\magnification=\magstephalf \voffset=30pt \hoffset=10pt \hsize=6.25 true in \vsize=9.25 true in \font\smallit=cmti9 \font\smalltt=cmtt9 \font\smallrm=cmr8 \font\bbld=msbm10 \font\bigbbld=msbm10 scaled \magstep1 \font\bigmath=cmsy10 scaled \magstep1 \font\Bigmath=cmsy10 scaled \magstep3 \font\bigrm=cmr10 scaled \magstep1 \font\bigbf=cmbx10 scaled \magstep1 \font\bigit=cmti10 scaled \magstep1 \font\smallbf=cmbx8 \font\smallit=cmti8 \font\smallrm=cmr8 \font\subbld=msbm8 \font\goth=eufm10 \font\subgoth=eufm8 \font\smc=cmcsc10%Large and small caps \font\eightrm=cmr8 \font\eightsy=cmsy8 \font\eighti=cmmi8 \font\sixrm=cmr6 \font\sixsy=cmsy6 \font\sixi=cmmi6 \font\eightit=cmti8 \font\eighttt=cmtt8 \font\eightbf=cmbx8 \font\eightsl=cmsl8 \font\sixbf=cmbx6 \font\fivebf=cmbx5 \def\eightpoint{\def\rm{\fam0\eightrm}% \textfont0=\eightrm \scriptfont0=\sixrm \scriptscriptfont0=\fiverm \textfont1=\eighti \scriptfont1=\sixi \scriptscriptfont1=\fivei \textfont2=\eightsy \scriptfont2=\sixsy \scriptscriptfont2=\fivesy \textfont\itfam=\eightit \def\it{\fam\itfam\eightit}% \textfont\slfam=\eightsl \def\sl{\fam\slfam\eightsl}% \textfont\ttfam=\eighttt \def\tt{\fam\ttfam\eighttt}% \textfont\bffam=\eightbf \scriptfont\bffam=\sixbf \scriptscriptfont\bffam=\fivebf \def\bf{\fam\bffam\eightbf}% \normalbaselineskip=9pt \setbox\strutbox=\hbox{\vrule height7pt depth2pt width0pt}% \let\sc=\sixrm \let\big=\eightbig \normalbaselines\rm} %End copied section \def\abstract#1{\bigskip\noindent\hskip .5 true in \vbox{\eightpoint \hsize 5.3 true in \noindent{\bf Abstract}. #1}} \def \subgc{\hbox{\subgoth c}} \def \bignearrow{\hbox{\Bigmath \char'45}} \def \bigsearrow{\hbox{\Bigmath \char'46}} \def \bignwarrow{\hbox{\Bigmath \char'55}} \def \bigswarrow{\hbox{\Bigmath \char'56}} \def \biguparrow{\hbox{\Bigmath \char'42}} \def \bigdownarrow{\hbox{\Bigmath \char'43}} \def \bigrightarrow{\hbox{\Bigmath \char'0 \hskip -4 pt\char'41}} \def \bigleftarrow{\hbox{\Bigmath \char'40\hskip -4 pt\char'0}} \def \cchi{\raise 2 pt \hbox{$\chi$}} \def \today{\ifcase\month\or January\or February\or March\or April\or May\or June\or July\or August\or September\or October\or November\or December\fi \ \number\day, \number\year} \def \gc{\hbox{\goth c}} \def \iff{if and only if } \def \ben{\hbox{\bbld N}} \def \ber{\hbox{\bbld R}} \def \beq{\hbox{\bbld Q}} \def \beqp{\hbox{$\beq^p$}} \def \bk{\backslash} \def \dbrk{\discretionary{}{}{}} \def \con{\hbox{${}^{\frown}$}} \def \bet{\hbox{\bbld T}} \def \bew{\hbox{\bbld W}} \def \bez{\hbox{\bbld Z}} \def \betaZ{\hbox{$\beta\!${\bbld Z}}} \def \betaS{\hbox{$\beta S$}} \def \subbea{\hbox{\subbld A}} \def \subbeb{\hbox{\subbld B}} \def \subbec{\hbox{\subbld C}} \def \subbed{\hbox{\subbld D}} \def \subbee{\hbox{\subbld E}} \def \subbef{\hbox{\subbld F}} \def \subbeg{\hbox{\subbld G}} \def \subbeh{\hbox{\subbld H}} \def \subbei{\hbox{\subbld I}} \def \subbej{\hbox{\subbld J}} \def \subbek{\hbox{\subbld K}} \def \subbel{\hbox{\subbld L}} \def \subbem{\hbox{\subbld M}} \def \subben{\hbox{\subbld N}} \def \subbeo{\hbox{\subbld O}} \def \subbep{\hbox{\subbld P}} \def \subbeq{\hbox{\subbld Q}} \def \subber{\hbox{\subbld R}} \def \subbes{\hbox{\subbld S}} \def \subbet{\hbox{\subbld T}} \def \subbeu{\hbox{\subbld U}} \def \subbev{\hbox{\subbld V}} \def \subbew{\hbox{\subbld W}} \def \subbex{\hbox{\subbld X}} \def \subbey{\hbox{\subbld Y}} \def \subbez{\hbox{\subbld Z}} \def \nstr{\hbox{$\hbox{\bbld N}^*$}} \def \betaN{\hbox{$\beta${\bbld N}}} \def \limit{\displaystyle\lim} \def \bigcuptxt{\textstyle\bigcup} \def \bigcaptxt{\textstyle\bigcap} \def\limst #1#2{\displaystyle \lim_{\hbox{$\buildrel{\hbox{${}_{#1}$}}\over{\hbox{${}_{#2}$}}$}}} %The following symbols (\Bigcup, \Bigcap,\Prod,\Sum) take two arguments. %The first argument is a subscript, which is placed below the symbol. %The second argument is the number of points right to shift the symbol. %For a subscript like {n\in H}, 5 is about right while for the %subscript {n\in H\cup K} 12 is about right. \def \Bigcup #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\textstyle\bigcup$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \Bigcap #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\textstyle\bigcap$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \Prod #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\Pi$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \Sum #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\Sigma$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \subsetneq{\hbox{\ \vtop{\hbox{$\subseteq$}\vskip-15pt \moveright2pt\hbox{${}_{{}_{\not}}$}}\ }} \def %\proclaim #1. #2\par{\medbreak\noindent% \proclaim #1. #2\par{\noindent% {\bf #1}.\enspace{\it #2\par}\medbreak} \def \definition #1. {\noindent{\bf#1}.\enspace} \def \section #1{\bigbreak\centerline{\bigbf #1}\nobreak\bigskip} \def \nhat #1{\{1,2,\discretionary{}{}{}\ldots,\discretionary{}{}{}#1\}} \def \ohat #1{\{0,1,\discretionary{}{}{}\ldots,\discretionary{}{}{}#1\}} %The following produces a labelled line. It takes two arguments. %The first of these is a label and the second is the line. \def \labline #1#2{\vskip 8 pt\line{\vrule height 0 pt depth 0 pt \hskip 20 pt #1\hfill #2\hfill}\vskip 8 pt\noindent\hskip 0 pt}%For some weird reason, the last is needed %to keep from having a small indentation. \def \proof{\medbreak\noindent{\it Proof}. } \def \qed{\hfill\vrule height 3pt depth -3 pt\nobreak \hfill \hbox{\vbox{\hrule width6pt height 6.4pt depth -6pt \hrule width.4pt depth6pt\hrule width6pt}\vbox{\hrule width.4pt height7pt}}\medskip} \def \qeddis{\vskip -22 pt\qed} %The following should be used instead of \qeddis. %Type the display without $$ ending with paragraph break \def \endproofdisplay #1\par{ \vskip 6pt plus 3pt minus 1pt \vbox{$$ #1 $$ \vskip -13 pt\qed} \medbreak} \def \bigtimes{\hbox{\bigmath \char'2}} \def \seqx{\hbox{$\langle x_n\rangle_{n=1}^\infty$}} \def \seqz{\hbox{$\langle z_n\rangle_{n=1}^\infty$}} \def \seqy{\hbox{$\langle y_n\rangle_{n=1}^\infty$}} \def \cl{c\ell} \def \FS{\hbox{$FS(\langle x_n\rangle_{n=1}^\infty)$}} \def \supp{\hbox{\rm supp}} \def \leqr{\hbox{$\leq_R$}} \def \leql{\hbox{$\leq_L$}} \def \mod{\hbox{\rm mod }} \def \scra{\hbox{${\cal A}$}} \def \scrb{\hbox{${\cal B}$}} \def \scrc{\hbox{${\cal C}$}} \def \scrd{\hbox{${\cal D}$}} \def \scre{\hbox{${\cal E}$}} \def \scrf{\hbox{${\cal F}$}} \def \scrg{\hbox{${\cal G}$}} \def \scrh{\hbox{${\cal H}$}} \def \scri{\hbox{${\cal I}$}} \def \scrj{\hbox{${\cal J}$}} \def \scrk{\hbox{${\cal K}$}} \def \scrl{\hbox{${\cal L}$}} \def \scrm{\hbox{${\cal M}$}} \def \scrn{\hbox{${\cal N}$}} \def \scro{\hbox{${\cal O}$}} \def \scrp{\hbox{${\cal P}$}} \def \scrq{\hbox{${\cal Q}$}} \def \scrr{\hbox{${\cal R}$}} \def \scrs{\hbox{${\cal S}$}} \def \scrt{\hbox{${\cal T}$}} \def \scru{\hbox{${\cal U}$}} \def \scrv{\hbox{${\cal V}$}} \def \scrw{\hbox{${\cal W}$}} \def \scrx{\hbox{${\cal X}$}} \def \scry{\hbox{${\cal Y}$}} \def \scrz{\hbox{${\cal Z}$}} \def \pf{\hbox{${\cal P}_f$}} \def \pfa{\hbox{${\cal P}_f({\cal A})$}} \def \pfn{\hbox{${\cal P}_f(\ben)$}} \def \pfb{\hbox{${\cal P}_f({\cal B})$}} \def \emp{\emptyset} \def \wtau{\widetilde\tau} \def \plim{\hbox{$p$-lim}} \def \nodiv{\hbox{$\not\hskip -.4 pt|$\hskip 2 pt}} \def \start{\hbox{\rm start}} \def \endd{\hbox{\rm end}} \newcount\refno \refno=0 \def \refsetup#1.