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\begin{center}
\vskip 1cm{\LARGE\bf 
Cyclic Products of Purely Periodic Irrationals 
}
\vskip 1cm
\large
C.~R.~Carroll\\
Department of Mathematics\\
Texas A\&M University\\
Kingsville, TX 78363 \\
USA\\
\href{maito:c-carroll@tamuk.edu}{\tt c-carroll@tamuk.edu}
\end{center}

\vskip .2 in


\begin{abstract}
Let  $\left(a_{0},  \dotsb  ,a_{k-1} \right)$ be a sequence of positive integers and  $m$ a positive integer.  We prove that   ``almost every'' real quadratic unit $\epsilon$ of norm $(-1)^k$  admits at least $m$ distinct factorizations into  a product of  purely periodic irrationals of the form
\begin{equation*}
{\epsilon=
[\overline{a_{0}; a_{1}, \ldots  ,a_{k-1},x,y}]\, \times \,}  { [\overline{a_{1}; a_{2}, \ldots, x,y  ,a_{0}}] }\,\times  \cdots \times  \,{[\overline{y;a_{0}, \ldots, a_{k-1},x}] }.
\end{equation*} 
Periods exhibited in this expression are not assumed minimal.  The analogous assertion holds for real quadratic units $\epsilon>1$ with prime trace and $m=1$.  The proofs are based on the fact that an  integral polynomial map of the form  $f(x,y)=axy+by+cx+d$, $\gcd(a,bc)=1$, $a>1$, $b,c>0$, assumes almost every positive integral value and almost every prime value when evaluated  over the positive integers.
\end{abstract}

  
  
\section{Introduction}\label{1}
To a sequence of positive integers $\nu=(a_0,a_1,a_2,\ldots, a_{k-1})$, $k\geq1$, we associate the real quadratic unit $\epsilon>1$  obtained by taking the following  product   of purely periodic quadratic  irrationals 
\begin{equation} \label{eq:aa}
\epsilon=\prod_{i=0}^{k-1} \ 
 [\overline{a_{i}; a_{i+1}, \ldots, a_{k-1}, a_{0}  ,a_{i-1}}]
.\end{equation}
Deviating from the standard convention we allow  periods exhibited in  this expression  to be multiples of the minimal period.  This convention is retained throughout.   An induction shows that  $\epsilon>1$ is a quadratic unit \cite{Y}.    Alternatively, this fact follows easily from the matrix approach  to the continued fraction algorithm (see van der Poorten  \cite{vdp} and \S \ref{4}); from this point of view (\ref{eq:aa})  corresponds to the matrix relation
\begin{equation} \label{eq:aaa}
\begin{pmatrix}
a_{0}& 1 \\ 1& 0
\end{pmatrix}
\begin{pmatrix}
a_{1}& 1 \\ 1& 0
\end{pmatrix} \dotsb
\begin{pmatrix}
a_{k-1}& 1 \\ 1& 0
\end{pmatrix}
\begin{pmatrix}
\alpha \\ 1
\end{pmatrix}
=\epsilon \begin{pmatrix}
\alpha \\ 1
\end{pmatrix}
\end{equation}
where
 \[\alpha=  [\overline{a_{0}; a_{1}, \ldots  ,a_{k-1}}].\]
Note from (\ref{eq:aaa})  that $\epsilon$ has norm  $(-1)^k$.   

Every  real quadratic unit $\epsilon>1$ has a  trivial factorization of the form given in (\ref{eq:aa}).  Let  $N$ denote the trace of  $\epsilon$.   For $\epsilon$ with norm $-1$ we have $\epsilon = [\overline{N;}]$; otherwise when the norm is  $+1$   we have  $\epsilon =[\overline{1;N-2}]\times [\overline{N-2;1}]$.


We call  a  product  of the form given in  (\ref{eq:aa}) a   \emph{cyclic factorization}  of the corresponding unit $\epsilon$.  Such factorizations are of interest from a variety of different  points of view. Distinct factorizations of  $\epsilon$  may be taken to represent distinct conjugacy classes of hyperbolic matrices in $GL(2,\mathbb{Z})$ with dominant  eigenvalue   $\epsilon$.   These classes, in the study of hyperbolic automorphisms of the torus,  correspond to  topological conjugacy classes of homeomorphisms, the invariants $\epsilon$ and $\nu$ having natural topological interpretations  \cite{adler}. By a result due to Latimer and MacDuffee  \cite{lm}  there is a further identification  with  ideal classes of the order $\mathbb{Z}[\epsilon]$. If $\epsilon_0$ is the fundamental unit of $ \mathbb{Q}[\sqrt{d}]$ ($d$ square-free),  then,   as explained by Yamamoto  \cite{Y}, the field class number  $h(d)$   can  be identified with  the number of cyclic factorizations of  $\epsilon_0$ satisfying the special condition that the factors have discriminant $d$.

