\LargeAdditive structure of $Z$(.) mod $m_k$ (squarefree)\\[1ex] and Goldbach's Conjecture
{\sc Nico F. Benschop} - {~~Amspade Research}
~~The Netherlands
The product $m_k$ of the first $k$ primes (2..$p_k$) has neighbours $m_k \pm 1$ with all prime divisors beyond $p_k$, implying there are infinitely many primes [Euclid]. All primes between $p_k$ and $m_k$ are in the group $G_1(k)$ of units in semigroup $Z_{m_k}$(.) of mutiplication mod $m_k$. Due to its squarefree modulus $Z_{m_k}$ is a disjoint union of $2^k$ groups, with as many idempotents - one per divisor of $m_k$, which form a Boolean lattice $BL$. The $additive$ properties of $Z_{m_k}$ and its lattice are studied. It is shown that each complementary pair in $BL$ adds to 1 mod $m_k$ , and each even idempotent $e$ in $BL$ has successor $e$+1 in $G_1$. It follows that $G_1(k)+G_1(k) \equiv E(k)$, the set of even residues in $Z_{m_k}$, so each even residue is the sum of two roots of unity, proving "Goldbach for Residues" mod $m_k ~(GR)$. The prime units in $G_1(k)$ have principle (natural) values in the corresponding set $G(k)$ of naturals $u<m_k$. A proof by contradiction and finite reduction using epimorphism $G_1(k+1) \rightarrow G_1(k)$ mod $m_k$, and verifying $GC$ for $4<2n<30 ~(k=3)$, yields a contradiction for $k$=3. This establishes $GC$: Each $2n>4$ is the sum of two odd primes. \\ The structure of $G_1(k)$ mod $m_k$ is illustrated by the next features. The smallest unit in $G_1(k)$ is $p_{k+1}$, so its smallest composite is $(p_{k+1})^2$. Hence its units between $p_{k+1}$ and $(p_{k+1})^2$ are all prime if considered as naturals (their principle values in set $G(k)$), to be used as summands for successive $2n < m_k$. For $k=3~(m_3$=30) it is shown by complete inspection that each $2n$ with $4<2n<30$ is indeed the sum of two odd primes. For $k>3$ the addition to obtain $2n < (p_{k+1})^2$ produces no carry, thus yielding a natural sum. The known Bertrand Postulate: $p_{k+1} < 2p_k$, implies overlapping intervals for successive $2n$, again yielding $GC$ by contradiction to $GR$.
math0103091.tex
 %============(normal LaTeX)=========ngb0907=====N.F.Benschop==========
% corr 9-7-09 Thm 4.1
% 21-7-09: prime sieve (niet nodig) sect9.4-9.5
% 22-7-09: constructive proof by union over k \ge 3 of 2n< (p_{k+1})^2.
% 29-7-09: proof by contradiction (after all): principle values in G(k)
%
\documentclass[11pt]{article}
\pagenumbering{arabic}
\onecolumn
%------- Text + Margins -----------------------------
\textheight     = 24cm
\textwidth      = 16cm
\topmargin      = -1cm
\oddsidemargin  = -0.5cm
\setlength{\parskip}{1ex}
\setlength{\parindent}{0pt}
%------- Lemma/Theorem/Corollary/Proof --------------------------------
\newtheorem{lem}{{\sc Lemma}}[section]              % refer:  (lem3.1)
\newtheorem{thm}{{\sc Theorem}}[section]            % refer:  (thm3.1)
\newtheorem{cor}{{\sc Corollary}}[section]          % refer:  (cor3.1)
\def\proof{\rm \trivlist \item[\hskip \labelsep{\sc Proof.}] }
\def\endproof{ {\large$\Box$} \endtrivlist }
\newcommand{\bc}[2]{\left(\!\!\begin{array}c#1\\#2\end{array}\!\!\right)}

\usepackage{latexsym} % for symbol \Box  in \endproof
\newcommand{\ul}{\underline}
\newcommand{\e}{\equiv}
\newcommand{\nqv}{\equiv \!\!\!\!\!\!\//~~}
%===================================================================

\title{ \Large\bf
Additive structure of $Z$(.) mod $m_k$ (squarefree)\\[1ex]
   and Goldbach's Conjecture}
\author{ {\sc Nico F. Benschop} -  {\it ~~Amspade Research}, ~~The Netherlands}
\date{}

\begin{document}
\maketitle

\begin{abstract}
The product $m_k$ of the first $k$ primes (2..$p_k$) has neighbours $m_k \pm 1$
with all prime divisors beyond $p_k$, implying there are infinitely many
primes [Euclid]. All primes between $p_k$ and $m_k$ are in the group $G_1(k)$
of units in semigroup $Z_{m_k}$(.) of mutiplication mod $m_k$. Due to its
squarefree modulus $Z_{m_k}$ is a disjoint union of $2^k$ groups, with as many
idempotents - one per divisor of $m_k$, which form a Boolean lattice $BL$.
The  $additive$ properties of $Z_{m_k}$ and its lattice are studied.
It is shown that each complementary pair in $BL$ adds to 1 mod $m_k$ ,
and each even idempotent $e$ in $BL$ has successor $e$+1 in $G_1$. It follows
that $G_1(k)+G_1(k) \equiv E(k)$, the set of even residues in $Z_{m_k}$, so
each even residue is the sum of two roots of unity, proving "Goldbach for Residues"
mod $m_k ~(GR)$. The prime units in $G_1(k)$ have principle (natural) values
in the corresponding set $G(k)$ of naturals $u<m_k$. A proof by contradiction
and finite reduction using epimorphism $G_1(k+1) \rightarrow G_1(k)$ mod $m_k$,
and verifying $GC$ for $4<2n<30 ~(k=3)$, yields a contradiction for $k$=3.
This establishes $GC$: Each $2n>4$ is the sum of two odd primes.
\\
The structure of $G_1(k)$ mod $m_k$ is illustrated by the next features. The
smallest unit in $G_1(k)$ is $p_{k+1}$, so its smallest composite is $(p_{k+1})^2$.
Hence its units between $p_{k+1}$ and $(p_{k+1})^2$ are all prime if considered as
naturals (their principle values in set $G(k)$), to be used as summands for successive
$2n < m_k$. For $k=3~(m_3$=30) it is shown by complete inspection that each
$2n$ with $4<2n<30$ is indeed the sum of two odd primes. For $k>3$ the addition
to obtain $2n < (p_{k+1})^2$ produces no carry, thus yielding a natural sum.
The known Bertrand Postulate: $p_{k+1} < 2p_k$, implies overlapping intervals
for successive $2n$, again yielding $GC$ by contradiction to $GR$.
\end{abstract}

{\bf Keywords}: ~Residue arithmetic, ring $Z$ mod $m$, squarefree modulus,
\\ \hspace*{2cm} Boolean lattice, Goldbach conjecture.