{\advance\refno by 1\edef#1{\number\refno}} \refsetup\refBH. \refsetup\refDHLL. \refsetup\refF. \refsetup\refH. \refsetup\refHS. \refsetup\refM. \def \bignearrow{\hbox{\Bigmath \char'45}} \def \bigsearrow{\hbox{\Bigmath \char'46}} \def \bignwarrow{\hbox{\Bigmath \char'55}} \def \bigswarrow{\hbox{\Bigmath \char'56}} \def \biguparrow{\hbox{\Bigmath \char'42}} \def \bigdownarrow{\hbox{\Bigmath \char'43}} \def \bigrightarrow{\hbox{\Bigmath \char'0 \hskip -4 pt\char'41}} \def \bigleftarrow{\hbox{\Bigmath \char'40\hskip -4 pt\char'0}} \def \cchi{\raise 2 pt \hbox{$\chi$}} \def \nhat #1{\{1,2,\discretionary{}{}{}\ldots,\discretionary{}{}{}#1\}} \def \seqx{\hbox{$\langle x_n\rangle_{n-1}^\infty$}} \def \seqy{\hbox{$\langle y_n\rangle_{n-1}^\infty$}} \def \seqz{\hbox{$\langle z_n\rangle_{n-1}^\infty$}} \def \gc{\hbox{\goth c}} \def \iff{if and only if } \def \ben{\hbox{\bbld N}} \def \ber{\hbox{\bbld R}} \def \beq{\hbox{\bbld Q}} \def \beqp{\hbox{$\beq^p$}} \def \bk{\backslash} \def \dbrk{\discretionary{}{}{}} \def \con{\hbox{${}^{\frown}$}} \def \bet{\hbox{\bbld T}} \def \bew{\hbox{\bbld W}} \def \bez{\hbox{\bbld Z}} \def \betaZ{\hbox{$\beta\!${\bbld Z}}} \def \betaS{\hbox{$\beta S$}} \def \subbea{\hbox{\subbld A}} \def \subbeb{\hbox{\subbld B}} \def \subbec{\hbox{\subbld C}} \def \subbed{\hbox{\subbld D}} \def \subbee{\hbox{\subbld E}} \def \subbef{\hbox{\subbld F}} \def \subbeg{\hbox{\subbld G}} \def \subbeh{\hbox{\subbld H}} \def \subbei{\hbox{\subbld I}} \def \subbej{\hbox{\subbld J}} \def \subbek{\hbox{\subbld K}} \def \subbel{\hbox{\subbld L}} \def \subbem{\hbox{\subbld M}} \def \subben{\hbox{\subbld N}} \def \subbeo{\hbox{\subbld O}} \def \subbep{\hbox{\subbld P}} \def \subbeq{\hbox{\subbld Q}} \def \subber{\hbox{\subbld R}} \def \subbes{\hbox{\subbld S}} \def \subbet{\hbox{\subbld T}} \def \subbeu{\hbox{\subbld U}} \def \subbev{\hbox{\subbld V}} \def \subbew{\hbox{\subbld W}} \def \subbex{\hbox{\subbld X}} \def \subbey{\hbox{\subbld Y}} \def \subbez{\hbox{\subbld Z}} \def \nstr{\hbox{$\hbox{\bbld N}^*$}} \def \betaN{\hbox{$\beta${\bbld N}}} \def \limit{\displaystyle\lim} \def \bigcuptxt{\textstyle\bigcup} \def \bigcaptxt{\textstyle\bigcap} \def\limst #1#2{\displaystyle \lim_{\hbox{$\buildrel{\hbox{${}_{#1}$}}\over{\hbox{${}_{#2}$}}$}}} \def \Bigcup #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\textstyle\bigcup$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \Bigcap #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\textstyle\bigcap$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \Prod #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\Pi$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \Sum #1#2{\hbox{\vtop{\moveright#2pt\hbox{$\Sigma$} \vskip-8pt\hbox{${}_{#1}$}}}} \def \subsetneq{\hbox{\ \vtop{\hbox{$\subseteq$}\vskip-15pt \moveright2pt\hbox{${}_{{}_{\not}}$}}\ }} \def \labline #1#2{\vskip 8 pt\line{\vrule height 0 pt depth 0 pt \hskip 20 pt #1\hfill #2\hfill}\vskip 8 pt\noindent\hskip 0 pt} %For some weird reason, the last is needed %to keep from having a small indentation. \def \proof{\medbreak\noindent{\it Proof}. } \def \qed{\hfill\vrule height 3pt depth -3 pt\nobreak \hfill \hbox{\vbox{\hrule width6pt height 6.4pt depth -6pt \hrule width.4pt depth6pt\hrule width6pt}\vbox{\hrule width.4pt height7pt}}\medskip} \def \qeddis{\vskip -22 pt\qed} %\def \bigtimes{\hbox{\bigmath \char'2}} %\def \seqx{\hbox{$\langle x_n\rangle_{n=1}^\infty$}} %\def \seqz{\hbox{$\langle z_n\rangle_{n=1}^\infty$}} %\def \seqy{\hbox{$\langle y_n\rangle_{n=1}^\infty$}} %\def \cl{c\ell} %\def \FS{\hbox{$FS(\langle x_n\rangle_{n=1}^\infty)$}} %\def \supp{\hbox{\rm supp}} %% Here is the macro definition \newcount\refno \refno=0 \def \refsetup#1.{\advance\refno by 1\edef#1{\number\refno}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% one sets up \refA by: \refsetup\refA. Note the period. \refsetup\refBH. \refsetup\refDHLL. \refsetup\refF. \refsetup\refH. \refsetup\refHS. \refsetup\refM. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%The reference list follows. They are printed at the end by "\references" %%One needs a closing } at the end of the list. \def \references{\vskip 20pt \line{\bf References \hfil}\nobreak \baselineskip=5pt \vskip 10pt \item{[\refBH]} {V.\ Bergelson and N.\ Hindman,} {\it Additive and multiplicative Ramsey Theorems in \ben ~--~ some elementary results}, Comb.\ Prob.\ and Comp.\ {\bf 2} (1993), 221-241. \vskip -5pt \item{[\refDHLL]} W.\ Deuber, N.\ Hindman, I.\ Leader, and H.\ Lefmann, {\it Infinite partition regular matrices\/}, Combinatorica {\bf 15} (1995), 333-355. \vskip -5pt \item{[\refF]} L. Fuchs, {\it Abelian groups\/}, Pergamon Press, Oxford, 1960. \vskip -5pt \item{[\refH]} N.\ Hindman, {\it Partitions and sums and products of integers\/}, Trans.\ Amer.\ Math.\ Soc.\ {\bf 247} (1979), 227-245. \vskip -5pt \item{[\refHS]} N.\ Hindman and D.\ Strauss, {\it Algebra in the Stone-\v Cech compactification -- theory and applications\/}, Walter de Gruyter, Berlin, 1998. \vskip -5pt \item{[\refM]} A.\ Maleki, {\it Solving equations in $\beta\ben$\/}, Semigroup Forum, to appear.} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%This is the theorem numbering section. Theorems, definitions, etc %%in Section 1 are labeled \tha* where * is a name begining with a %%capital letter. They should be entered in the list below %% in the same order as they appear in the text as %%\thasetup\tha*. Again note the period. %%Similar remarks apply to Section 2 (\th*), Section 3 (\tht*), %%Section 4(\thf*) and Section 5(thv*). (If longer than that, be %%inventive!) %%Note that if a space is wanted after the number %%it must be forced by \ . \newcount\thano \thano=0 \def \thasetup#1.{\advance\thano by 1\edef#1{1.\number\thano}} \newcount\thno \thno=0 \def \thsetup#1.{\advance\thno by 1\edef#1{2.\number\thno}} \newcount\thtno \thtno=0 \def \thtsetup#1.{\advance\thtno by 1\edef#1{3.\number\thtno}} \newcount\thfno \thfno=0 \def \thfsetup#1.{\advance\thfno by 1\edef#1{4.\number\thfno}} \newcount\thvno \thvno=0 \def \thvsetup#1.{\advance\thvno by 1\edef#1{5.