With the aid of a computer the complete family of cyclic factorizations of a real quadratic unit $\epsilon>1$, for $\epsilon$ not too large, can be determined.  Looking at such families  one observes  in general a profusion of  cyclic factorizations,   the associated sequences $\nu$ displaying a great deal of randomness.  A natural question is the extent to which these sequences $\nu$  may vary over the family of factorizations determined by a unit $\epsilon$.    We call   a sequence of integers whose terms are known with the exception of $d$  integers in fixed positions a \emph{d-pattern}.  We consider the extent to which   units $\epsilon$ admit cyclic factorizations  instantiating a given $d$-pattern.     If the  length of the pattern is odd, it may be assumed the corresponding units  have norm $-1$,  a condition noted above to be necessary.   
 
 
For a $1$-pattern  $\nu(x)=(  a_{0} ,\ldots  ,a_{k-1},x )$, $k\geq 2$,  it is not hard to see that  a significant proportion of  real quadratic units $\epsilon>1$  of norm $(-1)^{k+1}$  have no factorization  instantiating  $\nu(x)$.   This is a consequence of the fact that the existence of a corresponding  cyclic factorization would induce a constraint on  the congruence class, relative to certain  moduli, of the  trace of $\epsilon$.   The same  constraint arises also for certain $d$-patterns,  $d>1$, for instance for  the patterns  $(a,x,a-1,a+1,y)$,  $a>1$.    However, if  a $d$-pattern has at least two adjacent free variables,    a corresponding cyclic factorization almost always exits.  We prove
\begin{theorem}\label{main}
Let $\nu=(a_0,a_{1},\dotsb , a_{k-1})$ be a sequence of positive integers and  $m$ be an arbitrary positive integer.  Almost every real quadratic unit $\epsilon>1$ of norm $(-1)^k$ admits a factorization over  $\mathbb{Q}[\epsilon]$ of the form
\begin{equation}
\label{eq:cc}
 \begin{split}
{ \epsilon =
[\overline{a_{0}; a_{1}, \ldots  ,a_{k-1},x,y}]\, \times \,}  { [\overline{a_{1}; a_{2}, \ldots, x,y  ,a_{0}}] }\,\times  \cdots \times  \,{[\overline{y;a_{0}, \ldots, a_{k-1},x}] }.
 \end{split} 
\end{equation}
The integers $x$, $y$ may be selected in at least $m$ different ways.    Furthermore, if $\nu \neq (1)$ almost every unit whose trace belongs to a fixed integer translate of the primes has at least one such  factorization.
\end{theorem}
The argument given  extends to certain $2$-patterns whose free variables are not adjacent.  Moreover, Theorem \ref{main}  immediately generalizes to finite families of sequences.  Fix a finite family of sequences $\cal{F}$  with lengths of a single parity.  For example we may take $\cal{F}$ to  consist of all sequences of odd (resp., even) length  whose length and terms satisfy  fixed    upper bounds.  Then  almost every real quadratic unit $\epsilon>1$   of  norm $-1$ (resp.,  $+1$) has  a cyclic  factorization of the form given in (\ref{eq:cc}) for each   $ \nu \in \cal{F}$.  In the contrary direction,  we see  in \S \ref{2} that given a fixed sequence  $\nu$  of length $k$   there exist infinitely many real quadratic units $\epsilon>1$ of norm $(-1)^k$  that have no corresponding  cyclic  factorization.    If Dickson's conjecture is correct, then given a finite family  $ \cal{F}$ as above,  these units may be taken to have no cyclic factorization corresponding to any sequence $\nu \in \cal{F}$.

The sense in which units are to be considered generic can be made
precise as follows. Let
$\epsilon_{(i,+)}$, $i \geq 3$, (resp., $\epsilon_{(i,-)}$, $i\geq 1$) 
denote
the real quadratic  unit of magnitude greater than one with trace $i$ and  norm $+1$ (resp., $-1$).
The magnitude of such a unit is well approximated by its trace (since  $ \left| \epsilon_{{(i,\pm )}} -i    \right|<1$). Let $\cal{T_+}$ (resp., $\cal{T_-}$) be the set of traces $i$ of units of the form   $\epsilon_{{(i,+)}}$ (resp., $\epsilon_{{(i,-)}}$) that  satisfy a given statement.  The statement will be said to hold of \textit{almost every}  unit $\epsilon_k$ of norm $+1$ (resp., $-1$)  or, also, of a \textit{generic unit} of this type,  if and only if the corresponding  subset of   $\cal{T_+}$  (resp., of $\cal{T_-}$)  is of asymptotic density one in the integers.   The same approach can be employed to define genericity relative to any infinite subset of units   $\epsilon_{{(i,+)}}$ (resp., $\epsilon_{{(i,-)}}$); in particular, the trace $i$ may be taken to be prime. It is not hard to show that  almost every  real quadratic unit $\epsilon_{{(i,+)}}$ (resp., $\epsilon_{{(i,-)}}$) is a fundamental unit \cite{sprind}.    


