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The Feld Group Institute has created a unique program for the development of next-generation IT leaders. Our Technology Leadership Development (TLD) program starts with a class that's designed for enterprise-scale CIOs/CTOs and their direct reports. It's a very thorough 2-week class, with the curriculum drawing heavily upon The Feld Group Institute's leadership insights, operational experience and time-tested strategic framework for enterprise success. Completion of the program is then followed by an ongoing relationship with The Feld Group Institute and the community of peers from other companies that have also completed the program.
We started offering the program in 2016, and as of this post in November 2018, we are completing our 7th cohort. We're proud to have included CIOs, CTOs, VPs and Directors of IT. We've even included some business partners from great companies such as FedEx, Southwest Airlines, Bank of Montreal, Rent-A-Center, Vistra Energy, Baxter International, Volkswagen, York Risk Services, Vanguard, Fossil, Cummins, Schneider Electric and others. So far, we have over 100 graduates who are applying what they've learned in the program at their companies, and these folks are now part of our community of peers that share knowledge, challenges, successes and new discoveries.
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text:
Operational Situation Report USSR No. 97
The Chief of the Security Police and the SD, Berlin,
(36th copy)
Vorkommando 4a directly with the combat troops in Kiev since September 19. Einsatzgruppe HQ came up on September 24. Office Building NKVD, October 24th street, assigned and commandeered as the seat of Einsatzgruppe Headquarters. The building was evacuated this morning to move into emergency quarters in the one-time Tsar's castle.
The town was almost destroyed upon entry of troops, Numerous barricades and tank traps were placed in the main streets. In addition, other strong defensive installations were setup in the town area. On September 20th, the citadel blew up and the Artillery Commander and his chief of staff were killed.
On September 24, violent explosions in the quarters of the Feldkommandantur; the ensuing fire has not yet been extinguished. Fire also burns in the center of the town. Some very valuable buildings have been destroyed as a result. All fire-fighting efforts have thus far had no effect. Demolitions by blasting being carried out to bring the fire under control. Fire also rages in the immediate vicinity of this office and forced us to evacuate due to considerable damage from the blasting. The blasts continue.
Up to now, 670 mines have been detected in buildings, according to a mine-laying plan which was discovered; all public buildings and squares are mined, among them is stated the building appropriated for the future use of this office. The building was then search most assiduously and in the course of this search 60 Molotov cocktails, and various forms of explosives were detected.
In the Lenin Museum, 8000 lbs of dynamite were discovered which were to be detonated by a wireless control. It was repeatedly observed that fires broke out the moment the buildings were taken over by German forces. As has been proved, Jews have played a major role in this sabotage.
Allegedly 150,000 Jews living here. Verification of these statements has not been validated yet. 1,600 arrests were made in the course of the first action, measures underway to check the status of the Jewish population.
Execution of at least 50,000 Jews planned. The German Army welcomes measures and demands drastic procedure. Garrison commander advocates public execution of 20 Jews.
A larger number of NKVD officials, political commissars, partisan leaders and partisans have been arrested. According to reliable information, demolition battalion of the NKVD and considerable number of NKVD men in Kiev.
This morning, enemy plots detected. Contact established with German Army and authorities. Participated in setting up initial town administration. Informants posted, Vorkommando of the Higher SS and Police leaders have arrived.
Detailed reports to follow.
Translation by Hermann Feuer

redpajama_set_name: RedPajamaCommonCrawl


text:
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Start: 27th February, 2023 at 10:00am
End: 27th February, 2023 at 11:00am
Published: 3rd September, 2022
Author: Unverified User


text:
Charity worker dies
Anne Copplestone
By GUY WILLIAMS
A former Queenstown woman who was found dead in her apartment in Cambodia on Monday after a suspected heart attack was dedicated to helping underprivileged children, friends say.
Cambodian media reported the body of Anne Copplestone, 48, was discovered after she didn't turn up for work in Battambang, the country's second-largest city.
She worked as director of Hope for Cambodian Children, which provides healthcare and education support to impoverished children, for the past three years.
Copplestone's parents Graham and Sandra ran the Arthurs Point camping ground during the '80s and '90s, and she went to Queenstown Primary School and Wakatipu High School.
Tina Clyma (nee Brosig), of Auckland, says the news came as a "real, real shock" to her and a close circle of friends who graduated from the high school in 1989.
She last saw her friend in January after she'd spent Christmas with her mother and sister in Whangarei during a trip home from Cambodia, and spoke to her again only 10 days ago.
Clyma says after a long career as a travel agent, Copplestone decided to change direction and pursue a passion for working with underprivileged children, first in Australia's Northern Territory and then in Cambodia.
She loved Cambodia and its people.
"It was her place.
"Everyone just adored her."
Such was her dedication, she'd turned down her boss's offer to fly her home to New Zealand after the outbreak of Covid-19.
"She said, 'who's going to run the place and look after everyone?"'
Clyma understands Copplestone will be laid to rest in a Battambang temple.
Another friend, Victoria Robertson, of Queenstown, says she and other old resort friends kept in regular touch with Copplestone.
She had planned to visit her in Cambodia with her family this year.
Her friend will be remembered by many longtime resort residents from her years working as a travel agent at House of Travel in Shotover Street.
She had a gentle-natured, quiet personality.
"She had a really kind, caring nature."
guy.williams@scene.co.nz
Resort hoteliers feel the pain


text:
\section{Introduction}

The early developments in Ramsey theory focused mainly on partition
regularity of systems of linear equations. A system $\mathcal{L}$ of
linear homogeneous equations with coefficients in a ring $R$ is called
\emph{$r$-regular} over $R$ if, for every $r$-coloring of the nonzero
elements of $R$, there is a monochromatic solution to $\mathcal{L}$. A
system $\mathcal{L}$ of linear homogeneous equations is called
\emph{regular} over $R$ if it is $r$-regular over $R$ for all positive
integers $r$.

In 1916, Schur \cite{Sc} proved that the equation $x+y=z$ is regular
over $\mathbb{Z}$. In 1927, van der Waerden \cite{Wa} proved his
celebrated theorem that every finite coloring of the positive integers
contains arbitrarily long monochromatic arithmetic progressions. In his
1933 thesis, Rado \cite{Ra} generalized the theorems of Schur and van
der Waerden by classifying those systems of linear homogeneous equations
that are regular over $\mathbb{Z}$.  In particular, a linear homogeneous
equation with nonzero integer coefficients is regular over $\mathbb{Z}$
if and only if some nonempty subset of the coefficients sums to zero. In
1943, Rado \cite{Ra1} generalized the theorem further by classifying
those systems of linear equations that are regular over a subring of the
complex numbers. More recently, analogues of Rado's theorem have been
proven for abelian groups \cite{De1}, finite fields \cite{BeDeHi}, and
commutative rings \cite{BeDeHiLe}.

Some of the major remaining open problems on partition regularity
concern the properties of colorings that are free of monochromatic
solutions to a system of equations.

\begin{definition}
\label{mini} A coloring of the nonzero elements of a ring $R$ (or
more generally, a set of numbers $S$) is called \emph{minimal for a
  system $\mathcal{L}$ of linear homogeneous equations} if it is free of
monochromatic solutions to $\mathcal{L}$ and uses as few colors as
possible.
\end{definition}

Three basic questions arise for a given ring $R$ and system
$\mathcal{L}$ of linear homogeneous equations.

\begin{question}
What are the minimal colorings for $\mathcal{L}$?
\end{question}

\begin{question}\label{question:colors}
How many colors are used in a minimal coloring for $\mathcal{L}$?
\end{question}

\begin{question}\label{question:colorings}
How many minimal colorings, up to isomorphism, are there for
$\mathcal{L}$?
\end{question}

Rado made the following unresolved conjecture in his thesis on
Question~\ref{question:colors}.

\begin{conjecture} [Rado, \cite{Ra}] \label{RBC}
For all positive integers $m$ and $n$, there exists a positive integer
$k(m,n)$ such that if a system of $m$ linear equations in $n$ variables
is $k(m,n)$-regular over $\mathbb{Z}$, then the system is regular over
$\mathbb{Z}$.
\end{conjecture}

Conjecture~\ref{RBC} is commonly known as Rado's Boundedness Conjecture
\cite{HiLeSt}. Rado proved that Conjecture~\ref{RBC} is true if it is
true in the case when $m=1$, that is, for single linear equations. Rado
also settled his conjecture in the simple cases $n=1$ and
$n=2$. Kleitman and the second author \cite{FoKl} recently proved Rado's
Boundedness Conjecture for $n=3$. They proved that if a linear equation
in 3 variables is 36-regular over $\mathbb{Z}$, then it is regular over
$\mathbb{Z}$.

Rado also made the following conjecture in his thesis.

\begin{conjecture}[Rado, \cite{Ra}] \label{next}
For each positive integer $n$, there is a linear equation that is
$n$-regular over $\mathbb{Z}$ but not $(n+1)$-regular over $\mathbb{Z}$.
\end{conjecture}

While Conjecture~\ref{next} has remained open, Radoi\v{c}i\'c and the
second author found a family of linear homogeneous equations that they
conjecture verifies Conjecture~\ref{next}. For a rational number $q$ and
positive integer $n$, let $E(q,n)$ denote the equation
\[ x_0+qx_1+\dotsb+q^{n-2}x_{n-2}=q^{n-1}x_{n-1}. \]

\begin{definition}
For a nonzero rational number $q$ and prime number $p$, there is a
unique representation of $q$ as $q=p^va/b$ such that $a$, $b$, and $v$
are integers, $b$ is positive, $\gcd(a,b)=1$, and $p\nmid a,b$.  Define
$v_p(q)$ to be the integer $v$ and $w_p(q) \in \{1,\dotsc,p-1\}$ by
$w_p(q) \equiv ab^{-1} \pmod{p}$.
\end{definition}

For a prime $p$ and positive integer $n$, let $c_{p,n}\colon \mathbb{Q}
\setminus \{0\} \rightarrow \{0,1,\dotsc,n-1\}$ be the $n$-coloring of
the nonzero rational numbers defined by $c_{p,n}(q)\equiv v_p(q)
\pmod{n}$.

To avoid any possible confusion, we now define what it means for two
colorings of a set to be isomorphic.

\begin{definition}
  Two colorings $c_1\colon S \rightarrow C_1$ and $c_2\colon S
  \rightarrow C_2$ of the same set $S$ are \emph{isomorphic} if there is
  a bijection $\phi\colon C_2 \rightarrow C_1$ such that $c_1 =\phi
  \circ c_2$.
\end{definition}

Radoi\v ci\'c and the second author \cite{FoRa} proved that the
$n$-coloring $c_{p,n}$ is free of monochromatic solutions to
$E(p,n)$. Hence, the equation $E(p,n)$ is not $n$-regular over
$\mathbb{Z}$. They also conjecture that $E(2,n)$ is $(n-1)$-regular over
$\mathbb{Z}$, which would imply Conjecture~\ref{next}. We make the
following stronger conjecture.

\begin{conjecture} \label{stronger}
For $n>2$, $c_{2,n}$ is the only $n$-coloring, up to isomorphism, of the
nonzero rational numbers without a monochromatic solution to $E(2,n)$.
\end{conjecture}

In Section~\ref{section2}, we verify Conjecture~\ref{stronger} for $n=3$
and $n=4$. With a computer-generated proof, we have also verified
Conjecture~\ref{stronger} for $n=5$ and $n=6$, and that $E(2,7)$ is
$6$-regular. For brevity, we do not include these proofs. By the same
technique, it can be shown that $c_{3,3}$ is the only minimal coloring
of the nonzero rational numbers that is free of monochromatic solutions
to $E(3,3)$. We do, however, include a proof of the following result.

\begin{proposition} \label{proposition7}
The only $3$-colorings, up to isomorphism, of the nonzero rational
numbers without a monochromatic solution to $E(\frac{3}{2},3)$ are
$c_{2,3}$ and $c_{3,3}$.
\end{proposition}

As it pertains to Question~\ref{question:colorings}, the following
definition is natural.

\begin{definition}
\label{numb} For a system $\mathcal{L}$ of linear homogeneous
equations over a ring $R$, let $\Delta(\mathcal{L};R)$ denote the number
(as a cardinality) of minimal colorings, up to isomorphism, of the
nonzero elements of $R$ for $\mathcal{L}$.
\end{definition}

For example, we have $\Delta(E(2,n);\mathbb{Q})=1$ for $n \in
\{3,4,5,6\}$ and $\Delta(E(\frac{3}{2},3);\mathbb{Q})=2$.

The following conjecture would reduce the problem of finding
$\Delta(E(q,n);\mathbb{Q})$ to the case when $q$ is a prime power.

\begin{conjecture}
If $a$, $b$, and $n$ are integers, $n>2$, $a>1$, $\lvert b\rvert > 1$,
and $\gcd(a,b)=1$, then
\[ \Delta(E(a,n);\mathbb{Q})=\Delta(E(-a,n);\mathbb{Q}) \]
and
\[
\Delta(E(ab,n);\mathbb{Q})=\Delta(E(a/b,n);\mathbb{Q})=\Delta(E(a,n);\mathbb{Q})+\Delta(E(b,n);\mathbb{Q}). \]
\end{conjecture}

Let $\mathbb{Q}\setminus\{0\}=Q_0 \cup Q_1$ be the partition of the
nonzero rational numbers given by $q \in Q_0$ if and only if $v_2(q)$ is
even. Consider the $3$-coloring $c$ of $Q_0$ given by $c_0(q)=i$ if and
only if $v_2(q)\equiv 2i \pmod{6}$. For a permutation $\pi$ of the set
$\{0,1,2\}$, define the $3$-coloring $c_{\pi}$ of the nonzero rational
numbers by $c_{\pi}(q)=c(q)$ for $q \in Q_0$ and $c_{\pi(q)}=\pi(c(2q))$
for $q \in Q_1$. It is easy to check that the six colorings of the form
$c_{\pi}$ are each minimal for the equation $E(4,3)$. It can be shown
that these are all the minimal colorings of the nonzero rational numbers
for the equation $E(4,3)$, but we leave it out for brevity. This
construction can be easily generalized to give $(n!)^{r-1}$ different
$n$-colorings that are free of monochromatic solutions to $E(2^r,n)$. It
seems likely that these are the only minimal colorings for $E(2^r,n)$.