\number\thvno}} %%Do the theorem number setups below here. \thasetup\thaA. \thasetup\thaB. \thasetup\thaBB. \thasetup\thaC. \thasetup\thaD. \thasetup\thaE. \thasetup\thaF. \thasetup\thaG. \thasetup\thaH. \thasetup\thaI. \thasetup\thaJ. \thasetup\thaK. \thsetup\thA. \thsetup\thB. \thsetup\thC. \thsetup\thD. \thsetup\thE. \thsetup\thF. \thsetup\thG. \thsetup\thH. \thsetup\thI. \thsetup\thJ. \thsetup\thK. \thtsetup\thtAA. \thtsetup\thtAAA. \thtsetup\thtAAB. \thtsetup\thtAB. \thtsetup\thtA. \thtsetup\thtB. \thtsetup\thtC. \thtsetup\thtD. \thtsetup\thtH. \thtsetup\thtI. \thtsetup\thtN. \thtsetup\thtO. \thtsetup\thtP. \thtsetup\thtQ. \thtsetup\thtR. \thtsetup\thtS. \thtsetup\thtT. \thtsetup\thtMAIN. \thtsetup\thtW. \thtsetup\thtU. \thtsetup\thtV. \thtsetup\thtX. \thfsetup\thfA. \thfsetup\thfB. \thfsetup\thfC. \thfsetup\thfD. \thfsetup\thfMAIN. \thfsetup\thfE. \thfsetup\thfF. \thfsetup\thfG. \thfsetup\thfH. \thfsetup\thfI. \thfsetup\thfJ. \thfsetup\thfK. \thvsetup\thvA. \thvsetup\thvB. \thvsetup\thvC. \thvsetup\thvD. \thvsetup\thvE. \thvsetup\thvF. \thvsetup\thvG. \thvsetup\thvH. \thvsetup\thvI. \thvsetup\thvJ. \thvsetup\thvK. %End of theorem numbering section %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\apsum{a_1\cdot p+a_2\cdot p+\ldots+a_n\cdot p} \def\bpsum{b_1\cdot p+b_2\cdot p+\ldots+b_m\cdot p} \def\varlist{a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_m} \def\asum{a_1+a_2+\ldots+a_n} \def\bsum{b_1+b_2+\ldots+b_m} %The following definitions are added to %change the presentation in Section 3 to %something less clear, but more pleasing %to Dona. \def \oplus{+} \def \odot{\cdot} \parskip=10pt \parindent=0pt \headline={\ifnum\pageno<2 \hfil \else \smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 0 (2000), \#A02 \hfil \rm \folio \fi} \footline={\hfil} \centerline{\bf LINEAR EQUATIONS IN THE STONE-\v CECH COMPACTIFICATION OF \hbox{\bbld N}} \vskip 30pt \centerline{\bf Neil Hindman} \centerline{\smallit Department of Mathematics, Howard University, Washington, D.C. 20059, U.S.A.} \centerline{ \tt nhindman@howard.edu} \vskip 10pt \centerline{\bf Amir Maleki} \centerline{\smallit Department of Mathematics, Howard University, Washington, D.C. 20059, U.S.A.} \centerline{ \tt amaleki@howard.edu} \vskip 10pt \centerline{\bf Dona Strauss} \centerline{\smallit Department of Pure Mathematics, University of Hull, Hull HU6 7RX, U.K.} \centerline{ \tt d.strauss@maths.hull.ac.uk} \vskip 40pt \centerline{\smallit Received: 5/7/99, Revised: 12/7/99, Accepted: 12/14/99, Published: 5/19/00} \vskip 30pt \centerline{\bf Abstract} {Let $a$ and $b$ be distinct positive integers. We show that the equation $u+a\cdot p=v+b\cdot p$ has no solutions with $u,v\in\beta \subben$ and $p\in\beta \subben\bk \subben$. More generally, we show that if $(S,+)$ is any commutative cancellative semigroup and $S$ has no nontrivial solutions to $n\cdot s=n\cdot t$ for $n\in\subben$ and $s,t\in S$, then the equation $u+a\cdot p=v+b\cdot p$ has no solutions with $u,v\in\beta S$ and $p\in\beta S\bk S$. We characterize completely the Abelian groups for which such an equation can be satisfied. We also show that if $S$ can be embedded in the circle group $\subbet$, then the equation $a\cdot p+u=b\cdot p+v$ has no solutions with $u,v\in\beta S$ and $p\in\beta S\bk S$. Finally, we investigate solutions to the equation $a_1\cdot p+a_2\cdot p+\ldots +a_n\cdot p=b_1\cdot p+b_2\cdot p+\ldots +b_m\cdot p$ where $p\in\beta\subben\bk\subben$ and $a_1,a_2,\ldots ,a_n,b_1,b_2,\ldots ,b_m\in\subben$. } \baselineskip=15pt \parindent=15pt \vskip 30pt \line{\bf 1. Introduction \hfil} \vskip 10pt \noindent Given any discrete semigroup $(S,+)$ there is a unique extension of the operation in $S$ to its Stone-\v Cech compactification $\beta S$ making $\beta S$ a right topological semigroup with $S$ contained in its topological center. (That is, for each $p\in\beta S$, the function $\rho_p:\beta S\to\beta S$ defined by $\rho_p(q)=q+p$ is continuous. And for each $s\in S$, the function $\lambda_s:\beta S\to\beta S$ defined by $\lambda_s(p)=s +p$ is continuous.) We are denoting the operation by $+$ because we shall be concerned in this paper almost exclusively with commutative semigroups $S$. In fact, we are primarily concerned with the semigroup $(\ben,+)$. However, the reader should be cautioned that $\beta S$ is hardly ever commutative. (See [\refHS, Theorem 4.27].) (We are writing $\ben$ for the set of positive integers and write $\omega =\ben\cup \{0\}$.) For almost 25 years, the algebra of $\beta \ben$ has been a powerful tool in Ramsey Theory, beginning with the Galvin-Glazer proof of the Finite Sums Theorem. (See the notes to Chapter 5 of [\refHS] for a discussion of the history of this proof.) Conversely, the fact that there is a finite partition of $\ben$ with the property that no cell contains a sequence with all of its pairwise (distinct) sums and all of its pairwise (distinct) products shows that the equation $p+p=p\cdot p$ has no solutions in $\beta\ben$. The fact that there is an idempotent $p$ of $(\beta\ben,\cdot)$ in the topological closure of $\{q\in\beta\ben:q+q=q\}$ provided the first [\refH], and for fifteen years the only, proof of the following fact: {\it Let $r\in\ben$ and let $\ben=\bigcup_{i=1}^rA_i$. Then there exist $i\in\nhat{r}$ and sequences $\seqx$ and $\seqy$ in $\ben$ such that $FS(\seqx)\cup FP(\seqy)\subseteq A_i$\/} where $FS(\seqx)=\{\sum_{n\in F}\,x_n:F$ is a finite nonempty subset of $\ben\}$ and $FP(\seqy)=\{\prod_{n\in F}\,y_n:F$ is a finite nonempty subset of $\ben\}$. An elementary proof of this fact was eventually provided in [\refBH]. Other results, equally easy when done in terms of $\beta \ben$, seem unlikely to be provided with elementary proofs any time soon. For example, $\{p\in\beta\ben:$ every $A\in p$ has positive upper density$\}$ is a compact subsemigroup of $(\beta\ben,+)$, and consequently has an idempotent. Therefore, whenever \ben\ is finitely colored, one may inductively choose a sequence $\langle x_t\rangle_{t=1}^\infty$, with $FS(\langle x_t\rangle_{t=1}^\infty)$ monochrome, in such a way that, having chosen $\langle x_t\rangle_{t=1}^n$, the set of choices for $x_{n+1}$ has positive upper density. An alternative description of this result in the terminology of game theory was suggested to us recently by Tomasz \L uczak. At the start of the game, $\ben$ is finitely colored. At the $n^{\hbox{\smallrm th}}$ turn, player one chooses a subset $A_n$ of $\ben$ with positive upper density and player two chooses any element $x_n\in A_n$. Player one wins provided $FS(\seqx)$ is monochrome. The theorem says that player one has a winning strategy. See Part III of [\refHS] for numerous other examples of the applications of the algebra of $\beta\ben$ to