Subject msc: ~11P32

\section{Introduction} % sect1

Detailed analysis of the algebraic structure of modulo arithmetic is
persued, especially multiplication in relation to addition and exponentiation.
Addition and multiplication are associative operations, so semigroup structure
analysis provides a good perspective for basic problems in arithmetic [2,3,6,8]
such as {\it Goldbach's conjecture} of every even number $2n>4$ being the sum
of two odd primes.
The additive structure of multiplicative semigroups with squarefree moduli
is studied, in ring $Z(+, .)$ mod $m_k$. Choosing as modulus the product
$m_k$ of the first $k$ primes, all primes between $p_k$ and $m_k$ are in
the group of roots of 1 mod $m_k$, denoted as the group $G_1$ of units.
As shown (thm~\ref{thm:GR}), $G_1+G_1$ covers all even residues $2n$ in $Z$ mod $m_k$.

The direct product $Z_{rs} = Z_r \times Z_s$ of multiplications with
coprime component moduli $r$ and $s$, is represented by component-wise
multiplication [4]. Squarefree modulus $m_k$ implies $Z_{m_k}(.) =
Z_{p_1} \times~.~.~ \times Z_{p_k}$ is a direct product of multiplications
mod $p_i$. This direct product is analysed as an ordered {\it disjoint
union} of maximal $subgroups$ derived from the component semigroups
$Z_{p_i}$. The emphasis is on the $additive$ properties of idempotents,
and the "fine structure" of residue ring $Z(+,~.)$ mod $m_k$. Considering
the principle values of units in units group $G_1(k)$ transfer additive
results from residues to positive integers. So residues $u$ mod $m_k$ are
taken as naturals $u< m_k$, with upper bound $u+v < 2m_k$ so a
possible carry is at most 1, and for sum $u+v<m_k$ no carry is produced.
For instance in Lemma~\ref{lem:compl}: the sum of each pair of complementary
idempotents equals 1 mod $m_k$, yielding the natural sum $m_k+1$ for pairs
other than \{0,1\}.

{\bf Notation}:
The known number representation (base $m$) $n=c.m+r$ with carry $c$ and
rest $0 \leq r < m$ is used. Operation + is natural addition, which for
two summands $<m_k$ can produce a maximal carry of 1 (base $m_k$).
For residue arithmetic $c=0$. This in contrast to the usual
interpretation of a residue arithmetic closure as an image $Z/Z_m$ of the
integers $Z$, consisting of residue classes to express the irrelevance
of the carry in residue arithmetic (mod $m$). In short, these additive
and multiplicative interpretations, with residue values vs. residue classes,
correspond to $n=r+0.m$ resp. $n=r+Z.m$, resetting the carry to $0$ or to
set $Z$. The latter interpretation is a mix of numbers and sets, which is
cumbersome and not required in the present additive analysis. Furthermore,
$m$ will denote modulus $m_k$ if no confusion can arise, and ~$\equiv$~
denotes congruence mod $m_k$. Sections 4 and 5 interprete residues
$n$ mod $m_k$ as naturals $n < m_k$ by taking their principle value,
where addition is restricted to summands that produce no carry.

The idempotents ~$e^2 \equiv e$ of ~$Z_m$(.) play an essential role.
For prime modulus $p$ it is known that $Z_p$ has just two idempotents:
0 and 1 mod $p$. And all residues $1,~.~.~,p$-1, coprime to $p$, are in
some permutation generated as residues of powers $g^i$ of some
{\it primitive root} $g<p$ of unity [1]. They form an order $p-1$ cyclic
subgroup $G$ of $Z_p$, written $G = g^* \equiv \{g^i\} ~(i=1..p$-1), with
$g^{p-1} \equiv 1$. Hence $Z_p(.)$ is a cyclic group, adjoined to zero.

{\bf Summary}:
~The product $m_k$ of the first $k$ primes is used
for analysis of all primes and their additive properties. Each of the $2^k$
divisors $d$ of $m_k$ yields a maximal subgroup $G_d$ of $Z_{m_k}$ containing
all $n<m_k$ with the same set of prime divisors as $d$. The respective group
identities are the $2^k$ idempotents of $Z_{m_k}$, ordered as Boolean lattice
$BL$ ~[4][6]~ of which the additive properties are studied.

The additive properties of $Z_{m_k}$ are characterised by the successor $n+1$
of any $n$, especially of the idempotents. An essential additive property is
that each complementary pair of idempotents in $BL$ sums to 1 mod $m_k$ (lemma~\ref{lem:compl}),
and every even $e^2=e$ has successor $e+1$ in $G_1$, while $G_1+G_1$ covers
all $2n$ mod $m_k$. This residue version $GR$ of Goldbach's Conjecture ($GC$)
is extended, by considering the set of principle values (naturals) $G(k)$ of
the units in $G_1(k)$ for $k \ge 3$, to prove $GC$ for positive integers.
Results listed in the Conclusions may be new.

For completeness, these essential concepts [5][6] are reviewed in sections
1 and 2. Section 3 derives a 'Goldbach-for-Residues'($GR$) result. Sections
4 and 5 give the approach to Goldbach's conjecture, followed by conclusions.

\section{Lattice of groups}  %sect2

In modulus $m_k=\prod p_i ~(i=1 ~..~ k)$~ each prime factor has exponent one.
So $m_k$, having no square divisor, is called {\it square free}. The prime
divisors of $m_k$ are referred to as {\bf base primes}.

Residues $n$ with the same base-prime divisors as squarefree divisor $d~|~m_k$
form a {\it maximal subgroup} $G_d \subset Z_{m_k}$(.) with closure due to
all possible products having the same base primes. If $e$ is the identity
(idempotent) of $G_d$, then each $n$ in subgroup ~$G_d \e G_e$~ has a unique
$local$ inverse $n^{-1}$~ defined by ~$n.n^{-1} \e e$.

