have

$\displaystyle\Delta$
$\displaystyle\leq\delta_{a}+\delta_{b}+\delta_{c}-\delta_{a,b}-\delta_{b,c},$
(47) $\displaystyle\leq 1/m_{a}+1/m_{b}+1/m_{c}-1/(m_{a}\cdot
m_{b})-1/(m_{b}\cdot m_{c}),$ (48)

using the bound from Lemma 4.7. We now split into cases based on the parity of
$n$.

  1. (i)

We handle the case when $n$ is odd first. We see by our assumptions and Lemma
4.6, we have $m_{a},m_{b},m_{c}>1$. Since $m_{a}=[L_{a}:C]$ must divide $n$
and $n$ is odd, we have $m_{a},m_{b},m_{c}\geq 3$. So then $\Delta\leq
1/3+1/3+1/3-1/9-1/9=7/9<1$.

  2. (ii)

When $n$ is even, there are few more exceptional cases for
$(\delta_{a},\delta_{b},\delta_{c})$ which we handle now:

     * •

(1/$k$, 1/3, 1/2), $3\leq k\leq 6$: From our bound above, we get

$\Delta\leq\frac{1}{k}+\frac{1}{3}+\frac{1}{2}-\frac{1}{6}-\frac{1}{2k}=\frac{2}{3}+\frac{1}{2k}<1.$
(49)

     * •

(1/$k$, 1/2, 1/2), $k\geq 3$: From our bound above, we get

$\Delta\leq\frac{1}{k}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}-\frac{1}{2k}=3/4+\frac{1}{2k}<1.$
(50)

     * •

(1/2, 1/2, 1/2): This case is genuine cause for concern. For example, it is
possible to have 3 quadratic extensions of a field where every prime in the
base splits in at least one of them (consider
$\mathbb{Q}(\sqrt{a}),\mathbb{Q}(\sqrt{b})$, and $\mathbb{Q}(\sqrt{ab}$)).
However, we rule this out by the additional assumption on $a$. Suppose
$m_{a}=2$ so that $\delta_{a}=1/2$. Then since $x^{n}-a=(x^{2}-\alpha)f(x)$
for some $f(x)\in C[x]$, we are justified in writing $\alpha=a^{2/n}$. We have
$K((a^{2})^{1/n})/K$ is abelian since $K((a^{2})^{1/n})=K(\alpha)\subseteq C$,
and $C/K$ is abelian as it is a cyclotomic extension. Applying Lemma 4.5, we
see that

$a^{2\omega_{n}}=k^{n}$ (51)

for some $k\in K$, contradicting the original assumption on $a$.

     * •

Otherwise, we have already $\delta_{a}+\delta_{b}+\delta_{c}<1$, so we get
$\Delta<1$.

In all cases we obtain that the density of primes $p$ where at least one of
$a,b,c$ is an $n$th power modulo $p$ is less than 1, so there exists a set of
primes of positive density where none are $n$th powers. ∎

In the case $K=\mathbb{Q}$, this specializes to a more pleasant form. We
examine an equally pleasant generalization for
$\alpha,\beta,\gamma\in\mathbb{Q}$ rather than $a,b,c\in\mathbb{Z}$ as
demanded by the applications from Section 3. We remind the reader that if
$p\in\mathbb{N}$ is a prime and $r,s\in\mathbb{Z}$ are such that $p\nmid s$,
then $\frac{r}{s}\equiv rs^{-1}\pmod{p}$.

######  Corollary 4.9.

Let $\alpha,\beta,\gamma\in\mathbb{Q}$.

  1. (i)

Suppose $n$ is odd and $\alpha,\beta,\gamma$ are not $n$th powers; or

  2. (ii)

Suppose $n$ is even, $\alpha,\beta,\gamma$ are not $\frac{n}{2}$th powers, and
$\alpha$ is not an $\frac{n}{4}$th power if $4\mid n$.

Then there exists infinitely many primes $p\in\mathbb{N}$ modulo which none of
$\alpha,\beta,\gamma$ are $n$th powers.

###### Proof.

Let $c\in\mathbb{N}$ be such that for
$\alpha^{\prime}:=c^{n}\alpha,\beta^{\prime}:=c^{n}\beta,$ and
$\gamma^{\prime}:=c^{n}\gamma$ we have
$\alpha^{\prime},\beta^{\prime},\gamma^{\prime}\in\mathbb{Z}$. Since
$\omega_{n}=1$ when $n$ is odd and $\omega_{n}=2$ when $n$ is even we may
apply Lemma 4.8 to find infinitely many primes $p\in\mathbb{N}$ for which
$\alpha^{\prime},\beta^{\prime},$ and $\gamma^{\prime}$ are noth $n$th powers
modulo $p$. If $p\in\mathbb{N}$ is a prime for which $p\nmid c$, then
$\alpha^{\prime}$ is an $n$th power modulo $p$ if and only if $\alpha$ is an
$n$th power modulo $p$, and similarly for $\beta^{\prime}$ and
$\gamma^{\prime}$. It follows that there are infinitely many primes $p$ for
which $\alpha,\beta,$ and $\gamma$ are also not $n$th powers modulo $p$. ∎

######  Remark 4.10.

When working over $\mathbb{Q}$, as long as $n\neq 4,8$, when specialized to
our situation where $c=a+b$, the condition that one of the three numbers not
be an $\frac{n}{4}$th power is automatic by Fermat’s Last Theorem.

Now we give a few more remarks about the subtleties that arose in this proof,
and what obstacles prevent pushing it further, both over $\mathbb{Q}$ and for
general number fields $K$.

  1. (i)

It could happen that $x^{n}-a$ has a root mod all primes
$\mathfrak{p}\subset\mathcal{O}_{K}$ even though $a$ is not an $n$th power.
For example, this happens when $K=\mathbb{Q}$; the polynomial $x^{8}-16$ has a
root mod $p$ for all primes $p\in\mathbb{Z}$, but 16 is not an $8$th power.
The Grunwald-Wang theorem says that if $x^{n}-a$ has a root mod $p$ for all
primes $p$, then $n$ must be even of a special form, determined by $K$. This
is ruled out above since $16$ is a perfect $8/2=4$th power.

  2. (ii)

Even if one were to exclude the situation arising in the Grunwald-Wang
theorem, it is possible that $(x^{n}-a)(x^{n}-b)$ could have a root mod $p$
for all primes $p$, despite neither factor having this behavior. We consider
$3$ such examples when $K=\mathbb{Q}$, each of which also suggests the
necessity of the conditions in item (ii) of Lemma 4.8 and item (ii) or
Corollary 4.9.

    1. (a)

The polynomial $(x^{12}-3^{6})(x^{12}-\beta^{4})$ has this feature for any
$\beta\in\mathbb{N}$, because $(x^{2}+3)(x^{3}-\beta)$ has a root mod $p$ for
all $p\in\mathbb{Z}$. The problem here is that $x^{12}-3^{6}$ has a quadratic
factor, which lives inside $\mathbb{Q}(\zeta_{12})$. In the language above,
this means that $e=1$, so the density argument won’t work. (cf. [12])

    2. (b)

Since $x^{8}-16=(x^{4}-4)(x^{4}+4)$ and $x^{8}-16$ has a root modulo $p$ for
any prime $p\in\mathbb{N}$, one of $4$ and $-4$ is a $4$th power modulo $p$.

    3. (c)

Since $36$ is a fourth power modulo $p$ for any prime $p\not\equiv
13\pmod{24}$ and $9$ is a fourth power modulo $p$ for any prime $p\equiv
13\pmod{24}$, we see that $(x^{4}-36)(x^{4}-9)$ has a root modulo $p$ for any
prime $p\in\mathbb{N}$.

For general $K$, one imagines it only gets harder to determine whether
$x^{n}-a$ has a factor whose splitting field lies in a cyclotomic extension.
The proof above implies that if $K$ doesn’t have many roots of unity, then
there are mild conditions on $a,b,c$ to get the existence of the desired
prime.

  3. (iii)

When $K$ doesn’t have unique factorization, we lose some of the power afforded
by Lemma 4.6 because we cannot ensure that if $a^{x}=b^{y}$, then $a$ is a
perfect $y/(x,y)$th power. We can pass to ideals generated by $a$ and $b$, and
use the fact that $\mathcal{O}_{K}$ is Dedekind, so its ideals satisfy unique
factorization. But this only says that the ideal generated by $a$ is a perfect
$y/(x,y)$th power, and this need not imply $a$ has the same property. For
example, if $K=\mathbb{Q}(\sqrt{-5})$ and
$\mathfrak{p}=(2,1+\sqrt{-5})\subset\mathcal{O}_{K}$, then
$\mathfrak{p}^{2}=(2)$, but $2\neq\alpha^{2}$ for any
$\alpha\in\mathcal{O}_{K}$.

  4. (iv)

When $K$ has lots of units, it can also be difficult to deduce $a$ is a
perfect $y/(x,y)$th power given $a^{x}=b^{y}$. For example, the ideal
generated by $(-4)$ in $\mathbb{Z}$ is the square of the ideal $(2)$, but $-4$
is of course not a perfect square. It’s possible one could say more here by
trying to control the units of $\mathcal{O}_{K}$, but this seems a bit
intimidating and not of immediate interest.

We see that Corollary 4.9 is not useful when $n=2$ or $n=4$, so we will
address the case of $n=2$ separately and observe the implications that it has
for the case of $n=4$. We first require a lemma for constructing primes modulo
which certain numbers are not squares.

######  Lemma 4.11.

Suppose that $\alpha,\beta,\gamma\in\mathbb{Q}$ are not squares, and
$\alpha\beta\gamma$ is also not a square. There exists a prime
$p\in\mathbb{N}$ for which $\alpha,\beta,$ and $\gamma$ are not squares modulo
$p$.

###### Proof.

Suppose on the contrary that no such prime exists. Let
$\left(\frac{a}{p}\right)$ be the Legendre symbol and let $S_{\alpha,\beta}$
be the set of primes $p$ for which

$\left(\frac{\alpha}{p}\right)=\left(\frac{\beta}{p}\right)=-1.$ (52)

By our assumption, we have $\left(\frac{\gamma}{p}\right)=1$ for all $p\in
S_{\alpha,\beta}$ since otherwise we would have produced a prime with the
desired features. Using similar notation for the other pairs, we see that
$S_{\alpha,\beta}\cap S_{\beta,\gamma}=\varnothing$ because for primes $p$ in
the first set, we have $\left(\frac{\alpha}{p}\right)=-1$ but for $p$ in the
second set we have $\left(\frac{\alpha}{p}\right)=1$.

Thus the density of $S:=S_{\alpha,\beta}\cup S_{\beta,\gamma}\cup
S_{\gamma,\alpha}\geq 3/4$ as the density of each $S_{\alpha,\beta}$ is at
least $1/4$ by Quadratic Reciprocity. But for each $p\in S$, we have

$\left(\frac{\alpha\beta\gamma}{p}\right)=1.$ (53)

Hence a set of primes of density $>1/2$ split in the extension
$\mathbb{Q}(\sqrt{\alpha\beta\gamma})$. But by Quadratic Reciprocity, the set
of primes which split in a degree 2 extension has density 1/2. Thus we have
$[\mathbb{Q}(\sqrt{\alpha\beta\gamma}):\mathbb{Q}]=1$, so we must have
$\alpha\beta\gamma$ is a square in $\mathbb{Q}$, contradicting the original
assumption. ∎

######  Corollary 4.12 (cf. Theorem 1.1(iii)(c)).

If $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ are such that
$\frac{a}{c},\frac{b}{c},\frac{a+b}{c},$ and
$(\frac{a}{c})(\frac{b}{c})(\frac{a+b}{c})$ are not squares, then for any
$n\in\mathbb{N}$ the equation

$ax+by=cwz^{2n}$ (54)

is not partition regular over $\mathbb{Q}$.

###### Proof.

We want to use Theorem 3.8, so we produce a prime satisfying the conditions
there. In particular, we want a prime $p$ such that
$\alpha:=\frac{a}{c},\beta:=\frac{b}{c}$, and $\gamma:=\frac{a+b}{c}$ are not
perfect squares in $\mathbb{Z}/p\mathbb{Z}$. Noting that
$\alpha,\beta,\gamma,$ and $\alpha\beta\gamma$ are not squares in
$\mathbb{Q}$, we see that the existence of our desired prime $p$ is a
consequence of Lemma 4.11. ∎

We may also use Lemma 4.11 to obtain a strengthening of the special case of
Corollary 4.9 in which there are $2$ variables instead of $3$, which will be
of use in Section 7 when we determine necessary conditions for some systems of
equations to be partition regular.

######  Lemma 4.13.

Let $\alpha,\beta\in\mathbb{Q}$ and $n\in\mathbb{N}$. Suppose that one of the
following conditions holds:

  1. (i)

$4\nmid n$ and neither of $\alpha,\beta$ are $n$th powers.

  2. (ii)

$4|n$ and neither of $\alpha,\beta$ are $\frac{n}{2}$th powers.

Then there exists infinitely many primes $p\in\mathbb{N}$ modulo which neither
$\alpha,\beta$ are $n$th powers.

###### Proof.

We begin by proving the desired result for item (i) and observe that the only
case not handled by Corollary 4.9 is when $n=2m$ with $m$ odd and at least one
of $\alpha,\beta$ is an $m$th power. Suppose $\alpha=x^{m}$. Write
$\beta=y^{d}$ for $d$ maximal. By similar density arguments as before, the
only edge case is when $d=m$, otherwise the sum of densities of primes where
at least one is an $n$th power will be strictly less than 1. In the case
$d=m$, since $m$ is odd, we see that

$\left(\frac{\alpha}{p}\right)=\left(\frac{x}{p}\right)\text{ and
}\left(\frac{\beta}{p}\right)=\left(\frac{y}{p}\right).$

Letting $\gamma=\alpha=x$ and $\beta=y$, we see that none of
$\alpha,\beta,\gamma,$ or $\alpha\beta\gamma=\alpha^{2}\beta$ are squares in
$\mathbb{Q}$, so by Lemma 4.11 there exists infinitely many primes $p$ such
that $x,y$ are not squares mod $p$, and thus $\alpha,\beta$ are not $n$th
powers mod $p$.