combinatorics, specifically Ramsey Theory. In this paper, much of the interaction between algebra and Ramsey Theory occurs in the reverse direction. That is, we use combinatorial results involving partitions of $\ben$, $\ber$, or $\bet$ to conclude that certain equations in $\beta\ben$ cannot be solved. We take the points of $\beta S$ to be the ultrafilters on $S$ and identify the principal ultrafilters with the points of $S$. The topology of $\beta S$ is defined by choosing the sets of the form $\overline A=\{p\in\beta S:A\in p\}$, where $A\subseteq S$, as a base for the open sets. The set $\overline A$ is clopen in this topology and is equal to $\hbox{\rm cl}_{\beta S}(A)$. Given $p,q\in\beta S$ and $A\subseteq S$, one has that $A\in p+q$ \iff$ \{x\in S:-x+A\in q\}\in p$, where $-x+A=\{s\in S:x+s\in A\}$. Alternatively, $p+q$ can be defined topologically as $\limst{s\to p}{s\in S}\limst{t\to q}{t\in S}(s+t)$. The following fact is easy to verify. Suppose that $p,q\in\beta S$, that $P\in p$ and that $Q_s\in q$ for each $s\in P$. Then $\bigcup_{s\in P}(s+Q_s)\in p+q$. We shall use this fact several times. See [\refHS] for an elementary introduction to the properties of the semigroup $\beta S$, as well as for any unfamiliar algebraic assertions encountered here. If $X$ is a completely regular space, any function $f$ from $S$ to $X$, has a continuous extension mapping $\beta S$ to $\beta X$, which we shall denote by $\widetilde f$. In this paper, we investigate solutions to certain ``linear'' equations. Before we talk about these, we need to clarify what we mean, for example, by $2\cdot p$ where $p\in\beta S\bk S$. (We do {\it not\/} mean $p+p$.) We take our motivation from the situation in $\beta\ben$. Here $2\cdot p$ is naturally interpreted as multiplication in the semigroup $(\beta\ben,\cdot )$. It is defined to be $\widetilde{l_2}(p)$ where $\widetilde{l_2}:\beta\ben\to\beta\ben$ is the continuous extension of $l_2:\ben\to\ben$ defined by $l_2(x)=2\cdot x$. It is characterized by the fact that for any $A\subseteq\ben$, $A\in 2\cdot p$ \iff$2^{-1}A\in p$ where $2^{-1}A=\{x\in\ben:2\cdot x\in A\}$. We take a similar approach in an arbitrary semigroup. \noindent {\bf Definition} \definition\thaA\ \hskip -5pt . Let $(S,+)$ be a semigroup and let $a\in\ben$. Define $l_a:S\to S$ by $l_a(s)=s+s+\ldots +s$ ($ a$ times) and let $\widetilde{l_a}:\beta S\to\beta S$ be the continuous extension of $ l_a$. For $p\in\beta S$ define $a\cdot p=\widetilde{l_a}(p)$. Note that if $p\in S$ then, as usual, $a\cdot p$ is the sum of $p$ with itself $a$ times. It is immediate from the definition that $a\cdot p=\displaystyle\limst{s\to p}{s\in S}a\cdot s$. \noindent {\bf Lemma} \proclaim\thaB\ \hskip -5pt . Let $(S,+)$ be a semigroup, let $a\in\ben$, let $p\in\beta S$, and let $A\subseteq S$. Then $A\in a\cdot p$ \iff$a^{-1}A\in p$, where $a^{-1}A=\{x\in S:a\cdot x\in A\}$. In particular, if $B\in p$, then $a\cdot B\in a\cdot p$. \proof Necessity. We have that $\overline A$ is a neighborhood of $\widetilde{l_a}(p)$ so pick $B\in p$ such that $\widetilde{l_a}[\overline B]\subseteq\overline A$. Then $B\subseteq a^{-1}A$. Sufficiency. Suppose that $A\notin a\cdot p$. Then $S\bk A\in a\cdot p$ and so, by the sufficiency applied to $S\bk A$, $a^{-1}(S\bk A)\in p$. Since $a^{-1}A\cap a^{-1}(S\bk A)=\emptyset$, this is a contradiction. \qed \noindent {\bf Lemma} \proclaim \thaBB\ \hskip -5pt . Let $(S,+)$ be a commutative semigroup, let $a\in\ben$, and let $p,q\in\beta S$. Then $a\cdot(p+q)=a\cdot p+a\cdot q$. \proof Since $S$ is commutative, $l_a:S\to S$ is a homomorphism and thus, by [\refHS, Corollary 4.22], so is $\widetilde{l_a}: \beta S\to\beta S$.\qed In Sections 2 and 3 we investigate the equations $a\cdot p+u=b\cdot p+v$ and $u+a\cdot p=v+b\cdot p$, where $u,v\in \beta S$, $p\in S^{*}=\beta S\bk S$, and $a,b\in\ben$ with $a\neq b$. In Section 2, we show that the equation $a\cdot p+u=b\cdot p+v$ has no solution for any semigroup $S$ which is embeddable in the unit circle $\bet$. In Section 3, we show that the equation $u+a\cdot p=v+b\cdot p$ has no solution (again for $p\in S^*$) for a wide class of commutative cancellative semigroups. We in fact characterize those Abelian groups $S$ for which this equation can hold as exactly those for which $\{s\in S:ab|b-a|\cdot s=0\}$ is finite. In Section 4 we turn our attention to a particular class of linear equations in $\beta\ben$. (These equations were in fact our original motivation for the equations studied in Sections 2 and 3.) Consider the following result. \noindent {\bf Theorem} \proclaim\thaC\ \hskip -5pt . Let $a_1,a_2,\ldots ,a_n,b_1,b_2,\ldots ,b_m\in\ben$ so that for any $i\in\nhat{n-1}$ and any $j\in\nhat{m-1}$, $a_i\neq a_{i+1}$ and $ b_j\neq b_{j+1}$. Let $p$ be an idempotent in $(\beta\ben,+)$. Suppose that $$a_1\cdot p+a_2\cdot p+\ldots +a_n\cdot p=b_1\cdot p+b_2\cdot p+ \ldots +b_m\cdot p\,.$$ Then $m=n$ and for all $i\in\nhat{n}$, $a_i=b_i$. \proof This is [\refM, Theorem 2.19], where it was derived as an easy consequence of [\refDHLL, Theorems 2.10 and 3.3].\qed Notice that, if $a\in\ben$ and $p=p+p$, then by [\refHS, Lemma 13.1] $a\cdot p=a\cdot (p+p)=a\cdot p+a\cdot p$. Consequently, the restriction on repeating coefficients in Theorem \thaC\ is necessary. We were led then to ask the following question, asking essentially whether the idempotent requirement can be omitted from Theorem \thaC. (Originally, it was a conjecture, but we have lost faith because of the difficulties that we have encountered.) \noindent {\bf Question} \proclaim\thaD\ \hskip -5pt . Let $a_1,a_2,\ldots ,a_n,b_1,b_2,\ldots ,b_m\in\ben$ so that for any $i\in\nhat{n-1}$ and any $j\in\nhat{m-1}$, $a_i\neq a_{i+1}$ and $ b_j\neq b_{j+1}$. Suppose that there exists some $p\in\ben^{*}$ such that $$a_1\cdot p+a_2\cdot p+\ldots +a_n\cdot p=b_1\cdot p+b_2\cdot p+ \ldots +b_m\cdot p\,.