The $2^k$ divisors of $m_k$ correspond to as many subsets of the $k$
base primes. Each divisor $d$ of $m_k$ generates a finite cycle $d^*=\{d^i\}$
with an idempotent $\underline{d}$, the identity of subgroup $G_d$. Each
subgroup has just one idempotent as its identity. So ~$Z_{m_k}$ has $2^k$
disjoint subgroups $G_d$, one for each divisor $d$ of $m_k$, ordered in a
Boolean lattice as their identities are ordered, as follows.

\subsection{Ordering of commuting idempotents}  % sect2.1

$Z_{m_k}$ is a disjoint union of $2^k$ groups $G_d$, and the group identities,
the idempotents, form a {\it Boolean lattice}. In fact, {\it commuting
idempotents} ~$e^2=e,~f^2=f$ can be {\it ordered} $e \geq f$ whenever $ef=fe=f$,
in other words $e$ is identity for $f$. ~~This is readily verified to be an
ordering relation, being transitive, anti-symmetric and reflexive [4].

The lattice {\it meet} (greatest lower bound) operation  is modelled by
{\it multiplication}. The product of two commuting idempotents $e, f$ is
idempotent: $ef.ef = effe = efe = eef = ef$, while $e,f$ are left- and right-
identity for $ef$ since $e.ef = ef = fe= fe.e$, sothat $e \geq ef$, and
similarly $f \geq ef$. Also, $ef$ is the greatest idempotent ordered under $e$
and $f$, since ~$c \leq e$ and $c \leq f$ imply $c \leq ef$, which is easily
verified.

The {\it join} (least upper bound) of two idempotents is the idempotent
with the {\it intersection} of the corresponding baseprime sets. Idempotent
'1' at the top has the smallest base-prime set (empty), while '0' at the
bottom contains all base-primes since 0=$m$ mod $m$.

The sum of two idempotents is generally not an idempotent, nor is its
generated idempotent their lattice-join, except for complementary
idempotents, to be derived next.

\subsection{ Lattice of idempotents: add {\it vs} join }  %sect2.2

As shown earlier, the set of idempotents of $Z$ mod $m$ is closed under
multiplication, forming a lower semi-lattice [4,6]. {\it Multiplication} models
the {\bf meet} (glb: greatest lower bound) operation of two idempotents,
yielding an idempotent with the $union$ of the respective base-prime sets.

Notice that all primes $p: ~p_k<p<m_k$~ are 'units' in topgroup $G_1$. In the
base-prime set of any idempotent or subgroup they are considered equivalent
to 1 mod $m_k$. For instance, cycle 2* mod $m$ (in $G_2$) produces residues
$c.2^n$, where ~$c \in G_1$~ are relative prime to $m_k$, and $c$ has
prime divisors $p_r > p_k$. Residues in $G_1$ can occur as factor in
each $n \in Z_{m_k}$, according to their name of $units$ in $Z_{m_k}$.

The {\bf join} (least upper bound $lub$) of two idempotents follows by
$intersecting$ their baseprime sets, yielding an idempotent with their
common baseprimes. \\{\bf Def:} two idempotents $a,b$ are
{\it complementary} iff $ab \e 0$ and $lub(a,b) \e 1$.

The endomorphism '$.e$' ~for idempotents $e$ in commutative ~$Z_m(.)$~
models the lattice meet operation by multiplication, since for each
~$x,y \in Z_m: ~xy.e \e xy.e^2 \e xe.ye$ .\\
Although in general the sum of two idempotents is  not an idempotent,
the next exception is an essential additive property of $Z_m(.)$ :

\begin{lem} \label{lem:compl} %% lem2.1
~~For any squarefree $m>1$ with at least two prime divisors:\\
For each $complementary$ pair $\{a,b\} \neq \{0,1\}$ of idempotents
in $Z_m(.)$ holds~ $a+b=m+1$.
\end{lem}
\begin{proof}
The lattice of idempotents has order ~$2^k$, with~ $2^{k-1}$~ complementary
pairs. Consider a sublattice of order four: ${0,1}$ ~and any other
complementary pair ~${a,b}$. It must be shown that $a+b \e 1$ mod $m$.
Now idempotents ~$a,b$~ are complementary, ~so ~$ab \e 0$ mod $m$, implying :
\\$(a+b)^2 \e a^2+2ab+b^2 \e a+b$ (mod $m$), thus~ $a+b$~ is idempotent.
And ~$(a+b)a \e a^2+ba \e a$ mod $m_k$, ~so~ $a+b \geq a$, and similarly
~$a+b \geq b$. Hence $a+b \e 1$ mod $m$, because by $lub(a,b) \e 1$ the
only idempotent covering complementary $a$ and $b$ is 1. Clearly, for
$\{a,b\} \neq \{0,1\}$ holds $1<a+b<2m$ so $a+b=m+1$, with carry =1 (base $m$).
\end{proof}

In other words : complementary idempotents $a,~b$ have disjoint base-prime
sets $A$ and $B$, and union $A \cup B$ consists of all base-primes in
$m$. For square-free $m, ~a.b \e 0$ is the idempotent containing all base-primes.
And $join(a,b)$ has the trivial intersection $A \cap B =1$ as base-prime
set, relative prime to $m$, with corresponding idempotent '1' of $G_1$.

\begin{lem} \label{lem:auto} % lem2.2
For squarefree modulus $m=2.odd: ~h=m/2$ is the lowest odd idempotent in
$Z_m(.)$ and~ $a \rightarrow a+h$ is the only additive automorphism of $Z_m(.)$
\end{lem}
\begin{proof}
Notice that $2h \e 0$, so for each even or odd pair $a,b$ in $Z_m$ holds $(a+b)h \e 0$.
Hence :\\ $(a+h)(b+h)\e ab +(a+b)h + h^2 \e ab+h$ , and only if $h^2 \e h$
this yields $a \rightarrow a+h$ as additive automorphism of $Z_m(.)$.
Furthermore, $h=m/2$ is the lowest odd idempotent, namely the image under
$+h$ of the lowest even idempotent 0 in $Z_m$ (for squarefree $m:$ no
divisors of 0 exist). It is readily verified that this morphism is
1-1 onto, mapping $Z_m(even)$ and $Z_m(odd)$ into each other.
\end{proof}

Now consider product $m= m_k = \prod_{i=1}^k p_i$ of the first $k$ primes.
Unit 1 is ordered at the top of the lattice of idempotents, being the
identity for all idempotents in $Z_m = \times_i ~~Z_{p_i}$. Top group
$G_1$ of all residues relative prime to $m$ misses all base primes.
Thus ~$G_1 = \times_i ~~C(p_i$-1) ~$[i=2..k]$ is a direct product
of $k-1$ cycles of periods $p_i -1$.