The proof of Theorem \ref{main} rests on the following result  on the representation of integers by polynomials.    Let $f(x,y)=axy+by+cx+d$ be an integral  polynomial such that  $a>1$, $b,c\in \mathbb{Z}^+$, and $\gcd(a,bc)=1$. Let $P$ denote the set of  primes.   

\begin{theorem} \label{mainr}
 Let $R_f\, = \, \left\{ f(x,y):  x,y \in \mathbb{Z}^+ \right\}$. Fix $m \in \mathbb{Z}^+$.   Then:
(i) $R_f\cap \mathbb{Z}^+$
is of asymptotic density one within the set of positive integers, each element of the set being represented by at least $m$ distinct pairs of positive integers $(x,y)$; (ii)  $R_f\cap P$
is of relative asymptotic density one in $P$.
\end{theorem}
Assertion (i) follows quickly from the fact that, given a fixed  modulus $a$,   almost every integer has at least one divisor in each    residue class $r \bmod{a}$,  $\gcd(r,a)=1$  \cite{erdos2}.  S.K. Stein \cite{stein} obtains a more general factorization result from which he derives (i) with $m=1$ and $d=0$.  The proof of  (ii) relies on Dirichlet's and de la Vall\'ee Poussin's theorems, together with a modern variant of Euler's  product identity for arithmetic progressions.  The relative density of the set of primes that is represented   by a general  two-variable integral quadratic polynomial over the integers  was investigated in well-known work of Iwaniec \cite{iwaniec}.   Their results imply that the above polynomials $f(x,y)$  assume a set of prime values of positive relative density when evaluated over the integers.  We should note that in contrast with the  restricted class of polynomials considered here, which assume a set of integer values of asymptotic density one, the usual    number-theoretic focus is on polynomials that assume a sparse set of integer values.   It is possible to extend the argument given for (ii)  to a more general class of integral polynomials of the form $f(x,y)=(rx+ty)(ux+my)+bx+cy+d$, subject to various restrictions on the parameters, but this level of generality is not needed for our purposes.
  
 In \S \ref{2}-\ref{3} a proof of  Theorem \ref{mainr} is given.  The derivation of Theorem \ref{main}  is given in \S \ref{4}. 





\section{ The representation of positive integers }\label{2}
  

Let $f(x,y)=a\,xy+by+cx+d$ be an integral polynomial whose coefficients satisfy
\begin{equation}\label{cond} a>1,\  \ b,c >0,\  \ \gcd(a,bc)=1.\end{equation}
We consider the  positive integral values taken by $f(x,y)$ over the positive integers. This set may be identified with  $f(\mathbb{Z}^+  \times \mathbb{Z}^+) $ up to a possible finite number of non-positive values that may arise when $d<0$.


  
Given $N\in \mathbb{Z}^+$ we say  $f(x,y)$ \textit{represents} $N$ if  there exist  integers $x$, $y \in \mathbb{Z}^+$ such that $f(x,y)=N$; it should be emphasized that the representation we are considering is over the  positive integers, an assumption necessitated by our later arguments (see \S \ref{4}). We will say $f(x,y)$ represents a set  $B\subset \mathbb{Z}^+$ if it represents  each element of $ B$ in the above sense.
  

Fix an infinite set of positive integers $B$ and let $A\subset B$. Then the \textit{asymptotic density} (or\textit{ natural density}) of $A$ relative to $B$ is defined to be the limit
 \[ d_B(A)=\lim_{k\rightarrow \infty} \frac{\left| A\cap \left\{1,2,\ldots,k\right\}\right|}{\left|B\cap \left\{1,2,\ldots,k\right\}\right|} \] 
provided it exists.   If $d_B(A)=1$ we will say  that \textit{almost every element of $B$ lies in $A$} or, alternatively, that \textit{$A$ is of full density in $B$}.  On occasion the reference to $B$ is omitted, in which case it is to be assumed $B=\mathbb{Z}^+$ . 

In this section we show that any  polynomial $f(x,y)$ satisfying the  conditions stated in  (\ref{cond})  represents almost every positive integer, or $d_{\mathbb{Z}^+}(f(\mathbb{Z}^+ \times \mathbb{Z}^+)\cap \mathbb{Z}^+)=1.$


We recall some  basic facts. Let $B$ be an infinite set of integers and
let $\cal{C}$ be the collection of all subsets of  $B$ having
well-defined asymptotic density relative to $B$.  It is  well-known
that $\cal{C}$  is closed under finite disjoint unions and under
complementation but is not closed under intersection and hence does not
form an algebra.  A  useful property of $d_B$  is that it is  additive
over disjoint sets in $\cal{C}$.  Below we list several additional
useful properties \cite[pp.~79--80]{nark}.


\begin{lemma}
\label{eprop}
Let  $B$  be an infinite subset of positive integers and $A\subset B$.
\renewcommand{\labelenumi}{(\roman{enumi})} 
\begin{enumerate}
\item
If $A$ is finite then $d_B(A)=0$.
\item
Assume  $A$  has well-defined density, given by $d_B ( A) =d$.  If  $A^c$ is  the complement of $A$ in $B$ then $d_B(A^c)=1-d$.
\item
Let $A_i$, $1\leq i \leq k$, be a collection of subsets of $B$ for which $d_B ( A_i) =1$.  Then the asymptotic density of $\bigcap_{i=1}^{k} A_i $  exists and is given  by  $d_B(\bigcap_{i=1}^{k} A_i)=1$.
\item
Assume   $A_1\subset A_{2} $  are nested subsets of $B$   with well-defined asymptotic densities.        Then $ d_B( A_1)\leq  d_B( A_2)$.
\item
Let $A$ be any subset of positive integers. Fix $k_1 \in \mathbb{Z}^+$ and $k_2\in \mathbb{Z}$. Then  $d_{\mathbb{Z}^+}(A)=d$ if and only if  $d_{\mathbb{Z}^+}(k_1\,A+k_2 \, \cap \,\mathbb{Z}^+)=d/k_1$.  
\item
If $A_1,\, A_2 \subset B$ satisfy $d_B(A_1)=1$ and $d_B(A_2)=d$  then the density of $ A_1\cap A_2$ relative to $B$ is well-defined and given by $d_B(A_1\cap A_2)=d$.
\end{enumerate}
\end{lemma}