To see that the desired result holds for item (ii) we consider the cases of
$n\neq 4$ and $n=4$ separately. When $n\neq 4$, we let $\gamma\in\mathbb{Q}$
be any element that is not a $\frac{n}{4}$th power and apply Corollary 4.9.
When $n=4$, we see that neither of $\alpha$ or $\beta$ are squares, so by item
(i) there exists infinitely many primes $p\in\mathbb{N}$ modulo which neither
of $\alpha,\beta$ are squares. ∎

We observe that Item (ii)(b) of Remark 4.10 tells us that for any prime $p$
and odd number $m$, one of $4^{m}$ and $-4^{m}$ will be a $4m$th power modulo
$p$, which justifies our assumptions in item (ii) of Lemma 4.13.

##  5 Reduction to the case min(m,n) = 1

The main purpose of this section is to show that the equation
$ax+by=cw^{m}z^{n}$ is not partition regular over $\mathbb{Z}\setminus\\{0\\}$
if $a,b,c\in\mathbb{Z}\setminus\\{0\\}$, $a+b\neq 0$, and $n,m\geq 2$.
Afterwards, we will show that the equation $ax+by=cwz^{n}$ is not partition
regular over $\mathbb{Z}\setminus\\{0\\}$ for some values of $a,b,c,$ and $n$
that are not already addressed by Theorem 1.1.

######  Theorem 5.1 (cf. Theorem 1.1(i)(a)).

If $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $n,m\in\mathbb{N}$ are such that
$a+b\neq 0$ and $n,m\geq 2$, then the equation

$ax+by=cz^{n}w^{m}$ (55)

is not partition regular over $\mathbb{Z}\setminus\\{0\\}$.

In order to prove Theorem 5.1 we will use item $(1)$ of Theorem 2.19 of [1].
We now review the definitions necessary to state and use this result.

######  Definition 5.2.

Given a polynomial $P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$, we let Supp($P$)
denote the collection of $\alpha\in\mathbb{N}_{0}^{n}$ for which

$P(x_{1},\cdots,x_{n})=\sum_{\alpha\in\text{Supp}(P)}c_{\alpha}x^{\alpha}.$
(56)

######  Example 5.3.

If $P(x,y)=x^{2}+y^{2}$, we have $\text{Supp}(P)=\\{(2,0),(0,2)\\}$.

######  Example 5.4.

If $P(x,y,z)=xyz+1$, we have $\text{Supp}(P)=\\{(0,0,0),(1,1,1)\\}$.

######  Example 5.5.

If $P(w,x,y,z)=wx+y+z^{2}$, we have
$\text{Supp}(P)=\\{(1,1,0,0),(0,0,1,0),(0,0,0,2)\\}$.

######  Definition 5.6 (cf. Def. 2.6 in [1]).

Let $\phi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}$ be a positive linear map, i.e.

$\phi:(\alpha_{1},\cdots,\alpha_{n})\mapsto
t_{1}\alpha_{1}+\cdots+t_{n}\alpha_{n}$

with $t_{1},\cdots,t_{n}\in\mathbb{N}_{0}$.

  * •

If $c$ is a finite coloring of $\mathbb{N}$, then we say that $\phi$ is
$c$-monochromatic if $\\{t_{1},\cdots,t_{n}\\}$ is $c$-monochromatic;

  * •

If $P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$, and $(M_{0},\cdots,M_{\ell})$ is the
increasing enumeration of $\phi(\text{Supp}(P))$, then the partition of
$\text{Supp}(P)$ determined by $\phi$ is the ordered tupe
$(J_{0},\cdots,J_{\ell})$, where
$J_{i}=\\{\alpha\in\text{Supp}(P):\phi(\alpha)=M_{i}\\}$.

######  Definition 5.7 (cf. Def. 2.7 in [1]).

A Rado partition of $P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$ is an ordered tuple
$(J_{0},\cdots,J_{\ell})$ such that, for every finite coloring $c$ of
$\mathbb{N}$, there exist infinitely many $c$-monochromatic positive linear
maps $\phi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}$ such that
$(J_{0},\cdots,J_{\ell})$ is the partition of $\text{Supp}(P)$ determined by
$\phi$.

######  Definition 5.8 (cf. Def. 2.8 in [1]).

A Rado set for $P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$ is a set
$J\subseteq\text{Supp}(P)$ such that there exists a Rado partition
$(J_{0},\cdots,J_{\ell})$ for $P$ such that $J=J_{i}$ for some
$i\in\\{0,1,\cdots,\ell\\}$

######  Definition 5.9 (cf. Def. 2.13 in [1]).

An upper Rado functional of order $m$ for $P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$
is a tuple $(J_{0},\cdots,J_{\ell},d_{0},\cdots,d_{m-1})$ for some $\ell\geq
m$ and $d_{0},\cdots,d_{m-1}\in\mathbb{N}$ such that, for every finite
coloring $c$ of $\mathbb{N}$ and for every $r\in\mathbb{N}$, there exist
infinitely many $c$-monochromatic positive linear maps
$\phi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}$,
$(\alpha_{1},\cdots,\alpha_{n})\rightarrow
t_{1}\alpha_{1}+\cdots+t_{n}\alpha_{n}$ such that $(J_{\ell},\cdots,J_{0})$ is
the partition of $\text{Supp}(P)$ determined by $\phi$, and if
$(M_{\ell},\cdots,M_{0})$ is the increasing enumeration of
$\phi(\text{Supp}(P))$, then $M_{i}-M_{m}=d_{i}$ for
$i\in\\{0,1,\cdots,m-1\\}$, and $M_{m}-M_{i}\geq r$ for
$i\in\\{m+1,\cdots,\ell\\}$.

######  Definition 5.10 (cf. Def. 2.16 in [1]).

A polynomial
$P(x)=\sum_{\alpha}c_{\alpha}x^{\alpha}\in\mathbb{Z}[x_{1},\cdots,x_{n}]$
satisfies the maximal Rado condition if for every $q\geq 2$ there exists an
upper Rado functional $(J_{0},\cdots,J_{\ell},d_{0},\cdots,d_{m-1})$ for $P$
such that, setting $d_{m}=0$, the polynomial

$g(w)=\sum_{i=0}^{m}q^{d_{i}}\sum_{\alpha\in J_{i}}c_{\alpha}w^{|\alpha|}$
(57)

has a real root in $[1,q]$.

######  Theorem 5.11 (cf. Theorem 2.19 of [1]).

Fix $P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$. If the equation
$P(x_{1},\cdots,x_{n})=0$ is partition regular over $\mathbb{N}$, then $P$
satisfies the maximal Rado condition.

We are now ready to begin proving Theorem 5.1.

######  Lemma 5.12.

If $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $n,m\in\mathbb{N}$ are such that
$a+b\neq 0$ and $n,m\geq 2$, then the equation

$ax+by=cw^{m}z^{n}$ (58)

is not partition regular over $\mathbb{N}$.

###### Proof.

Let us fix $a,b,n,m\in\mathbb{N}$ with $n,m\geq 2$. We will begin by showing
that the polynomial $P(w,x,y,z)=cz^{n}w^{m}-ax-by$ does not have any upper
Rado functional of order $m\geq 1$. From there it will be relatively simple to
verify that $P$ does not satisfy the maximal Rado condition. To this end, let
us assume for the sake of contradiction that
$(J_{0},\cdots,J_{\ell},d_{0},\cdots,d_{m-1})$ is an upper Rado functional of
order $m$ for $P$. Note that
$\text{Supp}(P)=\\{(m,0,0,n),(0,1,0,0),(0,0,1,0)\\}=\\{M_{0}^{\prime},M_{1}^{\prime},M_{2}^{\prime}\\}$.
By considering all possible orderings of the set
$\\{M_{0}^{\prime},M_{1}^{\prime},M_{2}^{\prime}\\}$ and the definition of
$d_{0}$, we see that for any finite coloring $c$ of $\mathbb{N}$ there exists
infinitely many $c$-monochromatic positive linear maps

$\phi(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})=t_{1}\alpha_{1}+t_{2}\alpha_{2}+t_{3}\alpha_{3}+t_{4}\alpha_{4},$
(59)

for which at least one of equations (60)–(65) has a solution:

$\displaystyle\phi(m,0,0,n)$ $\displaystyle-$
$\displaystyle\phi(0,1,0,0)=d_{0}$
$\displaystyle\iff{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}-}mt_{1}$
$\displaystyle-$ $\displaystyle t_{2}$ $\displaystyle+$ $\displaystyle nt_{4}$
$\displaystyle=$ $\displaystyle d_{0},$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
$\displaystyle{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}a}$
(60) $\displaystyle\phi(0,1,0,0)$ $\displaystyle-$
$\displaystyle\phi(m,0,0,n)=d_{0}$ $\displaystyle\iff-mt_{1}$ $\displaystyle+$
$\displaystyle t_{2}$ $\displaystyle-$ $\displaystyle nt_{4}$ $\displaystyle=$
$\displaystyle d_{0},$ (61) $\displaystyle\phi(m,0,0,n)$ $\displaystyle-$
$\displaystyle\phi(0,0,1,0)=d_{0}$
$\displaystyle\iff{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}-}mt_{1}$
$\displaystyle-$ $\displaystyle t_{3}$ $\displaystyle+$ $\displaystyle nt_{4}$
$\displaystyle=$ $\displaystyle d_{0},$ (62) $\displaystyle\phi(0,0,1,0)$
$\displaystyle-$ $\displaystyle\phi(m,0,0,n)=d_{0}$ $\displaystyle\iff-mt_{1}$
$\displaystyle+$ $\displaystyle t_{3}$ $\displaystyle-$ $\displaystyle nt_{4}$
$\displaystyle=$ $\displaystyle d_{0},$ (63) $\displaystyle\phi(0,1,0,0)$
$\displaystyle-$ $\displaystyle\phi(0,0,1,0)=d_{0}$ $\displaystyle\iff$
$\displaystyle t_{2}$ $\displaystyle-$ $\displaystyle t_{3}$ $\displaystyle=$
$\displaystyle d_{0},$ (64) $\displaystyle\phi(0,0,1,0)$ $\displaystyle-$
$\displaystyle\phi(0,1,0,0)=d_{0}$ $\displaystyle\iff$ $\displaystyle t_{3}$
$\displaystyle-$ $\displaystyle t_{2}$ $\displaystyle=$ $\displaystyle d_{0}.$
(65)

Equivalently, for any finite coloring of $\mathbb{N}$ at least one of
equations (60)–(65) must possess infinitely many monochromatic solutions
$(t_{1},t_{2},t_{3},t_{4})$. By Theorem 2.5 we see that any one of equations
(60)–(65) is partition regular over $\mathbb{Z}$ (and hence over $\mathbb{N}$)
if and only if there exists a constant solution $t_{1}=t_{2}=t_{3}=t_{4}=t$.
It follows that equations (64) and (65) are not partition regular.
Furthermore, we see that one of equations (60)–(63) possess infinitely many
monochromatic solutions $(t_{1},t_{2},t_{3},t_{4})$ in any finite coloring of
$\mathbb{N}$ if and only if one of equations (66)–(69) below possess
infinitely many monochromatic solutions $(t_{1},t_{2},t_{3},t_{4})$ in any
finite coloring of $\mathbb{N}$:

$\displaystyle
m(t_{1}-\frac{d_{0}}{m+n-1})-(t_{2}-\frac{d_{0}}{m+n-1})+n(t_{4}-\frac{d_{0}}{m+n-1})$
$\displaystyle=0,$ (66)
$\displaystyle-m(t_{1}-\frac{d_{0}}{m+n-1})+(t_{2}-\frac{d_{0}}{m+n-1})-n(t_{4}-\frac{d_{0}}{m+n-1})$
$\displaystyle=0,$ (67) $\displaystyle
m(t_{1}-\frac{d_{0}}{m+n-1})-(t_{3}-\frac{d_{0}}{m+n-1})+n(t_{4}-\frac{d_{0}}{m+n-1})$
$\displaystyle=0,$ (68)
$\displaystyle-m(t_{1}-\frac{d_{0}}{m+n-1})+(t_{3}-\frac{d_{0}}{m+n-1})-n(t_{4}-\frac{d_{0}}{m+n-1})$
$\displaystyle=0.$ (69)

Since $m,n\geq 2$ we may repeatedly use Corollary 2.4 to create a finite
partition of $\mathbb{N}$ for which the only monochromatic solution to any of
equations (66)–(69) (considered separately, not as a system) is the solution
$t_{1}=t_{2}=t_{3}=t_{4}=\frac{d_{0}}{m+n-1}$. It follows that $P$ does not
have any upper Rado functionals of order $m\geq 1$. Noting that an upper Rado
functional of order $0$ is just a Rado partition, we note that the set of Rado
partitions of $P$ is

$\displaystyle\Big{\\{}\Big{(}\\{(m,0,0,n)\\},\\{(0,1,0,0)\\},\\{(0,0,1,0)\\}\Big{)},$
$\displaystyle\Big{(}\\{(m,0,0,n)\\},\\{(0,0,1,0)\\},\\{(0,1,0,0)\\}\Big{)},$
$\displaystyle\Big{(}\\{(0,1,0,0)\\},\\{(m,0,0,n)\\},\\{(0,0,1,0)\\}\Big{)},$
$\displaystyle\Big{(}\\{(0,1,0,0)\\},\\{(0,0,1,0)\\},\\{(m,0,0,n)\\}\Big{)},$
$\displaystyle\Big{(}\\{(0,0,1,0)\\},\\{(m,0,0,n)\\},\\{(0,1,0,0)\\}\Big{)},$
$\displaystyle\Big{(}\\{(0,0,1,0)\\},\\{(0,1,0,0)\\},\\{(m,0,0,n)\\}\Big{)},$
$\displaystyle\Big{(}\\{(0,1,0,0),(0,0,1,0)\\},\\{(m,0,0,n)\\}\Big{)},$
$\displaystyle\Big{(}\\{(m,0,0,n)\\},\\{(0,1,0,0),(0,0,1,0)\\}\Big{)}\Big{\\}}.$