For each odd prime $p$ and integer $n > 2$, we can find
$(n!)^{\frac{p-3}{2}}$ different $n$-colorings of the nonzero rational
numbers that are free of monochromatic solutions to $E(p,n)$. These
$n$-colorings $c$ are given by the following two properties:
\begin{enumerate}
\item For every nonzero rational number $q$, we have $c(q)=c(p^jq)$
if and only if $j$ is a multiple of $n$. \item If
$v_p(q_1)=v_p(q_2)$ and $w_p(q_1) \equiv \pm w_p(q_2)
\pmod{p}$, then $c(q_1)=c(q_2)$.
\end{enumerate}

For an odd prime $p$, positive integers $n$ and $r$ with $n>2$, the
observations above can be generalized to construct a family of
$(n!)^{r+\frac{p-5}{2}}$ different $n$-colorings of the nonzero rational
numbers that are free of monochromatic solutions to $E(p^r,n)$. It seems
plausible, though we have shown little evidence to support it, that
these are all the minimal colorings for $E(p^r,n)$. This would imply
that $\Delta(E(p^r;n);\mathbb{Q})=(n!)^{r+\frac{p-5}{2}}$ for $n>2$, $p$
an odd prime, and $r$ a positive integer.

The total number of colorings of the nonzero rational numbers is
$2^{\aleph_0}$. Hence, for every system $\mathcal{L}$ of equations, we
have $\Delta(\mathcal{L};\mathbb{Q}) \leq 2^{\aleph_0}$. This upper
bound can be achieved, as the following proposition demonstrates.

\begin{proposition} \label{proposition9}
We have
\[ \Delta(x_1+x_2+x_3=4x_4;\mathbb{Q})=2^{\aleph_0}. \]
\end{proposition}
In fact, we classify all of the $2^{\aleph_0}$ minimal colorings for
$x_1+x_2+x_3=4x_4$ in Section~\ref{section3}.

For a prime number $p$, let $C_p$ be the $(p-1)$-coloring of the nonzero
rational numbers defined by $C_p(q) = w_p(q)$. For any set
$A=\{a_1,\dotsc,a_n\}$ such that no non-empty subset of $A$ sums to
zero, Rado proved that the equation $a_1x_1+\dotsb+a_nx_n=0$ is not
regular by showing that if $p$ is a sufficiently large prime number,
then the coloring $C_p$ is free of monochromatic solutions to
$a_1x_1+\dotsb+a_nx_n=0$. It turns out that the $4$-coloring $C_5$ is
minimal for the equation $x_1+x_2+x_3=4x_4$.

Let $\Pi_p=(\pi_n)_{n \in \mathbb{Z}}$ be a sequence of permutations of
the set $\{1,\dotsc,p-1\}$ of nonzero elements of $\mathbb{Z}_p$ such
that $\pi_0$ is the identity permutation. For each such sequence
$\Pi_p$, we define the coloring $c_{\Pi_p}$ of the nonzero rational
numbers by $c_{\Pi_p}(q)=\pi_{v_p(q)}(w_p(q))$. In particular, the
coloring $c_{\Pi_p}$ is the same as $C_p$ if $\pi_n$ is the identity
permutation for all integers $n$. It is a straightforward exercise to
show that each of the $(p-1)$-colorings $c_{\Pi_p}$ is free of
monochromatic solutions to the equation
$x_1+\dotsb+x_{p-2}=(p-1)x_{p-1}$. For $p=5$, we can say even more.

\begin{proposition} \label{proposition10}
Each of the $4$-colorings $c_{\Pi_5}$ is minimal for $x_1+x_2+x_3=4x_4$,
and there are no other minimal colorings for $x_1+x_2+x_3=4x_4$.
\end{proposition}

There are exactly $2^{\aleph_0}$ colorings of the form $c_{\Pi_5}$,
which establishes Proposition~\ref{proposition9}.

In all the examples above, either an equation has finitely many or
$2^{\aleph_0}$ minimal colorings of the nonzero rational numbers.  It
seems likely that this is always the case.

\begin{conjecture}
For every nonregular finite system $\mathcal{L}$ of linear homogeneous
equations, either $\Delta(\mathcal{L},\mathbb{Q})$ is finite or
$2^{\aleph_0}$. In particular, there is no $\mathcal{L}$ satisfying
$\Delta(\mathcal{L},\mathbb{Q})=\aleph_0$.
\end{conjecture}

\subsection{Minimal colorings over the real numbers}

When studying colorings of the real numbers, we must be careful about
which axioms we choose for set theory. In this subsection, we assume
that $q$ is any rational number other than $-1$, $0$, or $1$.

Radoi\v ci\'c and the second author \cite{FoRa} showed that it is
independent of Zermelo-Fraenkel (ZF) set theory that the equation
$E(q,3)$ is $3$-regular over $\mathbb{R}$. They also showed that it is
independent of ZF that $E(2,4)$ is $4$-regular over $\mathbb{R}$. We
extend these results by showing that for $n=5$ and $n=6$, it is
independent of ZF that $E(2,n)$ is $n$-regular over $\mathbb{R}$. Also,
we show that it is independent of ZF that $x_1+x_2+x_3=4x_4$ is
$4$-regular over $\mathbb{R}$.

They also show that in the Zermelo-Fraenkel-Choice (ZFC) system of
axioms, if $r$ is a positive integer and $\mathcal{L}$ is a finite
system of linear homogeneous equations with rational coefficients, then
$\mathcal{L}$ is $r$-regular over $\mathbb{R}$ if and only if
$\mathcal{L}$ is $r$-regular over $\mathbb{Z}$. It follows, in ZFC, that
the equation $E(q,3)$ is not $3$-regular over $\mathbb{R}$. In
Section~\ref{section:real}, we prove in ZFC that for every integer $n
\geq 3$, there are $2^{2^{\aleph_0}}$ different $n$-colorings of the
nonzero real numbers without a monochromatic solution to $E(q,n)$.
Hence, in ZFC, we have $\Delta(E,\mathbb{R})=2^{2^{\aleph_0}}$ for
$E=E(q,3)$ or $E=E(2,n)$ with $n \in \{3,4,5,6\}$.

Solovay \cite{So} proved that the following axiom is relatively
consistent with ZF.

\begin{axiom}[LM]
Every subset of the real numbers is Lebesgue measurable.
\end{axiom}

Notice that the axiom LM is not consistent with ZFC because, with the
axiom of choice, there are subsets of $\mathbb{R}$ that are not Lebesgue
measurable.

The following lemma is useful in proving in ZF+LM that a given linear
equation is $r$-regular over $\mathbb{R}$ for appropriate $r$.

\begin{lemma} \label{lemma:lm}
Suppose $c$ is a coloring of the nonzero real numbers that uses at most
countably many colors and there are positive real numbers $a$, $b$, $d$
such that $\log_a b$ is irrational and $c(x)=c(ax)=c(bx)\ne c(dx)$ for
all nonzero real numbers $x$.  Then there is a color class of $c$ that
is not Lebesgue measurable.
\end{lemma}

In ZF+LM, if $n\in \{3,4,5,6\}$, then for every $n$-coloring $c$ of the
nonzero real numbers without a monochromatic solution to $E(2,n)$, we
have $c(x)=c(3x)=c(5x)\ne c(2x)$ for all non-zero rational numbers $x$,
and $\log_3 5$ is irrational. Hence, by Lemma~\ref{lemma:lm}, the
equation $E(2,n)$ is $n$-regular in ZF+LM for $n \in \{3,4,5,6\}$. For
every $4$-coloring $c$ of the nonzero real numbers without a
monochromatic solution to the equation $x_1+x_2+x_3=4x_3$, we have
$c(x)=c(6x)=c(11x)\ne c(2x)$ for all nonzero rational numbers $x$ and
$\log_6 11$ is irrational. Hence, by Lemma~\ref{lemma:lm}, the equation
$x_1+x_2+x_3=4x_4$ is $4$-regular in ZF+LM.

\section{Minimal colorings over the rationals} \label{section2}
The following straightforward lemma \cite{FoKl} is useful in proving
that certain colorings are free of monochromatic solutions to particular
linear equations.

\begin{lemma} \label{lemma:valuation}
Suppose $t_1,\dotsc,t_n$ are nonzero rational numbers and $p$ is a prime
number such that $v_p(t_1) \leq v_p(t_2) \leq \dotsb \leq v_p(t_n)$ and
$\sum_{i=1}^n t_i=0$.  Then $v_p(t_1)=v_p(t_2)$.
\end{lemma}

The following proposition is due to the second author and Radoi\v{c}i\'c
~\cite{FoRa}.

\begin{proposition} \label{proposition:E2n}
For each integer $n>1$, the $n$-coloring $c_{2,n}$ is free of
monochromatic solutions to $E(2,n)$.
\end{proposition}
\begin{proof}
For $n>1$, if $x_0+2x_1+\dotsb+2^{n-2}x_{n-2}-2^{n-1}x_{n-1}=0$ is a
solution to $E(2,n)$ in the nonzero rationals, then we have by
Lemma~\ref{lemma:valuation} that for some $i$ and $j$ with $0 \leq i <j
< n$,
\[ v_2(2^{i}x_i)=v_2(2^{j}x_{j}). \]
It follows that $v_2(x_i)=(j-i)+v_2(x_j)$ and $v_2(x_j)-v_2(x_i)
\in \{1,\dotsc,n-1\}$. Therefore, $v_2(x_j)-v_2(x_i)$ is not a
multiple of $n$. Hence, $x_i$ and $x_j$ are not the same color,
and the coloring $c_{2,n}$ is free of monochromatic solutions to
$E(2,n)$.
\end{proof}

They \cite{FoRa} also prove the following structural result for
$3$-colorings of the nonzero rational numbers without a monochromatic
solution to $E(q,3)$.

\begin{lemma} \label{lemma:fr}
Let $x$ and $q$ be nonzero rational numbers with $q \ne \pm 1$, $m$ and
$n$ be integers, and let $a=\frac{q+1}{q^2}$ and $b=q(q-1)$. For every
$3$-coloring $c$ of the nonzero rational numbers without a monochromatic
solution to $E(q,3)$, we have $c(x)=c(a^mb^nx)$ if and only if $m+n$ is
a multiple of $3$.
\end{lemma}
\begin{proof}
Since $x+q(x)=q^2(ax)$, $ax$ must be a different color than $x$.  Since
$bx+q(x)=q^2(x)$, then $bx$ must be a different color than $x$. Since
$x+q(abx)=q^2(x)$, then $abx$ must be a different color than $x$. Hence,
$c(x) \ne c(rx)$ for $r \in \{a,b,ab\}$.

Recall that for a group $G$ and subset $S \subset G$, the Cayley graph
$\Gamma(G,S)$ has vertex set $G$ and two elements $x$ and $y$ are
adjacent if there is an element $s \in S$ such that $x=sy$ or $y=sx$.

We associate each rational number $a^mb^nx$ (with $m$ and $n$ integers)
with the lattice point $(m,n)$. Let $S=\{(1,0),(0,1),(1,1)\}$ and
consider the Cayley graph $\Gamma(\mathbb{Z}^2,S)$. Define the
$3$-coloring $\chi$ of $\mathbb{Z}^2$ by $\chi(m,n)=c(a^mb^nx)$. Since
$c(rx) \ne c(x)$ for $r \in \{a,b,ab\}$, then $\chi$ is a proper
$3$-coloring of the vertices of $\Gamma(\mathbb{Z}^2,S)$. By induction,
it is a straightforward check that there is only one proper coloring of
$\Gamma(\mathbb{Z}^2,S)$ up to isomorphism and this coloring is given by
$\chi(m,n) \equiv m+n \pmod{3}$. Hence, $c(x)=c(a^mb^nx)$ if and only if
$m+n$ is a multiple of $3$.
\end{proof}

\subsection{The minimal coloring for $E(2,3)$}
We prove that $c_{2,3}$ is the only minimal coloring of the nonzero
rational numbers, up to isomorphism, for the equation $E(2,3)$. By
Proposition~\ref{proposition:E2n}, we know that the $3$-coloring
$c_{2,3}$ is free of monochromatic solutions to $E(2,3)$. By
Lemma~\ref{lemma:fr}, the numbers $2$, $3$, and $4$ must be all
different colors, so $E(2,3)$ is $2$-regular. Hence $c_{2,3}$ is a
minimal coloring of the nonzero rational numbers for $E(2,3)$.

\begin{proposition} \label{proposition:E23}
The only minimal coloring of the nonzero rational numbers for $E(2,3)$
is $c_{2,3}$.
\end{proposition}
\begin{proof}
Suppose $c$ is a $3$-coloring of the nonzero rational numbers without a
monochromatic solution to $E(2,3)$. By Lemma~\ref{lemma:fr}, for every
nonzero rational number $x$ and integers $m$ and $n$, we have
$c((\frac{3}{4})^m2^nx)=c(x)$ if and only if $m+n$ is a multiple of
$3$. Equivalently, for every nonzero rational number $x$ and integers
$m$ and $n$, $c(2^m3^nx)=c(x)$ if and only if $m$ is a multiple of $3$.