We   consider now the  positive  integers $N$  represented by $ f(x,y) $. A standard approach to finding solutions of  an equation of the form $a\,xy+by+cx+d =N$  is to rewrite it as 
\begin{eqnarray}
aN - (ad - bc)\, & = &\,(ax + b) (ay + c). \label{two}
\end{eqnarray}
From this equation we see that  $N$ is represented by $f(x,y)$ exactly when  $aN - (ad - bc)$ is the product of two   integers greater than $a$ which belong to  suitable  residue classes modulo $a$.   Of course, such a factorization need not exist; in particular $aN - (ad - bc)$ may be prime.   By Dirichlet's theorem
\begin{theorem}[Dirichlet] \label{dirichlet} Every arithmetic progression of the form $a\,x+b$, with $a,\, b$ non-zero integers satisfying $a >0$ and $\gcd(a,b)=1$, contains infinitely many primes.   In fact, the sum of the reciprocals of the primes generated by such a progression diverges.
\end{theorem}
Given that the fixed parameters of  (\ref{two})  satisfy $\gcd(a,ad - bc)=1$, by  Dirichlet's theorem   the integer $ aN - (ad - bc)$ is prime  for infinitely many positive integral values of $N$; hence these values cannot be  represented by $f(x,y)$.     On the other hand we will see now that $f(x,y)$ represents almost every positive  integer.   This is a consequence of the fact that the divisors of an integer are in general well-distributed over residue classes.   As stated by  Erd\H{o}s \cite{erdos2},
\begin{proposition}\label{erdos2}
Let $a\in \mathbb{Z}^+$, $a>1$.  Let $F(a)$ denote the set of positive integers that have at least one divisor congruent to $k$ modulo $a$ for every integer $k$ relatively prime to $a$.  Then $F(a)$ is of full asymptotic density in $\mathbb{Z}^+$.
\end{proposition}

By  Lemma~\ref{eprop} (vi) this observation extends to any subset $S\subset \mathbb{Z}^+ $  of positive asymptotic density; that is,  
$d_S( S\cap F(a))=1$.  With this fact in hand it is easy to see that for  almost every positive integer $N$ there  exist  non-negative integers  $x,\, y$ satisfying  (\ref{two}).



Let $T$ be the affine transformation given by  $T(N)=a\,N - (ad - bc)$.   The set $T(\mathbb{Z}^+)$ may include a finite number of  non-positive integers which can be ignored  since for any solution $x,y>0$ of    (\ref{two})  the right-hand side of the equation must  be positive.   Accordingly  let   $B= T( \mathbb{Z}^+)\cap \mathbb{Z}^+$.  Consider the subset $B_1= B \cap F(a)$.    Since by   Lemma \ref{eprop} (v)   $d_{\mathbb{Z}^+}(B)=1/a$, we have  $d_{\mathbb{Z}^+}(B_1)=1/a$ as well.

An  element  $m=T(N') $ of $B_1$ must satisfy the congruence $m\equiv bc \pmod{a}$.  Given that $\gcd(b,a)=1$, it follows  $m$ has  a factorization of the form  $m=m_1\, m_2$ with  $m_1 \equiv b \pmod{a}$ and hence with  $m_2 \equiv c \pmod{a}$.  Therefore  (\ref{two})  has a solution  in non-negative integers  $x,\, y$ for  $N=N'$  and indeed  for any $N \in T^{-1}(B_1)$.  


Consider now the   asymptotic density of  $T^{-1}({B}_1 ) $.  To simplify notation put $l=ad - bc$.   Let $D_k=T^{-1}({B}_1 )  \cap   \left\{1,2,\ldots,k\right\} $.   Then  $T(D_k)= B_1 \cap \left\{ 1,\, 2,\, \ldots, ak-l \right\}$ must be a set of the same cardinality, say $n_k $.  Hence we may express the asymptotic density of $B_1$  as  \[d_{\mathbb{Z}^+} (B_1) =\lim_{k\rightarrow \infty} \frac{n_k}{ak+l} =  \frac{1}{a} \lim_{k\rightarrow \infty} \frac{{n_k}}{k}. \] 
Recalling that $d_{\mathbb{Z}^+} (B_1)=  1/a$ this shows $  \lim_{k\rightarrow \infty} \frac{{n_k}}{k}=1$.   Therefore  $T^{-1}({B}_1 )$ is of full density. 

We now know that for almost every positive integer $N$  there exist non-negative integers $x,\, y$ satisfying (\ref{two}).  It remains to be shown that there exist multiple  positive solutions.