Finally, to see that $P$ does not satisfy the maximal Rado condition, we see
that for $q=2$ (or any other value of $q\geq 2$) we have

$g(w)=\sum_{i=0}^{m}q^{d_{i}}\sum_{\alpha\in
J_{i}}c_{\alpha}w^{|\alpha|}=\sum_{i=0}^{0}2^{0}\sum_{\alpha\in
J_{i}}c_{\alpha}w^{|\alpha|}=\sum_{\alpha\in
J_{i}}c_{\alpha}w^{|\alpha|}\in\\{cw^{n+m},-aw,-bw,-(a+b)w\\},$ (70)

so none of the polynomials that $g(w)$ could be contains a root in $[1,2]$. ∎

###### Proof of Theorem 5.1.

By Lemma 5.12 we see that

$ax+by=cw^{m}z^{n}$ (71)

and

$ax+by=(-1)^{n+m-1}cw^{m}z^{n}$ (72)

are not partition regular over $\mathbb{N}$. Let
$\mathbb{N}=\bigcup_{i=1}^{r_{1}}C_{i}$ be a partition for which no cell
contains a solution to equation (71) and let
$\mathbb{N}=\bigcup_{i=1}^{r_{2}}D_{i}$ be a partition for which no cell
contains a solution to equation (72). It now suffices to show that no cell of
the partition

$\mathbb{Z}\setminus\\{0\\}=\left(\bigcup_{i=1}^{r_{2}}(-D_{i})\right)\cup\bigcup_{i=1}^{r_{1}}C_{i}$
(73)

contains a solution to equation (71). It follows from the definition of the
$C_{i}$ that none of them contain a solution to equation (71), so let us
assume for the sake of contradiction that for some $1\leq i\leq r_{2}$ there
exist $w^{\prime},x^{\prime},y^{\prime},z^{\prime}\in-D_{i}$ which satisfy
equation (71). Letting $w=-w^{\prime},x=-x^{\prime},y=-y^{\prime},$ and
$z=-z^{\prime}$, we see that $w,x,y,z\in D_{i}$ and

$a(-x)+b(-y)=ax^{\prime}+by^{\prime}=c(w^{\prime})^{m}(z^{\prime})^{n}=c(-w)^{m}(-z)^{n}=(-1)^{m+n}cw^{m}z^{n}$
(74)

$\Rightarrow ax+by=(-1)^{m+n-1}cw^{m}z^{n},$ (75)

which yields the desired contradiction. ∎

We now describe another condition for a polynomial to be partition regular of
a particular flavor, involving lower Rado functionals.

######  Definition 5.13 (cf. Def. 2.12 in [1]).

A lower Rado functional of order $m\in\mathbb{N}\cup\\{0\\}$ for
$P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$ is a tuple
$(J_{0},\cdots,J_{\ell},d_{1},\cdots,d_{m})$ for some $\ell\geq m$ and
$d_{1},\cdots,d_{m}\in\mathbb{N}$ such that, for every finite coloring $c$ of
$\mathbb{N}$ and for every $N\in\mathbb{N}$, there exist infinitely many
$c$-monochromatic positive linear maps

$\displaystyle\phi:\mathbb{Z}^{n}$ $\displaystyle\rightarrow\mathbb{Z},$
$\displaystyle(\alpha_{1},\cdots,\alpha_{n})$
$\displaystyle\mapsto(t_{1}\alpha_{1}+t_{2}\alpha_{2}+\cdots+t_{n}\alpha_{n}),$

such that $(J_{0},\cdots,J_{\ell})$ is the partition of Supp$(P)$ determined
by $\phi$ and, if $(M_{0},\cdots,M_{\ell})$ is the increasing enumeration of
$\phi(\text{Supp}(P))$, then $M_{i}-M_{m}=d_{i}$ for $i\in\\{1,2,\cdots,m\\}$,
and $M_{m+1}-M_{m}\geq N$.

######  Definition 5.14 (cf. Def. 2.2 in [1]).

For $P\in\mathbb{Z}[x_{1},x_{2},\cdots,x_{n}]$ and $q\in\mathbb{N}$, the
equation

$P(x_{1},x_{2},\cdots,x_{n})=0$

is $q$-partition regular if for any $k\in\mathbb{N}$ and any partition
$q^{k}\mathbb{N}=\bigcup_{i=1}^{v}A_{i}$, there exists $1\leq i_{0}\leq v$ and
$y_{1},\cdots,y_{n}\in A_{i_{0}}$ for which $P(y_{1},\cdots,y_{n})=0$.

######  Theorem 5.15 (cf. Theorem 3.3 of [1]).

Suppose that $p\in\mathbb{N}$ is a prime. If
$P\in\mathbb{Z}[x_{1},\cdots,x_{n}]$ is $p$-partition regular, then there
exists a lower Rado functional $(J_{0},\cdots,J_{\ell},d_{1},\cdots,d_{m})$
for $P$ such that setting $d_{0}=0$, the equation

$\sum_{i=0}^{m}p^{d_{i}}\sum_{\alpha\in
J_{i}}\frac{1}{\alpha!}\frac{\partial^{\alpha}P}{\partial
x^{\alpha}}(0,0,\cdots,0)w^{|\alpha|}=0$ (76)

has an invertible solution in the ring $\mathbb{Z}_{p}$ of $p$-adic integers.

We provide a lemma on a condition in order for our polynomial to be partition
regular.

######  Lemma 5.16.

Suppose that $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $n\in\mathbb{N}$ are
such that the equation

$ax+by=cwz^{n}$ (77)

is partition regular. If $p$ is a prime for which $v_{p}(\frac{a+b}{c})\notin
n\mathbb{N}\cup\\{0\\}$, then equation (77) is $p$-partition regular.

###### Proof.

We will use induction on $k$ to show that equation (77) is partition regular
over $p^{k}\mathbb{N}$ for each $k\geq 0$. The base case of $k=0$ holds by
assumption, so let us proceed to the inductive step and show that the desired
result holds for $k+1$ if it holds for $k$. Let $m_{1}=v_{p}(a+b)$,
$m_{2}=v_{p}(c)$, and $M=\text{max}(m_{1},m_{2})$. Consider the partition

$p^{k}\mathbb{N}=\bigcup_{j=1}^{p^{M+1}}C_{j}\text{ where
}C_{j}=\\{n\in\mathbb{N}\ |\ n\equiv p^{k}j\pmod{p^{k+M+1}}\\}.$ (78)

Since equation (77) is partition regular over $p^{k}\mathbb{N}$, let
$w,x,y,z\in C_{j_{0}}$ satisfy equation (77). We see that

$\displaystyle ax+by=cwz^{n}\rightarrow ax+by\equiv cwz^{n}\pmod{p^{k+M+1}}$
(79) $\displaystyle\implies$ $\displaystyle aj_{0}+bj_{0}\equiv
cj_{0}^{n+1}p^{nk}\equiv 0\pmod{p^{M+1}},$ (80) $\displaystyle\implies$
$\displaystyle j_{0}(a+b-cj_{0}^{n}p^{nk})\equiv 0\pmod{p^{M+1}}.$ (81)

Since $v_{p}(\frac{a+b}{c})\notin n\mathbb{N}\cup\\{0\\}$, we see that
$v_{p}(a+b-cj_{0}^{n}p^{nk})=\text{min}(v_{p}(a+b),v_{p}(cj_{0}^{n}p^{nk}))\leq
M$, so we must have that $j_{0}\equiv 0\pmod{p}$. We now see that for any
partition

$p^{k+1}\mathbb{N}=\bigcup_{j=1}^{r}B_{j},$ (82)

we may let $B_{r+1}=p^{k}\mathbb{N}\setminus p^{k+1}\mathbb{N}$ and construct
the partition

$p^{k}\mathbb{N}=\bigcup_{\underset{\scriptstyle 1\leq j_{2}\leq
p^{M+1}}{\scriptstyle 1\leq j_{1}\leq
r+1{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}^{\prime\prime\prime}}}}(B_{j_{1}}\cap
C_{j_{2}}).$ (83)

Since equation (77) is partition regular over $p^{k}\mathbb{N}$, let
$w,x,y,z\in B_{J_{1}}\cap C_{J_{2}}$ satisfy equation (77). We have already
shown that since $w,x,y,z\in C_{J_{2}}$, we have $w,x,y,z\in
p^{k+1}\mathbb{N}$. Since $w,x,y,z\notin B_{r+1}$, we have shown that equation
(77) is also partition regular over $p^{k+1}\mathbb{N}$ as desired. ∎

This leads us to the following useful criterion.

######  Theorem 5.17.

Suppose that $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $n\in\mathbb{N}$ are
such that the equation

$ax+by=cwz^{n}$ (84)

is partition regular. If $p$ is a prime for which $v_{p}(\frac{a+b}{c})\notin
n\mathbb{N}\cup\\{0\\}$, then one of $\frac{a}{c},\frac{b}{c},\frac{a+b}{c}$
must be an $n$th power in $\mathbb{Q}_{p}$.

###### Proof.

We will begin by determining all of the lower Rado functionals for
$P(x_{1},x_{2},x_{3},x_{4})=ax_{1}+bx_{2}-cx_{3}x_{4}^{n}$. Since the system
of equations

$\begin{array}[]{ccccccc}\phi(1,0,0,0)&=&\alpha_{1}&=&\alpha_{2}&=&\phi(0,1,0,0)\\\
\phi(0,1,0,0)&=&\alpha_{2}&=&\alpha_{3}+n\alpha_{4}&=&\phi(0,0,1,n)\end{array}$
(85)

is partition regular, we see that

$\displaystyle\Big{\\{}\Big{(}\\{(1,0,0,0),(0,1,0,0),(0,0,1,n)\\}\Big{)},$
$\displaystyle\Big{(}\\{(1,0,0,0),(0,1,0,0)\\},\\{(0,0,1,n)\\}\Big{)},$
$\displaystyle\Big{(}\\{(0,1,0,0),(0,0,1,n)\\},\\{(1,0,0,0)\\}\Big{)},$
$\displaystyle\Big{(}\\{(1,0,0,0),(0,0,1,n)\\},\\{(0,1,0,0)\\}\Big{)}\Big{\\}},$

is the collection of nontrivial222A lower Rado functional or order $0$
$(J_{0},\cdots,J_{\ell})$ is trivial if $J_{0}$ is a singleton, as such a
functional will never yield an invertible solution to equation (76). lower
Rado functionals of order $0$. We now proceed to determine all lower Rado
functionals of order $1$. Let $(J_{0},\cdots,J_{\ell},d_{1})$ be a lower Rado
functional of order $1$. Since $d_{1}>0$ we may use Theorem 2.5 to see that
neither of the equations

$\displaystyle d_{1}=\phi(1,0,0,0)-\phi(0,1,0,0)$
$\displaystyle=\alpha_{1}-\alpha_{2}\text{, and}$ (86) $\displaystyle
d_{1}=\phi(0,1,0,0)-\phi(1,0,0,0)$ $\displaystyle=\alpha_{2}-\alpha_{1}$ (87)

are partition regular. It follows that any lower Rado functional of order $1$
must have $(0,0,1,n)\in J_{0}\cup J_{1}$. We also note by Theorem 2.5 that the
equations

$\displaystyle d_{1}=\phi(1,0,0,0)-\phi(0,0,1,n)$
$\displaystyle=\alpha_{1}-\alpha_{3}-n\alpha_{4},$ (88) $\displaystyle
d_{1}=\phi(0,1,0,0)-\phi(0,0,1,n)$
$\displaystyle=\alpha_{2}-\alpha_{3}-n\alpha_{4},$ (89) $\displaystyle
d_{1}=\phi(0,0,1,n)-\phi(1,0,0,0)$
$\displaystyle=-\alpha_{1}+\alpha_{3}+n\alpha_{4}$ (90) $\displaystyle
d_{1}=\phi(0,0,1,n)-\phi(0,1,0,0)$
$\displaystyle=-\alpha_{2}+\alpha_{3}+n\alpha_{4},$ (91)

are partition regular over $\mathbb{Z}$ if and only if $n|d_{1}$. This results
in the following list of lower Rado functionals of order $1$:

$\displaystyle\Big{\\{}\Big{(}\\{(1,0,0,0)\\},\\{(0,0,1,n)\\},\\{(0,1,0,0)\\},nd\Big{)},$
$\displaystyle\Big{(}\\{(0,1,0,0)\\},\\{(0,0,1,n)\\},\\{(1,0,0,0)\\},nd\Big{)},$
$\displaystyle\Big{(}\\{(0,0,1,n)\\},\\{(1,0,0,0)\\},\\{(0,1,0,0)\\},nd\Big{)},$
$\displaystyle\Big{(}\\{(0,0,1,n)\\},\\{(0,1,0,0\\},\\{(1,0,0,0)\\},nd\Big{)},$
$\displaystyle\Big{(}\\{(1,0,0,0),(0,1,0,0)\\},\\{(0,0,1,n)\\},nd\Big{)},$
$\displaystyle\Big{(}\\{(0,0,1,n)\\},\\{(1,0,0,0),(0,1,0,0)\\},nd\Big{)}\Big{\\}}.$

Lastly, we recall that the only lower Rado functionals of order $2$ are of the
form $(J_{0},J_{1},J_{2},d_{1},d_{2})$, but such a lower Rado functional
cannot exist since equations (86) and (87) are not partition regular. We have
now determined all of the lower Rado functionals for
$P(x_{1},x_{2},x_{3},x_{4})$. By Lemma 5.16 we see that equation (84) is
$p$-partition regular, so we may apply Theorem 5.15 to see that at least $1$
of equations (92)-(101) has an invertible solution in $\mathbb{Z}_{p}$.