For nonzero rational numbers $x$ and $r$ with $r$ positive, we now show
that $c(x)=c(rx)$ if and only if $v_2(r) \equiv 0 \pmod{3}$. By
induction, it suffices to prove that $c(x)=c(px)$ for every odd positive
integer $p$ and nonzero rational number $x$. For $p=1$ or $3$, we have
already established that $c(x)=c(px)$ for every nonzero rational number
$x$.

So let $p$ be an odd integer greater than $3$ and suppose that for every
positive odd integer $p'<p$ and rational number $x$, $c(p'x)=c(x)$. The
numbers $px$ and $4x$ are different colors because otherwise
$(x_0,x_1,x_2)=(4(p-2)x,4x,px)$ is a monochromatic solution to
$E(2,3)$. We have $px$ and $2x$ are different colors because otherwise
$(16x,2(p-4)x,px)$ is a monochromatic solution to $E(2,3)$. Since
$c(x)$, $c(2x)$, and $c(4x)$ are all different colors, then
$c(px)=c(x)$. By induction, for nonzero rational numbers $x$ and $r$
with $r$ positive, we have $c(x)=c(rx)$ if and only if $v_2(r) \equiv 0
\pmod{3}$.

To finish the proof, we need to show that $c(x)=c(-x)$ for every nonzero
rational number $x$. Since $(10x,-x,2x)$ is a solution to $E(2,3)$ and
$10x$ and $2x$ are the same color, then $-x$ and $2x$ are different
colors. Since $(12x,-8x,-x)$ is a solution to $E(2,3)$, $-8x$ and $-x$
are the same color, and $12x$ is the same color as $4x$, then $-x$ and
$4x$ are differently colored. Since $x$, $2x$, and $4x$ are all
different colors, then $c(-x)=c(x)$. Therefore, $c_{2,3}$ is the only
minimal coloring, up to isomorphism, of the nonzero rational numbers
without a monochromatic solution to $E(2,3)$.
\end{proof}

Using a very similar argument to the proof of
Proposition~\ref{proposition:E23}, it is not difficult to show that
there are only two minimal colorings of the positive integers for the
equation $E(2,3)$. These two minimal colorings consist of $c_{2,3}$ with
its domain restricted to the positive integers and a coloring $c_{2,3}'$
that is identical to $c_{2,3}$ with its domain restricted to the
positive integers except that the color of $1$ is different. Formally,
$c_{2,3}'\colon \mathbb{N} \rightarrow \{0,1,2\}$ is the coloring of the
positive integers such that $c_{2,3}'(1)=2$ and $c_{2,3}'(n)=c_{2,3}(n)$
for $n>1$.

\begin{proposition}
The only two minimal colorings of the positive integers for the equation
$E(2,3)$ are the colorings $c_{2,3}$ with its domain restricted to the
positive integers and $c_{2,3}'$.
\end{proposition}

\subsection{The minimal coloring for $E(2,4)$}

In this subsection we prove that the only minimal coloring of the
nonzero rational numbers, up to isomorphism, for $E(2,4)$ is
$c_{2,4}$. The proof uses the following lemma from \cite{FoRa}, proven
below.

\begin{lemma} \label{lemma:muliple4}
For every $4$-coloring $c$ of the nonzero rational numbers without a
monochromatic solution to $E(2,4)$ and for nonzero rational number $x$
and integers $m$ and $n$, we have $c(x)=c(2^m3^nx)$ if and only if $m$
is a multiple of $4$.
\end{lemma}

In particular, Lemma~\ref{lemma:muliple4} implies that $E(2,4)$ is
$3$-regular since in every $4$-coloring of the nonzero rational numbers
without a monochromatic solution to $E(2,4)$, the numbers $1$, $2$, $4$,
and $8$ are different colors. It follows that $c_{2,4}$ is minimal for
$E(2,4)$.  Moreover, we have the following proposition.

\begin{proposition} \label{proposition:E24}
The coloring $c_{2,4}$ is the only minimal coloring of the nonzero
rational numbers without a monochromatic solution to $E(2,4)$.
\end{proposition}
\begin{proof}[Proof of Lemma~\ref{lemma:muliple4}.]
By induction on $m$ and $n$, it suffices to prove that for all nonzero
rational numbers $q$, we have $c(q)=c(3q)=c(16q)$ and $c(q) \notin
\{c(2q),c(4q),c(8q)\}$.  By considering all solutions to $E(2,4)$ with
exactly two distinct variables, for $r \in \{\frac{n+1}{n} : n \in
\mathbb{Z}~\textrm{and}~ 1 \leq n \leq 7\}$, we have that $q$ and $rq$
are different colors (call these ratios $r$ \emph{forbidden}). Note
immediately that $c(x) \ne c(2x)$ by the forbidden ratio $2$.

We begin by showing that $c(x) \ne c(4x)$.  Proceeding by contradiction,
suppose $c(x) = c(4x)$ instead for some $x$.  By forbidden ratios
amongst themselves, we have that $c(x) = c(4x)$, $c(2x)$, $c(3x)$, and
$c(\frac{3}{2}x)$ are distinct colors.  Then $c(\frac{9}{4}x) = c(2x)$
by forbidden ratios from $3x$ and $\frac{3}{2}x$ and because of the
solution $(x,4x,\frac{9}{4}x,\frac{9}{4}x)$.  Similarly, $c(6x) =
c(\frac{3}{2}x)$ because of forbidden ratios from $4x$ and $3x$ and
because of the solution $(6x,2x,2x,\frac{9}{4}x)$.  Finally,
$\frac{18}{7}x$ cannot be colored with any colors because of forbidden
ratios from $\frac{9}{4}x$ and $3x$ as well as the solutions
$(\frac{18}{7}x, x, 4x, \frac{18}{7}x)$ and $(\frac{18}{7}x, 6x,
\frac{3}{2}x, \frac{18}{7}x)$.

We now show that $c(x) = c(3x)$.  Assume otherwise, so that $c(x) \ne
c(3x)$ for some $x$.  Once again, by forbidden ratios amongst
themselves, we have that $c(x)$, $c(3x)$, $c(2x)$, and $c(4x)$ are
distinct; furthermore, $c(\frac{3}{2}x) = c(4x)$ also by forbidden
ratios.  We have $c(\frac{4}{3}x) = c(3x)$ by forbidden ratios from $x$
and $2x$ as well as the solution $(4x, \frac{4}{3}x, \frac{4}{3}x,
\frac{3}{2}x)$.  Next, $c(\frac{5}{3}x) = c(x)$ by forbidden ratios from
$\frac{4}{3}x$ and $2x$ as well as the solution $(4x, \frac{5}{3}x,
\frac{3}{2}x, \frac{5}{3}x)$.  Similarly, $c(6x) = c(2x)$ by forbidden
ratios from $3x$ and $\frac{3}{2}x$ and the solution $(6x, \frac{5}{3}x,
x, \frac{5}{3}x)$.  Finally, $\frac{9}{4}x$ cannot be colored with any
colors, because of forbidden ratios from $3x$ and $\frac{3}{2}x$ as well
as the solutions $(x, \frac{5}{3}x, \frac{9}{4}x, \frac{5}{3}x)$ and
$(6x, 2x, 2x, \frac{9}{4}x)$.

Clearly, $c(x) \ne c(8x)$ since otherwise $(8x, \frac{8}{3}x,
\frac{8}{3}x, 3x)$ would be a monochromatic solution.  Completing the
proof, both $c(x)$ and $c(16x)$ must be different from all of $c(2x)$,
$c(4x)$, and $c(8x)$ (which are distinct), so $c(x) = c(16x)$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{proposition:E24}.]
The proof is similar to the proof of
Proposition~\ref{proposition:E23}. Suppose $c$ is a $4$-coloring of the
nonzero rational numbers without a monochromatic solution to
$E(2,4)$. By Lemma~\ref{lemma:muliple4}, for every nonzero rational
number $x$ and integers $m$ and $n$, we have $c(x)=c(2^m3^nx)$ if and
only if $m$ is a multiple of $4$. We now show that for every positive
rational number $r$ with $v_2(r) \equiv 0 \pmod{4}$, we have
$c(x)=c(rx)$. By induction, it suffices to prove that $c(x)=c(px)$ for
every positive odd integer $p$ and nonzero rational number $x$. For
$p=1$ or $3$, Lemma~\ref{lemma:muliple4} implies that $c(x)=c(px)$ for
every nonzero rational number $x$.

For $p=5$, we have $(18x,5x,5x,6x)$ is a solution to $E(2,4)$ and
$c(18x)=c(2x)=c(6x)$, so $5x$ and $2x$ are different colors.  Also,
$(12x,4x,5x,5x)$ is a solution to $E(2,4)$ and $c(12x)=c(4x)$, so $5x$
and $4x$ are different colors.  Finally, $(5x,\frac{3}{2}x,8x,5x)$ is a
solution to $E(2,4)$ and $c(\frac{3}{2}x)=c(8x)$, so $c(5x)$ and $c(8x)$
are different colors. Since $x$, $2x$, $4x$, and $8x$ are all different
colors, then $c(5x)$ must be the color of $c(x)$.

The rest of the proof is by induction on $p$.  Suppose $p$ is an odd
integer greater than $5$ and that for all positive odd $p' < p$ and
rational $x$, $c(p'x) = c(x)$.  Then for all $x$, we know that $px$ and
$2x$ are different colors since $(\frac{32}{3}x, \frac{32}{3}x,
2(p-4)x,px)$ would otherwise be a monochromatic solution. Similarly,
$px$ and $4x$ are different colors because of $(\frac{64}{5}x, 4(p-2)x,
\frac{4}{5}x, px)$.  Finally, $px$ and $8x$ are different colors because
of $(8(p-2)x, \frac{8}{3}x, \frac{8}{3}x, px)$. Since $c(x)$, $c(2x)$,
$c(4x)$, and $c(8x)$ are all distinct, it follows that $c(px) = c(x)$ as
desired.

To finish the proof, we need to show that $c(x)=c(-x)$ for every nonzero
rational number $x$. Since $(10x,-x,2x,2x)$ is a solution to $E(2,4)$
and $c(10x)=c(2x)$, then $c(-x)$ and $c(2x)$ are different colors. Since
$(28x,-16x,-x,-x)$ is a solution to $E(2,4)$, $c(-16x)=c(-x)$, and
$c(28x)=c(4x)$, then $-x$ and $4x$ are different colors. Since
$(88x,-16x,-16x,-x)$ is a solution to $E(2,4)$, $c(-16x)=c(-x)$, and
$c(88x)=c(8x)$, then $-x$ and $8x$ are different colors. Since $x$,
$2x$, $4x$, and $8x$ are all different colors, then
$c(-x)=c(x)$. Therefore, $c_{2,4}$ is the only minimal coloring, up to
isomorphism, of the nonzero rational numbers without a monochromatic
solution to $E(2,4)$.
\end{proof}

\subsection{The minimal colorings for $E(\frac{3}{2},3)$}

In this subsection we prove that the two minimal colorings of the
nonzero rational numbers for $E(\frac{3}{2},3)$ are $c_{2,3}$ and
$c_{3,3}$. By a proof similar to Proposition~\ref{proposition:E2n}, it
is clear that $c_{2,3}$ and $c_{3,3}$ are free of monochromatic
solutions to $E(\frac{3}{2},3)$. The equation $E(\frac{3}{2},3)$ is
$2$-regular since in any coloring of the nonzero rational numbers
without a monochromatic solution to $E(\frac{3}{2},3)$, the numbers $9$,
$10$, and $12$ are all different colors (by Lemma~\ref{lemma:fr}). Hence
$c_{2,3}$ and $c_{3,3}$ are minimal colorings for
$E(\frac{3}{2},3)$. Again from Lemma~\ref{lemma:fr}, it follows that for
every nonzero rational number $x$ and integers $m$ and $n$, we have
$c(x)=c((\frac{6}{5})^m(\frac{10}{9})^nx)$ if and only if $m+n$ is a
multiple of $3$. In particular, $c(x)=c(rx)$ for $r \in \{\frac{8}{5},
\frac{64}{27}, \frac{40}{27}\}$ and $c(x) \ne c(rx)$ for $r \in
\{\frac{6}{5}, \frac{10}{9},\frac{4}{3}\}$.