Let $\mathbb{Z}(a;s)$ denote   the set of all positive integers $m$ that are congruent to $s$  modulo $a$.
\begin{proposition} \label{corerd}
Let $k$ be an arbitrary positive integer and let $a,b,c\in \mathbb{Z}^+$ such that $a>1$ and $\gcd(a,bc)= 1$.  Then almost every integer $m\in\mathbb{Z}(a;bc)$ admits at least $k$ distinct factorizations of the form $m=(xa+b)(ya+c)$,  
where $x,\,y\in \mathbb{Z}^+$. 
\end{proposition}
\begin{proof}   
Let  $s,\, k \in \mathbb{Z}^+$  and $q$ be a prime  such that  $s>2k+\left\lfloor {\frac{c}{a}} \right\rfloor $ and  $q> b+s\,a$.   Since $\gcd(a,b)=1$  the integers $b+i\,a$,  $0< i \leq s$, are  relatively prime to $qa$.  Note that they represent $s$ distinct reduced residue classes modulo $qa$. 
  By the  remark following Proposition \ref{erdos2},  $\mathbb{Z}(a;bc)$ contains a subset $\mathbb{Z}_1(a;bc)$
of full relative density   whose elements have divisors in each of these residue classes.  Fix   $m^\prime\in\mathbb{Z}_1(a;bc)$.  Then, given any  residue class $ b+i_0\,a \bmod{qa}$, $0< i_0 \leq s$,   we may write $m^\prime$  as a product of positive integers  $m^\prime=m_1 m_2$ where \[m_1\equiv b+i_0\,a\pmod{qa}.\]    Since $m_1 m_2\equiv b m_2\equiv b\,c \pmod{a}$  it follows that $m_2\equiv c \pmod a$.  
Put  $\overline{c}= c-\left\lfloor \frac{c}{a}\right\rfloor$.  
We may write
\begin{eqnarray}
m^\prime\,= \,m_1\,m_2 \, = \,( b+i_0\,a+i_1\,qa)\,( \overline{c}+j_0\,a) \nonumber
\end{eqnarray} 
for  non-negative   integers $ i_1   $ and $j_0$.
Notice that  $m_1 = b + xa$ with $ x = i_0 + i_1q > 0$.
When $i_0$ varies the factor $m_1$  determines at least $s$ distinct residue classes modulo $qa$.  We have  $ j_0\leq\left\lfloor {\frac{c}{a}} \right\rfloor$  for at most $\left\lfloor {\frac{c}{a}} \right\rfloor+1$ values.  In the remaining cases   $m_2$ is of the form $c+ya$ with $y>0$.  
Thus we have products  $m^\prime\,=\,( b + xa)(c+ya)$ with $x,\,y$ positive and with at least  $2\,k$  choices for the first factor.   These  factorizations are distinct unless  $b= c$, in which case there are still at least $k$ distinct factorizations.
\end{proof}

We now are in a position to conclude
\begin{corollary}  \label{int}
Fix $k\in \mathbb{Z}^+$.  Let $f(x,y)=axy+by+cx+d$ be an integral polynomial such that $a,b,c>0$, $a>1$, and $\gcd(a,bc)= 1$. Then there exists a set of positive integers $N$ of full asymptotic density such that $f(x,y)=N$ for at least $k$ distinct pairs of positive integers $(x,y)$.
\end{corollary} 


An alternative proof of Corollary \ref{int}  can  be given starting from  the following  reformulation of
 $a\,xy+by+cx+d=N$, $a\neq 0$:
\begin{eqnarray} \nonumber
N\,=\,(ax+b)y+cx+d. \label{one}
\end{eqnarray}
The idea, roughly, is as follows.    Since $\gcd(a,b)=1$  Dirichlet's theorem yields an infinite sequence of distinct primes $p_i = a\,x_i+b  .$   Each value of the index $i$  in turn determines a corresponding arithmetic progression $p(x_i,y)=p_i\,y + (c\,x_i +d)$ in the variable $y$.   
By a result due to C. A. Rogers 
\cite[p.~242]{halberstam} the asymptotic density of the positive integers that are  realized by at least one of  these progressions is bounded below by the asymptotic density of the set of positive integers that are  divisible by at least one of the primes $p_i$.   This density is known to be $1$ provided the sum $\sum_{j=1}^\infty \frac{1}{p_i}$ diverges \cite{nark}, a fact which in the case at hand is a consequence of Dirichlet's theorem.  Consequently $T(x,y)$ represents almost every positive integer  over $\mathbb{Z}^+ \times \mathbb{Z}^+$.  A trick similar to that used in the proof of Proposition \ref{corerd} is needed to  obtain multiple representations.



  
  

\section{The representation of primes}\label{3}

Corollary \ref{int} gives no information about the representation of sets of integers of asymptotic density zero.  In this section we consider the representation of primes.  We  show
\begin{proposition} \label{adp} 
Let $f(x,y)\,=\,a\,xy+by+cx+d$ be an  integral polynomial whose coefficients satisfy $a>1$, $b,c>0$, and $\gcd(a,bc)=1$.   Then  $f(x,y)$ represents  (over the positive integers) a set of primes $p$  of full relative density.
\end{proposition}

\begin{remark}
Since the integer  $d$ is arbitrary, $f(x,y)$ also  represents a subset of full  relative density of any fixed integer translate of the primes.
\end{remark}

The proof of Proposition \ref{adp}  is based on  Dirichlet's theorem and   the following additional classical results.    De la Vall\'ee Poussin established that the  primes  generated by an arithmetic progression are equidistributed among the possible residue classes.  Let  $P$ be the set of  primes and  $P_{a,b}$  the set of primes generated by the arithmetic progression $a\,x+b$,   $a>0,\ \gcd(a,b)=1$.  We have
\begin{theorem}[de la Vall\'ee Poussin] \label{dv} \[ \lim_{k\rightarrow \infty} \frac{\left| P_{a,b}\cap \left\{1,2,\ldots,k\right\}\right|}{\left|P\cap \left\{1,2,\ldots,k\right\}\right|}=\frac{1}{\phi(a) }\],
\end{theorem} 
\noindent where $\phi$ denotes Euler's totient function. 
 