$\displaystyle(a+b)w-cw^{n}$ $\displaystyle=0\left(\text{from the lower Rado
functional}\Big{(}\\{(1,0,0,0),(0,1,0,0),(0,0,1,n)\\}\Big{)}\right),$ (92)
$\displaystyle(a+b)w-cp^{nd}w^{n}$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(1,0,0,0),(0,1,0,0)\\},\\{(0,0,1,n)\\},nd\Big{)}\right),$ (93)
$\displaystyle(a+b)p^{nd}w-cw^{n}$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(0,0,1,n)\\},\\{(1,0,0,0),(0,1,0,0)\\},nd\Big{)}\right),$ (94)
$\displaystyle aw-cw^{n}$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(1,0,0,0),(0,0,1,n)\\},\\{(0,1,0,0)\\}\Big{)}\right),$ (95)
$\displaystyle aw-cp^{nd}w$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(1,0,0,0)\\},\\{(0,0,1,n)\\},\\{(0,1,0,0)\\},nd\Big{)}\right),$
(96) $\displaystyle ap^{nd}w-cw^{n}$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(0,0,1,n)\\},\\{(1,0,0,0)\\},\\{(0,1,0,0)\\},nd\Big{)}\right),$
(97) $\displaystyle bw-cw^{n}$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(0,1,0,0),(0,0,1,n)\\},\\{(1,0,0,0)\\}\Big{)}\right),$ (98)
$\displaystyle bw-cp^{nd}w$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(0,1,0,0)\\},\\{(0,0,1,n)\\},\\{(1,0,0,0)\\},nd\Big{)}\right),$
(99) $\displaystyle bp^{nd}w-cw^{n}$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(0,0,1,n)\\},\\{(0,1,0,0\\},\\{(1,0,0,0)\\},nd\Big{)}\right),$
(100) $\displaystyle(a+b)w$ $\displaystyle=0\left(\text{from the
LRF}\Big{(}\\{(1,0,0,0),(0,1,0,0)\\},\\{(0,0,1,n)\\}\Big{)}\right).$ (101)

The desired result follows after noting that one of equations (92)-(101) has
an invertible solution in $\mathbb{Z}_{p}$ if and only if one of
$\frac{a}{c},\frac{b}{c},\frac{a+b}{c}$ is an $n$th power in $\mathbb{Q}_{p}$.
∎

Before using Theorem 5.17, let us recall when $a\in\mathbb{Z}_{2}$ is an $n$th
power. If $a=2^{k}m$ with $m$ odd, it is a well-known consequence of Hensel’s
lemma that $a$ is a $2^{n}$th power in $\mathbb{Z}_{2}$ if and only if
$2^{n}\mid k$ and $m\equiv 1\pmod{2^{n+2}}$.

######  Corollary 5.18.

The following equations are not partition regular as seen by an application of
Theorem 5.17 with $p=2$ for items (i)-(iii), $p=3$ for item (iv), and $p=5$
for item (v).

  * (i)

$3x+13y=wz^{8}.$

Observe that $16$ is an $8$th power modulo $p$ for every prime $p$, so this
equation is not susceptible to Theorem 3.8.

  * (ii)

$16x+16y=wz^{8}$.

Observe that $16$ is an $8$th power modulo $p$ for every prime $p$, so this
equation is not susceptible to Theorem 3.8.

  * (iii)

$3\cdot 5\cdot 2^{2}x+2\cdot 5\cdot 3^{2}y=wz^{2}.$

Observe that at least one of $3\cdot 5\cdot 2^{2},2\cdot 5\cdot 3^{2}$, or
$3\cdot 5\cdot 2^{2}+2\cdot 5\cdot 3^{2}=2\cdot 3\cdot 5^{2}$ is a square
modulo $p$ for every prime $p$, so this equation is not susceptible to Theorem
3.8.

  * (iv)

$3^{4}x+3^{6}y=wz^{12}.$

Observe that $-3$ is a square modulo $p$ if $p\equiv 1\pmod{3}$ and $3$ is a
cube modulo $p$ if $p\equiv 2\pmod{3}$, so either $3^{4}$ or $3^{6}$ is a
$12$th power modulo any prime $p$ (cf. Item (ii)(a) or Remark 4.10), so this
equation is not susceptible to Theorem 3.8.

  * (v)

$(3^{2}\cdot 4\cdot 5)^{2}x+(3\cdot 4^{2}\cdot 5)^{2}y=wz^{4}$

Observe that $\alpha=(3^{2}\cdot 4\cdot 5)^{2},\beta=(3\cdot 4^{2}\cdot
5)^{2}$, and $\gamma=\alpha+\beta=(3\cdot 4\cdot 5^{2})^{2}$ are all squares
but are not fourth powers. Since one of $3^{2}\cdot 4\cdot 5,3\cdot 4^{2}\cdot
5$, and $3\cdot 4\cdot 5^{2}$ will be a perfect square modulo any prime $p$,
we see that one of $\alpha^{2},\beta^{2},$ or $\gamma^{2}$ will be a perfect
fourth power modulo any prime $p$, so this equation is not susceptible to
Theorem 3.8.

##  6 On the Partition Regularity of $ax+by=cwz^{n}$ over Integral Domains

The purpose of this section is to try and generalize as much of Theorem 1.1 as
we can to the more general setting of integral domains instead of just
$\mathbb{Z}$ or $\mathbb{N}$.

######  Theorem 6.1 (cf. Theorem 9.18).

Let $R$ be an integral domain. There exists an ultrafilter $p\in\beta
R\setminus\\{0\\}$ with the following properties.

  1. (i)

For any $A\in p$ and $\ell\in\mathbb{N}$, there exists $b,g\in A$ with
$\\{bg^{j}\\}_{j=0}^{\ell}\subseteq A$.

  2. (ii)

For any $A\in p$, $\ell\in\mathbb{N}$, and $h,s\in R\setminus\\{0\\}$, there
exists $a,d\in R$ for which $\\{hd,ha,ha+sd\\}\subseteq A$.

  3. (iii)

For every $\alpha\in R\setminus\\{0\\}$, we have $\alpha R\in p$.

######  Lemma 6.2.

Let $R$ be an integral domain and let $p\in\beta R\setminus\\{0\\}$ be an
ultrafilter satisfying the conditions of Theorem 6.1. For any $A\in p$,
$a,b\in R\setminus\\{0\\}$ and $n\in\mathbb{N}$, the equation

$ax+by=cwz^{n}$ (102)

has a solution in $A$ if $c\in\\{a,b,a+b\\}$.

###### Proof.

Let

$A^{\prime}=\\{v\in A\ |\ v=wz^{n}\text{ for some }z,w\in A\\}.$ (103)

Since $A\in p$, to see that $A^{\prime}=A\setminus(A\setminus A^{\prime})\in
p$ it suffices to observe that $A\setminus A^{\prime}\notin p$ because
$A\setminus A^{\prime}$ does not satisfy condition $(i)$ of Theorem 6.1. Our
first case is when $c=a+b$, and in this case we let $x\in A^{\prime}$ be
arbitrary and let $w,z\in A$ be such that $x=wz^{n}$. Since

$ax+bx=cx=cwz^{n},$ (104)

we see that $x,x,w,z$ is a solution to equation (102) coming from $A$. For our
second case it suffices to consider $c=a$ since the case of $c=b$ is handled
similarly. Observe that $A^{\prime}_{a}:=A^{\prime}\cap aR\in p$ since
$A^{\prime},aR\in p$ and consider

$A^{\prime\prime}=\\{x_{1}\in A^{\prime}_{a}\ |\ \text{there exists }x_{2}\in
A^{\prime}_{a}\text{ satisfying }x_{1}+b\frac{x_{2}}{a}\in A^{\prime}_{a}\\}.$
(105)

Since $A^{\prime}_{a}\in p$, to see that
$A^{\prime\prime}=A^{\prime}_{a}\setminus(A^{\prime}_{a}\setminus
A^{\prime\prime})\in p$ it suffices to observe that $A^{\prime}_{a}\setminus
A^{\prime\prime}\notin p$ because $A^{\prime}_{a}\setminus A^{\prime\prime}$
does not satisfy condition $(ii)$ of Theorem 6.1 with $(h,s)=(a,b)$. Now let
$x_{1}\in A^{\prime\prime}$ be arbitrary, let $x_{2}\in A^{\prime}_{a}$ be as
in equation (105), and observe that

$ax_{1}+bx_{2}=a(x_{1}+b\frac{x_{2}}{a}).$ (106)

Since $x_{1}+b\frac{x_{2}}{a}\in A^{\prime}$, we may pick $w,z\in A$ for which
$x_{1}+b\frac{x_{2}}{a}=wz^{n}$. In this case we observe that

$ax_{1}+bx_{2}=c(x_{1}+b\frac{x_{2}}{a})=cwz^{n},$ (107)

so $x_{1},x_{2},w,z$ is a solution to equation (102) coming from $A$. ∎

Before proceeding further let us recall some notation. If $R$ is an integral
domain, then for $u,v\in R\setminus\\{0\\}$ and $A\subseteq R$ we have

$\frac{v}{u}A=\\{r\in R\ |\ \frac{u}{v}r\in A\\}=\\{\frac{v}{u}a\ |\ a\in
A\cap uR\\}.$ (108)

Similarly, if $p\in\beta R$ is an ultrafilter, then we have

$\frac{u}{v}\cdot p=\\{A\subseteq R\ |\ \frac{v}{u}A\in p\\}=\\{\frac{u}{v}A\
|\ A\in p\\}.$ (109)

######  Theorem 6.3.

Let $R$ be an integral domain with field of fractions $K$ and let $p\in\beta
R\setminus\\{0\\}$ be an ultrafilter satisfying the conditions of Theorem 6.1.
If $a,b,c\in R\setminus\\{0\\}$ and $n\in\mathbb{N}$ are such that one of
$\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ is of the form
$(\frac{u}{v})^{n}$ for some $u,v\in R$, then the equation

$ax+by=cwz^{n}$ (110)

contains a solution for any $A\in q$ where

$q=\begin{cases}p&\text{if }u=0\\\ \frac{u}{v}\cdot p&\text{else}\end{cases}.$
(111)

In particular, equation (110) is partition regular over $R\setminus\\{0\\}$.

###### Proof.

We see that if $u=0$ then $a+b=0$, so the desired result in this case follows
from Lemma 6.2. Now let us assume that $u\neq 0$ and let $d\in\\{a,b,a+b\\}$
be such that $\frac{d}{c}=(\frac{u}{v})^{n}$. Let $A\in\frac{u}{v}\cdot p$ be
arbitrary and note that $\frac{v}{u}A\in p$. By Lemma 6.2 there exists
$w,x,y,z\in A$ for which
$\frac{v}{u}w,\frac{v}{u}x,\frac{v}{u}y,\frac{v}{u}z\in\frac{v}{u}A$ and

$a(\frac{v}{u}x)+b(\frac{v}{u}y)=d(\frac{v}{u}w)(\frac{v}{u}z)^{n}\Rightarrow
ax+by=d(\frac{v}{u})^{n}wz^{n}=cwz^{n}.$ (112)

For the latter half of the Theorem, it suffices to note that if
$R\setminus\\{0\\}=\bigcup_{i=1}^{r}C_{i}$ is a partition, then there exists
$1\leq i_{0}\leq r$ for which $C_{i_{0}}\in q$, hence $C_{i_{0}}$ contains a
solution to equation (110). ∎

We recall that if $R$ is an Dedekind domain and $\mathfrak{p}\subseteq R$ is a
prime (hence maximal) ideal, then for any $u\in R$ and $v\in
R\setminus\\{0\\}$ we have $\frac{u}{v}\equiv uv^{-1}\pmod{\mathfrak{p}}$.

######  Theorem 6.4.

Let $R$ be a Dedekind domain with field of fractions $K$. Let $a,b,c\in
R\setminus\\{0\\}$ and $n\in\mathbb{N}$ be such that none of
$\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$ are $n$th powers in
$R/\mathfrak{p}$ for some prime ideal $\mathfrak{p}\subseteq R$ satisfying
$a,b,a+b,c\notin\mathfrak{p}$ and $[R:\mathfrak{p}]<\infty$. Let
$K_{\mathfrak{p}}$ denote the completion of $K$ at $\mathfrak{p}$. The
equation

$ax+by=cwz^{n}$ (113)

is not partition regular over $K_{\mathfrak{p}}\setminus\\{0\\}$.