For the rest of this subsection, we suppose that $c$ is a $3$-coloring
that is free of monochromatic solutions to $E(\frac{3}{2},3)$ and we
will deduce that $c$ is either $c_{2,3}$ or $c_{3,3}$. We first build up
structural properties about $c$ if there is a nonzero rational number
$x$ such that $c(x)=c(2x)$ and deduce that $c$ is the coloring
$c_{3,3}$. We then prove properties about $c$ if $c(x) \ne c(2x)$ for
every nonzero rational number $x$.  We deduce in this case that $c$ is
the coloring $c_{2,3}$.

\begin{lemma} \label{lemma:2powers}
If $x$ is a nonzero rational number such that $c(2x)=c(x)$, then
$c(2^nx)=c(x)$ for every nonnegative integer $n$.
\end{lemma}

\begin{lemma} \label{lemma:3multiples}
If $x$ is a nonzero rational number such that $c(2x)=c(x)$, then for all
nonnegative integers $k$, $m$, and $n$ we have $c(2^k3^m5^nx)=c(x)$ if
and only if $m$ is a multiple of $3$.
\end{lemma}

\begin{lemma} \label{lemma:36}
If $x$ is a nonzero rational number such that $c(2x)=c(x)$, then
$c(3x)=c(6x)$.
\end{lemma}

From Lemma~\ref{lemma:3multiples} and Lemma~\ref{lemma:36}, we have the
following result.

\begin{lemma}  \label{lemma:235}
If $c$ is a $3$-coloring of the nonzero rational numbers without a
monochromatic solution to $E(\frac{3}{2},3)$ and $x$ is a rational
number such that $c(2x)=c(x)$, then for all integers $m_1$, $n_1$,
$p_1$, $m_2$, $n_2$, $p_2$, we have
$c(2^{n_1}3^{m_1}5^{p_1}x)=c(2^{n_2}3^{m_2}5^{p_2}x)$ if and only if
$m_1 \equiv m_2 \pmod{3}$.
\end{lemma}

The following lemma gets us much closer to proving that $c_{3,3}$ is the
only $3$-coloring of the nonzero rational numbers for which there is a
nonzero rational number $x$ such that $c(x)=c(2x)$.

\begin{lemma} \label{lemma:3multiples2}
If $x$ is a nonzero rational number such that $c(x)=c(2x)$, then for
every positive integer $n$, we have $c(nx)=c(x)$ if and only if $v_3(n)$
is a multiple of $3$.
\end{lemma}

Finally, to finish the proof that $c_{3,3}$ is the only $3$-coloring $c$
of the nonzero rational numbers without a monochromatic solution to
$E(\frac{3}{2},3)$ and for which there is a rational number $x$ such
that $c(x)=c(2x)$, it suffices to prove that $c(y)=c(-y)$ for all $y$.

\begin{lemma} \label{lemma:onlyc33}
The only $3$-coloring $c$ of the nonzero rational numbers for which
there is a rational number $x$ such that $c(x)=c(2x)$ is $c_{3,3}$.
\end{lemma}

Having completed the case when there is a nonzero rational number $x$
such that $c(x)=c(2x)$, we now look at those $3$-colorings for which
$c(x) \ne c(2x)$ for all nonzero rational numbers $x$.

\begin{lemma} \label{lemma:iff3multiple}
If $c(x) \ne c(2x)$ for every nonzero rational number $x$, then
$c(2^ny)=c(y)$ holds for integer $n$ and nonzero rational number $y$ if
and only if $n$ is a multiple of $3$.
\end{lemma}

\begin{lemma} \label{lemma:iff3multiple2}
If $c(x) \ne c(2x)$ for every nonzero rational number $x$, then
$c(2^m3^n5^py)=c(y)$ holds for nonzero rational number $y$ and integers
$m$, $n$, and $p$ if and only if $m$ is a multiple of $3$.
\end{lemma}

\begin{lemma} \label{lemma:neg}
If $c(x) \ne c(2x)$ for all nonzero rational numbers $x$, then
$c(y)=c(-y)$ for all nonzero rational numbers $y$.
\end{lemma}

Finally, to finish the proof of Proposition~\ref{proposition7} that
$c_{2,3}$ and $c_{3,3}$ are the only $3$-colorings of the nonzero
rational numbers without a monochromatic solution to $E(\frac{3}{2},3)$,
it suffices to prove the following lemma.

\begin{lemma} \label{lemma:3multiple3}
If no nonzero rational number $x$ satisfies $c(x)=c(2x)$, then
$c(nx)=c(x)$ if and only if $v_2(n)$ is a multiple of $3$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lemma:2powers}.]
By induction on $n$, it suffices to prove that $c(4x)=c(x)$ if
$c(x)=c(2x)$.  Assume for contradiction that $c(x) = c(2x) \ne c(4x)$.
Then we know that $c(3x)$ must further be distinct from $c(x)$ and
$c(4x)$ because of the forbidden ratio from $4x$ and the solution
$(3x,x,2x)$.  It follows that $c(\frac{10}{3}x) = c(x)$ because of
forbidden ratios from $3x$ and $4x$.  Similarly, $c(\frac{5}{2}x) =
c(4x)$ because of forbidden ratios from $3x$ and $\frac{10}{3}x$.  It
follows that $c(\frac{13}{3}x) = c(3x)$ because of the solutions $(x,
\frac{13}{3}x, \frac{10}{3}x)$ and $(\frac{5}{2}x, \frac{13}{3}x, 4x)$.
Similarly, $c(\frac{9}{2}x) = c(4x)$ because of the solutions
$(\frac{9}{2}x, 2x, \frac{10}{3}x)$ and $(3x, \frac{9}{2}x,
\frac{13}{3}x)$.  We have $c(6x) = c(3x)$ because of a forbidden ratio
from $\frac{9}{2}x$ and the solution $(6x, x, \frac{10}{3}x)$.  Finally,
the number $\frac{3}{4}x$ cannot be colored with any colors, because of
a forbidden ratio from $x$ as well as the solutions $(\frac{9}{2}x,
\frac{3}{4}x, \frac{5}{2}x)$ and $(\frac{3}{4}x, 6x, \frac{13}{3}x)$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:3multiples}.]
By the previous lemma, we have $c(2^nx)=c(x)$ for all nonnegative
integers $n$. Since $c(\frac{5}{8}y)=c(y)$ for every nonzero rational
number $y$, then $c(2^n5^px)=c(x)$ for all nonnegative integers $n$ and
$p$. To finish the proof, it suffices, by induction, to prove that
neither $3x$ nor $9x$ is the same color as $x$, and $27x$ is the same
color as $x$. Since $3x$ and $4x$ have ratio $\frac{3}{4}$, then $c(3x)
\ne c(4x)=c(x)$. Since $9x$ and $16x$ have ratio
$(\frac{6}{5})^{-2}(\frac{10}{9})^{-2}$, then $c(9x) \ne
c(16x)=c(x)$. Since $27x$ and $64x$ have ratio $\frac{27}{64}$, then
$c(27x)=c(64x)=c(x)$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:36}.]
Assume for contradiction that $c(3x) \ne c(6x)$. By
Lemma~\ref{lemma:3multiples}, $c(x)=c(4x)=c(8x)$. Since
$4x=\frac{4}{3}(3x)$ and $8x=\frac{4}{3}(6x)$, then $x$, $3x$, and $6x$
are all different colors. Since $\frac{64}{27}(6x)=\frac{32}{9}x$, then
$\frac{32}{9}x$ is the same color as $6x$. Since
$(6x,\frac{4}{3}x,\frac{32}{9}x)$ is a solution to $E(\frac{3}{2},3)$,
then $c(\frac{4}{3}x)$ and $c(6x)$ are different colors. Since
$\frac{4}{3}x$ and $x$ have ratio $\frac{4}{3}$, then $\frac{4}{3}x$ and
$x$ are different colors. Hence, $\frac{4}{3}x$ is the same color as
$3x$. Since $\frac{3}{4}x$, $x$, and $\frac{4}{3}x$ are all different
colors, then $\frac{3}{4}x$ is the same color as $6x$. Since
$\frac{3}{4}(2x)=\frac{3}{2}x$, then $\frac{3}{2}x$ and $x$ are
different colors. If $\frac{3}{2}x$ and $3x$ are the same color, then by
Lemma~\ref{lemma:3multiples}, $6x$ would also be the same color as $3x$,
contradicting the assumption that $3x$ and $6x$ are different
colors. Hence, $\frac{3}{2}x$ is the same color as $6x$. But then
$\frac{3}{2}x$ and $\frac{3}{4}x$ are the same color, which implies, by
Lemma~\ref{lemma:3multiples}, that $3x$ is the same color as $6x$, a
contradiction.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:235}.]
Clearly, from Lemma~\ref{lemma:3multiples} and Lemma~\ref{lemma:36}, we
have the result for \emph{nonnegative} integers.

Recall that, for every nonzero rational number $y$, we have
$c(y)=c(\frac{8}{5}y)$ and we also have $y$, $\frac{4}{3}y$, and
$\frac{16}{9}y$ are all different colors. Therefore, it suffices, by the
remark above and induction to prove that $c(\frac{x}{2})=c(x)$. Since
$c(3x)=c(6x)$ and $(6x,\frac{x}{2},3x)$ is a solution to
$E(\frac{3}{2},3)$, then $\frac{x}{2}$ is a different color from
$3x$. We have $\frac{80}{3}x=\frac{40}{27}(18x)$, so $18x$ and
$\frac{80}{3}x$ are the same color as $9x$. Since
$(\frac{x}{2},\frac{80}{3}x,18x)$ is a solution to $E(\frac{3}{2},3)$,
then $\frac{x}{2}$ and $9x$ are different colors. Hence, $\frac{x}{2}$
is the same color as $x$, which completes the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:3multiples2}.]
Suppose $c$ is a $3$-coloring of the nonzero rational numbers without a
monochromatic solution to $E(\frac{3}{2},3)$ and $x$ is a nonzero
rational number such that $c(x)=c(2x)$. By Lemma~\ref{lemma:235}, for
integers $m_1$, $n_1$, $p_1$, $m_2$, $n_2$, and $p_2$ we have
$c(2^{n_1}3^{m_1}5^{p_1}x)=c(2^{n_2}3^{m_2}5^{p_2}x)$ if and only if
$m_2-m_1$ is a multiple of $3$. By induction, it suffices to prove for
every prime $p>3$, that $c(x)=c(px)$. For $p=5$, Lemma~\ref{lemma:235}
implies that $c(x)=c(px)$.

The proof is by induction on the size of $p$. Suppose $p$ is prime with
$p>5$. We write $p=6a+b$, where $a$ and $b$ are nonnegative integers and
$b \in \{1,5\}$. The induction hypothesis is that $c(q)=c(p'q)$ for
every nonzero rational number $q$ satisfying $c(q)=c(2q)$ and prime $p'$
satisfying $3<p'<p$. The induction hypothesis implies that $c(q)=c(p'q)$
for every nonzero rational number $q$ satisfying $c(q)=c(2q)$ and
positive odd integer $p'$ which is not a multiple of $3$ and satisfies
$p'<p$. Then we have for all $p$ that $c(px) \ne c(9x) =
c(\frac{9}{4}x)$ because of the solution $(\frac{9}{4}(p-6)x, 9x, px)$.
It suffices to show that in the two cases below, $c(px) \ne c(3x)$ since
$c(x)$, $c(3x)$, and $c(9x)$ are distinct.

\textbf{Case 1:} $p=6a+1$. For $a=1$, we have $p=7$, and $(3x,7x,6x)$ is
a solution to $E(\frac{3}{2},3)$, so $7x$ and $3x$ are different
colors. For $a>1$, we have $0<3a+5<6a+1$, so by the induction
hypothesis, we have $c((3a+5)q)=c(q)$ for every rational number $q$
satisfying $c(q)=c(2q)$, and in particular, for $q=\frac{8}{9}x$. The
numbers $6x$ and $\frac{8}{9}x$ are the same color as the color of $3x$
by Lemma~\ref{lemma:235}. Since $(px,6x,(3a+5)\frac{8}{9}x)$ is a
solution to $E(\frac{3}{2},3)$, then $px$ and $3x$ are different
colors. Hence, $c(px)=c(x)$.

\textbf{Case 2:} $p=6a+5$. For each prime $p>5$ of the form $p=6a+5$ we
have $3a+7<6a+5$, so by the induction hypothesis we have
$c((3a+5)q)=c(q)$ for every rational number $q$ satisfying $c(q)=c(2q)$,
and in particular, for $q=\frac{8}{9}x$. The numbers $6x$ and
$\frac{8}{9}x$ are the same color as the color of $3x$ by
Lemma~\ref{lemma:235}. Since $(px,6x,(3a+7)\frac{8}{9}x)$ is a solution
to $E(\frac{3}{2},3)$, then $px$ and $3x$ are different colors. Hence,
$c(px)=c(x)$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:onlyc33}.]
By Lemma~\ref{lemma:3multiples2}, it suffices to prove that $c(-y)=c(y)$
for all nonzero rational numbers $y$. By Lemma~\ref{lemma:3multiples2},
we have $c(\frac{85}{6}y)=c(9y)$. Since $(-y,\frac{85}{6}y,9y)$ is a
solution to $E(\frac{3}{2},3)$, then $-y$ and $9y$ are different
colors. By Lemma~\ref{lemma:3multiples2}, we have
$c(\frac{14}{9}y)=c(3y)$. Since $(-y,3y,\frac{14}{9}y)$ is a solution to
$E(\frac{3}{2},3)$, then $-y$ and $3y$ are different colors. Therefore,
we have $c(-y)=c(y)$, completing the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:iff3multiple}.]
It suffices to prove that $c(y) \ne c(4y)$ for all nonzero rational
numbers $y$. So suppose for contradiction that there is a nonzero
rational number $y$ such that $c(y)=c(4y)$. Let red be the color of $y$,
blue be the color of $2y$, and green be the remaining color.  Since
$3y=\frac{4}{3}(4y)$, then $3y$ is green or blue.

\textbf{Case 1:} $3y$ is green. Since $3y$, $4y$, and $\frac{16}{3}y$
are all different colors, then $\frac{16}{3}$y is blue. Since
$\frac{9}{4}y=\frac{27}{64}(\frac{16}{3}y)$, then $\frac{9}{4}y$ is
blue. Since $\frac{9}{2}y=2(\frac{9}{4}y)$, then $\frac{9}{2}y$ is not
blue. Since $(9y,2y,\frac{16}{3}y)$ is a solution to $E(\frac{3}{2},3)$,
then $9y$ is not blue. Since $9y=2(\frac{9}{2}y)$, then $\frac{9}{2}y$
and $9y$ are different colors. Therefore, either $\frac{9}{2}y$ is red
and $9y$ is green or $\frac{9}{2}y$ is green and $9y$ is red.