We also  rely on the following variant of Euler's famous product identity for  primes generated by an arithmetic progression  \cite{lang}:
\begin{theorem}\label{eul}
Let $a,\,b$ be integers such that $a>0$ and $\gcd(a,b)=1$.   Then
\[ \prod_{q\in P_{a,b}} \, \left(1-\frac{1}{q} \right) =0 . \]
\end{theorem}
 
To establish  Proposition \ref{adp},  we  consider the families of primes generated by arithmetic progressions obtained by evaluating the polynomials $f(x,y)\,=\,(a\,x+b)\,y+cx+d$ at selected  values of  $x$.


To fix ideas  let us consider an  arbitrary pair of arithmetic progressions $q_1\,y+r_1$, $q_2\,y+r_2$,   with   $\ q_1,\, q_2$  distinct odd primes and $r_1,\,  r_2$ relatively prime to $q_1,\, q_2$, respectively. We wish to determine the relative asymptotic density of the set of primes $A$ generated by these progressions,   $A=P_{q_1,r_1}\cup P_{q_2,r_2}$.  This is most conveniently accomplished by computing the relative asymptotic density of the complementary set of primes $P-A$.   Notice that only finitely many primes $q_c\in P-A$  can belong to one of the four residue classes $0 \pmod{q_i}$, $r_i \pmod{q_i}$, $i=1,2$.    The remaining primes $q_c$  are distributed across $ (q_1-2)(q_2-2)$  possible pairs of residue classes.  Let   $ k_1\pmod{q_1}$,  $ k_2 \pmod{q_2}$  be such a pair, with the residues $k_i$ taken to be reduced modulo $q_i, \ i=1,2$.   
By the Chinese remainder theorem the condition that the prime  $q_c$ belong to this pair of  classes is equivalent to a condition of the form
\begin{equation}\label{eq:ch}
q_c\equiv {s}_3 \pmod{q_1 q_2},
\end{equation}
with  $s_3$ a fixed integer such that $1\leq s_3 \leq q_1 q_2-1$.   Since  $k_1,\,k_2 \neq 0$, we have  $\gcd({s}_3,q_1\,q_2) =1$.  Applying  Theorem \ref{dv} we see that the set of all primes satisfying (\ref{eq:ch}) has relative density
\begin{equation*} d_P( P_{q_1q_2,s_3})\,=\,
 \lim_{k\rightarrow \infty} \frac{\left| P_{q_1q_2,s_3}\cap \left\{ 1,2,\ldots, k\right\}\right|}{\left|P\cap \left\{1,2,\ldots, k\right\}\right|}=\frac{ 1}{\phi(q_1\,q_2)}=\frac{1}{(q_1-1) (q_2-1) }.
\end{equation*}
Since this computation does not depend on which of the $ ( q_1-2 )(q_2-2 )$ pairs of residue classes under consideration is selected,  the total number of primes  in  $P-A$ is asymptotic to 
\[  ( q_1-2 )\,(q_2-2 )\,\frac{1}{(q_1-1) (q_2-1)}=\left(1- \frac{1}{q_1-1} \right)\left(1- \frac{1}{q_2-1}\right)\]   

The  argument generalizes to  any number $k$ of progressions.   Thus we have
\begin{lemma}\label{eugn2}
Let $q_j$,  $1\leq j\leq k$, be  $k$ distinct primes  and  $r_j$  corresponding  integers such that   $\gcd(q_j,r_j) = 1$.  Let $A=\bigcup_{j=1}^k P_{q_j,r_j}$.
Then the asymptotic density  of  $A$ within the set of primes is given by \[ 1-\prod_{j=1}^k \,\left( 1-\frac{1}{q_j-1}\right)\]  
\end{lemma}

This fact can be extended to the case of an infinite union  $\bigcup_{j=1}^\infty P_{q_j,r_j}$   by applying Lemma \ref{eprop} (iv) to the nested sequence of  sets \[A_i=\bigcup_{j=1}^i P_{q_j,r_j},\ \ i\in \mathbb{Z}^+  .\]    We obtain

\begin{lemma}\label{prodinf} 
Let $q_j$ be an infinite sequence of distinct primes and  $r_j$ a corresponding sequence of integers satisfying  $\gcd(q_j,r_j)=1 $.  Then   $\bigcup_{j=1}^\infty P_{q_j,r_j}$ is a set of full density within the set of primes  provided 
 \begin{equation*}
\prod_{j=1}^\infty \,\left( 1-\frac{1}{q_j - 1}\right)=0.
\end{equation*}
\end{lemma} 