###### Proof.

Since $R$ is a Dedekind domain we see that $R_{\mathfrak{p}}$ is a discrete
valuation ring under the valuation $v_{\mathfrak{p}}$, so let $\pi$ be a
generator of the maximal ideal of $R_{\mathfrak{p}}$. Let $F\subseteq R$ be a
set of coset representatives of $(\pi)$ such that $\bigcup_{f\in
F}\left(f+(\pi)\right)=R\setminus\mathfrak{p}$ and
$\left(f_{1}+(\pi)\right)\cap\left(f_{2}+(\pi)\right)=\emptyset$ whenever
$f_{1}\neq f_{2}$. We note that $|F|=[R:\mathfrak{p}]-1<\infty$. Let
$\chi:K_{\mathfrak{p}}\setminus\\{0\\}\rightarrow F$ be given by

$\frac{x}{\pi^{v_{\mathfrak{p}}(x)}}\equiv\chi(x)\pmod{\mathfrak{p}}.$ (114)

Note that $\chi(rs)=\chi(r)\chi(s)\pmod{\pi}$ for all $r,s\in
R\setminus\\{0\\}$. We also see that

$\chi(r+s)\equiv\begin{cases}\chi(r)+\chi(s)\hfill\pmod{\pi}&\text{ if
}v_{\mathfrak{p}}(r)=v_{\mathfrak{p}}(s)\text{ and }r+s\not\equiv
0\pmod{\pi}\\\ \chi(s)\hfill\pmod{\pi}&\text{ if
}v_{\mathfrak{p}}(r)>v_{\mathfrak{p}}(s)\\\ \chi(r)\hfill\pmod{\pi}&\text{ if
}v_{\mathfrak{p}}(s)>v_{\mathfrak{p}}(r)\end{cases}.$ (115)

Let $K_{\mathfrak{p}}\setminus\\{0\\}=\bigcup_{f\in F}C_{f}$ be the partition
given by $C_{f}=\chi^{-1}(\\{f\\})$. Let us assume for the sake of
contradiction that there exists $d\in F$ for which $w,x,y,z\in C_{d}$ and
equation (113) is satisfied. We now have $3$ cases to consider. If
$v_{\mathfrak{p}}(x)=v_{\mathfrak{p}}(y)$, then we see that

$0\not\equiv(a+b)d\equiv\chi(a)\chi(x)+\chi(b)\chi(y)\equiv\chi(ax+by)\equiv\chi(cwz^{n})\equiv
cd^{n+1}\pmod{\mathfrak{p}}$ (116)

$\implies(a+b)c^{-1}\equiv d^{n}\pmod{\mathfrak{p}},$ (117)

which yields the desired contradiction. For our next case we assume that
$v_{\mathfrak{p}}(x)<v_{\mathfrak{p}}(y)$ and note that

$0\not\equiv ad\equiv\chi(a)\chi(x)\equiv\chi(ax+by)\equiv\chi(cwz^{n})\equiv
cd^{n+1}\pmod{\mathfrak{p}}$ (118)

$\implies ac^{-1}\equiv d^{n}\pmod{\mathfrak{p}},$ (119)

which again yields the desired contradiction. Similarly, in our final case
when $v_{\mathfrak{p}}(x)>v_{\mathfrak{p}}(y)$ we have

$0\not\equiv bd\equiv\chi(b)\chi(y)\equiv\chi(ax+by)\equiv\chi(cwz^{n})\equiv
cd^{n+1}\pmod{\mathfrak{p}}$ (120)

$\implies bc^{-1}\equiv d^{n}\pmod{\mathfrak{p}},$ (121)

which once more yields the desired contradiction. ∎

######  Corollary 6.5.

Let $K$ be a number field and let $\omega_{m}$ be the number of $m$th roots of
unity in $K$. Let $a,b,c\in\mathcal{O}_{K}$ and let $n\in\mathbb{N}$. Let
$d_{a}$ be the largest integer for which
$a^{\frac{1}{d_{a}}}\in\mathcal{O}_{K}$, and define $d_{b}$ and $d_{c}$
similarly. Let $m_{a}=\frac{n}{d_{a}},m_{b}=\frac{n}{d_{b}},$ and
$m_{c}=\frac{n}{d_{c}}$.

  1. (i)

Suppose $n$ is odd, and none of $a^{\omega_{m_{a}}},b^{\omega_{m_{b}}},$ and
$c^{\omega_{m_{c}}}$ are an $n$th power in $\mathcal{O}_{K}$; or

  2. (ii)

Suppose $n$ is even, and $a,b,c$ satisfy the same conditions as in (i), but
$a^{2\omega_{m_{a}}}$ is also not an $n$th power.

Then the equation

$ax+by=cwz^{n}$ (122)

is not partition regular over $K\setminus\\{0\\}$.

###### Proof.

The given assumptions are precisely what we need to use Lemma 4.8 and obtain a
prime ideal $\mathfrak{p}\subseteq\mathcal{O}_{K}$ for which none of $a,b,$
and $c$ are $n$th powers modulo $\mathfrak{p}$. After noting that $K$ embeds
in $K_{\mathfrak{p}}$, we see that the desired result follows from Theorem
6.4. ∎

######  Remark 6.6.

Consider the equation

$2x+3y=wz^{2}.$ (123)

Since $2,3,$ and $5$ are not squares modulo $43$, Theorem 6.5 tells us that
equation (123) is not partition regular over
$\mathbb{Q}_{43}\setminus\\{0\\}$. Since equation (123) is partition regular
over the countable set $\mathbb{Z}[\sqrt{2}]$ as a consequence of Theorem 6.3
but not over the uncountable set $\mathbb{Q}_{43}$, we see that the algebraic
properties of the underlying set $S$ have a stronger influence on the
partition regularity of equations of the form $ax+by=cwz^{n}$ than the
cardinality of $S$.

##  7 Systems of equations

######  Theorem 7.1.

Let $R$ be an integral domain with field of fractions $K$ and let $p\in\beta
R\setminus\\{0\\}$ be an ultrafilter satisfying the conditions of Theorem 6.1.
If $a_{1},\cdots,a_{k},b_{1},\cdots,b_{k},c_{1},\cdots,c_{k}\in
R\setminus\\{0\\}$ and $n_{1},\cdots,n_{k}\in\mathbb{N}$ are such that

$I:=\bigcap_{i=1}^{k}\left\\{\sqrt[n_{i}]{\frac{a_{i}}{c_{i}}},\sqrt[n_{i}]{\frac{b_{i}}{c_{i}}},\sqrt[n_{i}]{\frac{a_{i}+b_{i}}{c_{i}}}\right\\}\neq\emptyset,$
(124)

then the system of equations

$\begin{array}[]{ccccc}a_{1}x_{1}&+&b_{1}y_{1}&=&c_{1}w_{1}z_{1}^{n}\\\
a_{2}x_{2}&+&b_{2}y_{2}&=&c_{2}w_{2}z_{2}^{n}\\\ &&\vdots&&\\\
a_{k}x_{k}&+&b_{k}y_{k}&=&c_{k}w_{k}z_{k}^{n}\end{array}$ (125)

contains a solution in every $A\in q$, where we may take

$q=\begin{cases}p&\text{if }0\in I\\\ i\cdot p&\text{if there exists }i\in
I\setminus\\{0\\}\end{cases}.$ (126)

In particular, the system of equations in (125) is partition regular over
$R\setminus\\{0\\}$.

###### Proof.

Since none of the equations in the system of equations in (125) share any
variables, the desired result follows from Theorem 6.3 ∎

######  Theorem 7.2.

Let $R$ be a Dedekind domain with field of fractions $K$. Let
$a_{1},\cdots,a_{k},b_{1},\cdots,b_{k},c_{1},\cdots,c_{k}\in
R\setminus\\{0\\}$, let $n\in\mathbb{N}$, and let

$I:=\bigcap_{i=1}^{k}\left\\{\frac{a_{i}}{c_{i}},\frac{b_{i}}{c_{i}},\frac{a_{i}+b_{i}}{c_{i}}\right\\}$
(127)

Suppose that there exists a prime ideal $\mathfrak{p}\subseteq R$ satisfying:

  1. (i)

$a_{1},\cdots,a_{i},b_{1},\cdots,b_{i},c_{1},\cdots,c_{i},a_{1}+b_{1},\cdots,a_{i}+b_{i}\notin\mathfrak{p}$.

  2. (ii)

If
$v_{1},v_{2}\in\bigcup_{i=1}^{k}\\{\frac{a_{i}}{c_{i}},\frac{b_{i}}{c_{i}},\frac{a_{i}+b_{i}}{c_{i}}\\}$
are distinct, then $v_{1}\not\equiv v_{2}\pmod{\mathfrak{p}}$.

  3. (iii)

No element of $I$ is an $n$th power modulo $\mathfrak{p}$.

  4. (iv)

$[R:\mathfrak{p}]<\infty$.

The system of equations

$\begin{array}[]{ccccc}a_{1}x_{1}&+&b_{1}y_{1}&=&c_{1}w_{1}z_{1}^{n}\\\
a_{2}x_{2}&+&b_{2}y_{2}&=&c_{2}w_{2}z_{2}^{n}\\\ &&\vdots&&\\\
a_{k}x_{k}&+&b_{k}y_{k}&=&c_{k}w_{k}z_{k}^{n}\end{array}$ (128)

is not partition regular over $K_{\mathfrak{p}}\setminus\\{0\\}$, where
$K_{\mathfrak{p}}$ is the localization of $K$ at $\mathfrak{p}$.

###### Proof.

We begin the proof similarly to that of Theorem 6.4. Since $R$ is a Dedekind
domain we see that $R_{\mathfrak{p}}$ is a discrete valuation ring under the
valuation $v_{\mathfrak{p}}$, so let $\pi$ be a generator of the maximal ideal
of $R_{\mathfrak{p}}$. Let $F\subseteq R$ be a set of coset representatives of
$(\pi)$ such that $\bigcup_{f\in
F}\left(f+(\pi)\right)=R\setminus\mathfrak{p}$ and
$\left(f_{1}+(\pi)\right)\cap\left(f_{2}+(\pi)\right)=\emptyset$ whenever
$f_{1}\neq f_{2}$. We note that $|F|=[R:\mathfrak{p}]-1<\infty$. Let
$\chi:K_{\mathfrak{p}}\setminus\\{0\\}\rightarrow F$ be given by

$\frac{x}{\pi^{v_{\mathfrak{p}}(x)}}\equiv\chi(x)\pmod{\mathfrak{p}}.$ (129)

Observe that $\chi(r)\equiv r\pmod{\mathfrak{p}}$ for all
$r\notin\mathfrak{p}$. Let $K_{\mathfrak{p}}\setminus\\{0\\}=\bigcup_{f\in
F}C_{f}$ be the partition given by $C_{f}=\chi^{-1}(\\{f\\})$. Let us assume
for the sake of contradiction that there exists $d\in F$ and
$x_{1},x_{2},y_{1},y_{2},w_{1},w_{2},z_{1},z_{2}\in C_{d}$ for which the
system of equations in (128) is satisfied. We see that for $1\leq i\leq k$ we
have

$0\not\equiv\chi(c_{i}w_{i}z_{i}^{n})\equiv\chi(a_{i}x_{i}+b_{i}y_{i})\equiv\begin{cases}\chi(a_{i})d+\chi(b_{i})d\hfill\pmod{\pi}&\text{
if }v_{\mathfrak{p}}(x_{i})=v_{\mathfrak{p}}(y_{i})\\\
\chi(a_{i})d\hfill\pmod{\pi}&\text{ if
}v_{\mathfrak{p}}(y_{i})>v_{\mathfrak{p}}(x_{i})\\\
\chi(b_{i})d\hfill\pmod{\pi}&\text{ if
}v_{\mathfrak{p}}(x_{i})>v_{\mathfrak{p}}(x_{i})\end{cases}$ (130)

$\Rightarrow d^{n}\equiv
d^{-1}\chi(w_{i}z_{i}^{n})=\begin{cases}(a_{i}+b_{i})c^{-1}\hfill\pmod{\pi}&\text{
if }v_{\mathfrak{p}}(x_{i})=v_{\mathfrak{p}}(y_{i})\\\
a_{i}c^{-1}\hfill\pmod{\pi}&\text{ if
}v_{\mathfrak{p}}(y_{i})>v_{\mathfrak{p}}(x_{i})\\\
b_{i}c^{-1}\hfill\pmod{\pi}&\text{ if
}v_{\mathfrak{p}}(x_{i})>v_{\mathfrak{p}}(x_{i})\end{cases}.$ (131)

For $1\leq i\leq k$ let
$v_{i}\in\\{a_{i}c_{i}^{-1},b_{i}c_{i}^{-1},(a_{i}+b_{i})c_{i}^{-1}\\}$ be
such that $d^{n}\equiv v_{i}\pmod{\mathfrak{p}}$. Since we must have that
$v_{i}\equiv d^{n}\equiv v_{j}\pmod{\mathfrak{p}}$ for all $1\leq i<j\leq k$,
we see that there is some $v\in K$ for which $v=v_{i}$ for all $1\leq i\leq
k$, hence $v\in I$. The desired contradictions follows after recalling that no
element of $I$ is an $n$th power modulo $\mathfrak{p}$. ∎

We observe that the conditions of Theorem 7.2 are vacuously fulfilled if
$I=\emptyset$.

######  Corollary 7.3.

Let
$a_{1},\cdots,a_{k},b_{1},\cdots,b_{k},c_{1},\cdots,c_{k}\in\mathbb{Z}\setminus\\{0\\}$
and $n\in\mathbb{N}$ be such that the system of equations

$\begin{array}[]{ccccc}a_{1}x_{1}&+&b_{1}y_{1}&=&c_{1}w_{1}z_{1}^{n}\\\
a_{2}x_{2}&+&b_{2}y_{2}&=&c_{2}w_{2}z_{2}^{n}\\\ &&\vdots&&\\\
a_{k}x_{k}&+&b_{k}y_{k}&=&c_{k}w_{k}z_{k}^{n}\end{array}$ (132)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$. Let

$I:=\bigcap_{i=1}^{k}\left\\{\frac{a_{i}}{c_{i}},\frac{b_{i}}{c_{i}},\frac{a_{i}+b_{i}}{c_{i}}\right\\}.$
(133)

  1. (i)

If $4\nmid n$ then $I$ contains an $n$th power.

  2. (ii)

If $4|n$ then $I$ contains an $\frac{n}{2}$th power.

###### Proof.

Let us assume for the sake of contradiction that one of items (i) and (ii)
were false. Since $|I|\leq 2$, we may invoke Lemma 4.13 to find a prime $p$
satisfying the conditions of Theorem 7.2 to attain the desired contradiction.
∎

######  Remark 7.4.

The following systems of equations are not partition regular as seen by an
application of Corollary 7.3.

  1. (i)

$\begin{array}[]{rcrcc}3^{4}\cdot 4^{2}\cdot 5^{2}x_{1}&+&3^{2}\cdot
4^{4}\cdot 5^{2}y_{1}&=&w_{1}z_{1}^{4}\\\ 5^{4}\cdot 12^{2}\cdot
13^{2}x_{2}&+&5^{2}\cdot 12^{4}\cdot
13^{2}y_{2}&=&w_{2}z_{2}^{4}\end{array},n\in\mathbb{N}$

.

  2. (ii)

$\begin{array}[]{rcrcc}16x_{1}&+&17y_{1}&=&w_{1}z_{1}^{8}\\\
33x_{2}&+&(2^{12}-33)y_{2}&=&w_{2}z_{2}^{8}\end{array}$

.