$$ Must it then be true that $m=n$ and for all $i\in\nhat{n}$, $a_i= b_i$? We investigate this question in Section 4, obtaining some partial results. \vskip 30pt \line{ \bf {2. The Equation} {\bigit a} {\bigmath\char'001} {\bigit p} {\bigrm+} {\bigit u} {\bigrm=} {\bigit b} {\bigmath\char'001} {\bigit p} {\bigrm+} {\bigit v} \hfil} \vskip 20pt \noindent Let $\bet_d$ be the circle group with the discrete topology (while $\bet$ denotes the circle group with its usual topology). We show here that if $a,b\in\ben$ with $a\neq b$, then the equation $a\cdot p+u=b\cdot p+v$ has no solution with $u,v\in\beta\bet_d$ and $p\in\bet_d^{*}$. As a consequence, if $S$ is a semigroup which can be algebraically embedded in $\bet$, such as the semigroup $(\ben,+)$ and the group $(\bez,+)$, then the equation $a\cdot p+u=b\cdot p+v$ has no solution with $u,v\in\beta S$ and $p\in S^{*}$. We take $\bet=\ber/\bez$ and let $\pi :\ber\to\bet$ be the usual projection. Notice that, since $\pi$ is a homomorphism, if $a\in\ben$ and $r\in\ber$, then $a\cdot\pi (r)=\pi (a\cdot r)$. We observe that $\pi$ is an open mapping and so are the maps $l_a$ and $l_b$, considered as maps from \bet\ to itself. We define $\gamma :\beta\bet_d\to\bet$ to be the continuous extension of the identity map from $\bet_d$ to \bet, and note that this is a homomorphism by [\refHS, Theorem 4.8]. For any $p\in\beta\bet_d$, $\gamma (p)$ is the point of \bet\ to which $p$ converges. To see this, suppose that $U$ is a neighborhood of $\gamma (p)$ in \bet. Then $U\in p$, because otherwise we should have $p\in\overline {\bet\setminus U}$ and hence $\gamma (p)\in\cl_{\subbet}\big(\gamma (\bet\setminus U)\big)=\cl_{\subbet}(\bet\setminus U)$. \noindent {\bf Lemma} \proclaim\thA\ \hskip -5pt . Let $a\in\ben$ and $q\in\beta\bet_d$. Then $a\cdot\gamma (q)=\gamma (a\cdot q)$. \proof Since $l_a:\bet\to\bet$, $\gamma :\beta\bet_d\to\bet$, and $\widetilde{ l_a}:\beta\bet_d\to\beta\bet_d$ are continuous functions, and $l_a\circ\gamma$ agrees with $\gamma\circ\widetilde{l_a}$ on $\bet$, it follows that $l_a\circ\gamma (q)=\gamma\circ\widetilde{l_a}(q)$. \qed \noindent {\bf Lemma} \proclaim\thB\ \hskip -5pt . Let $U$ be an open subset of $\bet$. If $q,r\in\beta\bet_d$ and $U\in q$, then $U+\gamma (r)\in q+r$. \proof For every $x\in U$, $-x+U+\gamma (r)$ is a neighborhood of $\gamma (r)$ and so $-x+U+\gamma (r)\in r$. Thus $\{x\in\bet:-x+U+\gamma (r)\in r\}\in q$. \qed \noindent {\bf Lemma} \proclaim\thC\ \hskip -5pt . Let $p\in\beta\bet_d$, let $a,b\in\ben$ with $a\neq b$, and let $V$ be an open subset of $\bet$ such that $a\cdot V\cap b\cdot V=\emptyset$. If there exist $u,v\in\beta\bet_d$ such that $a\cdot p+u=b\cdot p+v$, then $\gamma (p)+V\notin p$. \proof Put $p'=-\gamma (p)+p$, $u'=-\gamma (u)+u$ and $v'=-\gamma (v)+v$. Since $\gamma$ is a homomorphism, $\gamma(a\cdot p)+\gamma(u)= \gamma(b\cdot p)+\gamma(v)$. By Lemma \thA, $a\cdot\gamma (p)+\gamma (u)=b\cdot \gamma (p)+\gamma (v)$, and so $$\matrix{\hfill a\cdot p'+u'&= &a\cdot\big(-\gamma(p)\big)+a\cdot p+(-\gamma(u)+u)\hfill &\hbox{(by Lemma \thaBB)}\hfill\cr &=&-a\cdot\gamma(p)+\big(-\gamma(u)\big)+a\cdot p+u\hfill &\hbox{by [\refHS, Theorem 6.54]}\hfill\cr &=&-b\cdot\gamma(p)+\big(-\gamma(v)\big)+b\cdot p+v\hfill\cr &=&b\cdot p'+v'\,.\hfill\cr}$$ Notice that $\gamma (u')=\gamma (v') =0$. If $V\in p'$, then $a\cdot V\in a\cdot p'+u'$ and $b\cdot V\in b\cdot p'+v'$ (by Lemma \thB) -- a contradiction. \qed \noindent {\bf Lemma} \proclaim\thD\ \hskip -5pt . Let $a,b\in\ben$ with $a0$. Pick $m\in\bez$ such that $a\cdot r=b\cdot s+m$. Now $$00$ be a cardinal and let $\{T_{\iota}:\iota <\kappa \}$ be a set of semigroups such that for each $\iota <\kappa$, $\Psi (T_{\iota})$ and let $S=\bigoplus_{\iota <\kappa}T_{\iota}$. If each of the sets $$\{\iota <\kappa :\hbox{\rm \ there exist }s\neq t\hbox{\rm \ in } T_{\iota}\hbox{\rm \ such that either }a\cdot s=a\cdot t\hbox{\rm \ or } b\cdot s=b\cdot t\}$$ and $$\{\iota <\kappa :\hbox{\rm \ there exists }t\in T_{\iota}\bk\{0 \}\hbox{\rm \ such that }a\cdot t=b\cdot t\}$$ is finite, then $\Psi (S)$. \proof Suppose the theorem is false, and choose the smallest cardinal $\kappa$ for which a counterexample exists. We consider first the possibility that $\kappa <\omega$. For each $\iota <\kappa$, $\widetilde\pi_{\iota}$ is a homomorphism by [\refHS, Corollary 4.22]. So $\widetilde{\pi_{\iota}}(u)+a\cdot\widetilde{\pi_{\iota}}(p)=\widetilde{ \pi_{\iota}}(v)+b\cdot\widetilde{\pi_{\iota}}(p)$ (by Lemma \thtAAA). Since $\Psi (T_{\iota})$ holds, this is impossible if $\widetilde{\pi_{\iota}}(p)\in T_{\iota}^{*}$ and so one has some $y_{\iota}\in T_{\iota}$ such that $\{s\in S:\pi_{\iota}(s)=y_{\iota}\}\in p$. Let $y\in S$ such that for each $\iota <\kappa$, $\pi_{\iota}(y)=y_{\iota}$. But then, $\{y\}=\bigcap_{\iota <\kappa}\{s\in S:\pi_{\iota}(s)=y_{ \iota}\}\in p$, a contradiction. We thus assume that $\kappa\geq\omega$. Define $\psi :S\to\kappa$ by $\psi (0)=0$ and for $s\in S\bk\{0\}$, $\psi (s)=\max\{\iota <\kappa :\pi_{\iota}(s)\neq 0\}$. We claim that for each $\gamma <\kappa$, $\{s\in S:\psi (s)\geq\gamma \}\in p$. So suppose instead that we have some $\gamma <\kappa$ such that $\{s\in S:\psi (s)<\gamma \}\in p$. Let $f:S\to\bigoplus_{\iota <\gamma}T_{\iota}$ be the projection onto the first $\gamma$ coordinates and let $\widetilde f:\beta S\to\beta (\bigoplus_{\iota <\gamma}T_{\iota} )$ be its continuous extension. Then $\widetilde f(u)+a\cdot\widetilde f(p)=\widetilde f(u+a\cdot p)=\widetilde f(v+b\cdot p)=\widetilde f(v)+b\cdot\widetilde f(p)$, so by the induction hypothesis, $\widetilde f(p)$ is principal. So pick $x\in\bigoplus_{\iota <\gamma}T_{\iota}$ such that $\{x\}\in\widetilde f(p)$. Define $y\in S$ by $$y_\iota=\left\{\matrix{x_\iota&\hbox{if }\iota<\gamma\hfill\cr 0&\hbox{if }\iota\geq\gamma\,.}\right.$$ Pick $V\in p$ such that $\widetilde f[V]=\{x\}$. Then $\{y\}=V\cap\{s\in S:\psi(s)<\gamma\}\in p$, a contradiction. Thus $\{s\in S:\psi (s)\geq\gamma \}\in p$ for each $\gamma <\kappa$ as claimed. Pick $\gamma <\kappa$ such that, whenever $\gamma\leq\iota <\kappa$, we have $a\cdot s\neq a\cdot t$ and $b\cdot s\neq b\cdot t$ whenever $s$ and $t$ are distinct members of $T_{\iota}$ and $a\cdot t\neq b\cdot t$ whenever $t\in T_{\iota}\bk\{0\}$. (We can do this because the set of $\iota$'s violating these conditions is finite.) Define $g:S\to S$ by $$g(s)_\iota=\left\{\matrix{0&\hbox{if }\iota\neq\psi(s)\hfill\cr s_\iota&\hbox{if }\iota=\psi(s)\,.\hfill\cr}\right.$$ Let $P\in p$. We note that $\big\{s+a\cdot t:s\in S$, $t\in P$, and $\psi (t)>\max\{\psi (s),\gamma\}\big\}\in u+a\cdot p$ and $\big\{s'+b\cdot t':s'\in S$, $t'\in P$, and $\psi (t')>\max\{\psi (s'),\gamma\}\big\}\in v+b\cdot p$. Thus we have $s+a\cdot t=s'+b\cdot t'$ for some $s,s'\in S$ and $t,t'\in P$, with $\psi (t)>\max\{\psi (s),\gamma \}$ and $\psi (t' )>\max\{\psi (s'),\gamma \}$. Now $g(s+a\cdot t)=a\cdot g(t)$ and $g(s'+b\cdot t')=b\cdot g(t')$. So $a\cdot\widetilde g(p)=b\cdot\widetilde g(p).