The proof of Proposition \ref{adp} is now straightforward.  Assume the polynomial  \begin{equation} \nonumber f(x,y)=axy+by+cx+d =(ax+b)y+cx+d \label{eq:f}\end{equation} satisfies the hypotheses of Proposition \ref{adp}.   By Dirichlet's theorem the values of $ax+b$ are prime   for an increasing sequence of  positive integers $x_i$.  We write $p_i = a\,x_i+b $, $r_i =cx_i+d$.  Consider  the arithmetic progression in $y$ given by $f(x_i,y)=p_i\,y +r_i$.  If $\gcd{(p_i,r_i)}=1$,  Dirichlet's theorem  can be applied a second time.  
 \begin{lemma}\label{pred}
There exists an integer $i_0$ such that for $i \geq i_0$  \[\gcd(p_i,r_i)=\gcd(ax_i+b,c\,x_i+d)=1  .   \]
 \end{lemma}
\begin{proof} 
Since $a\neq 1$ and $\gcd(a,c)=1$, it follows $a \neq c$.

\medskip

\textit{Case (i):}  $a>c$.  For $x_i$ sufficiently large  necessarily  $0<cx_i+d<ax_i+b$.   Since $ p_i=ax_i+b$ is prime  $\gcd(p_i,cx_i+d)=1$.  

\medskip

 \textsl{Case (ii):}  $a< c$.    Since $\gcd(a,c)=1$ we have $c=k\,a+ c^\prime$ for positive integers $k$ and $c^\prime$ with $0< c^\prime<a$.  Thus we may write 
\begin{eqnarray*} 
f(x_i,y)&=&(ax_i+b)y+(k\,a+ c^\prime)x_i+d\\
&=&(ax_i+b)(y+k)+c^\prime x_i+d^\prime 
\end{eqnarray*}
 where $d^\prime=  d-k\,b$.   Again,  for $x_i$ sufficiently large
$0<c^\prime x_i+d^\prime<ax_i+b$.   Hence $\gcd(ax_i+b,cx_i+d)= \gcd(ax_i+b,(c x_i+d) -k(ax_i+b))= \gcd(ax_i+b,c^\prime x_i+d^\prime) =1.$  
 \end{proof}
  
Hence we may apply Dirichlet's theorem  to the progressions $p_iy+r_i$ for sufficiently large values of the index $i$, say, $i\geq i_0$; accordingly,   $f(x,y)$ represents the  primes   $P_{p_i,r_i }   $  for $i\geq i_0$.


Consider $\bigcup_{i=i_0}^\infty P_{p_i,r_i }   $.   Theorem \ref{eul} yields \[ \prod_{i=i_0}^\infty \, \left(1-\frac{1}{p_i-1} \right) \ \leq \ \prod_{i=i_0}^\infty \, \left(1-\frac{1}{p_i} \right)\, =\, 0 .\] It follows by  Lemma \ref{prodinf}  that  the  set of primes $\bigcup_{i=i_0}^\infty P_{p_i,r_i } $  is of full density in the primes. Therefore $f(x,y)$ represents,  over the positive integers, a set of primes of  full relative density, completing the proof of Proposition \ref{adp}.


 
 
 


\section{Periods prescribed up to two adjacent integers}\label{4}


There exists  a  well-known correspondence, mentioned already  in \S {1}, between certain matrix  products  and continued fractions.   In particular, the  following two statements may be regarded as equivalent: 
\begin{enumerate}
\renewcommand{\labelenumi}{(\roman{enumi})} 
\item 
$\alpha=  [\overline{a_0;a_{1},\ldots ,a_{k-1}}]$   with $k$ a fixed  multiple of the minimal period).
\item
For some unit     $\epsilon \in \mathbb{Q}[\alpha]$,  $\epsilon >1$,
\begin{equation} \label{eq:a}
\begin{pmatrix}
a_{0}& 1 \\ 1& 0
\end{pmatrix}
\begin{pmatrix}
a_{1}& 1 \\ 1& 0
\end{pmatrix} \dotsb
\begin{pmatrix}
a_{k-1}& 1 \\ 1& 0
\end{pmatrix}
\begin{pmatrix}
\alpha \\ 1
\end{pmatrix}
= \epsilon \begin{pmatrix}
\alpha \\ 1
\end{pmatrix}.
\end{equation}
\end{enumerate}

From the second equation we   obtain a  corresponcing factorization of $\epsilon$,  as follows.  Let $\alpha_m$, $m \geq 0$ denote    the purely periodic irrational  obtained by cyclically permuting the period of the expansion of   $\alpha$  so that $a_m$ occurs as the initial element, i.e.,
\[
\alpha_{m} =[\overline{{{{a_{m}}; a_{m+1},\dotsb,a_{k-1},a_0, \dotsb,a_{m-1} }}}].
\]
Notice that $\alpha_{i+k}=\alpha_i$.  A computation yields
\begin{equation} \label{e}
\begin{pmatrix}
a_{m}& 1 \\ 1& 0
\end{pmatrix}^{-1}  
\begin{pmatrix}
\alpha_m \\ 1
\end{pmatrix}
= \frac{1}{\alpha_{m+1}} \,\begin{pmatrix}
\alpha_{m+1} \\ 1
\end{pmatrix}.
\end{equation}
The  $k$ matrices occurring in   (\ref{eq:a}) may be eliminated  by repeatedly multiplying each side of  the equation   by the inverse of the left-most matrix and applying the  equality of Equation (\ref{e}) to the right-hand side of the equation.  After the $k$-th multiplication we are left with
\[  
 \begin{pmatrix}
\alpha \\ 1
\end{pmatrix}
= \frac{\epsilon}{\alpha_0\alpha_{k-1} \cdots   \alpha_1} \,
\begin{pmatrix}
\alpha\\ 1
\end{pmatrix}.\]
Hence
\begin{equation} \label{cfprod}
\epsilon \, = \, 
\alpha_0\, \alpha_1 \ldots \alpha_{k-1}.
\end{equation}