%cor2.1
\begin{cor} \label{cor:G1=G2}
~In $Z$(.) mod $m$ with square-free $m= 2.odd$, and let $h=m/2$ then:\\ \hspace*{.5cm}
$Odd$ and $even$ top-groups are isomorphic ~$G_1 \cong G_2$ under additive
automorphism $+h$.\\
{\rm Note:} isomorphic max cycles~ $(2+h)^* \cong 2^*$ in $G_1$ and $G_2$
(e.g. $5<primes<25$ are $15 \pm 2^i$)
\end{cor}

\section{ Primes, composites and neighbours}  % sect3

{\bf Equivalent sum and difference :}
~$(-1)^2$=1 implies $-1 \in G_1$, ~so~ $G_1 \e -G_1$ ~hence :

{\bf (1)}~~~~~~~~~~~~$G_1 + G_1 \equiv G_1 - G_1$

So sums and differences of pairs in $G_1$ yield the same set of
residues mod $m$. ~~Notice that:\\~$(-n)^2=n^2$, ~so ~$n$ and $-n$
generate the same idempotent, thus are in the same subgroup:

{\bf (2)}~~~~For every group $G_d \subset Z_m :$
~if $n \in G_d$~ then so is ~$-n$, while ~$G_d+G_d \equiv G_d-G_d$.

{\bf Neighbours $n$+1 and $n$-1 in the lattice of $Z_m$ :}

For integers and residues: $~n$ and $n$+1 are coprime for each $n$ so their
prime divisors form disjoint sets. The same holds for $n$ and $n-1$. Then
one would expect $n$ and $n$+1 to be in complementary subgroups of $Z_m$.
More precisely, the subgroup ordering of their idempotents implies:

\begin{lem}~ %lem3.1
For each $n \in Z_m$~ and base-prime complementary ~$\overline{n}:$~~
 $G_{n \pm 1} \geq G_{\overline{n}}$
\end{lem}
\begin{proof} Due to the subgroup ordering, a {\it subset} of baseprimes
disjoint from (complementary to) those in $n$ defines a subgroup ordered
above or equal to $G_{\overline{n}}$.
\end{proof}

Hence $e$+1 for any {\bf even} idempotent $e$ must be in an odd subgroup
$G_d$ that is ordered $G_d \geq G_{\overline{e}}$, with $\overline{e}$ the
complement of $e$ in the lattice of $Z_m$. In fact, as shown next: ~$e$+1
is in topgroup $G_1$.

\subsection{Each idempotent's successor is in $G_1$ or $G_2$}  %sect2.1

The sum of two complementary idempotents yields an idempotent namely 1
(lemma~\ref{lem:compl}), which is their join or least upper bound. This is an
exception, and in general idempotents do not sum to an idempotent, let alone
their join. For instance, in $Z_{10}$ with idempotents 1, 5, 6, 0 : $5+1=6$
is idempotent, but join(5,1)= 1. And join(6,1)= 1 while $6+1=7$ is not
idempotent, although 7 does generate the proper idempotent 1, due to:

 %lem3.2
\begin{lem} \label{lem:e+1}
~In $Z$(.) mod $m$, with square-free $m=2.odd$:\\
(a) \hspace{1in} Each {\bf even} idempotent $e$ has $e$+1 in $G_1$, and\\
(b) \hspace{1in} each {\bf odd} idempotent $d$ has $d$+1 in $G_2$. \\
(c) \hspace{1cm}
  For period $n$ of $e+1$ in $G_1$ mod $m_k$ holds: ~$e.(2^n-1) \equiv 0$.
\end{lem}
\begin{proof} (a,c):
~Given $e^2=e$, notice that ~$(e+1)(e-1) \equiv e^2-1 \equiv e-1$, so $e$+1
is identity for $e-1$, hence $G_{e+1} \geq G_{e-1}$ for $every$ idempotent $e$.
~Now $(e+1)^2 \equiv e^2+2e+1 \equiv 3e+1$, and in general expanding $(e+1)^n$,
with $e^i \equiv e$ for all $i>0$ and factoring out $e$, yields:

~~~~~~~$(e+1)^n \equiv 1 + \sum_{i=1}^n \bc{n}{i} e^i \equiv 1 + (2^n-1)e$

We need to show $c=(2^n-1)e \equiv 0$ for every even idempotent $e$, where
$n$ is the period of $e+1$, with corresponding odd idempotent $d=(e+1)^n=c+1$,
which equals 1 iff $c \equiv 0$. ~In fact it would suffice if $2^n-1$ is in
a group complementary to $G_e$ in the lattice of $Z_m$. The baseprimes in
$2^n-1$, which are all necessarily odd, would then complement those in
even idempotent $e$.

This can be seen as follows: $d^2=d$ implies $(c+1)^2 \equiv c+1$, hence
$c^2+c \equiv 0$,\\ so:~~~~
$(2^n-1)^2~e +(2^n-1)e \equiv (2^n-1)(2^n-1+1)e \equiv (2^n-1)2^n e \equiv 0$.
\\ Apparently, the odd baseprimes in $2^n-1$ complement at least those
in $e$ because their union is complete (product 0). This implies
$(2^n-1)e = c \equiv 0$, independent of the extra factor $2^n$. So :

{\bf (3)} ~~~~~$(e+1)^n \equiv 1+(2^n-1)e \equiv 1$, ~where $n$ is the
period of $e$+1 in $G_1$.

Part (b) is dual to (a), proven similarly by using $G_1 \cong G_2$ (lemma~\ref{lem:G1=G2})
\end{proof}

\begin{thm}~( {\it Goldbach for Residues} ~~$GR$ ): \label{thm:GR} \\   %thm3.1
\hspace*{.5cm} For squarefree $m_k= \prod p_i ~(i=1...k)$ with $p_1$=2,
 and $E$ the set of even residues mod $m_k$:\\
\hspace*{1cm}
  In $Z$ mod $m_k:~~E \equiv ~\{2n\}~ \equiv G_1+G_1 \equiv G_1-G_1$, ~so :
\\ \hspace*{.5cm} Each even residue in ~$Z_{m_k}$ is a sum or difference
  of two units.
\end{thm}
\begin{proof}
~In short write $G$ for $G_1$. Let $e$ be any even idempotent, then multiply
$e \in G-1$ (lem~\ref{lem:e+1}) on both sides by $G$. On the lefthand side this
yields $G.e = G_e$ which is the max-subgroup on $e$, and on the righthand side
$G(G-1)= G^2-G = G-G$, sothat $G_e \subseteq G-G$. ~Using (1) yields:
~$G_e \subseteq G-G=G+G$ for all even $G_e$, so $G+G$ covers all even
residues.
\end{proof}

This also holds for any even squarefree modulus $m=2.odd$. Theorem 3.1 can
be generalized to hold for naturals which are the principle values of the
units in groups $G_1(k)$, as shown next.

\section{Prime units and carry extension} \label{def:principle}           %sect4

{\bf Define} ~$G_1(k)$~ as group of units mod $m_k$, and the correponding
set $G(k)$ of principle values (naturals) $\{1,u\}$~ where $p_k<u<m_k$
with $u$ coprime to base primes $p \leq p_k$. Use set $P(k)$ of all primes
in $G(k)$. The emphasis in the sequel is on the principle (natural) values
in $G(k)$ of units in group $G_1(k)$.

The primes $p>p_k$ are congruent mod $m_k$ to units in $G_1(k)$, and all
those $p<m_{k+1}$ in $G_1(k+1)$ are covered by $G(k) + a~m_k ~(carry~a~:
~0 \leq a<p_{k+1})$. \\An example for $k$=3 with all units in $G_1(4)$ follows
(table 1). It illustrates the relation between the prime structures of $G_1(k)$
and $G_1(k+1)$, which is a generalization of the known fact that all primes are
congruent to $G_1(2)=\{1,5\}$ mod $m_2$=6: remove the numbers that have a base
prime as divider (re Eratosthenes' ~prime sieve). Here : $G_1(4) \cong G_1(3)$
mod $m_3$=30.

\begin{verbatim}
 Units:  1     7<   11    13    17    19    23    29 :   mod m3 = 30

 +30:   31    37    41    43    47   7^2<   53    59     p_{k+1} = 7

 +60:   61    67    71    73  7.11<   79    83    89     7.n (8x '<')
                                                         not in G(4)
 +90:  7.13<  97   101   103   107   109   113   7.17<  (baseprime 7)

+120: 11^2#  127   131  7.19<  137   139  11.13# 149     Composites #
                                                         Smallest 11^2
+150:  151   157  7.23<  163   167  13^2#  173   179      in G(4)

+180:  181 11.17#  191   193   197   199  7.29< 11.19#   mod m4= 210
\end{verbatim}
{\bf Table 1} Unit extensions: $G(k+1) = \{~u+a.m_k~\}~:~unit~u \in G(k),~carry~0<a<p_{k+1}$\\

The set of all units in $G_1(k+1)$ is generated as illustrated for $G_1(4)$ in
table 1, including all primes in $P(k+1)$. Each natural unit $u \in G(k)$
generates at most $p_{k+1}-1$~ primes ~$p=u+a.m_k \in P(k+1)$~, with
~$p_k< p <m_{k+1}$ and carry $0 \le a < p_{k+1}$. For large enough $2n$
there are several prime pairsums in $GC$ format (see diagonals of equal
carry-sum in table 2). Clearly $p_{k+1}$ is the smallest unit in $G_1(k)$,
so ~$(p_{k+1})^2$~ is its smallest composite, hence:\\

{\bf (4)} ~~All (natural) units $u$ with $p_{k+1} \leq u<(p_{k+1})^2$ in $G(k)$ are prime.

Moreover, principle values of composite units in $G_1(k)$ are necessarily
generated under multiplication by the corresponding prime principle values $>p_k$
of units in $G_1(k)$. The reverse process of unit reduction by multiples of $m_k$
yields the next lemma:

\begin{lem} \label{lem:GRepi}  % lem4.1
~~$G_1(k+1) \rightarrow G_1(k)$ mod $m_k$, symbolizes that group $G_1(k)$ is
an epimorphic image of $G_1(k+1)$ with $v=t+c.m_k$, relating each principle
value $t \in G_1(k)$ to $p_{k+1}$ principle values $v \in G_1(k+1)$ with
a carry $c$.
\end{lem}
\begin{proof}
The mappings $v-c.m_k \longrightarrow t$ form a morphism because
$v.w \equiv (t+c.m_k)(u+d.m_k) \equiv (t.u)+e.m_k \equiv t.u $ mod $m_k$,
where $e = (td+cu)+cd.m_k$.
\end{proof}

Notice that each natural $n<m_k$~ is represented uniquely by $k$ digits
of a multi base code using the successive baseprimes: $p_1~.~.~.~p_k$.
The $k-1$ lower significant digits are extended with a most
significant digit or carry $a<p_k$, of weight $m_k$.

This in contrast to the usual single base code, e.g. decimal, using
powers of ten.  The successive bases 2, 6, 30, 210, ... have maximal
digit values $p_k$-1:~ 1, 2, 4, 6, ... respectively. For instance decimal
$331=210+11^2=1.210+4.30+0.6+0.2+1$ ~yields 5-digit code 1 4 0 0 1.

All primes $p>$3 are congruent to \{1, 5\} mod 6, while primes $p>p_3=5$
are congruent to the eight prime residues \{1,7,~...~,23,29\} mod 30 in
$G_1(3)$, obtained from ~$G(2)=\{1,5\}$~ by $p_3$-1=5-1=4 extensions with
increment $m_2=6$, namely \{7, 11\} ; \{13, 17\} ; \{19, 23\} ; \{25, 29\}.\\

Composite $25=5^2$ is not coprime to 30, hence is not in $G(3)$. The other
seven extensions are all primes $p_3<p<30=m_3$, forming with 1 the 8 units
in $G_1(3)=C_2 \times C_4$, ~in fact of form $15 \pm 2^i$ (cor~\ref{cor:G1=G2}).
The 7-1=6 extensions $G(3)+a.m_3$ generate all 2.4.6 = 48 units in $G_1(4):$
the 5 composites with prime divisors $p>7$ (exclude eight non-units of form
7.n in table 1), identity 1 and all ~$48-6=42$~ primes in open interval ~$(7,210)$.

\subsection{Pair sums of carry extended units}   %sub4.1

{\bf Define} ~set $S_0(k)=G(k)+G(k)$ ~of pair sums of (natural) units.
\\Denote even numbers interval by set ~$E(k)=\{2n~|~4<2n<m_k\}$, and the set
of natural carry-extended units: $T_a(k)=G(k)+a.m_k~~(a<p_{k+1})$ in $G(k+1)$.
The set of baseprimes is extended by $p_{k+1}$, so its multiples in $G(k+1)$
are not units (see table 1 for expanding G(3) to G(4) by baseprime 7).
All other extended units of $G(k)$ are  units of $G(k+1)$ since none is
divisible by a baseprime $p \leq p_{k+1}$.

Table 2 shows these sums for $k$=2 and 3 (by commutation half an
array suffices). Notice that ~$G(2)=\{1,5\}$ with pair sums $S_0(2)=
\{2,6,10\}$, while pair sums $2n$ in $S_0(3)= G(3)+G(3)$~ covers all $2n$ with
$2p_4 \leq 2n<m_3=30$, where ~$G(3)=\{1,7,~.~.~,29\}$ coprime to ~$2.3.5=30=m_3$.
For 6, 8, 10 use 3 and 5 to avoid non-prime 1.
In fact all ~$2n>16$~ have several ~$GC$ pair sums, e.g. each ~$2n$~ in
~$S_0(2)+6c=\{2,6,10\}+6c$~ for $1<c<p_3=5$~ has distinct unit pair sums,
all of which are prime pair sums.

\subsection{Pair sums of primes in ~$G(3)$}  %sub4.2

{\bf Define} ~Set ~$S_{a+b}(k) = S_0(k)+(a+b)m_k$ of pairsums in
$T_a(k)+T_b(k)$ of extended (natural) units in $G(k+1)$, except multiples
of $p_{k+1}$ (e.g. $5^2 \notin G(3)$), with carrysum ~$0 \leq a+b < p_{k+1}$.

{\bf Table 2} Extension sums:
carry sum diagonals ~$a+b=c<5$~ cover $2n$ in $E(3)$ by $S_0(2)+6c$
\begin{verbatim}
 Ta+Tb |    0        1        2        3        4    : carry b (wgt 6)
_______#__1___5 #  7__11   13__17   19__23   25__29  : translations Tb
     1 | .2.  6 |<- 6= 3+3                   xx
 0   5 |  6 .10.|<- S0(2)= {2,6,10}                25=5^2: no unit in G(3)
       #--------#   S0(3)= {2,6,..2n..,28)          6=3+3, 8=3+5, 10=5+5
     7 |  8  12  .14. 18
 1  11 | 12 >16<  18 .22.
 a  Ta  ........ ........+--------*
    13 | 14  18   20  24 |.26. 30 | <- S4(2)= S0(2)+4.6
 2  17 | 18  22   24  28 | 30 .34.|         ={26,30,34}
                +--------*--------+
    19 | 20  24 | 26  30 |
 3  23 | 24  28 | 30  34 |
       +--------*--------+
   5.5 x 26  30 |
 4  29 | 30  34 |
 ------*--------+
 \end{verbatim}

For instance extend ~$m_2$=6 ~to~ $m_3$=30 ~($p_3=5$) then
translations ~$S_{a+b}$~ of ~$S_0(2)$=\{2,6,10\} yield ~$5-1=4$
diagonals ~of~ 2 $\times$ 2 sums with carries ~$a+b<p_3$~ (table 2):