  3. (iii)

$\begin{array}[]{rcrcc}2^{n}x_{1}&+&3^{n}y_{1}&=&w_{1}z_{1}^{n}\\\
3^{n}x_{2}&+&7^{n}y_{2}&=&w_{2}z_{2}^{n}\\\
7^{n}x_{3}&+&2^{n}y_{3}&=&w_{3}z_{3}^{n}\end{array},n\in\mathbb{N}$

.

  4. (iv)

$\begin{array}[]{rcrcc}9x_{1}&+&16y_{1}&=&w_{1}z_{1}^{2}\\\
25x_{2}&-&9y_{2}&=&w_{2}z_{2}^{2}\\\ 25x_{3}&-&16y_{3}&=&w_{3}z_{3}^{2}\\\
9x_{4}&+&7y_{4}&=&w_{4}z_{4}^{2}\end{array}$

.

A few remarks are in order regarding these examples. In example (i) neither of
the constituent equations of the system are individually partition regular.
This fact can be determined through the use of Lemma 5.17, but not through the
use of Theorem 3.8 alone, despite the similarity of the proofs of Theorem 3.8
and 7.2. In example (ii) we do not currently know whether either of the
constituent equations of the system are individually partition regular (cf.
Section 8). In example (iii) any proper subsystem of equations is partition
regular as a consequence of Theorem 7.1. Theorem 7.1 also shows us that in
example (iv) the system of equations becomes partition regular if any equation
other than the first equation is removed from the system.

##  8 Conjectures and Concluding Remarks

Theorem 1.1 and Corollary 5.18 naturally lead us to the following conjecture:

######  Conjecture 8.1.

Given $a,b,c\in\mathbb{Z}\setminus\\{0\\}$ and $n\in\mathbb{N}$, the equation

$ax+by=cwz^{n}$ (134)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$ if and only if one of
$\frac{a}{c},\frac{b}{c},\frac{a+b}{c}$ is an $n$th power in $\mathbb{Q}$.

We see that the situation in which we have yet to resolve Conjecture 8.1 fully
is when $n$ is even and one of $\frac{a}{c},\frac{b}{c},$ or $\frac{a+b}{c}$
is an $\frac{n}{2}$th power in $\mathbb{Q}$. Since some special instances of
this situation have been resolved in Corollary 5.18, we list here some
equations whose partition regularity remains unknown. Firstly, the equation

$33x+(2^{12}-33)y=wz^{8}$ (135)

is not expected to be partition regular since $33,2^{12}-33,$ and $2^{12}$ are
not $8$th powers, but $2^{12}$ is an $8$th power modulo $p$ for every prime
$p$, and $33$ is an $8$th power in $\mathbb{Z}_{2}$, so we are unable to apply
Theorem 3.8 or Theorem 5.17. Similarly, the equation

$16x+17y=wz^{8}$ (136)

is not expected to be partition regular since $16,17,$ and $33$ are not $8$th
powers, but $16$ is an $8$th power modulo $p$ for every prime p, and $16$ is
also an $8$th power in $\mathbb{Z}_{3}$ and $\mathbb{Z}_{11}$, so we are once
again unable to apply Theorem 3.8 or Theorem 5.17. Next, we see that for any
coprime $a,b\in\mathbb{N}$ for which $a,b,$ and $a+b$ are not squares, the
equation

$a^{2}b(a+b)x+ab^{2}(a+b)y=wz^{2}$ (137)

is not expected to be partition regular since none of
$a^{2}b(a+b),ab^{2}(a+b),$ and $a^{2}b(a+b)+ab^{2}(a+b)=ab(a+b)^{2}$ are
squares. However, at least one of $a^{2}b(a+b),ab^{2}(a+b),$ and $ab(a+b)^{2}$
is a square modulo $p$ for any prime $p$, so we cannot make use of Theorem
3.8. We have seen in item (iii) of Corollary 5.18 that Theorem 5.17 can be
used in some cases, but it is unclear to the authors as to whether or not
Lemma 5.17 can be used in all cases. A similar difficulty arises in the case
of $n=4$. Recalling that for any $m,n,k\in\mathbb{N}$ we have
$(2mnk)^{2}+(km^{2}-kn^{2})^{2}=(km^{2}+kn^{2})^{2}$, we take
$k=2mn(m^{2}-n^{2})(m^{2}+n^{2})$ and consider the equation

$\left((2mn)^{2}(m^{2}-n^{2})(m^{2}+n^{2})\right)^{2}x+\left(2mn(m^{2}-n^{2})^{2}(m^{2}+n^{2})\right)^{2}y=wz^{4}.$
(138)

We see that for any prime $p$ at least one of
$\alpha:=(2mn)^{2}(m^{2}-n^{2})(m^{2}+n^{2}),\beta:=2mn(m^{2}-n^{2})^{2}(m^{2}+n^{2}),$
or $\gamma:=2mn(m^{2}-n^{2})(m^{2}+n^{2})^{2}$ will be a perfect square modulo
$p$, hence one of $\alpha^{2},\beta^{2},$ or $\gamma^{2}$ will be a fourth
power modulo $p$, so equation (138) is not susceptible to Theorem 3.8. We saw
in item (v) of Corollary 5.18 that Lemma 5.17 can occasionally be of use in
this situation, but it is once again unclear to the authors whether or not
Lemma 5.17 can always be used in this situation.

While the methods of this paper are not strong enough to fully resolve
Conjecture 8.1, they are strong enough to prove the following Theorem:

######  Theorem 8.2.

Fix $a_{1},\cdots,a_{r},b_{1},\cdots,b_{s},c\in\mathbb{Z}\setminus\\{0\\}$.

  * (i)

If min$(b_{1},\cdots,b_{s})\geq 2$ then the equation

$\sum_{i=1}^{r}a_{i}x_{i}=c\prod_{j=1}^{s}y_{j}^{b_{j}}$ (139)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$ if and only if there
exists $F\subseteq[1,r]$ for which $\sum_{i\in F}a_{i}=0$.

  * (ii)

If $s\geq 2$, then the equation

$\sum_{i=1}^{r}a_{i}x_{i}=cy_{1}\prod_{j=2}^{s}y_{j}^{b_{j}}$ (140)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$ if for some
$\emptyset\neq F\subseteq[1,r],s_{F}:=\sum_{i\in F}\frac{a_{i}}{c}$ is an
$n$th power in $\mathbb{Q}$, where $n=\sum_{j=2}^{s}b_{j}$. Furthermore, if
there exists $F\subseteq[1,r]$ for which $s_{F}$ is a $n$th power in
$\mathbb{Q}_{\geq 0}$, then equation (140) is partition regular over
$\mathbb{N}$.

  * (iii)

The equation

$\sum_{i=1}^{r}a_{i}x_{i}=cy_{1}\prod_{j=2}^{s}y_{j}^{b_{j}}$ (141)

is not partition regular over $\mathbb{Q}\setminus\\{0\\}$ if there exists a
prime $p$ such that for any $\emptyset\neq F\subseteq[1,r],\sum_{i\in
F}\frac{a_{i}}{c}$ is not an $n$th power modulo $p$, where
$n=\sum_{j=2}^{s}b_{j}$.

This naturally leads to the following generalization of Conjecture 8.1.

######  Conjecture 8.3.

Given $a_{1},\cdots,a_{r},c\in\mathbb{Z}\setminus\\{0\\}$ and
$b_{2},\cdots,b_{s}\in\mathbb{N}$, the equation

$\sum_{i=1}^{r}a_{i}x_{i}=cy_{1}\prod_{j=2}^{s}y_{j}^{b_{j}}$ (142)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$ if and only if there is
some $\emptyset\neq F\subseteq[1,r]$ for which $\sum_{i\in F}\frac{a_{i}}{c}$
is an $n$th power in $\mathbb{Q}$, where $n=\sum_{j=2}^{s}b_{j}$.

We have already seen that the methods of this paper cannot be extended to
prove Conjecture 8.3 even when $r=2$. When $r>2$, there are even more
problematic cases to consider. For example, if $r=4$ then at least one of
$2,5,10,$ or $20$ is a perfect cube modulo $p$ for any prime $p$, so we are
unable to use item (iii) of Theorem 8.2 to show that the equation

$2x_{1}+5x_{2}+10x_{3}+20x_{4}=y_{1}y_{2}^{3}$ (143)

is not partition regular over $\mathbb{Z}$. Furthermore, since $7^{3}\equiv
10\pmod{37}$, we may use Hensel’s lemma to see that $10$ is a perfect cube in
$\mathbb{Z}_{37}$, so analogues of Lemma 5.16 will be of no use here.

In light of Theorem 6.3 and Corollary 6.5 it is natural to pose the following
generalization of Conjecture 8.3.

######  Conjecture 8.4.

Let $K=\mathbb{Q}(\alpha)$ be a number field with ring of integers
$\mathcal{O}_{K}$ and let $\mathfrak{p}\subseteq\mathcal{O}_{K}$ be a prime
ideal. Given $a_{1},\cdots,a_{r},c\in\mathcal{O}_{K}\setminus\\{0\\}$,
$b_{2},\cdots,b_{s}\in\mathbb{N}$, and
$\mathcal{K}\in\\{K,K_{\mathfrak{p}}\\}$, the equation

$\sum_{i=1}^{r}a_{i}x_{i}=cy_{1}\prod_{j=2}^{s}y_{j}^{b_{j}}$ (144)

is partition regular over $\mathcal{O}_{\mathcal{K}}\setminus\\{0\\}$, where
$\mathcal{O}_{\mathcal{K}}$ is the ring of integers of $\mathcal{K}$, if and
only if there is some $\emptyset\neq F\subseteq[1,r]$ for which $\sum_{i\in
F}\frac{a_{i}}{c}$ is an $n$th power in $\mathcal{K}$, where
$n=\sum_{j=2}^{s}b_{j}$.

An analog of Theorem 5.11 would have to be proven for polynomial equations
over $\mathcal{O}_{K}$ instead of $\mathbb{N}$ in order to prove an analog of
item (i) of Theorem 8.2, which would nicely complement Conjecture 8.4 to
provide a complete classification of the
$a_{1},\cdots,a_{r},c\in\mathcal{O}_{K}\setminus\\{0\\}$ and
$b_{1},\cdots,b_{s}\in\mathbb{N}$ for which the equation

$\sum_{i=1}^{r}a_{i}x_{i}=c\prod_{j=1}^{s}y_{j}^{b_{j}}$ (145)

is partition regular over $\mathcal{O}_{\mathcal{K}}\setminus\\{0\\}$.

######  Remark 8.5.

In light of Remark 3.7 we are led to ask about sign obstructions to partition
regularity of polynomial equations in rings of integers of totally real number
fields. Let us consider for example the number field $K=\mathbb{Q}[\sqrt{2}]$,
which is totally real since all of its embeddings into $\mathbb{C}$ turn out
to be embeddings into $\mathbb{R}$. We recall that
$\mathcal{O}_{K}=\mathbb{Z}[\sqrt{2]}$ is the ring of integers of $K$. It is
clear that the equation

$x+y=-wz$ (146)

is not partition regular regular over $\mathbb{Z}[\sqrt{2}]_{>0}$ due to sign
obstructions despite being partition regular over
$\mathbb{Z}[\sqrt{2}]\setminus\\{0\\}$ as a consequence of Theorem 6.3. We are
unable to determine whether or not the equation

$2x-2\sqrt{2}y=wz^{3}$ (147)

is partition regular over $\mathbb{Z}[\sqrt{2}]_{>0}$ since there are no sign
obstructions but we are unable to apply Theorem 6.3 and we are unable to apply
the methods of Theorem 3.5 since $-2\sqrt{2}=(-\sqrt{2})^{3}$ and
$-\sqrt{2}\notin\mathbb{Z}[\sqrt{2}]_{>0}$. Unlike Remark 3.7, we may take
this line of inquiry a step further by examining the semiring $P$ of totally
positive elements of $\mathbb{Z}[\sqrt{2}]$, which are those elements that
remain positive under every embedding of $\mathbb{Q}[\sqrt{2}]$ into
$\mathbb{R}$. In this case we can directly determine $P$ to be given by
$P=\\{a+b\sqrt{2}\ |\ a>|b|\sqrt{2}\\}$. It is clear that equation (146) is
not partition regular over $P$ since $P\subseteq\mathbb{Z}[\sqrt{2}]_{>0}$,
but can we determine whether or not equation (147) is partition regular over
$P$? Furthermore, it can be shown using the techniques of this paper that the
equation

$\sqrt{2}x+2\sqrt{2}y=wz^{3}$ (148)

is partition regular over $\mathbb{Z}[\sqrt{2}]_{>0}$ since
$2\sqrt{2}=(\sqrt{2})^{3}$ and $\sqrt{2}\in\mathbb{Z}[\sqrt{2}]_{>0}$.
However, $\sqrt{2}$ is a positive element of $\mathbb{Z}[\sqrt{2}]$ that is
not totally positive, so equation (148) is not partition regular over $P$
since $w,x,y,z\in P$ woud result in the left hand side of the equation being
positive but not totally positive, while the right hand side of the equation
would be totally positive.

Now let us consider the equation

$2x+2\sqrt{2}y=wz^{3}.$ (149)

We can show that equation (149) is partition regular over
$\mathbb{Z}[\sqrt{2}]_{>0}$ using the considerations from above, but now that
there are no “generalized sign obstructions” can we also determine whether or
not equation (149) is partition regular over $P$?

In light of Theorem 7.1 and Corollary 7.3 we are led to the Conjecture 8.6.