$ Now $l_a$ is injective on $g[\{s\in S:\psi (s)>\gamma \}]\in\widetilde g(p)$. It follows from Lemma \thtAAB\ that $\{s\in S:a\cdot s=b\cdot s\}\in\widetilde g(p)$ and hence that $\{s\in S:a\cdot g(s)=b\cdot g(s)\}\in p$. This is a contradiction, because $a\cdot g(s)\neq b\cdot g(s)$ if $\psi (s) >\gamma$. \qed We now need to establish that the statement $\Psi (T)$ of Theorem \thtAB\ holds for some specific groups. We shall use a real number $x$ to denote the element $\pi (x)$ of \bet\ in order to avoid equations in \bet\ that are too cumbersome. Of course, with this notation, the equation $x=y$ in \bet\ is equivalent to the relation $x\in y+\bez$ in \ber. \noindent {\bf Definition} \definition\thtA\ \hskip -5pt . Let $r$ be a prime number. Then $\bez_r^{\infty}=\displaystyle\{{m\over {r^n}}\in\bet:m\in\ben$ and $ n\in\omega \}$. The subgroups of \bet\ of this form are called quasicyclic. When using the notation $\beta \bez_r^{\infty}$, we shall assume that $\bez_r^{\infty}$ has the discrete topology. \noindent {\bf Definition} \definition\thtB\ \hskip -5pt . Let $r$ be a prime number. For each $s\in\bez_r^{\infty}$, we define $o(s)$ to be the order of $s$, that is the least $k\in\ben$ such that $k\cdot s=0$, and we put $\tau (s)=\log_r\big(o(s)\big)$. Let $\widetilde\tau :\beta \bez_r^{\infty}\to\beta\bez$ be the continuous extension of $\tau$. \noindent {\bf Lemma} \proclaim\thtC\ \hskip -5pt . Let $r$ be a prime, let $S=\bez_r^{\infty}$, let $p\in S^{*},$ let $u\in\beta S$ and let $ a\in\ben$. If $r^d$ is the highest power of $ $$r$ which divides $a$, then $\widetilde\tau (u+a\cdot p)=-d+\widetilde\tau (p)$. \proof We note that every $s\in S\setminus \{0\}$ can be expressed as $s={m\over {r^n}}$, where $m,n\in\ben$ and $(m,r)=1$. Then $o(s)= r^n$ and $\tau (s)=n$. It is not hard to show that, for any $s,t\in S$, $o(t)>o(s)$ implies that $o(s+t)=o(t)$, and that $o(t)>r^d$ implies that $o(a\cdot t)=r^{-d}o(t)$ and thus $\tau (a\cdot t)=-d+\tau (t)$. Now, for any $ s\in S$, $\big\{t\in S:o(t)>\max\{o(s),r^d\}\big\}\in p$, because this set has a finite complement in $S$. Thus $\widetilde\tau (u+a\cdot p)= \displaystyle\lim_{s\to u}\displaystyle\lim_{ t\to p}\tau (s+a\cdot t)=\displaystyle\lim_{t\to p}(-d+\tau (t))= -d+\widetilde\tau (p)$. \qed \noindent {\bf Lemma} \proclaim\thtD\ \hskip -5pt . Let $r$ be a prime, let $S=\bez_r^{\infty}$, let $p\in S^{*}$, let $u,v\in\beta S$, and let $a,b\in\ben$. If $u+a\cdot p=v+b\cdot p$, then the highest power of $r$ which divides $a$ is equal to the highest power of $r$ which divides $b$. \proof Let $r^d$ and $r^e$ denote the highest powers of $r$ which divide $a$ and $b$ respectively. By Lemma \thtC, $-d+\widetilde\tau (p)=-e+\widetilde\tau (p)$. This implies that $d=e$ by [\refHS, Lemma 6.28]. \qed \noindent {\bf Theorem} \proclaim\thtH\ \hskip -5pt . Let $r$ be a prime, let $S=\bez_r^{\infty}$, let $ a,b\in\ben$, let $u,v\in\beta S$, and let $p\in S^{*}$. If $a\neq b$, then $u +a\cdot p\neq v+b\cdot p$. \proof Suppose that $u+a\cdot p=v+b\cdot p$. Then, by Lemma \thtD, we can write $a=a_1r^d$ and $b=b_1r^d$, where $a_1,b_1\in\ben$, $d\in\omega$ and $(a_1,r)=(b_1,r)=1$. We choose $k\in\ben$ satisfying $r^k>\vert a_1-b_1\vert .$ Each $s\in S\bk \{0\}$ can be expressed as $\displaystyle{{{m(s)}\over {r^{n(s)}}}}$, where $m(s),n(s)\in\ben$ and $(m(s),r)=1$. Thus we have $\displaystyle s={{x(s)r^k+y(s)}\over {r^{n(s)}}}$, for some $x(s)\in\omega$ and some $y(s)\in \{1,2,\ldots ,r^k-1\}$ with $(y(s),r)=1$. We can choose $y\in \{1,2,\ldots ,r^k-1\}$, with $(y,r)=1$, and a set $P\in p$ such that $y(t)=y$ for every $t\in P$. Let $s\in S$. Then $\{t\in P:\tau (t)>k+d+\tau (s)\}\in p$. It follows that the set of elements of \bez$_r^{\infty}$ of the form $\displaystyle{{z\over {r^l}}}+a\cdot{{xr^k+y}\over {r^n}}$, with $x,z,k,l,n\in\omega$ and $n>k+d+l$, is a member of $u+a\cdot p$. The corresponding statement also holds for $v+b\cdot p$. Thus we have an equation in \bez$_r^{\infty}$ of the form: $${z\over {r^l}}+a\cdot{{xr^k+y}\over {r^n}}={{z'}\over {r^{l'}}} +b\cdot{{x'r^k+y}\over {r^{n'}}}\,,$$ in which all the symbols denote non-negative integers and $n>k+d+l$ and $n'>k+d+l'$. We observe that the left hand side of this equation has order $r^{n-d}$ and the right hand side has order $r^{n'-d}$, and so $n=n'$. Thus we have the following relation in \ber: $${z\over {r^l}}+a_1\cdot{{xr^k+y}\over {r^{n-d}}}\in{{z'}\over { r^{l'}}}+b_1\cdot{{x'r^k+y}\over {r^{n-d}}}+\bez.$$ Multiplying by $r^{n-d}$ shows that $a_1y\equiv b_1y$ (mod $r^k$) and hence that $a_1\equiv b_1$ (mod $r^k$). Since $\vert a_1-b_1\vert \displaystyle\max\left\{{{-x\cdot\alpha}\over { \alpha -1}},{x\over {\alpha -1}}\right\}+1$, then there is some $k\in \{-1,0,1\}$ such that $f(x+y)=f(y)+k$. To see this, assume first that $x\geq 0$. Then $y>\displaystyle{{x\over {\alpha -1}}}$ so $\alpha y>x+y$ and thus $\alpha^{f(y)}\leq y\leq y+x<\alpha y<\alpha^{f(y)+2}$. Next assume that $x<0$. Then $y>\displaystyle{{{-x\cdot\alpha}\over { \alpha -1}}}$ and thus $y+x>\displaystyle{{y\over {\alpha}}}$. Then $\alpha^{f(y)-1}\leq\displaystyle{{y\over {\alpha}}}r^ko(s)$, then $\tau (s+a\cdot t)= \tau (a\cdot t)=-k+\tau (t)$. Thus $\widetilde\tau (u+a\cdot p)= \displaystyle\lim_{s\to u}\displaystyle\lim_{ t\to p}\tau (s+a\cdot t)=-k+\widetilde\tau (p)$. Similarly, since $r\nodiv b$, $\widetilde\tau (v+bp)=\widetilde\tau (p)$. By [\refHS, Lemma 6.28], this is a contradiction. We may thus suppose that $r\nodiv a$ and $r\nodiv b$. This implies that, for every $s\in S$, $o(a\cdot s)=o(b\cdot s)=o(s)$. If $s\in S\bk \{0\}$, let $l(s)=\max\{\iota <\kappa :\pi_{\iota}(s)\neq 0\}$ and let $\mu(s)=\max\{\iota <\kappa :o\big(\pi_{\iota}(s)\big)=o(s)\}$. Define $g:S\to S$ by $g(0)=0$ and for $s\in S\bk\{0\}$ and $\iota<\kappa$, $$g(s)(\iota)=\left\{\matrix{ 0&\hbox{if }\iota\neq\mu(s)\hfill\cr \pi_{\mu(s)}(s)&\hbox{if }\iota=\mu(s)\,.\hfill\cr}\right.$$ We claim that $a\cdot \widetilde g(p)=b\cdot \widetilde g(p)$, so suppose instead that $a\cdot \widetilde g(p)\neq b\cdot \widetilde g(p)$ and pick $P\in p$ such that for all $t,t'\in P$, $a\cdot g(t)\neq b\cdot g(t')$. Notice that for each $\lambda<\kappa$, $\{t\in S:\mu(t)>\lambda\}\in p$. To see this, suppose instead that $\{t\in S:\mu(t)\leq \lambda\}\in p$. Let $\sigma$ denote the projection of $S$ onto $\bigoplus_{\iota\leq\lambda}\bez_r^\infty$. Since $\{t\in S:\mu(t)\leq \lambda\}\in p$, we have that $\{t\in S:o\big(\sigma(t)\big)=o(t)\}\in p$ and thus $\widetilde o\big(\widetilde\sigma(p)\big)=\widetilde o(p) \in\ben^*$, contradicting our inductive assumption. Consequently, $$\matrix{\{s+a\cdot t:s\in S,\,t\in P,\, \mu (t)>l(s),\hbox{ and }o(t)>o(s)\}\in u+a\cdot p\hbox{\rm \ and}\hfill \cr \{s'+b\cdot t':s'\in S,\,t'\in P, \,\mu (t')>l(s'),\hbox{ and }o(t'>o(s')\}\in v+b\cdot p\, .\hfill \cr} $$ Thus we may choose $s,s'\in S$ and $t,t'\in P$ with $\mu (t)>l(s)$, $o(t)>o(s)$, $\mu (t')>l(s')$, and $o(t')>o(s')$ such that $s+a\cdot t=s'+b\cdot t'$. Since $\mu (a\cdot t)=\mu (t)>l(s)$ and $o(t)>o(s)$, we have $\mu(s+a\cdot t)=\mu(a\cdot t)$ and $\pi_{\mu(a\cdot t)}(s+a\cdot t)=a\cdot\pi_{\mu(t)}(t)$ and so $g(s+a\cdot t)=a\cdot g(t)$ and similarly $g(s'+b\cdot t')=b\cdot g(t')$. Since $t,t'\in P$, this is a contradiction and thus $a\cdot \widetilde g(p)=b\cdot \widetilde g(p)$ as claimed. Since $l_a$ is injective on $S$, it follows from Lemma \thtAAB\ that $\{t\in S:a\cdot t=b\cdot t\}\in\widetilde g(p)$. Let $r^m$ be the largest power of $r$ which divides $b-a$. Now $\{t\in S:o(t)>r^m\}\in p$ and $o\big(g(t)\big)=o(t)$ for all $t\in S$. So $\{t\in S:o(t)>r^m\}\in\widetilde g(p)$. This is a contradiction, because if $o(t)>r^m$, then $a\cdot t\neq b\cdot t$.