\textbf{Case 1a:} $\frac{9}{2}y$ is red and $9y$ is green. Since
$6y=\frac{4}{3}(\frac{9}{2}y)$, then $6y$ is not red. Since $6y=2(3y)$,
then $6y$ is not green. Hence, $6y$ is blue. Since $\frac{9}{2}y$, $6y$,
and $8y$ are all different colors, then $8y$ is green. Since
$12y=2(6y)$, then $12y$ is not blue. Since $12y=\frac{4}{3}(9y)$, then
$12y$ is red. Since $\frac{15}{2}y=\frac{5}{8}(12y)$, then
$\frac{15}{2}y$ is red. Then $(\frac{15}{2}y,y,4y)$ is a monochromatic
solution to $E(\frac{3}{2},3)$, a contradiction.

\textbf{Case 1b:} $\frac{9}{2}y$ is green and $9y$ is red. Since
$\frac{10}{3}y=\frac{5}{8}(\frac{16}{3}y)$, then $\frac{10}{3}y$ is
blue. Since $3y=2(\frac{3}{2}y)$ and $2y=\frac{4}{3}(\frac{3}{2}y)$,
then $\frac{3}{2}y$ is red. Since
$\frac{5}{3}y=\frac{10}{9}(\frac{3}{2}y)$ and
$\frac{5}{3}y=\frac{5}{6}(2y)$, then $\frac{5}{3}y$ is green.  Finally,
$\frac{28}{9}y$ can not be colored with any colors because of the
solutions $(y,4y,\frac{28}{9}y)$, $(2y,\frac{10}{3}y,\frac{28}{9}y)$,
and $(\frac{9}{2}y,\frac{5}{3}y,\frac{28}{9}y)$.

\textbf{Case 2:} $3y$ is blue. Since $3y$, $4y$, $\frac{16}{3}y$ are all
different colors, then $\frac{16}{3}y$ is green. Since
$\frac{16}{3}y=2(\frac{8}{3}y)$, then $\frac{8}{3}y$ is not green.
Since $\frac{8}{3}y=\frac{4}{3}(2y)$, then $\frac{8}{3}y$ is not
blue. Hence, $\frac{8}{3}y$ is red. Since $\frac{3}{2}y$, $2y$,
$\frac{8}{3}y$ are all different colors, then $\frac{3}{2}y$ is
green. Since $\frac{9}{4}=\frac{27}{64}(\frac{16}{3}y)$, then
$\frac{9}{4}y$ is green. Since $\frac{9}{2}y=2(\frac{9}{4}y)$, then
$\frac{9}{2}y$ is not green. Since $(\frac{9}{2}y,y,\frac{8}{3}y)$ is a
solution to $E(\frac{3}{2},3)$, then $\frac{9}{2}y$ is not
red. Therefore, $\frac{9}{2}y$ is blue. Since
$8y=(\frac{6}{5})^2(\frac{10}{9})^2(\frac{9}{2}y)$, then $8y$ is not
blue. Since $8y=2(4y)$, then $8y$ is not red. Hence, $8y$ is
green. Since $5y=\frac{5}{8}(8y)$, then $5y$ is green.  Since
$\frac{32}{9}y=\frac{64}{27}(\frac{3}{2}y)$, then $\frac{32}{9}$ is
green. Since $(\frac{y}{2},5y,\frac{32}{9}y)$ is a solution to
$E(\frac{3}{2},3)$, then $\frac{y}{2}$ is not green. Since
$y=2(\frac{y}{2})$, then $\frac{y}{2}$ is not red. Hence, $\frac{y}{2}$
is blue.

We found a contradiction in Case 1, so if $y$ and $4y$ are the same
color, then $\frac{y}{2}$, $2y$, $3y$ are the same color.  Therefore,
$\frac{y}{4}$, $y$, $\frac{3}{2}y$ must be the same color. But
$\frac{3}{2}y$ is green and $y$ is red, a contradiction.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:iff3multiple2}.]
It suffices, by induction and Lemma~\ref{lemma:iff3multiple}, to prove
that $c(y)=c(3y)=c(5y)$ for every $y$. By
Lemma~\ref{lemma:iff3multiple}, $c(y)=c(8y)$ for every $y$. Since
$8y=\frac{8}{5}(5y)$, then $5y$ is the same color as $8y$, so
$c(y)=c(5y)$ for all $y$. So suppose for contradiction that there is a
nonzero rational number $x$ such that $c(x) \ne c(3x)$. So, by
Lemma~\ref{lemma:iff3multiple}, $x$, $2x$, and $4x$ are all different
colors. Let red be the color of $x$, blue be the color of $2x$, and
green be the color of $4x$. Since $3x$ is a different color from $x$ and
$4x$, then $3x$ is blue. Since $24x=8(3x)$, then $24x$ is blue. Since
$50x=25(2x)$, then $50x$ is blue. Since
$\frac{64}{9}x=\frac{64}{27}(3x)$, then $\frac{64}{9}x$ is blue. Since
$(3x,24x,\frac{52}{3}x)$, $(3x,50x,\frac{104}{3}x)$, and
$(3x,\frac{26}{3}x,\frac{64}{9}x)$ are solutions to $E(\frac{3}{2},3)$,
then none of the numbers $\frac{26}{3}x$, $\frac{52}{3}x$,
$\frac{104}{3}x$ is blue. But $\frac{26}{3}x$, $\frac{52}{3}x$,
$\frac{104}{3}x$ are all different colors, so one of them has to be
blue, a contradiction.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:neg}.]
By Lemma~\ref{lemma:iff3multiple2}, the numbers $2y$ and $\frac{2}{9}y$
are the same color and the numbers $4y$, $\frac{4}{3}y$, $\frac{4}{9}y$
are the same color. Since $(2y,-y,\frac{2}{9}y)$ and
$(-y,\frac{4}{3}y,\frac{4}{9}y)$ are solutions to $E(\frac{3}{2},3)$ and
$y$, $2y$, $4y$ are all different colors, then $c(y)=c(-y)$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:3multiple3}.]
By Lemma~\ref{lemma:iff3multiple2} and Lemma~\ref{lemma:neg}, for
integers $l$, $m$, $n$, and $p$ and nonzero rational number $y$, we have
$c((-1)^{l}2^{m}3^{n}5^{p}y)=c(y)$ if and only if $m$ is a multiple of
$3$. By induction, it suffices to prove for every odd $p \geq 3$, that
$c(x)=c(px)$. For $p=3$ or $p=5$, Lemma~\ref{lemma:iff3multiple2}
implies that $c(x)=c(px)$.

The proof is by induction on the size of $p$. Suppose $p$ is an odd
number with $p>5$. The induction hypothesis is that $c(q)=c(p'q)$ for
every nonzero rational number $q$ and odd number $p'$ that is less than
$p$.  We know that $c(px) \ne c(2x) = c(\frac{9}{4}x)=c(6x)$ because of
the solution $(\frac{9}{4}(p-4)x, 6x, px)$.  Similarly, $c(px) \ne c(4x)
= c(\frac{3}{2}x)=c(\frac{9}{2}x)$ because of the solution
$(\frac{9}{2}x, \frac{3}{2}(p-2)x, px)$.  Since $c(x)$, $c(2x)$, and
$c(4x)$ are distinct, $c(px) = c(x)$.
\end{proof}
\section{Minimal colorings for $x_1+x_2+x_3=4x_4$} \label{section3}
In this section we prove that the minimal colorings for
$x_1+x_2+x_3=4x_4$ are those of the form $c_{\Pi_5}$. It is a
straightforward check that each of the colorings of the form $c_{\Pi_5}$
is minimal for $x_1+x_2+x_3=4x_4$. We suppose for the rest of this
subsection that $c$ is a $4$-coloring of the nonzero rational numbers
without a monochromatic solution to $x_1+x_2+x_3=4x_4$.

\begin{lemma} \label{lemma:forb1}
If $x$ is a nonzero rational number and $r \in
\{\frac{4}{3},\frac{3}{2},2\}$, then $c(x) \ne c(rx)$.
\end{lemma}

\begin{lemma} \label{lemma:forb2}
For every nonzero rational number $x$, we have $c(x) \ne c(3x)$.
\end{lemma}

\begin{lemma} \label{lemma:forb3}
For every nonzero rational number $x$, we have $c(x) \ne c(4x)$.
\end{lemma}

\begin{lemma} \label{lemma:235-2}
For every nonzero rational number $x$ and integers $m$ and $n$, we have
$c(x) = c(2^m3^nx)$ if and only if $w_5(2^m3^n) \equiv 1 \pmod{5}$.
\end{lemma}

\begin{lemma} \label{lemma:-4}
For every nonzero rational number $x$, we have $c(-x)=c(4x)$.
\end{lemma}

To finish the proof of Proposition~\ref{proposition10}, it suffices by
Lemma~\ref{lemma:-4} to prove the following lemma.

\begin{lemma} \label{lemma:mod5}
If $x$ is a nonzero rational number and $n$ is a positive integer
satisfying $v_5(n)=0$ and $n \equiv d \pmod{5}$ with $d\in \{1,2,3,4\}$,
then $c(nx)=c(dx)$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lemma:forb1}.]
Since $(\frac{4}{3}x,\frac{4}{3}x,\frac{4}{3}x,x)$,
$(\frac{3}{2}x,\frac{3}{2}x,x,x)$, and $(2x,x,x,x)$ are solutions to
$x_1+x_2+x_3=4x_4$, then $c(x)\ne c(rx)$ for $r \in
\{\frac{4}{3},\frac{3}{2},2\}$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:forb2}.]
  We suppose for contradiction that there is a nonzero rational number
  $x$ such that $c(x)=c(3x)$. Without loss of generality, we may take
  $x=1$. The previous lemmas imply that the ratios $2$, $\frac 32$, and
  $\frac 43$ are \emph{forbidden ratios}, that is, $c(x) \ne c(rx)$ for
  $r\in \{2, \frac 32, \frac 43\}$. Using these forbidden ratios, we see
  that $c(1)=c(3)$, $c(2)$, and $c(4)$ must be different colors.