######  Conjecture 8.6 (cf. Conjecture 1.3).

Let
$a_{1},\cdots,a_{k},b_{1},\cdots,b_{k},c_{1},\cdots,c_{k}\in\mathbb{Z}\setminus\\{0\\}$
and $n\in\mathbb{N}$. The system of equations

$\begin{array}[]{ccccc}a_{1}x_{1}&+&b_{1}y_{1}&=&c_{1}w_{1}z_{1}^{n}\\\
a_{2}x_{2}&+&b_{2}y_{2}&=&c_{2}w_{2}z_{2}^{n}\\\ &&\vdots&&\\\
a_{k}x_{k}&+&b_{k}y_{k}&=&c_{k}w_{k}z_{k}^{n}\end{array}$ (150)

is partition regular over $\mathbb{Z}\setminus\\{0\\}$ if and only if

$I:=\bigcap_{i=1}^{k}\left\\{\frac{a_{i}}{c_{i}},\frac{b_{i}}{c_{i}},\frac{a_{i}+b_{i}}{c_{i}}\right\\}$
(151)

contains an $n$th power in $\mathbb{Q}$.

We conclude with some examples of systems of equations whose partition
regularity remains unknown. We are unable to apply Theorem 7.2 to the system
of equations

$\begin{array}[]{rcrcc}16x_{1}&+&17y_{1}&=&w_{1}z_{1}^{8}\\\
33x_{2}&-&17y_{2}&=&w_{2}z_{2}^{8}\end{array}$ (152)

since $I=\\{16,33\\}$ and $16$ is an $8$th power modulo every prime $p$. Since
$33$ is an $8$th power in $\mathbb{Z}_{2}$ we also cannot expect a
generalization of Lemma 5.17 to systems of equations to help determine the
partition regularity of the system in (152). We are also unable to apply
Theorem 7.2 to the system of equations

$\begin{array}[]{rcrcc}5^{4}x_{1}&+&3^{6}y_{1}&=&w_{1}z_{1}^{12}\\\
(5^{4}-3^{6})x_{2}&+&3^{6}y_{2}&=&w_{2}z_{2}^{12}\end{array},$ (153)

since $I=\\{5^{4},3^{6}\\}$, $5$ is a cube modulo $p$ when $p\equiv
2\pmod{3}$, and $-3$ is a square modulo $p$ when $p\equiv 1\pmod{3}$, so one
of $5^{4}$ and $3^{6}$ will be a $12$th power modulo any prime $p$ (cf. Item
(ii)(a) of Remark 4.10).

##  9 The Existence of Special Ultrafilters

In this section we will review some knowledge about $\beta S$, the space of
ultrafilters over a set $S$, so that we can provide a thorough proof of
Theorems 2.8 and 6.1 as Theorems 9.12 and 9.18 respectively. After proving
Theorem 9.18, we give a brief discussion comparing the methods that we use to
show that certain equations are partition regular with the methods used in [3]
and [10]. As a result of this discussion (cf. Remark 9.13), we decide to prove
a generalization of Theorem 9.12 as Corollary 9.26. For a more comprehensive
study of ultrafilters and their usage in the study of semigroups the reader is
referred to [11].

Let us recall some notation before proceeding further. We let $\mathcal{P}(S)$
denote the collection of subsets of $S$ and $\mathcal{P}_{f}(S)$ denotes the
collection of finite subsets of $S$. If $(S,\diamond)$ is a semigroup, then
for $s\in S$ and $A\subseteq S$ we define $sA=\\{sa\ |\ a\in A\\}$ and
$s^{-1}A=\\{x\in S\ |\ sx\in A\\}$.

######  Definition 9.1 (cf. Def. 2.7).

Let $S$ be a set. $p\subseteq\mathcal{P}(S)$ is an ultrafilter if it satisfies
the following properties:

  1. (i)

$\emptyset\notin p$.

  2. (ii)

If $A\in p$ and $A\subseteq B$ then $B\in p$.

  3. (iii)

If $A,B\in p$ then $A\cap B\in p$.

  4. (iv)

For any $A\subseteq S$, either $A\in p$ or $A^{c}\in p$.

$\beta S$ denotes the space of all ultrafilters over $S$.

It is often useful to think about $\beta S$ as the set of finitely additive
$\\{0,1\\}$-valued measures on the set $S$. The topology of $\beta S$ is
generated by the basis of open sets $\\{\hat{A}\\}_{A\subseteq S}$, where

$\hat{A}:=\\{p\in\beta S\ |\ A\in p\\}.$ (154)

Since $(\hat{A})^{c}=\widehat{A^{c}}$ for all $A\subseteq S$, we see that each
$\hat{A}$ is also a closed set, and it is a fact that
$\\{\hat{A}\\}_{A\subseteq S}$ also generates the topology of $\beta S$ as a
basis of closed sets. We note that for any $s\in S$, the collection of sets
given by $e_{s}:=\\{A\subseteq S\ |\ s\in A\\}$ is an ultrafilter over $S$.
Let $e:S\rightarrow\beta S$ be given by $e(s)=e_{s}$ and observe that $e$ is
an injective map that naturally embeds $S$ inside of $\beta S$ as a dense
subset. Furthermore, when we endow $S$ with the discrete topology, which will
always be the case for the rest of this section, $e$ is a homeomorphism onto
its image. An ultrafilter $p$ is a principal ultrafilter if $p\in e(S)$, and a
nonprincipal ultrafilter otherwise. Since we naturally identify each $s\in S$
with the principal ultrafilter $e_{s}$, it is common to write $s$ in place of
$e_{s}$, and we will be using this convention for the rest of this section.
Theorem 9.2 is a universal property that can be used to characterize $\beta
S$.

######  Theorem 9.2 (cf. Theorem 3.28 in [11]).

Let $S$ be an infinite set with the discrete topology and let $\beta S$ denote
the space of ultrafilters over $S$. Given any compact space $Y$ and any
function $f:S\rightarrow Y$ there exists a unique continuous function
$\tilde{f}:\beta S\rightarrow Y$ such that $\tilde{f}|_{S}=f$.

A careful and repeated application of Theorem 9.2 also allows one to extend
binary operations from $S\times S$ to $\beta S\times\beta S$, which is of
great use when $S$ is a semigroup.

######  Theorem 9.3 (cf. Theorem 4.1 in [11]).

Let $S$ be a set and let $\diamond$ be a binary operation defined on $S$.
There is a unique binary operation $*:\beta S\times\beta S\rightarrow\beta S$
satisfying the following three conditions:

  1. (a)

For every $s,t\in S$, $s*t=s\diamond t$.

  2. (b)

For each $q\in\beta S$, the function $\rho_{q}:\beta S\rightarrow\beta S$ is
continuous, where $\rho_{q}(p)=p*q$.

  3. (c)

For each $s\in S$, the function $\lambda_{s}:\beta S\rightarrow\beta S$ is
continuous, where $\lambda_{s}(q)=s*q$.

######  Theorem 9.4 (cf. Theorems 4.4 and 4.12 in [11]).

If $(S,\diamond)$ is a semigroup and $*:\beta S\times\beta S\rightarrow\beta
S$ is the operation given by Theorem 9.3, then $*$ is an associative
operation. Furthermore, for $p,q\in\beta S$ the ultrafilter $p*q$ is given by

$A\in p*q\Leftrightarrow\\{s\in S\ |\ s^{-1}A\in q\\}\in p.$ (155)

It is customary to denote the operation $*$ that is produced by Theorem 9.3 by
$\diamond$ so that $\diamond$ represents an operation on $S\times S$ as well
as $\beta S\times\beta S$, and we shall adopt this practice. In light of
Theorem 9.4 We see that if $(S,\diamond)$ is a semigroup, then $(\beta
S,\diamond)$ is also a semigroup. Since $\beta S$ is a compact Hausdorff space
(cf. Theorem 3.18 in [11]), and for each $q\in\beta\mathbb{N}$ the map
$\rho_{q}:\beta\mathbb{N}\rightarrow\beta\mathbb{N}$ given by
$\rho_{q}(p)=p\diamond q$ is continuous (cf. Theorem 9.3), we see that $(\beta
S,\diamond)$ is a compact right topological semigroup.333A compact right
topological semigroup is a semigroup $(S,\diamond)$ that is also a compact
Hausdorff space for which each of the maps $\rho_{s}:S\rightarrow S$ given by
$\rho_{s}(t)=t\diamond s$ are continuous. Note that other sources may use
$\rho_{s}(t)=s\diamond t$ in their definition and that we base our definition
off of Definition 2.1 in [11]. We apologize to the reader for our seemingly
excessive emphasis on the operation $\diamond$ of our semigroup
$(S,\diamond)$, but we choose to do this to avoid confusion later on when we
work with rings $(R,+,\cdot)$ and are forced to consider the semigroups
$(R,+)$ and $(R,\cdot)$ separately. We will now collect some facts from the
theory of semigroups.

######  Definition 9.5.

Let $(S,\diamond)$ be a semigroup.

  * •

We say $e\in S$ is an idempotent if $e\diamond e=e$. We let $E(S,\diamond)$
denote the set of idempotents of $(S,\diamond)$.

  * •

We say $L\subseteq S$ is a left ideal if for any $s\in S$ and $\ell\in L$ we
have $s\diamond\ell\in L$. We say $R\subseteq S$ is a right ideal if for any
$s\in S$ and $r\in R$ we have $r\diamond s\in R$. In general, $I\subseteq S$
is an ideal if it is a left ideal and a right ideal.

  * •

We call $L\subseteq S$ a minimal left ideal if $L$ is a left ideal that does
not properly contain any other left ideal. Similarly, the smallest ideal of S,
if it exists, is an ideal $I$ that is contained in every other ideal of
$S$.444Note that not every semigroup possesses a smallest ideal. In the
semigroup $(\mathbb{N},+)$ each of the sets $I_{n}:=\\{m\in\mathbb{N}\ |\
m\geq n\\}$ is an ideal. Since there is no set that is contained in every
$I_{n}$, we see that $(\mathbb{N},+)$ does not have a smallest ideal. If
$(S,\diamond)$ does possess a smallest ideal, then we denote it by
$K(S,\diamond)$ and observe that $K(S,\diamond)$ is also a semigroup.

######  Theorem 9.6 (cf. Corollary 2.6 in [11]).

Let $(S,\diamond)$ be a compact right topological semigroup. Then every left
ideal of $S$ contains a minimal left ideal. Minimal left ideals are closed,
and each minimal left ideal has an idempotent.

######  Theorem 9.7 (cf. Theorem 1.51 in [11]).

Let $(S,\diamond)$ be a semigroup. If $S$ has a minimal left ideal, then
$K(S,\diamond)$ exists and $K(S,\diamond)=\bigcup\\{L\ |\ L\text{ is a minimal
left ideal of }S\\}$.

######  Lemma 9.8.

Let $(S,\diamond)$ be a semigroup and let $\leq$ be the partial ordering on
the set of idempotents of $S$ given by $f\leq e$ if and only if $fe=ef=f$.
Assume that $S$ has a minimal left ideal that has an idempotent.

  1. (i)

If $e\in S$ is an idempotent that is minimal with respect to $\leq$, then $e$
is a member of some minimal left ideal of $S$. Such an idempotent is a minimal
idempotent. It follows that $E(K(S,\diamond))$ is the set of minimal
idempotents of $(S,\diamond)$. (cf. Theorem 1.38(d) in [11])

  2. (ii)

If $f\in S$ is an idempotent then there exists a minimal idempotent $e$ such
that $e\leq f$. (cf. Theorem 1.60 in [11])

######  Definition 9.9.

Let $(S,\diamond)$ be a semigroup and let $A\subseteq S$. Then $A$ is central
if there exists an ultrafilter $p\in E(K(\beta S,\diamond))$ for which $A\in
p$.

######  Theorem 9.10 (cf. Theorem 3.5 in [4]).

Let $(S,\diamond)$ be a commutative semigroup, let $\ell\in\mathbb{N}$, and
let $A\subseteq\mathbb{N}$ be a central set.

  1. (i)

There exists $b,g\in A$ such that

$b,b\diamond g,b\diamond g\diamond g,\cdots,b\diamond\underbrace{g\diamond
g\diamond\cdots\diamond g}_{\ell}\in A.$ (156)

  2. (ii)

If $(S,\diamond)$ is a commutative group and $c:S\rightarrow S$ is a
homomorphism for which $[S:c(S)]<\infty$, then there exists $b,g\in S$ such
that

$\\{c(g)\\}\cup\\{c(b)\diamond g^{j}\\}_{j=-\ell}^{\ell}\subseteq A.$ (157)

######  Lemma 9.11 (cf. Lemma 17.2 and Theorem 17.3 in [11]).

There exists an ultrafilter $p\in\beta\mathbb{N}$ such that every $A\in p$ is
a central subset of $(\mathbb{N},+)$ and a central subset of
$(\mathbb{N},\cdot)$.

We are now ready to prove Theorem 2.8.

######  Theorem 9.12 (cf. Theorem 2.8).

Let $p\in\beta\mathbb{N}$ be such that every $A\in p$ is a central subset of
$(\mathbb{N},+)$ and a central subset of $(\mathbb{N},\cdot)$. $p$ satisfies
the following properties:

  1. (i)

For any $A\in p$ and $\ell\in\mathbb{N}$, there exists $b,g\in A$ with
$\\{bg^{j}\\}_{j=0}^{\ell}\subseteq A$.

  2. (ii)

For any $A\in p$ and $h,\ell\in\mathbb{N}$, there exists $a,d\in\mathbb{N}$
for which $\\{hd\\}\cup\\{ha+id\\}_{i=-\ell}^{\ell}\subseteq A$.

  3. (iii)

For any $s\in\mathbb{N}$, we have $s\mathbb{N}\in p$.