\qed \noindent {\bf Theorem} \proclaim\thtV\ \hskip -5pt . Let $S$ be an Abelian group. Suppose that $u+ap=v+bp$ for some $u,v\in\beta S$, some $p\in S^{*}$, and some $a\neq b$ in \ben. Then there exist $s\in S$ and $k\in\ben$ such that $\big(ab(b-a)\big)^kp\in S$. \proof By Lemmas \thtS\ and \thtT, there exist $s\in S$ such that $s+\bigoplus_{r\in R}S_r\in p$ and an embedding $h:\bigoplus_{r\in R}S_r\to D=\bigoplus_{r\in R}D_r$, where each $ D_r$ is a direct sum of copies of \bez$_r^{\infty}$. By [\refF, Theorem 16.1], $h$ extends to a homomorphism $h'$ from $S$ to $D$. Since $\pi_r\circ h'$ is a homomorphism, so is $\widetilde{\pi_r}\circ\widetilde{h'}$ by [\refHS, Corollary 4.22]. Thus $\widetilde{\pi_r}\circ\widetilde{h'}(u)+a\cdot\widetilde{\pi_r}\circ\widetilde{ h'}(p)=\widetilde{\pi_r}\circ\widetilde{h'}(v)+b\cdot\widetilde{\pi_ r}\circ\widetilde{h'}(p)$. So, by Lemma \thtU, for each $r\in R$, $\widetilde o\circ\widetilde{\pi_r}\circ\widetilde{h'}(p)\in\ben$ and hence $\widetilde o\circ\widetilde{\pi_r}\circ\widetilde{h'}(-s+p)\in\ben$. So $r^{n_r}\cdot\big(\widetilde{\pi_r}\circ\widetilde{h'}(-s+p)\big )=0$ for some $n_r\in\ben$. Let $n=\prod_{r\in R}r^{n_r}$. Then $\widetilde{h'}\big(n\cdot(-s+p)\big)=n\cdot \widetilde {h'}(- s+p)=0$. Now $h'$ is injective on $\bigoplus_{ r\in R}S_r$, so that (see [\refHS, Exercise 3.4.1]) $\widetilde{h'}$ is injective on $\cl(\bigoplus_{r\in R}S_r)$. Since $n\cdot (-s+p)\in\cl(\bigoplus_{r\in R}S_r)$ it follows that $n\cdot(-s+p)=0$. Our claim follows from the fact that we can choose $k\in\ben$ such that $\big(ab(b-a)\big)^k$ is a multiple of $n$. \qed Our results depended in an essential way on the assumption that $S$ is Abelian. However, our next corollary shows that we can obtain analogous results for some non-Abelian semigroups. If $S$ is an arbitrary semigroup and $a\in\ben$, we shall use $\pi_a:S\to S$ for the mapping $s\mapsto s^a$ and $\widetilde\pi_a:\beta S\to\beta S$ for its continuous extension. \noindent {\bf Corollary} \proclaim\thtX\ \hskip -5pt . Let $S$ denote the free semigroup on a finite set of generators. If $u,v\in\beta S$, $p\in S^{*}$, and $a,b\in\ben$ with $a\neq b$, then $u\widetilde\pi_a(p)\neq v\widetilde\pi_b(p)$. \proof For $s\in S$, let $\ell (s)$ be the length of $s$. We note that $\ell$ is a homomorphism and that $\ell \big(\pi_n(s)\big)=n\cdot\ell (s)$ for every $n\in\ben$ and every $s\in S$. Thus $\widetilde\ell :\beta S\to\beta\ben$ is a homomorphism by [\refHS, Corollary 4.22], and it follows from the continuity of the maps involved that $\widetilde\ell \big(\pi_n(x)\big)= n\cdot\widetilde\ell(x)$ for every $x\in\beta S$. So $\widetilde\ell (u)+a\cdot\widetilde\ell (p)=\widetilde\ell (v)+b \cdot\widetilde\ell (p)$. By Corollary \thtW, $\widetilde\ell (p)\in\ben$. However, if $\widetilde\ell (p)=n$, the fact that $\ell^{-1}[\{n\}]$ is finite implies that $p\in S$, a contradiction. \qed \vskip 30pt \line{\bf {4. The Equation} $\hbox{\bigit a}_{\hbox{\smallrm \hskip 1 pt 1}}$% {\bigmath\char'001}{\bigit p} {\bigrm +} $\hbox{\bigit a}_{\hbox{\smallrm \hskip 1 pt 2}}$% {\bigmath\char'001}{\bigit p} {\bigrm +} {\bigrm . . .} {\bigrm +} $\hbox{\bigit a}_{\hbox{\smallit \hskip 1 pt n}}$% {\bigmath\char'001}{\bigit p} {\rm=} $\hbox{\bigit b}_{\hbox{\smallrm \hskip 1 pt 1}}$% {\bigmath\char'001}{\bigit p} {\bigrm +} $\hbox{\bigit b}_{\hbox{\smallrm \hskip 1 pt 2}}$% {\bigmath\char'001}{\bigit p} {\bigrm +} {\bigrm . . .} {\bigrm +} $\hbox{\bigit b}_{\hbox{\smallit \hskip 1 pt m}}$% {\bigmath\char'001}{\bigit p} \hfil} \nobreak \vskip 10pt \noindent Recall from Theorem \thaC\ that if $p+p=p$, $a_1,a_2,\ldots,a_n\in\ben$ with $a_i\neq a_{i+1}$ for all $i\in\nhat{n-1}$, $b_1,b_2,\ldots,b_m\in\ben$ with $b_i\neq b_{i+1}$ for all $i\in\nhat{m-1}$, and $\apsum=\bpsum$, then $m=n$ and $a_i=b_i$ for all $i\in\nhat{n}$. Recall also that the restriction on repeated coefficients is necessary. For example, for any idempotent $p\in\beta\ben$, one has $p+p+2\cdot p+3\cdot p= p+2\cdot p+3\cdot p+3\cdot p$. In this section we investigate solutions to the equation $\apsum =\bpsum$ for $p\in\ben^*$. As an immediate consequence of Corollary \thG\ and Theorem \thtQ\ we have that necessarily $a_1=b_1$ and $a_n=b_m$. Beyond that, the information we are able to obtain is quite limited. In particular, our main results are restricted to the case when there exists some $d\in\ben\bk\{1\}$ such that $\varlist\in\{1,d\}$. (By [\refHS, Lemma 13.1] the restriction that $\varlist\in\{1,d\}$ is the same as requiring that $\varlist\in\{k,l\}$ where $k$ divides $l$.) Some special cases of our results include the facts that the equations $p+3\cdot p+p=p$ and $3\cdot p+p+3\cdot p+p= 3\cdot p+p$ have no solutions. The nature of the arguments is such that we cannot even determine whether the equation $2\cdot p+3\cdot p+2\cdot p=2\cdot p$ can be solved. (We conjecture very strongly that it cannot be solved.) The key to the amount of success that we have had is the following lemma, which allows us to work in base $d$ arithmetic and add numbers with no carrying. \noindent {\bf Lemma} \proclaim \thfA\ \hskip -5pt . Let $p\in\ben^*$ and let $\varlist\in\ben$. If $\apsum=\bpsum$ and $a_1+a_2+\ldots+a_n\neq b_1+b_2+\ldots+b_m$, then for all $d,l\in\ben$, $\ben d^l\in p$. \proof It suffices to show that for every prime $q$ and every $l\in\ben$, $\ben q^l\in p$. So let a prime $q$ be given. Without loss of generality assume that $c=\asum-(\bsum)>0$. Let $l\in\ben$ be given and pick $k\in\ben$ such that $q^k\geq c$. Pick $i\in\ohat{q^{k+l}-1}$ such that $A=\ben q^{k+l}+i\in p$. Then $a_1A+a_2A+\ldots+a_nA\in\apsum$ (as can be established by an easy induction) and $b_1A+b_2A+\ldots+b_mA\in\bpsum$. Consequently $a_1A+a_2A+\ldots+a_nA\cap b_1A+b_2A+\ldots+b_mA\neq \emp$. Thus $a_1i+a_2i+\ldots+a_ni\equiv b_1i+b_2i+\ldots+b_mi\ (\mod q^{k+l})$. That is, $c\cdot i\equiv 0\ (\mod q^{k+l})$. Since $q^{k+1}\nodiv c$, we have $q^l|i$. Then $\ben q^{k+l}+i\subseteq\ben q^l$ so that $\ben q^l\in p$ as required.