  \newcommand{\footnotesize }{\footnotesize }
  \newcommand{\hspace{1em}}{\hspace{1em}}
  \newcommand{\myfrac}[2]{\ensuremath{\frac{#1}{#2}}}
  \renewcommand{\arraystretch}{0.7}
  \begin{table}
    \begin{tabular}{|l|r@{}c@{}l|l|l|l|l|} \hline \footnotesize
      Assumptions &\multicolumn{3}{|l|}{\footnotesize  Claim} &\footnotesize   Why not 0 &\footnotesize   Why not 1 &\footnotesize   Why not 2 &\footnotesize   Why not 3 \\\hline \footnotesize
      $c(1)=c(3)=0, c(2) = 1, c(4) = 2$ &\footnotesize  $c(6)$&\footnotesize  ${}\in{}$&\footnotesize  $\{1,3\}$&\footnotesize  $3\cdot2 = 6$&\footnotesize  &\footnotesize  $4\cdot\myfrac{3}{2} = 6$&\footnotesize  \\ \hline \footnotesize
      $c(6) = 1$&\footnotesize  $c(8)$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $(3,1,8,3)$&\footnotesize  $6\cdot\myfrac{4}{3} = 8$&\footnotesize  $4\cdot2 = 8$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em} $c(8) = 3$&\footnotesize  $c(12)$&\footnotesize  ${}\in{}$&\footnotesize  $\{0,2\}$&\footnotesize  &\footnotesize  $6\cdot2 = 12$&\footnotesize  &\footnotesize  $8\cdot\myfrac{3}{2} = 12$\\ \hline \footnotesize
      \hspace{1em}\tab $c(12) = 0$&\footnotesize  $c(16)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $12\cdot\myfrac{4}{3} = 16$&\footnotesize  $(6,2,16,6)$&\footnotesize  &\footnotesize  $8\cdot2 = 16$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em} $c(16) = 2$&\footnotesize  $c(9)$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $12\cdot\myfrac{3}{4} = 9$&\footnotesize  $6\cdot\myfrac{3}{2} = 9$&\footnotesize  $(16,16,4,9)$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab $c(9) = 3$&\footnotesize  $c(\myfrac{9}{2})$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $3\cdot\myfrac{3}{2} = \myfrac{9}{2}$&\footnotesize  $6\cdot\myfrac{3}{4} = \myfrac{9}{2}$&\footnotesize  &\footnotesize  $9\cdot\myfrac{1}{2} = \myfrac{9}{2}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{9}{2}) = 2$&\footnotesize  $c(10)$&\footnotesize  ${}\in{}$&\footnotesize  $\{1,3\}$&\footnotesize  $(1,1,10,3)$&\footnotesize  &\footnotesize  $(4,4,10,\myfrac{9}{2})$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(10) = 1$&\footnotesize  $c(24)$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $12\cdot2 = 24$&\footnotesize  $(10,6,24,10)$&\footnotesize  $16\cdot\myfrac{3}{2} = 24$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(24) = 3$&\footnotesize  $c(32)$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $(6,2,32,10)$&\footnotesize  $16\cdot2 = 32$&\footnotesize  $24\cdot\myfrac{4}{3} = 32$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(32) = 0$&\footnotesize  $c(11)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $(32,1,11,11)$&\footnotesize  $(2,11,11,6)$&\footnotesize  &\footnotesize  $(24,9,11,11)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(11) = 2$&\footnotesize  $c(5)$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $(12,3,5,5)$&\footnotesize  $10\cdot\myfrac{1}{2} = 5$&\footnotesize  $(11,\myfrac{9}{2},\myfrac{9}{2},5)$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(5) = 3$&\footnotesize  $c(\myfrac{11}{2})$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(1,\myfrac{11}{2},\myfrac{11}{2},3)$&\footnotesize  $(10,10,2,\myfrac{11}{2})$&\footnotesize  $11\cdot\myfrac{1}{2} = \myfrac{11}{2}$&\footnotesize  $(5,9,8,\myfrac{11}{2})$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(10) = 3$&\footnotesize  $c(24)$&\footnotesize  ${}={}$&\footnotesize  $1$&\footnotesize  $12\cdot2 = 24$&\footnotesize  &\footnotesize  $16\cdot\myfrac{3}{2} = 24$&\footnotesize  $(8,8,24,10)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(24) = 1$&\footnotesize  $c(7)$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $(24,2,2,7)$&\footnotesize  $(\myfrac{9}{2},\myfrac{9}{2},7,4)$&\footnotesize  $(10,10,8,7)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(7) = 0$&\footnotesize  $c(18)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $12\cdot\myfrac{3}{2} = 18$&\footnotesize  $24\cdot\myfrac{3}{4} = 18$&\footnotesize  &\footnotesize  $9\cdot2 = 18$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(18) = 2$&\footnotesize  $c(\myfrac{27}{2})$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(1,\myfrac{27}{2},\myfrac{27}{2},7)$&\footnotesize  $(24,24,6,\myfrac{27}{2})$&\footnotesize  $18\cdot\myfrac{3}{4} = \myfrac{27}{2}$&\footnotesize  $9\cdot\myfrac{3}{2} = \myfrac{27}{2}$\\ \hline \footnotesize
      \hspace{1em}\tab $c(12) = 2$&\footnotesize  $c(16)$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $(6,2,16,6)$&\footnotesize  $12\cdot\myfrac{4}{3} = 16$&\footnotesize  $8\cdot2 = 16$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em} $c(16) = 0$&\footnotesize  $c(5)$&\footnotesize  ${}\in{}$&\footnotesize  $\{1,3\}$&\footnotesize  $(16,3,1,5)$&\footnotesize  &\footnotesize  $(12,4,4,5)$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab $c(5) = 1$&\footnotesize  $c(7)$&\footnotesize  ${}\in{}$&\footnotesize  $\{0,3\}$&\footnotesize  &\footnotesize  $(6,7,7,5)$&\footnotesize  $(12,12,4,7)$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(7) = 0$&\footnotesize  $c(9)$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $(16,3,9,7)$&\footnotesize  $6\cdot\myfrac{3}{2} = 9$&\footnotesize  $12\cdot\myfrac{3}{4} = 9$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(9) = 3$&\footnotesize  $c(14)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $7\cdot2 = 14$&\footnotesize  $(5,5,14,6)$&\footnotesize  &\footnotesize  $(9,9,14,8)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(14) = 2$&\footnotesize  $c(20)$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(7,1,20,7)$&\footnotesize  $(2,2,20,6)$&\footnotesize  $(14,14,20,12)$&\footnotesize  $(8,8,20,9)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(7) = 3$&\footnotesize  $c(10)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $(1,1,10,3)$&\footnotesize  $5\cdot2 = 10$&\footnotesize  &\footnotesize  $(8,10,10,7)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(10) = 2$&\footnotesize  $c(14)$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $(5,5,14,6)$&\footnotesize  $(12,14,14,10)$&\footnotesize  $7\cdot2 = 14$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(14) = 0$&\footnotesize  $c(20)$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $(16,20,20,14)$&\footnotesize  $(2,2,20,6)$&\footnotesize  $10\cdot2 = 20$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(20) = 3$&\footnotesize  $c(9)$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $6\cdot\myfrac{3}{2} = 9$&\footnotesize  $12\cdot\myfrac{3}{4} = 9$&\footnotesize  $(20,7,9,9)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(9) = 0$&\footnotesize  $c(13)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $(9,14,13,9)$&\footnotesize  $(5,6,13,6)$&\footnotesize  &\footnotesize  $(7,8,13,7)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(13) = 2$&\footnotesize  $c(17)$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(16,3,17,9)$&\footnotesize  $(5,2,17,6)$&\footnotesize  $(13,10,17,10)$&\footnotesize  $(7,8,17,8)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab $c(5) = 3$&\footnotesize  $c(\myfrac{9}{2})$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $3\cdot\myfrac{3}{2} = \myfrac{9}{2}$&\footnotesize  $6\cdot\myfrac{3}{4} = \myfrac{9}{2}$&\footnotesize  &\footnotesize  $(5,5,8,\myfrac{9}{2})$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{9}{2}) = 2$&\footnotesize  $c(10)$&\footnotesize  ${}={}$&\footnotesize  $1$&\footnotesize  $(1,1,10,3)$&\footnotesize  &\footnotesize  $(4,4,10,\myfrac{9}{2})$&\footnotesize  $5\cdot2 = 10$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(10) = 1$&\footnotesize  $c(7)$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $(10,7,7,6)$&\footnotesize  $(\myfrac{9}{2},\myfrac{9}{2},7,4)$&\footnotesize  $(5,8,7,5)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(7) = 0$&\footnotesize  $c(\myfrac{15}{2})$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(7,7,16,\myfrac{15}{2})$&\footnotesize  $10\cdot\myfrac{3}{4} = \myfrac{15}{2}$&\footnotesize  $(\myfrac{9}{2},4,\myfrac{15}{2},4)$&\footnotesize  $5\cdot\myfrac{3}{2} = \myfrac{15}{2}$\\ \hline \footnotesize
      $c(6) = 3$&\footnotesize  $c(8)$&\footnotesize  ${}={}$&\footnotesize  $1$&\footnotesize  $(3,1,8,3)$&\footnotesize  &\footnotesize  $4\cdot2 = 8$&\footnotesize  $6\cdot\myfrac{4}{3} = 8$\\ \hline \footnotesize
      \hspace{1em} $c(8) = 1$&\footnotesize  $c(\myfrac{9}{2})$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $3\cdot\myfrac{3}{2} = \myfrac{9}{2}$&\footnotesize  $(8,8,2,\myfrac{9}{2})$&\footnotesize  &\footnotesize  $6\cdot\myfrac{3}{4} = \myfrac{9}{2}$\\ \hline \footnotesize
      \hspace{1em}\tab $c(\myfrac{9}{2}) = 2$&\footnotesize  $c(9)$&\footnotesize  ${}\in{}$&\footnotesize  $\{0,1\}$&\footnotesize  &\footnotesize  &\footnotesize  $\myfrac{9}{2}\cdot2 = 9$&\footnotesize  $6\cdot\myfrac{3}{2} = 9$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em} $c(9) = 0$&\footnotesize  $c(12)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $9\cdot\myfrac{4}{3} = 12$&\footnotesize  $8\cdot\myfrac{3}{2} = 12$&\footnotesize  &\footnotesize  $6\cdot2 = 12$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab $c(12) = 2$&\footnotesize  $c(\myfrac{3}{2})$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $3\cdot\myfrac{1}{2} = \myfrac{3}{2}$&\footnotesize  $2\cdot\myfrac{3}{4} = \myfrac{3}{2}$&\footnotesize  $(12,\myfrac{9}{2},\myfrac{3}{2},\myfrac{9}{2})$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{3}{2}) = 3$&\footnotesize  $c(\myfrac{9}{4})$&\footnotesize  ${}={}$&\footnotesize  $1$&\footnotesize  $3\cdot\myfrac{3}{4} = \myfrac{9}{4}$&\footnotesize  &\footnotesize  $\myfrac{9}{2}\cdot\myfrac{1}{2} = \myfrac{9}{4}$&\footnotesize  $\myfrac{3}{2}\cdot\myfrac{3}{2} = \myfrac{9}{4}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(\myfrac{9}{4}) = 1$&\footnotesize  $c(\myfrac{27}{8})$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $\myfrac{9}{4}\cdot\myfrac{3}{2} = \myfrac{27}{8}$&\footnotesize  $\myfrac{9}{2}\cdot\myfrac{3}{4} = \myfrac{27}{8}$&\footnotesize  $(\myfrac{3}{2},6,6,\myfrac{27}{8})$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{27}{8}) = 0$&\footnotesize  $c(\myfrac{9}{8})$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $(\myfrac{27}{8},9,\myfrac{9}{8},\myfrac{27}{8})$&\footnotesize  $\myfrac{9}{4}\cdot\myfrac{1}{2} = \myfrac{9}{8}$&\footnotesize  &\footnotesize  $\myfrac{3}{2}\cdot\myfrac{3}{4} = \myfrac{9}{8}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(\myfrac{9}{8}) = 2$&\footnotesize  $c(\myfrac{3}{4})$&\footnotesize  ${}={}$&\footnotesize  $1$&\footnotesize  $1\cdot\myfrac{3}{4} = \myfrac{3}{4}$&\footnotesize  &\footnotesize  $\myfrac{9}{8}\cdot\myfrac{2}{3} = \myfrac{3}{4}$&\footnotesize  $\myfrac{3}{2}\cdot\myfrac{1}{2} = \myfrac{3}{4}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{3}{4}) = 1$&\footnotesize  $c(\myfrac{15}{2})$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $(3,3,\myfrac{15}{2},\myfrac{27}{8})$&\footnotesize  $(\myfrac{3}{4},\myfrac{3}{4},\myfrac{15}{2},\myfrac{9}{4})$&\footnotesize  $(\myfrac{9}{2},4,\myfrac{15}{2},4)$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(\myfrac{15}{2}) = 3$&\footnotesize  $c(\myfrac{21}{4})$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(\myfrac{27}{8},\myfrac{27}{8},\myfrac{21}{4},3)$&\footnotesize  $(\myfrac{3}{4},2,\myfrac{21}{4},2)$&\footnotesize  $(12,\myfrac{9}{2},\myfrac{9}{2},\myfrac{21}{4})$&\footnotesize  $(\myfrac{15}{2},\myfrac{15}{2},6,\myfrac{21}{4})$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em} $c(9) = 1$&\footnotesize  $c(10)$&\footnotesize  ${}\in{}$&\footnotesize  $\{1,3\}$&\footnotesize  $(1,1,10,3)$&\footnotesize  &\footnotesize  $(4,4,10,\myfrac{9}{2})$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab $c(10) = 1$&\footnotesize  $c(7)$&\footnotesize  ${}\in{}$&\footnotesize  $\{0,3\}$&\footnotesize  &\footnotesize  $(10,10,8,7)$&\footnotesize  $(\myfrac{9}{2},\myfrac{9}{2},7,4)$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(7) = 0$&\footnotesize  $c(\myfrac{8}{3})$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $(7,1,\myfrac{8}{3},\myfrac{8}{3})$&\footnotesize  $2\cdot\myfrac{4}{3} = \myfrac{8}{3}$&\footnotesize  $4\cdot\myfrac{2}{3} = \myfrac{8}{3}$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(\myfrac{8}{3}) = 3$&\footnotesize  $c(\myfrac{4}{3})$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $1\cdot\myfrac{4}{3} = \myfrac{4}{3}$&\footnotesize  $2\cdot\myfrac{2}{3} = \myfrac{4}{3}$&\footnotesize  &\footnotesize  $\myfrac{8}{3}\cdot\myfrac{1}{2} = \myfrac{4}{3}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{4}{3}) = 2$&\footnotesize  $c(\myfrac{16}{3})$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $8\cdot\myfrac{2}{3} = \myfrac{16}{3}$&\footnotesize  $4\cdot\myfrac{4}{3} = \myfrac{16}{3}$&\footnotesize  $\myfrac{8}{3}\cdot2 = \myfrac{16}{3}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(\myfrac{16}{3}) = 0$&\footnotesize  $c(\myfrac{32}{3})$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $\myfrac{16}{3}\cdot2 = \myfrac{32}{3}$&\footnotesize  $8\cdot\myfrac{4}{3} = \myfrac{32}{3}$&\footnotesize  $(\myfrac{4}{3},4,\myfrac{32}{3},4)$&\footnotesize  $(\myfrac{8}{3},\myfrac{32}{3},\myfrac{32}{3},6)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(7) = 3$&\footnotesize  $c(5)$&\footnotesize  ${}\in{}$&\footnotesize  $\{0,2\}$&\footnotesize  &\footnotesize  $10\cdot\myfrac{1}{2} = 5$&\footnotesize  &\footnotesize  $(7,7,6,5)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(5) = 0$&\footnotesize  $c(12)$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $(5,3,12,5)$&\footnotesize  $9\cdot\myfrac{4}{3} = 12$&\footnotesize  &\footnotesize  $6\cdot2 = 12$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(12) = 2$&\footnotesize  $c(16)$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(3,1,16,5)$&\footnotesize  $8\cdot2 = 16$&\footnotesize  $12\cdot\myfrac{4}{3} = 16$&\footnotesize  $(6,6,16,7)$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(5) = 2$&\footnotesize  $c(12)$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $9\cdot\myfrac{4}{3} = 12$&\footnotesize  $(4,4,12,5)$&\footnotesize  $6\cdot2 = 12$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(12) = 0$&\footnotesize  $c(\myfrac{7}{2})$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $(12,1,1,\myfrac{7}{2})$&\footnotesize  $(10,2,2,\myfrac{7}{2})$&\footnotesize  $(5,5,4,\myfrac{7}{2})$&\footnotesize  $7\cdot\myfrac{1}{2} = \myfrac{7}{2}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab $c(10) = 3$&\footnotesize  $c(\myfrac{16}{3})$&\footnotesize  ${}={}$&\footnotesize  $0$&\footnotesize  &\footnotesize  $8\cdot\myfrac{2}{3} = \myfrac{16}{3}$&\footnotesize  $4\cdot\myfrac{4}{3} = \myfrac{16}{3}$&\footnotesize  $(10,6,\myfrac{16}{3},\myfrac{16}{3})$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{16}{3}) = 0$&\footnotesize  $c(\myfrac{8}{3})$&\footnotesize  ${}={}$&\footnotesize  $3$&\footnotesize  $\myfrac{16}{3}\cdot\myfrac{1}{2} = \myfrac{8}{3}$&\footnotesize  $2\cdot\myfrac{4}{3} = \myfrac{8}{3}$&\footnotesize  $4\cdot\myfrac{2}{3} = \myfrac{8}{3}$&\footnotesize  \\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab $c(\myfrac{8}{3}) = 3$&\footnotesize  $c(\myfrac{4}{3})$&\footnotesize  ${}={}$&\footnotesize  $2$&\footnotesize  $1\cdot\myfrac{4}{3} = \myfrac{4}{3}$&\footnotesize  $2\cdot\myfrac{2}{3} = \myfrac{4}{3}$&\footnotesize  &\footnotesize  $\myfrac{8}{3}\cdot\myfrac{1}{2} = \myfrac{4}{3}$\\ \hline \footnotesize
      \hspace{1em}\tab\hspace{1em}\tab\hspace{1em}\tab\hspace{1em} $c(\myfrac{4}{3}) = 2$&\footnotesize  $c(\myfrac{32}{3})$&\footnotesize  !?&\footnotesize  $ $&\footnotesize  $\myfrac{16}{3}\cdot2 = \myfrac{32}{3}$&\footnotesize  $8\cdot\myfrac{4}{3} = \myfrac{32}{3}$&\footnotesize  $(\myfrac{4}{3},4,\myfrac{32}{3},4)$&\footnotesize  $(\myfrac{8}{3},\myfrac{32}{3},\myfrac{32}{3},6)$\\ \hline
    \end{tabular}
    \caption{The computer-generated proof of Lemma~\ref{lemma:forb2}.}
    \label{table:proof}
  \end{table}
  \normalsize