###### Proof.

By Lemma 9.11 let $p\in\beta\mathbb{N}$ be an ultrafilter such that every
$A\in p$ is a central subset of $(\mathbb{N},+)$ and a central subset of
$(\mathbb{N},\cdot)$. To see that $p$ satisfies (i), it suffices to observe
that each $A\in p$ is a central subset of $(\mathbb{N},\cdot)$, so Theorem
9.10(i) applies. To see that $p$ satisfies (ii) we first note that any central
subset of $(\mathbb{N},+)$ is also a central subset of $(\mathbb{Z},+)$ (cf.
Theorem 2.9 in [16]), so we may apply Theorem 9.10(ii) with the homomorphism
$c$ given by $c(x)=hx$ for all $x\in\mathbb{Z}$ to find some
$a,d\in\mathbb{Z}$ for which
$\\{hd\\}\cup\\{ha+id\\}_{i=-\ell}^{\ell}\subseteq A$. Since $ha,hd\in
A\subseteq\mathbb{N}$, we see that $a,d\in\mathbb{N}$. To see that $p$
satisfies $(iii)$, let $s\in\mathbb{N}$ be arbitrary and let us assume for the
sake of contradiction that $s\mathbb{N}\notin p$. It follows that
$(s\mathbb{N})^{c}\in p$, so by (ii) let $a,d\in\mathbb{N}$ be such that
$\\{sd\\}\cup\\{sa+id\\}_{i=-1}^{1}\subseteq(s\mathbb{N})^{c}$ to obtain the
desired contradiction. ∎

######  Remark 9.13.

In [3] and [10] the authors also used an ultrafilter $p$ for which every $A\in
p$ is a central subset of $(\mathbb{N},+)$ and a central subset of
$(\mathbb{N},\cdot)$ in order to show that the equation $x+y=wz$ is partition
regular over $\mathbb{N}$, so it is unsurprising that we have managed to use
such an ultrafilter to obtain our positive results over $\mathbb{N}$ and
$\mathbb{Z}\setminus\\{0\\}$. Unfortunately, if $R$ is a general integral
domain, then there may not exist an ultrafilter $p\in\beta R$ for which every
$A\in p$ is a central subset of $(R,+)$ and a central subset of
$(R\setminus\\{0\\},\cdot)$. Thankfully, we will see as a consequence of
Theorem 9.18 that we only need to work with central subsets of
$(R\setminus\\{0\\},\cdot)$ in order to get our desired results for a general
integral domain. In particular, the ultrafilter from Theorems 6.1 and 9.18 is
just a minimal idempotent in $(\beta R\setminus\\{0\\},\cdot)$, so it is a
corollary of Lemma 6.2 that any central subset of
$(\mathbb{Z}\setminus\\{0\\},\cdot)$ (and consequently of
$(\mathbb{N},\cdot)$) contains a solution to the equation $x+y=wz$. For the
sake of completeness, we will still examine rings $R$ for which there exists
an ultrafilter $p\in\beta R$ such that every $A\in p$ is a central subset of
$(R,+)$ and a central subset of $(R\setminus\\{0\\},\cdot)$ after we prove
Theorem 9.18.

We would also like to point out to the reader that we work with central
subsets of $(R\setminus\\{0\\},\cdot)$ instead of central subsets of
$(R,\cdot)$ because $K(\beta R,\cdot)=0$, so $\\{0\\}$ is the only central
subset of $(R,\cdot)$. For central subsets of $(R\setminus\\{0\\},\cdot)$ to
be defined, we need $(R\setminus\\{0\\},\cdot)$ to be a semigroup, which is
why we will only work with division rings and integral domains for the rest of
this section. We also observe that the natural inclusion map
$\iota:\beta(R\setminus\\{0\\})\rightarrow(\beta R)\setminus\\{0\\}$ given by

$\iota(p)=\\{A\subseteq R\ |\ A\setminus\\{0\\}\in p\\}=p\cup\\{A\cup\\{0\\}\
|\ A\in p\\}$ (158)

is a homeomorphism. Furthermore, we see that for any
$p_{1},p_{2}\in\beta(R\setminus\\{0\\})$ we have that $\iota(p_{1}\cdot
p_{2})=\iota(p_{1})\cdot\iota(p_{2})$, so $\iota$ is a semigroup isomorphism
as well. Since $(\beta R)\setminus\\{0\\}$ and $\beta(R\setminus\\{0\\})$ are
naturally isomorphic as compact right topological semigroups, we will write
$\beta R\setminus\\{0\\}$ for $(\beta R)\setminus\\{0\\}$ without any worry
for the potential confusion with $\beta(R\setminus\\{0\\})$.

######  Theorem 9.14 (cf. Theorem A in [5]).

Let $R$ be an infinite integral domain and let ${\bf M}$ be a matrix with
entries in $R$. Then the system ${\bf M}\vec{x}=\vec{0}$ is partition regular
over $R\setminus\\{0\\}$ if and only if ${\bf M}$ satisfies the columns
condition (cf. Definition 2.2).555In [5] the statements of the results discuss
partition regularity over $R$, not $R\setminus\\{0\\}$, but the definition of
partition regularity that is used in [5] explicitly forbids trivial solutions,
which is why we may modify their statement to mirror our previous statements.

######  Lemma 9.15.

Let $R$ be an infinite integral domain and let ${\bf M}$ be a matrix with
entries in $R$ for which the system ${\bf M}\vec{x}=\vec{0}$ is partition
regular over $R\setminus\\{0\\}$. Let $I_{\bf M}\subseteq\beta
R\setminus\\{0\\}$ denote the collection of ultrafilters $p$ such that every
$A\in p$ there exists $\vec{x}$ with entries from $A$ such that ${\bf
M}\vec{x}=\vec{0}$. $I_{\bf M}$ is an ideal of $(\beta
R\setminus\\{0\\},\cdot)$.

###### Proof.

First let us show that $I_{\bf M}$ is nonempty. To this end, let us assume for
the sake of contradiction that for each $p\in\beta R\setminus\\{0\\}$ there
exists $A_{p}\in p$ such that there is no $\vec{x}$ with entries in $A_{p}$
satisfying ${\bf M}\vec{x}=\vec{0}$. Since $\\{\widehat{A_{p}}\\}_{p\in\beta
R\setminus\\{0\\}}$ is an open cover of the compact space $\beta
R\setminus\\{0\\}$, let $\\{\widehat{A_{p_{i}}}\\}_{i=1}^{r}$ be a finite
subcover. The desired contradiction follows from the observation that
$R\setminus\\{0\\}=\bigcup_{i=1}^{r}A_{p_{i}}$ is a partition in which no cell
yields a solution to the equation ${\bf M}\vec{x}=\vec{0}$. Now let us show
that $I_{F}$ is a left ideal. To this end, let $p\in I_{F}$ and $q\in\beta
R\setminus\\{0\\}$ both be arbitrary. We see that for $A\in q\cdot p$ we have

$\\{r\in R\setminus\\{0\\}\ |\ r^{-1}A\in p\\}\in q,$ (159)

so let $r\in R\setminus\\{0\\}$ be such that $r^{-1}A\in p$. Since $r^{-1}A\in
p$, let $x_{1},\cdots,x_{k}\in r^{-1}A$ be a solution to $F$. Since $F$ is
homogeneous, we see that $rx_{1},\cdots,rx_{k}\in A$ is also a solution to
$F$, which completes the proof that $I_{F}$ is a left ideal. Now let us show
that $I_{F}$ is a right ideal. To this end, let $p\in I$ and $q\in\beta
R\setminus\\{0\\}$ both be arbitrary. We see that for $A\in p\cdot q$ we have

$\\{r\in R\setminus\\{0\\}\ |\ r^{-1}A\in q\\}\in p,$ (160)

so let $x_{1},\cdots,x_{k}\in\\{r\in R\setminus\\{0\\}\ |\ r^{-1}A\in q\\}$ be
a solution to $F$. Since $\bigcap_{i=1}^{k}x_{i}^{-1}A\in q$, let
$y\in\bigcap_{i=1}^{k}x_{i}^{-1}A$ be arbitrary and note that
$x_{1}y,\cdots,x_{k}y\in A$ is a solution to $F$ since $F$ is a homogeneous
system of equations. ∎

######  Corollary 9.16.

Let $R$ be an infinite integral domain. Let $I\subseteq\beta
R\setminus\\{0\\}$ denote the collection of ultrafilters $p$ such that for
every $A\in p$ and every finite system of homogeneous linear equations $F$
that is partition regular over $R\setminus\\{0\\}$, there is a solution to $F$
contained in $A$. Then $I$ is a nonempty ideal of $(\beta
R\setminus\\{0\\},\cdot)$.

###### Proof.

Let $\mathcal{M}$ denote the collection of matrices ${\bf M}$ with entries in
$R$ for which the system ${\bf M}\vec{x}=\vec{0}$ is partition regular. We see
that

$K(\beta R\setminus\\{0\\},\cdot)\subseteq\bigcap_{{\bf
M}\in\mathcal{M}}I_{\bf M}=I.$ (161)

We see that $I$ is an ideal since it is an intersection of ideals, and $I$ is
nonempty since it contains $K(\beta R\setminus\\{0\\},\cdot)$. ∎

######  Lemma 9.17.

Let $R$ be an infinite integral domain. Let $H\subseteq\beta
R\setminus\\{0\\}$ denote the collection of ultrafilters $p$ such that for
every $r\in R\setminus\\{0\\}$ we have $rR\in p$. Then $H$ is an ideal of
$(\beta R\setminus\\{0\\},\cdot)$.

###### Proof.

Let $p\in H,q\in\beta R\setminus\\{0\\},$ and $r\in R\setminus\\{0\\}$ all be
arbitrary. Let us first show that $H$ is a left ideal. We note that for any
$s\in R$, we have $s^{-1}(rR)\supseteq rR$. Since $rR\in p$, we see that

$R\subseteq\\{s\in R\setminus\\{0\\}\ |\ s^{-1}(rR)\in p\\}\Rightarrow\\{s\in
R\setminus\\{0\\}\ |\ s^{-1}(rR)\in p\\}\in q.$ (162)

It follows that $rR\in q\cdot p$, so $q\cdot p\in H$ and $H$ is indeed a left
ideal. Now let us show that $H$ is a right ideal. We note that for any $s\in
rR$ we have $s^{-1}(rR)\supseteq R$. Since $R\in q$, we see that

$rR\subseteq\\{s\in R\setminus\\{0\\}\ |\ s^{-1}(rR)\in q\\}\Rightarrow\\{s\in
R\setminus\\{0\\}\ |\ s^{-1}(rR)\in q\\}\in p.$ (163)

It follows that $rR\in p\cdot q$, so $p\cdot q\in H$. ∎

######  Theorem 9.18 (cf. Theorem 6.1).

Let $R$ be an infinite integral domain. If $p\in E(K(\beta
R\setminus\\{0\\},\cdot))$, then $p$ satisfies the following properties:

  1. (i)

For any $A\in p$ and $\ell\in\mathbb{N}$, there exists $b,g\in A$ with
$\\{bg^{j}\\}_{j=0}^{\ell}\subseteq A$.

  2. (ii)

For any $A\in p$ and any finite system of homogeneous linear equations $F$
that is partition regular over $R\setminus\\{0\\}$, there is a solution to $F$
contained in $A$.

  3. (iii)

For every $r\in R\setminus\\{0\\}$, we have $rR\in p$.

  4. (iv)

For any $A\in p,\ell\in\mathbb{N}$, and $h,j_{1},j_{2},\cdots,j_{\ell}\in
R\setminus\\{0\\}$, there exists $a,d\in R$ for which
$\\{hd,ha\\}\cup\\{ha+j_{i}d\\}_{i=0}^{\ell}\subseteq A$.

###### Proof.

Since $p$ is a minimal idempotent in $(\beta R\setminus\\{0\\},0)$ we see that
each $A\in p$ is a central subset of $(R\setminus\\{0\\},\cdot)$, so Theorem
9.10(i) shows us that $p$ satisfies condition (i). Since $I$ and $H$ (cf.
Corollary 9.16 and Lemma 9.17 respectively) are ideals of $(\beta
R\setminus\\{0\\},\cdot)$, $I\cap H$ is also an ideal. Hence $K(\beta
R\setminus\\{0\\},\cdot)\subseteq I\cap H$. Since $p\in I$, we see that
condition (ii) is satisfied. Since $p\in H$, we see that condition (iii) is
satisfied. The fact that $p$ satisfies (iv) when $R$ is an integral domain is
a corollary of the fact that $p$ satisfies (ii). We give a proof of this
implication for the sake of completeness since property (iv) is used in the
earlier sections of this paper.

To see that $p$ satisfies condition (iv), we first consider the system of
equations

$\begin{array}[]{lcrcrcc}hx_{3}&-&hx_{2}&-&j_{1}x_{1}&=&0\\\
hx_{4}&-&hx_{2}&-&j_{2}x_{1}&=&0\\\ &&&\vdots&&&\\\
hx_{\ell+2}&-&hx_{2}&-&j_{\ell}x_{1}&=&0\\\
hx_{\ell+3}&-&x_{2}&-&hx_{\ell+5}&=&0\\\
hx_{\ell+4}&-&hx_{2}&-&x_{1}&=&0\end{array}$ (164)

Let us assume that $\\{x_{i}\\}_{i=1}^{\ell+5}\subseteq R\setminus\\{0\\}$ is
a solution to the system of equations in (164). Since
$x_{1}=h(x_{\ell+5}-x_{2})$ and $x_{2}=h(x_{\ell+3}-x_{\ell+4})$, we may write
$x_{1}=hd$ and $x_{2}=ha$ for some $a,d\in R\setminus\\{0\\}$. It follows that
$x_{i}=ha+j_{i}d$ for $3\leq i\leq\ell+2$, so it suffices to show that each
$A\in p$ contains a solution to the system of equations in (164). To this end,
let ${\bf M}\in M_{\ell+3,\ell+5}(R)$ be the matrix such that the equation
${\bf M}\vec{x}=\vec{0}$ represents the system of equations in (164), and we
will proceed to show that ${\bf M}$ satisfies the columns condition. Let
$\\{\vec{c}_{i}\\}_{i=1}^{\ell+5}$ denote the columns of ${\bf M}$, with
$\vec{c}_{i}$ representing the column corresponding to $x_{i}$. We see that

$\vec{c}_{1}=\begin{pmatrix}-j_{1}\\\ -j_{2}\\\ \vdots\\\ -j_{\ell}\\\ 0\\\