\qed We now adopt some special notation to be used in our proof of the main theorem of this section, Theorem \thfMAIN. \noindent {\bf Definition} \definition \thfB\ \hskip -5pt . Let $d\in\ben\bk\{1\}$, let $x\in\ben$, and write $x$ in its base $d$ expansion as $\sum_{i=1}^ld^{t_i}\cdot g_i$ where $0\leq t_11$ and assume that the statement is true for $n-1$. Let $F=\{x\in\ben:f_{j+1-i}(x)\equiv n\cdot \alpha +c_1\ (\mod r)\}$. We consider four cases. \underbar{Case 1}: $a_{n-1}=1$ and $a_n=1$. Then by the induction hypothesis, $$E=\{x\in\ben:f_{j+1-i}(x)\equiv (n-1)\cdot \alpha +c_1\ (\mod r)\}\in a_1\cdot p+a_2\cdot p+\ldots+a_{n-1}\cdot p\,.$$ Also, by Lemma \thfC, $\{x\in\ben:\start(x)\equiv i\ (\mod 3)\}\in a_1\cdot p+a_2\cdot p+\ldots+a_{n-1}\cdot p$. We claim that $E\cap\{x\in\ben:\start(x)\equiv i\ (\mod 3)\}\subseteq \{x\in\ben:-x+F\in p\}$ so let $x\in E$ such that $\start(x)\equiv i\ (\mod 3)$ and let $l=\start(x)+1$. Then $B\cap\ben d^l\cap A\subseteq -x+F$. \underbar{Case 2}: $a_{n-1}=1$ and $a_n=d$. Then by the induction hypothesis, $$E=\{x\in\ben::f_{j+1-i}(x)\equiv (n-1)\cdot \alpha +c_1-1\ (\mod r)\}\in a_1\cdot p+a_2\cdot p+\ldots+a_{n-1}\cdot p\,.$$ Also, by Lemma \thfC, $\{x\in\ben:\start(x)\equiv i\ (\mod 3)\}\in a_1\cdot p+a_2\cdot p+\ldots+a_{n-1}\cdot p$. We claim that $E\cap\{x\in\ben:\start(x)\equiv i\ (\mod 3)\}\subseteq \{x\in\ben:-x+F\in d\cdot p\}$ so let $x\in E$ such that $\start(x)\equiv i\ (\mod 3)$ and let $l=\start(x)+1$. Then $B\cap\ben d^l\cap A\subseteq d^{-1}(-x+F)$. Cases 3 ($a_{n-1}=d$ and $a_n=1$) and 4 ($a_{n-1}=d$ and $a_n=d$) are handled similarly.\qed \noindent {\bf Theorem} \proclaim\thfMAIN\ \hskip -5pt . Let $d\in\ben\bk\{1\}$, let $p\in\bigcap_{l=1}^{\infty}\overline {\ben d^l}$, let $\varlist\in \{1,d\}$ and assume that $\apsum=\bpsum$. Let $c=|\{t\in\nhat{n-1}:a_t\neq a_{t+1}\}$ and let $e=|\{t\in\nhat{m-1}:b_t\neq b_{t+1}\}$. Then $a_1=b_1$, $a_n=b_m$, and $\displaystyle(n-m)|{{e-c}\over 2}$. \proof That $a_1=b_1$ and $a_n=b_m$ follows from Corollary \thG\ and Theorem \thtMAIN. Let $$\matrix{\hfill c_1&=&|\{t\in\nhat{n-1}:a_t=1\hbox{\rm \ and }a_{ t+1}=d\}\,,\hfill \cr \hfill c_2&=&|\{t\in\nhat{n-1}:a_t=d\hbox{\rm \ and }a_{t+1}=1\}\, ,\hfill \cr \hfill e_1&=&|\{t\in\nhat{m-1}:b_t=1\hbox{\rm \ and }b_{t+1}=d\}\, ,\hbox{\rm \ and }\hfill \cr \hfill e_2&=&|\{t\in\nhat{m-1}:b_t=d\hbox{\rm \ and }b_{t+1}=1\}\, .\hfill \cr} $$ Notice that \itemitem{(1)} If $a_1=a_n$, then $c_1=c_2$. \itemitem{(2)} If $a_1=1$ and $a_n=d$, then $c_1=c_2+1$. \itemitem{(3)} If $a_1=d$ and $a_n=1$, then $c_2=c_1+1$. \noindent Since similar statements hold for $e_1$ and $e_2$, we have in any event that $c_2-c_1=e_2-e_1$. Fix $i,j\in \{0,1,2\}$ such that $$\{x\in\ben:\start(x)\equiv i\ (\mod3)\hbox{\rm \ and }\endd(x)\equiv j\ (\mod3)\}\in p\,.$$ If $m=n$, pick $r\in\ben$ such that $r>|e_1-c_1|$. If $m\neq n$, let $r=|m-m|$. Pick $\alpha\in\ohat{r-1}$ such that $\{x\in\ben:f_{j+1-i}(x)\equiv\alpha\ (\mod r)\}\in p$. By Lemma \thfD, we have that $\{x\in\ben:f_{j+1-i}(x)\equiv n\cdot \alpha +c_1\ (\mod r)\}\in\apsum$ and $\{x\in\ben:f_{j+1-i}(x)\equiv m\cdot\alpha +e_1\ (\mod r)\}\in\bpsum$. Consequently $ n\cdot\alpha +c_1\equiv m\cdot\alpha +e_1\ (\mod r)$. If $m=n$, we have $c_1\equiv e_1\ (\mod r)$ so, since $r>|c_1-e_1|$, $c_1=e_1$. If $m\neq n$, then $r=|m-n|$ so that $c_1\equiv e_1\ (\mod r)$ and thus $(n-m)|(e_1-c_1)$. Therefore in any case $(n-m)|(e_1-c_1)$. We have observed that $c_2-c_1=e_2-e_1$ and thus $e-c=(e_2-c_2)+(e_1-c_1)=2(e_1-c_1)$ so that $\displaystyle(n-m)|{{e-c}\over 2}$ as required.\qed \noindent {\bf Corollary} \proclaim \thfE\ \hskip -5pt . Let $d\in\ben\bk\{1\}$, let $p\in\ben^*$, let $\varlist\in\{1,d\}$, and assume that $\apsum=\bpsum$. Let $c=|\{t\in\nhat{n-1}:a_t\neq a_{t+1}\}$ and let $e=|\{t\in\nhat{m-1}:b_t\neq b_{t+1}\}$. Then $a_1=b_1$, $a_n=b_m$, and either $\asum=\bsum$ or $\displaystyle(n-m)|{{e-c}\over 2}$. \proof By Lemma \thfA, if $\asum\neq\bsum$, then $p\in\bigcap_{l=1}^\infty\overline{\ben d^l}$ so that Theorem \thfMAIN\ applies.\qed We see in particular that if the coefficients alternate, they must match exactly. \noindent {\bf Corollary} \proclaim \thfF\ \hskip -5pt . Let $d\in\ben\bk\{1\}$, let $p\in\ben^*$, let $\varlist\in\{1,d\}$, such that for all $i\in\nhat{n-1}$ and all $j\in\nhat{m-1}$, $a_i\neq a_{i+1}$ and $b_j\neq b_{j+1}$. If $\apsum=\bpsum$, then $n=m$ and for all $i\in\nhat{n}$, $a_i=b_i$. \proof Let $c=|\{t\in\nhat{n-1}:a_t\neq a_{t+1}\}$ and let $e=|\{t\in\nhat{m-1}:b_t\neq b_{t+1}\}$. Then $c=n-1$ and $e=m-1$ and so $e-c=m-n$. By Corollary \thfE, either $\asum=\bsum$ or $(n-m)|\displaystyle{{{n-m}\over 2}}$. In either case we conclude that $n=m$. Since also $a_1=b_1$ we have that for all $i\in\nhat{n}$, $a_i=b_i$. \qed Notice also that Corollary \thfE\ tells us that many equations whose coefficients do not alternate have no solutions. For example, there is no $p\in\ben^*$ such that $p+p=p+2\cdot p+p+2\cdot p+p$. On the other hand we no not know whether there exist solutions to $p+p+p=p+2\cdot p+p+2\cdot p+p$. Nor do we know whether $p+p+2\cdot p+2\cdot p=p+2\cdot p+p+2\cdot p$ has any solutions (although Theorem \thfMAIN\ tells us there are no solutions with $p\in\bigcap_{l=1}^{\infty}\overline {\ben d^l}$). There are other equations which we know can be solved by idempotents such as $p+p+2\cdot p=p+2\cdot p$ for example, and we would conjecture that these are the only solutions. We know, of course, that there are equations in $\beta\ben$ only solvable by idempotents. Trivially $p+p=p$ is one such. Much less trivial is the fact that $p+p+p=p$ implies that $p$ is an idempotent. (Indeed, if $p+p+p=p$, then $\{p,p+p\}$ forms a subgroup of $\ben^*$ and the very difficult Zelenuk's Theorem [\refHS, Theorem 7.17] asserts that the only finite subgroups of $\ben^*$ are singletons.) It is unknown whether there exists any $p\neq p+p$ such that $p+p+p=p+p$. The existence of such $p$ is equivalent to the existence of a nontrivial continuous homomorphism from $\beta\ben$ to $\ben^*$ [\refHS, Corollary 10.20]. Similarly, we do not know whether the equation $p+p+2\cdot p=p+2\cdot p$ has any solutions in $\ben^*$ besides idempotents. It is a consequence of Theorem \thfG\ that there are no such solutions with $p\in K(\beta\ben)$, the smallest ideal of $\beta\ben$. \noindent {\bf Theorem} \proclaim \thfG\ \hskip -5pt . Let $p\in K(\beta\ben)$ and let $q\in\beta\ben$. If $p+p+q=p+q$, then $p+p=p$. \proof Pick by [\refHS, Theorem 2.8] a minimal left ideal $L$ and a minimal right ideal $R$ of $\beta\ben$ with $p\in R\cap L$. By [\refHS, Theorem 2.7], $R\cap L$ is a group. Let $e$ be the identity of $R\cap L$ and let $r$ be the inverse of $p$ in $R\cap L$. By [\refHS, Theorem 1.46], $L+q$ is a minimal left ideal of $\beta\ben$ and $e+q\in L+q$ so by [\refHS, Theorem 2.11(c)], $\rho_{e+q|L}$ is a homeomorphism from $L$ onto $L+q$. Since $\rho_{e+q}(p)=p+e+q=e+p+q=r+p+p+q=r+p+q=e+q=e+e+q= \rho_{e+q}(p)$, we have that $p=e$.\qed \references \bye