  We present a computer-generated proof of the rest of this lemma in
  Table~\ref{table:proof}.  Without loss of generality, assume that our
  set of colors is $\{0,1,2,3\}$ and that $c(1)=c(3)=0$, $c(2)=1$, and
  $c(4)=2$; we must derive a contradiction.

  We describe briefly how to read Table~\ref{table:proof}.  In the
  left-most column, the assumptions that we make are listed.  They are
  structured in a tree-like fashion, reflecting the trial-and-error
  argument.  The next column lists the current claim.  If this claim is
  ``$c(x)$!?'', then this means that $c(x)$ cannot be colored with any
  color and a contradiction has been obtained (thus we must backtrack);
  if the claim is $c(x) = a$, then this will be the next assumption.  If
  the claim is $c(x) \in \{a,b\}$, then we must consider the cases $c(x)
  = a$ and $c(x) = b$ separately.  Finally, the last four columns
  describe why $c(x) \ne 0$, \emph{etc.}, if this is needed to support
  the claim. An equation of the form $y\cdot r = x$ means that the
  forbidden ratio $r$ forbids $x$ and $y$ from being the same color; a
  $4$-tuple $(x_1, x_2, x_3, x_4)$ means that if we colored $x$ the
  color in question, then this would be a monochromatic solution to $x_1
  + x_2 + x_3 = 4 x_4$.

  For example, the first line of the proof can be read as follows:
  ``$c(6)$ is either $1$ or $3$ because of forbidden ratios from $3$ and
  $4$; we consider these possibilities separately.''
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:forb3}.]
  For the sake of contradiction, assume that $c(x)=c(4x)$ for some $x$.
  As in the previous lemma, we will use forbidden ratios; here, they are
  $2$, $\frac 32$, $\frac 43$, and $3$.  To begin, note that $c(x) =
  c(4x)$, $c(2x)$, $c(3x)$, and $c(6x)$ must be different colors by
  these forbidden ratios.  Now, $c(\frac 32x) = c(6x)$ by forbidden
  ratios from $x$, $2x$, and $3x$.  It follows that $c(\frac 94x) =
  c(2x)$ by forbidden ratios from $3x$ and $\frac 32x$ as well as the
  solution $(4x,4x,x,\frac 94x)$.  Then $c(\frac 92x) = c(x)$ by
  forbidden ratios from $\frac 94x$, $3x$, and $\frac 32x$.  Next,
  $c(9x) = c(2x)$ by forbidden ratios from $\frac 92x$, $3x$, and $6x$.
  Further, $c(12x)=c(3x)$ by forbidden ratios from $4x$, $9x$, and $6x$.
  Also, $c(8x) = c(2x)$ by forbidden ratios from $4x$, $12x$, and $6x$.
  Nearing the end, $c(\frac 12x) = c(3x)$ by forbidden ratios from $x$
  and $\frac 32x$ as well as the solution $(8x,\frac 12x,\frac 12x,
  \frac94x)$.  Finally, there are no possibilities left for $c(\frac
  34x)$ by forbidden ratios from $x$, $\frac 94x$, $\frac 12x$, and
  $\frac 32x$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:235-2}.]
By induction, it suffices to prove that $c(x)=c(6x)=c(16x)$ for every
nonzero rational number $x$. We have $x$, $2x$, $3x$, and $4x$ are all
different colors by the previous three lemmas. Also, $2x$, $3x$, $4x$,
and $6x$ are all different colors by Lemma~\ref{lemma:forb1} and
Lemma~\ref{lemma:forb2}. Hence, $c(x)=c(6x)$. By the previous three
lemmas, the numbers $2x$, $4x$, $6x$, and $8x$ are all different
colors. Hence, $c(8x)=c(3x)$. By the previous three lemmas, $3x$, $4x$,
$6x$, and $12x$ are all different colors, so $c(12x)=c(2x)$. By the
previous three lemmas, $4x$, $8x$, $12x$, and $16x$ are all different
colors, so $c(16x)=c(x)$, completing the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:-4}.]
Since $c(x)=c(6x)$ and $(-x,-x,6x,x)$ is a solution to
$x_1+x_2+x_3=4x_4$, then $-x$ and $x$ are different colors. Since
$c(2x)=c(\frac{3}{4}x)$ and $(-x,2x,2x,\frac{3}{4}x)$ is a solution to
$x_1+x_2+x_3=4x_4$, then $-x$ and $2x$ are different colors.

By Lemma~\ref{lemma:235-2}, $5x$, $10x$, $15x$, and $20x$ are all
different colors, so $c(3x) \in \{5x,\penalty0 10x,\penalty0
15x,\penalty0 20x\}$. By Lemma~\ref{lemma:235-2}, we have
$c(3x)=c(8x)=c(18x)=c(\frac{27}{4}x)$. Since $(5x,-x,8x,3x)$,
$(10x,-x,3x,3x)$, $(15x,-x,18x,8x)$, and $(20x,-x,8x,\frac{27}{4}x)$ are
solutions to $x_1+x_2+x_3=4x_4$, then $-x$ and $3x$ are different
colors. Hence, $c(-x)=c(4x)$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma:mod5}.]
By Lemma~\ref{lemma:235-2}, for integers $m$ and $n$ and nonzero
rational number $x$, we have $c(2^{m}3^{n}x)=c(x)$ if and only if
$w_5(2^m3^n)\equiv 1 \pmod{5}$. By Lemma~\ref{lemma:-4}, we have
$c(-x)=c(4x)$ for all nonzero rational numbers $x$. By induction, it
suffices to prove for every positive integer $p$ that is not a multiple
of $5$, we have, $c(px)=c(dx)$ for $d \in \{1,2,3,4\}$ if and only if $p
\equiv d \pmod{5}$. By Lemma~\ref{lemma:iff3multiple2}, we have already
established this for $p=2^m3^n$ and $n$ and $m$ are integers.

The proof is by induction on the size of $p$. Suppose $p>5$ is an
integer that is not a multiple of $2$, $3$, or $5$. We write $p=10a+b$,
where $a$ and $b$ are nonnegative integers and $b \in \{1,3,7,9\}$. The
induction hypothesis is that, for $d \in \{1,2,3,4\}$, we have
$c(p'q)=c(dq)$ for every nonzero rational number $q$ and positive
integer $p'$ such that $p' \equiv d \pmod{5}$, $p'<p$, and $p'$ is not a
multiple of $2$, $3$, or $5$.

By Lemma~\ref{lemma:235-2}, the four colors $c(5x), c(10x), c(15x),
c(20x)$ are distinct for every nonzero rational number $x$. Hence,
$c(x), c(2x), c(3x), c(4x) \in \{c(5x), c(10x), c(15x), c(20x)\}$ for
every nonzero rational number $x$.

\textbf{Case 1:} $p=10a+1$ with $a \geq 1$.

We have $c(3x)=c(8x)$.  Since $0<5a+6<p$, then by the induction
hypothesis, we have $c(3x)=c((\frac{5a+6}{2})x)$. Since
$(px,3x,8x,(\frac{5a+6}{2})x)$ is a solution to $x_1+x_2+x_3=4x_4$, then
$px$ and $3x$ are different colors.

We have $c(14x)=c(4x)$. Since $0<5a+4<p$, then by the induction
hypothesis, we have $c(4x)=c((5a+4)x)$. Since $(px,px,14x,(5a+4)x)$ is a
solution to $x_1+x_2+x_3=4x_4$, then $px$ and $4x$ are different colors.

By the induction hypothesis, Lemma~\ref{lemma:235-2}, and
Lemma~\ref{lemma:mod5}, we have $c(-3x)=c(2x)=c(7x)=c(12x)$. Since
$0<5a+9<p$ for $a>1$ and $\frac{5a+9}{2}=7<11$ for $a=1$, then by the
induction hypothesis, we have $c((\frac{5a+9}{2})x)=c(2x)$. Since
$(5x,12x,px,(\frac{5a+9}{2})x)$, $(10x,7x,px, (\frac{5a+9}{2})x)$,
$(15x,2x,px,(\frac{5a+9}{2})x)$, $(20x,-3x,px,(\frac{5a+9}{2})x)$ are
solutions to $x_1+x_2+x_3=4x_4$ and $c(2x) \in
\{c(5x),c(10x),c(15x),c(20x)\}$, then $px$ and $2x$ are different
colors. Hence, $c(px)=c(x)$.

\textbf{Case 2:} $p=10a+3$ with $a \geq 1$.

Since $0<5a+2<p$, then by the induction hypothesis, we have $(5a+2)x$
and $2x$ are the same color. Since $(px,px,2x,(5a+2)x)$ is a solution to
$x_1+x_2+x_3=4x_4$, then $px$ and $2x$ are different colors.

Since $p=10a+3 \geq 13$, then by the induction hypothesis, we have
$c(4x)=c(9x)=c(14x)$. For $a=1$, we have $p=13$, and $(13x,14x,9x,9x)$
is a solution to $x_1+x_2+x_3=4x_4$, so $13x$ and $4x$ are different
colors. For $a>1$, we have $0<5a+8 < p$ , so by the induction
hypothesis, we have $(\frac{5a+8}{2})x$ and $4x$ are the same
color. Since $(px,9x,4x,(\frac{5a+8}{2})x)$ is a solution to
$x_1+x_2+x_3=4x_4$, then $px$ and $4x$ are different colors.

Since $p=10a+3 \geq 13$ and $0<5a+7<p$, then by the induction
hypothesis, we have
$c(x)=c(6x)=c(\frac{7}{2}x)=c(\frac{72}{7}x)=c((\frac{5a+7}{2})x)$,
$c(\frac{5}{7}x)=c(15x)$, and $c(\frac{15}{2}x)=c(20x)$. Since the
tuples $(5x,px,6x,(\frac{5a+7}{2})x)$, $(10x,px,x,(\frac{5a+7}{2})x)$,
$(\frac{5}{7}x,px,\frac{72}{7}x,(\frac{5a+7}{2})x)$, and
$(\frac{15}{2}x,px,\frac{7}{2}x,(\frac{5a+7}{2})x)$ are solutions to
$x_1+x_2+x_3=4x_4$, and $c(x) \in \{c(5x),\penalty0 c(10x),\penalty0
c(15x),\penalty0 c(20x)\}$, then $px$ and $x$ are different
colors. Hence, $c(px)=c(3x)$.

\textbf{Case 3:} $p=10a+7$ with $a \geq 0$.

We have $c(-4x)=c(x)$. Since $0<5a+2<10a+7$, then by the induction
hypothesis, we have $c((\frac{5a+2}{2})x)=c(x)$. Since
$(px,-4x,x,(\frac{5a+2}{2})x)$ is a solution to $x_1+x_2+x_3=4x_4$, then
$px$ and $x$ are different colors.

We have $c(-2x)=c(3x)$. Since $0<5a+3<10a+7$, then by the induction
hypothesis, we have $c(5a+3)x)=c(3x)$. Since $(px,px,-2x,(5a+3)x)$ is a
solution to $x_1+x_2+x_3=4x_4$, then $px$ and $3x$ are different colors.

Note that $c(-16x)=c(-6x)=c(-\frac{9}{4}x)=c(-\frac{8}{3}x)=c(4x)$,
$c(\frac{5}{3}x)=c(10x)$, and $c(\frac{5}{4}x)=c(20x)$. Since
$0<5a+3<10a+7$, then by the induction hypothesis, we have
$(\frac{5a+3}{2})x$ and $4x$ are the same color. Since
$(5x,px,-6x,(\frac{5a+3}{2})x)$,
$(\frac{5}{3}x,px,-\frac{8}{3}x,(\frac{5a+3}{2})x)$,
$(15x,px,-16x,(\frac{5a+3}{2})x)$, and
$(\frac{5}{4}x,px,-\frac{9}{4}x,(\frac{5a+3}{2})x)$ are solutions to
$x_1+x_2+x_3=4x_4$ and $c(4x) \in \{c(5x),\penalty0 c(10x),\penalty0
c(15x),\penalty0 c(20x)\}$, then $px$ and $4x$ are different colors.
Hence, $c(px)=c(2x)$.