~~~~$S_1\{8,12,16\} ,~~S_2\{14,18,22\} ,~~S_3\{20,24,28\} ,~~S_4\{26,30,34\}$

Extending ~$G(2)=\{1,5\}$ ~yields~ $G(3)=\{~G(2)+6a~|~0<a<5~\}$,
containing prime set ~$P(3)=\{15 \pm 2^i,~29\}<m_3$ ~($i$=1,2,3)
~where $5^2$ is not coprime to 30, so not in $G(3)$.

\begin{lem}  \label{lem:base} % lem4.2
~(basis $k$=3): Let $P'(3)=P(3)\cup \{3,5\}$, then ~$P'(3)+P'(3)$ ~covers ~$E(3)$.
\end{lem}
\begin{proof}
By complete inspection. Increments 4 in ~$S_0(2)=\{2,6,10\}$~ cause
successive $S_c(2)=S_0(2)+c.m_2$ with carry increment $m_2$=6 to
$interlace$ for $2n<m_3$. Exclude non-prime 1 by including primes 3 and 5.
Then $E(3)=\{2n \in [~6,30)~\}$~ is covered by pair sums of primes
$p<m_3$ in $G(3)$, extended with primes 3,~5 in $G(1) \cup G(2)$.
\end{proof}

So pair sum set $S_0(3)$, adapted for the interlacing edge-effect by
including $\{3,5\}$, covers adjacent $2n$ in $E(3)$. Hence interlacing
does not occur for $k>$3, and a unique carry sum $a+b=c$ suffices for
covering successive $2n$ by unit pair sums, in adjacent and disjoint
extension sum intervals $S_c(k)$, while:

~~~~Each $2n$ in $E(k+1)$ ~has a unique carry sum c with
  $0 \leq c < p_k$, such that $2n \in S_c(k)$.

This is to be used as basis for $k>3$, ~first for unit pair sum sets ~$S_c(k)$.

{\bf Define:} The set $S_0(k)=G(k)+G(k)$ of pair sums of principle values of
units is called {\it complete} if it covers $E(k)$, otherwise it is $incomplete$.

\begin{lem} \label{lem:Sc(k)}          %lem4.3
~For ~$k \geq 3$:

~~~~Extended pairsum sets $S_c(k)$ ~for~ $0 \leq c < p_k$
       ~partition $E(k+1)$ ~iff~ $S_0(k)$ covers $E(k)$.
\end{lem}
\begin{proof}
 ~Extension sets $S_c(k)=S_0(k)+c.m_k$ are disjoint for different carries
$c< p_{k+1}$, and $\{x,y\}$ in distinct extension sum sets remain so under
any shift~ $s=c.m_k:~~x \neq y \iff x+s \neq y+s$. For distinct carrysums
$c<c'$ with $c'-c=d:~ S_c(k) ~\cap~ S_c'(k) = S_c(k) ~\cap~ (S_c(k)+d.m_{k-1})=
\emptyset$. Their union covers $E(k) ~only$ if pair sums $S_0(k)=G(k)+G(k)$~
cover $E(k)$. Because some $2n$ missing from $S_0(k)$ implies its translations
$2n'=2n+c.m_k$~ are also missing from all $S_c(k)$ with $c>0$.
\end{proof}

% sub4.3
\subsection{Excluding composites in $G(k)$, baseprimes and 1 as summands}

The set $G(k)$ of principle values (naturals) of units in group $G_1(k)$
coprime to baseprimes $2~..~p_k$, contains $p_{k+1}$ as smallest prime,
so the smallest composite in $G(k)$ is $(p_{k+1})^2$. Notice that $G(3)$
has no composites since $(p_4)^2=49>30=m_3$. Furthermore, the natural units
$u \in G(4)$ are in interval $(7<u<210)$ with smallest prime $p_5=11$,
hence minimal composite $11^2=121$, so all units of $G(4)$ in $[11,11^2)$
(coprime to 2.3.5.7=210) are prime. By inspection all $2n$ in interval
[22~..~222] are covered by prime pair sums, of which those $2n<210$
involve no carry.

The known Bertrand's Postulate is useful (Chebyshev 1850, simplified
by S.Pillai 1944) to prove a complete cover of even naturals:

{\bf BP} ~{\it (Bertrand's Postulate)}:
   ~~For each $n>1$ there is at least one prime between $n$ and $2n$.

Notice that  Pillai's proof [7] has an induction base of $2n \leq 60$
(see present lemma~\ref{lem:base}). In order to guarantee prime summands,
consider only pair sums of units $u<(p_{k+1})^2$, the smallest composite
in $G(k)$. In fact using $p_{k+1}<2p_k$ by Bertrands Postulate (BP),
the smaller interval $p_k < u< 2(p_{k+1})$ already suffices.
Successive $k$ yield $2n$ in overlapping intervals by $BP$, thus
covering all $2n$ beyond the induction base $k$=3. The next lemma is
readily verified, regarding the absence of a carry for $k > 3$.

\begin{lem} \label{lem:nocarry}  % lem4.4
For (natural) units in $G(k)$ and prime pairsums 2n in
$2~p_{k+1} \leq 2n < (p_{k+1})^2$ : no carry is produced for $k \geq 4$
since sum upperbound $(p_{k+1})^2 < m_k ~(~(p_{4+1})^2 =121 < m_4=210$).
\end{lem}

Notice that for initial $G(2)=\{1,5\}$ mod 6 (table 2) the baseprimes
2 and 3 are not used in pair sum residues $G(2)+G(2)= \{2,6,10\}$.
Considering $2n>4$ (re Goldbach's conjecture): non-prime 1 is avoided
by $6=3+3$ and $8=5+3$, the only $2n$ requiring 3. Moreover, 12=5+7
and 16=5+11 are the only extension pair sums $<30$ with one summand
of carry=0, thus requiring baseprime 5 of $G(3)$.


\section{Proving $GC$ by induction, or by reduction and contradiction}   %sect5

{\bf Approach} :~
Consider $G(k)$ (sect.~\ref{def:principle}) as set of 'natural units' $<m_k$
congruent to the units in group $G_1(k)$ of residues mod $m_k$, as defined in
the previous section. In other words, consider only the principle values in $G(k)$
\index{principle values} of the residue units in $G_1(k)$. Let $2n$ be small
enough, namely $2~p_{k+1} \leq 2n < (p_{k+1})^2$ with necessarily only prime
unit summands. There are two ways of proving $GC$: either a direct proof by
induction over $k \ge 3$ of primesums $2n<(p_{k+1})^2$ in $G(k)+G(k)$, or
an indirect proof by finite reduction of $GR(k)$ (thm~\ref{thm:GR}) and
contradiction to $S_0(3)$ (lemma~\ref{lem:base}). A direct proof, restricting
$GR(k)$ to the following primepair sums (item 3), would consist of the next
five steps:\\[1ex]
1. As summands use the principle value set $G(k)$ of the prime units in $G_1(k)$.\\
2. Complete inspection of $GC$ for $k=3$ (lemma~\ref{lem:base}).\\
3. Primesums $2n<(p_{k+1})^2$ in $G(k)+G(k)$ for $k>$3 yield no carry (lemma~\ref{lem:nocarry}).\\
4. Successive such restricted intervals of $2n$ intersect (Bertrand's Postulate).\\
5. The union over $k \ge 3$ of such primesums $2n$ in $G(k)+G(k)$ yield $GC$.

Using theorem~\ref{thm:GR} ($GR$) and lemma~\ref{lem:GRepi} an indirect proof
by \index{reduction} finite reduction: $S_0(k+1) \longrightarrow S_0(k)$ and
contradiction to lemma~\ref{lem:base} [~$S_0(3)$ covers $E(3)$~] runs as follows.