We call a matrix  of the type arising  in  Equation (\ref{eq:a})---that is, either a single matrix of the form $\begin{pmatrix}a & 1 \\ 1& 0\end{pmatrix}  $, $a \in \mathbb{Z}^+$, or  a finite product of such matrices---a \textit{CF-matrix} of \textit{depth} $k$.  Such matrices have a simple  characterization \cite[Proposition 3]{vdp}: namely, they are the  non-negative elements $A=\begin{pmatrix} s_1& s_2 \\ s_3& s_4\end{pmatrix}$ of  $ GL(2,\mathbb{Z})$  for which $s_1 \geq \max{(s_2,s_3)}.$
By an elementary induction  if $s_1 \neq 1$ the inequality is strict, $s_1 > \max{(s_2,s_3)}$.  

One may easily verify that an arbitrary CF-matrix $A$ of depth $k$ in addition satisfies the following  properties: $\texttt{Det}(A) = (-1)^{k}$;  $s_1,s_2,s_3>0$;  $\gcd(s_1,s_2)=\gcd(s_1,s_3)=1$.


Given  a sequence of positive integers   $\nu=(a_0, a_{1},\dotsb , a_{k-1})$,  let 
 \begin{equation}\label{eq:e}
M_\nu(x,y) =
\begin{pmatrix}
a_{0}& 1 \\ 1& 0
\end{pmatrix}
 \dotsb
\begin{pmatrix}
a_{k-1}& 1 \\ 1& 0
\end{pmatrix}
\begin{pmatrix}
x& 1 \\ 1& 0 
\end{pmatrix}
\begin{pmatrix}
y& 1 \\ 1& 0
\end{pmatrix}
\end{equation}
 and let $\epsilon_\nu(x,y)>1$  denote the  dominant eigenvalue of $M_\nu(x,y)$. Writing
\[\begin{pmatrix}
b_1& b_2 \\ b_3& b_4
\end{pmatrix} = \prod_{0 \leq i \leq k-1}\begin{pmatrix}
a_{i}& 1 \\ 1& 0
\end{pmatrix}, \]
we obtain the following expression for  the  trace of $M_\nu(x,y)$  (which  by the Cayley-Hamilton theorem  also gives the trace of its eigenvalues)
\[T_\nu(x,y)=b_1xy+b_2y+b_3x+b_1+b_4.\]



Thus if $N \in \mathbb{Z}^+$ is represented by the polynomial $T_\nu (x,y)$---that is, $ N=T_\nu (x_0,y_0)$ for some choice of positive integers  $x_0,\ y_0$---then $\epsilon_\nu(x_0,y_0)$  is the dominant eigenvalue of  the matrix  $M_\nu(x_0,y_0)$ and by  (\ref{cfprod}) it follows that    $\epsilon_\nu(x_0,y_0)$   admits
 the cyclic factorization
\begin{equation} \nonumber
{
[\overline{a_{0}; a_{1}, \ldots  ,a_{k-1},x_0,y_0}]\, \times \,}  { [\overline{a_{1}; a_{2}, \ldots, x_0,y_0  ,a_{0}}] }\,\times  \cdots \times  \,{[\overline{y_0;a_{0}, \ldots, a_{k-1},x_0}] }.
\end{equation}


By the characterization of  CF-matrices provided above  the polynomial  $T_\nu(x,y)$ satisfies the hypotheses of Theorem \ref{mainr} provided  $\nu \neq (1)$.   When   $\nu = (1)$,  $ T_\nu(x,y)=(x+1)(y+1)$. In the former case, applying  Theorem \ref{mainr}, and in the latter,  noting that composite integers form a set of integers of asymptotic density one, we may conclude that $T_\nu(x,y)$ represents almost every positive integer $N$  in at least $m$ different ways.   Turning to the representation of primes,  assuming   $\nu \neq(1)$,  by Theorem \ref{mainr}  almost every prime is represented as well.  This yields Theorem \ref{main}.

\section{Acknowledgement}

The author is grateful to an anonymous referee  for helpful comments and  corrections.


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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A55; Secondary 11N32, 37D20.

\noindent \emph{Keywords: } 
simple continued fraction, purely periodic irrational, prime
represented by quadratic polynomial, linear automorphism of the torus,
Anosov automorphism.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 3 2013;
revised version received   March 21 2014.
Published in {\it Journal of Integer Sequences},
March 22 2014.

\bigskip
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\noindent
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