\textbf{Case 4:} $p=10a+9$ with $a \geq 0$.

We have $c(x)=c(6x)$. Since $0<5a+6<p$, then by the induction
hypothesis, we have $(5a+6)x$ and $x$ are the same color. Since
$(px,px,6x,(5a+6)x)$ is a solution to $x_1+x_2+x_3=4x_4$, then $px$ and
$x$ are different colors.

We have $c(-3x)=c(2x)$. Since $0<5a+4<p$, then by the induction
hypothesis, we have $(\frac{5a+4}{2})x$ and $2x$ are the same
color. Since $(px,2x,-3x,(\frac{5a+4}{2})x)$ is a solution to
$x_1+x_2+x_3=4x_4$, then $px$ and $2x$ are different colors.

Since $p=10a+9 \geq 19$, then by the induction hypothesis, we have
$c(3)=c(-2)=c(-7)=c(-12)=c(-17)$. Since $0<5a+6<p$, then
$(\frac{5a+6}{2})x$ and $3x$ are the same color. Since
$(5x,px,-2x,(\frac{5a+6}{2})x)$, $(10x,px,-7x,(\frac{5a+6}{2})x)$,
$(15x,px,-12x,(\frac{5a+6}{2})x)$, and $(20x,px,-17x,(\frac{5a+6}{2})x)$
are solutions to $x_1+x_2+x_3=4x_4$, then $px$ and $3x$ are different
colors.  Hence, $px$ and $4x$ are the same color.
\end{proof}

\section{Minimal colorings of the nonzero reals}
\label{section:real} Let $A=\{a_1,\dotsc,a_n\}$ be a nonempty finite
multiset of nonzero real numbers. For example, $\{2,3\}$ and $\{2,2,3\}$
are considered to be different multisets. We define a coloring $c$ to be
\emph{strongly free} of monochromatic solutions to $a_1x_1+\dotsb+a_nx_n
= 0$ if for every nonempty subset $A' \subseteq A$, the coloring $c$ has
no monochromatic solutions to the equation $E(A')$ defined by
\[ E(A')\colon \sum_{a_i \in A'} a_ix_i=0. \]

\begin{proposition}
Suppose $c\colon \mathbb{Q}\setminus \{0\}\rightarrow \{1,\dotsc,r\}$ is
an $r$-coloring of the nonzero rational numbers with $r \not =1$ a
positive integer, and $c$ is \emph{strongly free} of monochromatic
solutions to $a_1x_1+\dotsb+a_nx_n=0$. Then in ZFC, there exists
$2^{2^{\aleph_0}}$ $r$-colorings of the nonzero real numbers that are
strongly free of monochromatic solution to $a_1x_1+\dotsb+a_nx_n=0$.
\end{proposition}
\begin{proof}
In ZFC, every vector space has a basis. Viewing $\mathbb{R}$ as a
$\mathbb{Q}$-vector space, there is a well-ordered basis
$B=\{b_j\}_{j<2^{\omega}}$ for $\mathbb{R}$ as a $\mathbb{Q}$-vector
space. So every real number $x$ has a unique representation as
$x=\sum_{j<2^{\omega}} q_jb_j$, where each $q_j$ is rational and $q_j
\ne 0$ for only finitely many $j$. For a nonzero real number $x$, let
$j(x)$ be the least ordinal $j$ such that $q_j$ is nonzero. For a finite
set $S$ of nonzero real numbers, let $j(S)=\min_{x \in S}j(x)$.

Let $\Pi=(\pi_j)_{j<2^{\omega}}$ be any sequence of permutations $\pi_j$
of the set $\{1,\dotsc,r\}$ with $\pi_0$ being the identity
permutation. Define the coloring $C_{\Pi}$ by
$C_{\Pi}(x)=\pi_{j(x)}(c(q_{j(x)}))$.

Suppose $x_{1},\dotsc,x_{n}$ are nonzero real numbers and $A' \subseteq
A$ is a nonempty subset such that $\sum_{a_i \in A'} a_ix_i=0$. Let
$q_{i,j}$ be the coefficient of $b_j$ in the representation of
$x_i$. Since $\sum_{a_i \in A'} a_ix_i=0$, then $\sum_{a_i \in A'}
a_iq_{i,j(A')}=0$. Letting $A'' \subset A'$ be those $a_i \in A'$ such
that $q_{i.j(A')} \ne 0$, we see that $A''$ is a nonempty subset of $A$
and $\sum_{a_i \in A''} a_iq_{i,j(A')}=0$. Since
$C_{\Pi}(x_i)=\pi_{j(A')}(c(q_{i,j(A')}))$ for $a_i \in A''$ and $c$ is
strongly free of monochromatic solutions to $a_1x_1+\dotsb+a_nx_n=0$,
then the set $\{x_i\}_{a_i \in A''}$ is not monochromatic and $C_{\Pi}$
is strongly free of monochromatic solutions to
$a_1x_1+\dotsb+a_nx_n=0$. Since there are $2^{2^{\aleph_0}}$
nonisomorphic $r$-colorings of the form $C_{\Pi}$ and there are a total
of $2^{2^{\aleph_0}}$ nonisomorphic $r$-colorings of the real numbers,
then there are $2^{2^{\aleph_0}}$ colorings of the nonzero real numbers
that are strongly free of monochromatic solutions to
$a_1x_1+\dotsb+a_nx_n=0$.
\end{proof}

For $p$ a prime number and $n \geq 3$, each of the colorings $c_{p,n}$
is strongly free of monochromatic solutions to $E(p,n)$, so there are
$2^{2^{\aleph_0}}$ different $n$-colorings of the nonzero real numbers
without a monochromatic solution to $E(p,n)$.  In general, define the
$n$-coloring $c_{p,v,n}\colon \mathbb{Q}\setminus\{0\} \rightarrow
\{0,\dotsc,n-1\}$ by $c_{p,v,n}(x)\equiv \bigl \lfloor \frac{v_p(x)}{v}
\bigr \rfloor \pmod{n}$.  Notice that if $q$ is a nonzero rational
number and $v_p(q) \ne 0$, then $c_{p,v,n}$ with $v=v_p(q)$ is strongly
free of monochromatic solutions to $E(q,n)$.

We now turn our attention to the ZF+LM system of axioms. We prove
Lemma~\ref{lemma:lm} using the following multiplicative version of a
theorem of Steinhaus \cite{St}. For a set $A$ of real numbers, define
$A/A=\{a/a':a,a' \in A, a' \ne 0\}$.

\begin{theorem}[Steinhaus's Theorem]
If $A$ is a set of real numbers with positive Lebesgue measure, then
$A/A$ contains an entire interval $(1-\epsilon,1+\epsilon)$ for some
$\epsilon>0$.
\end{theorem}

\begin{proof}[Proof of Lemma~\ref{lemma:lm}]
Suppose for contradiction that all color classes of $c$ are Lebesgue
measurable. Since $c$ partitions $\mathbb{R}$ into countably many color
classes, and Lebesgue measure is countably additive, then at least one
color class $C$ has positive Lebesgue measure. Since $x,da^mb^nx$ are
different colors for all integers $m$ and $n$, then $C/C$ and
$\{da^mb^n:m,n \in \mathbb{Z}\}$ are disjoint sets. Since $\log_a b$ is
irrational, then the set $\{da^mb^n:m,n \in \mathbb{Z}\}$ contains
numbers arbitrarily close to $1$. But by Steinhaus's theorem, $C/C$
contains an entire interval around $1$, so $C/C$ and $\{da^mb^n:m,b \in
\mathbb{Z}\}$ are not disjoint, a contradiction.
\end{proof}

\section{Acknowledgements}
The authors would like to thank Steven Butler, Daniel Kleitman, Rados
Radoi\v ci\'c, and Richard Stanley for interesting discussions
concerning the paper.

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redpajama_set_name: RedPajamaArXiv


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Jogja is mostly known as the City of Gudeg, but the truth is, in addition to the food dominated with savory and sweet taste, Jogja has a million of foods that you simply cannot resist. One of the legendary foods in the City of Gudeg is soto. You can even find one stall that has been selling soto since half a century ago.
Going through the history of soto, this one particular food is so full of culture and taste. As we all know, there are soto Madura, soto Lamongan, soto Kudus, soto Betawi, and many others. Furthermore, soto is actually a product of acculturation between Indonesian, Chinese, and even Indian culture. The book "Nusa Jawa: Silang Budaya" mentioned that soto originally came from China. In the beginning it was called caudo, then tauto, saoto, and finally sroto or soto to fit Indonesian tongue. The use of turmeric, one of the many empon-empon (a Javanese term used to refer to the various spices used in cooking) as an ingredient is assumed to follow Indian recipes. On the other hand, Indonesian culture's contribution is the addition of local spices, which differ from one region to another, and thus resulting in a large variety of soto. You can enjoy some of them when visiting Jogja.
If you head west from Wirobrajan intersection, you can taste one legendary soto in Jogja known with the name Soto Kadipiro. It has been running its business since 1921, and nowadays it is recognized as an icon of soto in Jogja. Soto Kadipiro is located in the center of the city, making it easy to find. What makes it more exceptional is because we can order a beverage named Limun Sarsaparilla (Sarsaparilla Lemonade), a rare Jogjanese soda beverage, for dessert. Check the address and the map of the location on Soto Kadipiro.
There is another soto widely known amongst food hunters in Jogja. Soto Pak Marto Taman Sari, located 200 meter to the west of Plengkung Taman Sari, never lacks customers. The main appeal of this stall, established by Mr. Martodimedjo in 1960, is the babat sapi (tripe-edible offals from the stomachs of various farm animals) cooked using sweet seasoning similar to the one used in making bacem (a way of cooking using a lot amount of sugar, or salt, water) foods. Check the address and the map of the location on Soto Pak Marto Taman Sari.
People would be surprised when they hear the name (the word sampah is literally translated as 'garbage' or 'junk') for the first time because it is not a word normally used to call a food. Going against the horrid name, Soto Sampah, which is located near Tugu Jogja (Jogja's iconic historical monument), is worth to try. To you who are wondering about the name, here is the answer: it is named Soto Sampah because instead of using chicken meat or beef, like people normally do, they use gajih or cow's fat. Furthermore, you can add any available side dishes into the mix, turning the food's appearance into something similar to a pile of 'junk'. Check the adress and the map of the location on Soto Sampah.
Some paintings, wood carvings, and scraps of car parts are scattered on some corners, both inside and outside. Nobody would have thought that the building, appear more like an artist's workshop, is in fact a soto stall. This stall serving soto Blitar has a very unique name. People know this place by the name Soto Djiancuk. The owner came from East Java and he deliberately used the swear word (djiancuk is a swear word normally used by the native people of East Java) to make it clear that he is selling East Java-style soto. Rumor has it that the taste of the soto here often provokes people to curse because it is so tasty and will be sold out in just a blink of the eye. Check the adress and the map of the location on Soto Djiancuk.
Enjoying fried chicken with basil is very common, but what about eating soto sprinkled with basil? That's exactly what this soto stall in Kemasan is offering. The exact name of the stall is Soto Kemasan. In addition to having basil sprinkled on the food, the stall, established in 1963, also mixes slices of tahu bacem (tofu cooked using bacem method) to replace the more commonly used meat. It is a very unique mixture that will surely attract food lovers' curiosity. Check the address and the map of the location on Soto Kemasan.
The monetary crisis striking Indonesia in the 90s became the philosophy behind the establishment of the soto stall, located north of Kotagede's ringroad intersection, Jogja. People identify this humble stall by the name Warung Soto Moneter. The signature dish of this stall is a bowl of beef soto with crystal clear broth. Soto Moneter also provides a unique sensation of enjoying soto with basil and lenthok (mashed fried cassava), as the side dish. Check the address and the map of the location on Soto Moneter.
Betawinese (Derived from Batavia, the old name of Jakarta; Betawinese is the native people of Jakarta) Soto is not as popular as soto Madura, soto Lamongan, and soto Kudus. Still, Soto Betawi Bang H. Pitung is one of the best places to use as reference if you're looking for a place to eat soto in Jogja. Soto Betawi is distinctly known with its coconut milk broth, which is thick and yellowish in color. The manner of serving is extraordinary, using a small aluminium stove with a candle to keep the ordered soto warm. Check the address and the map of the location on Soto Betawi Bang H. Pitung.
Many say that Soto Pak Sholeh Al-Barokah is a soto that has the most 'Jogjanese taste' due to the mixture of savory and sweet taste, a combination found in most Jogjanese foods. The process of cooking, which uses traditional charcoal stove instead of gas stove, makes the taste of this soto more exceptional. Back in 1952, it was sold door to door, but now it has turned into a very famous soto in Jogja. Check the address and the map of the location on Soto Pak Sholeh Al-Barokah.
Not far from the site of Sambisari Temple, there are several bamboo huts looking so conspicuous amongst the rice fields. That is the exact place where we can enjoy the tastiness of Saoto Bathok Mbah Katro. Saoto. is a term normally used to refer to soto in the region of Solo (as you have read in the beginning of this article, soto has a lot of names; in Makassar, for example, soto is called coto). The combination of remotely located compound far from the noisy town and the lovely rural atmosphere has made Saoto Bathok Mbah Katro the best option for breakfast in Kalasan. As the name suggests, Saoto Bathok Mbah Katro is served in a bathok-a bowl made out of coconut shell. Read the full review in Saoto Bathok Mbah Katro.
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