% thm5.1
\begin{thm}~(Goldbach's Conjecture)~~
    Each $2n>4$ is a sum of two odd primes. \label{thm:GC}
\end{thm}
\begin{proof}
By inspection (lemma~\ref{lem:base}) $GC$ holds for $4<2n<m_3=30$ ($k$=3).
Now assume $GC$ to fail for an even primepair sum $2n<(p_{k+1})^2$ in
$S_c(k) \subset G(k)+G(k)$, to guarentee prime summands (by eqn(4)).
Then lemma~\ref{lem:GRepi}: $G(k+1) \rightarrow G(k)$ mod $m_k$
implies $2n-c.m_k$ to be missing as primepair sum from $S_0(k)$, making it
incomplete by lemma~\ref{lem:Sc(k)}. This in turn, by $G(k) \rightarrow G(k-1)$
mod $m_{k-1}$, reduces to incomplete $S_0(k-1)$, etcetera, down to incomplete
$S_0(3)$, contradicting lemma~\ref{lem:base}. Notice that $S_0(3)$ contains
only primepair sums, by eqn(4). Combining this with the overlap (by Bertrand's
Postulate) of prime summand intervals for successive $k$, hence also of pairsum
intervals $2p_{k+1} \leq 2n<(p_{k+1})^2$ for $k \geq 3$, it follows that Goldbach's
Conjecture holds. \end{proof}

Regarding the values of prime summands that suffice to cover all even
naturals, the following can be said. Notice that in table 2 (for $2n<30$)
only primes $p \geq p_k$ are required to represent $2n \geq 2p_k$ in most
cases. However, exeptions occur if prime gap $p_{k+1}-p_k>2$. Then $2n+2$
requires a prime $p_{~k-1}$ or smaller: the larger the gap the smaller
$p_{~k-i}$ is required. See for instance (table 2): $2n = 2p_k +2=
16,~28,~40$ ~for $p_k=7,~13,~19$ respectively, which require $p_{k-1}$
as Goldbach summand, due to a gap $p_{~k+1} - p_k=4$ (versus gap 2 in
cases $p_k=5,~11,~17$).


\section{Conclusions}

Balanced analysis of multiplication and addition in relation to each
other, with finite square-free moduli $2...p_k$ yields a fruitful
analysis of prime sums (Goldbach), similar to that with prime power
moduli mod $p^k$ for $p$-th power sums (Fermat [3], Waring [8]).
In both  approaches the careful extension of residues with a $carry$
is essential for transferring additive structural results to integers.
This 'residue-and-carry' method, as used for proving FLT [3] and Goldbach's
Conjecture, is based on unique number representation by residue and carry:
using the associative (semigroup) properties of the residue closure,
combined with an induction proof by carry extension. As such it could
well serve as a generic method to solve other hard problems in
elementary number theory [6].

In fact, the semigroup $Z_m$(.) of multiplication mod $m$ is formed by the
{\it endomorphisms} of the additive cyclic group $Z_m$(+) generated by 1.
 So ~$Z_m(.)=endo[~Z_m(+)~]$ where (.) distributes over (+), suggesting
a strong link between these two operations, evident from the derived
additive fine structure of $Z_{m_k}$ for squarefree modulus $m_k$.
A two-dimensional table of prime pair sums revealed additive properties
of ~$2n<m_3=30$ as basis for the analysis, hard to find otherwise.

The product $m_k$ of the first $k$ primes as modulus restricts all
primes between $p_k$ and $m_k$ to the group $G_1$ of units. The additive
structure of $Z$(.) mod $m_k$ was analysed, and extended to positive
integers by considering the principle values (naturals) of residues,
starting with $k$=3 ($Z_{30}$). Units group ~$G_1(k)$, and the additive
properties of the Boolean lattice $BL$ of idempotents of $Z_{m_k}$(.)
play an essential role.

The lower semilattice of $BL$ is multiplicative, since the $meet$ glb($a,b$)
of two idempotents is their product. The additive properties of $BL$ were
analysed, regarding the $join$ lub($a,$b) in the upper semilattice. Although
$BL$ is not closed under (+) mod $m_k$, this yields the next main results :

Lem~\ref{lem:compl}:
  ~Each complementary pair of idempotents in $Z_{m_k}$(.) sums to 1 mod $m_k$

Cor~\ref{cor:G1=G2}:
   ~Congruent max cycles $2^* \cong (2+h)^*$ ~in~ $G_2 \cong G_1$ , with $h^2 \e h=m_k/2$

Lem~\ref{lem:e+1}:
   ~Each even [odd] idempotent $e^2 \e e$ ~has~ $e$+1 in $G_1$ ~[in $G_2$]

Thm~\ref{thm:GR}: ~Each residue $2n$ mod $m_k$ is a sum of two units :
           {\it Goldbach for Residues} ~ $GR(k)$

Consider principle value set $G(k)$ of units mod $m_k$ in group $G_1(k)$.

Eqn(4):
    Restrict $2n \in G(k)+G(k)$ to prime sums $2p_{k+1} \leq 2n < (p_{k+1})^2$.

Lem~\ref{lem:GRepi}:~~Epimorphism ~$G_1(k+1) \rightarrow G_1(k)$ mod $m_k$.

Thm~\ref{thm:GC}: Goldbach Conjecture $GC$ holds, via a proof by finite reduction:\\ \hspace*{.5cm}
  $incomplete~S_0(k+1) \longrightarrow incomplete~S_0(k)$ and contradiction to complete $S_0(3)$.

\newpage
\section*{References}

\begin{enumerate}
\item T.Apostol: "{\it Introduction to Analytical Number Theory}"
  thm 10.4-6, Springer Verlag 1976.
\item N.Benschop: "The semigroup of multiplication mod $p^k$,
  an extension of Fermat's Small Theorem, and its additive structure",
 {\it Semigroups and Applications} p7, Prague, July'96.
\item N.Benschop: "Additive structure of the Group of units mod $p^k$,
 with Core and Carry concepts for extension to integers", Acta Mathematica
 Univ. Bratislava (nov.2005) \\ http://pc2.iam.fmph.uniba.sk/amuc/\_vol74n2.html
  (pp169-184, incl. direct FLT proof)
\item G.Birhoff, T.Bartee: "{\it Modern Applied Algebra}", McGraw-Hill, 1970.
\item A.Clifford, G.Preston: "{\it The Algebraic Theory of Semigroups}",~
   Vol.I, \\AMS survey \#7, p130-135, 1961.
\item S.Schwarz: "The Role of Semigroups in the Elementary Theory of
   Numbers", \\{\it Math.Slovaca} V31, N4, pp369-395, 1981.
\item K.Chandrasekharan: "{\it Introduction to Analytic Number Theory}"
   (Ch.7 - Thm 4), \\Springer Verlag, 1968.
\item N.Benschop:  "Powersums representing residues mod $p^k$, from Fermat to Waring",
   \\ {\it Computers and Mathematics, with Applications}, V39 (2000) N7-8 pp253-261.
\end{enumerate}

\end{document}
