\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf On Some Properties of Two Simultaneous \\ \vskip .2cm Polygonal Sequences} \vskip 1cm \large Joseph C. Su\\ P. O. Box 710\\ Alhambra, CA 91802\\ USA\\ \href{mailto:jsu800@alum.mit.edu}{\tt jsu800@alum.mit.edu} \\ \end{center} \begin{abstract} This paper focuses on the properties of integers that are simultaneously representable as both $m$-gonal and $n$-gonal numbers. Employing hyperbolic geometry, we ascertain both lower and upper bounds on these numbers when there is a finite set of these integers. We then consider the fundamental class of solutions to a generalized Pell-Diophantine equation that relates to these integers, and derive a fast algorithm that can be used to generate them. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{definition}{Definition}[section] \newtheorem{example}[theorem]{Example} \newtheorem{ex}[corollary]{Example} \newtheorem{ex-def}[definition]{Example} \newenvironment{pf}{\noindent{\bf Proof. }}{\hfill $\Box$ \\} \begin{section}{Introduction} Polygonal numbers can be represented by a group of dots in arithmetic progression. The $r$-th $m$-gonal number can be expressed as \begin{equation} \label{polygonal-equation} f_r^m=\frac{(m-2)r^2-(m-4)r}2. \end{equation} \noindent For example, the triangular numbers are $1, 3, 6, 10, \ldots $, \seqnum{A000217} and the square numbers are $1, 4, 9, 16, \ldots$, \seqnum{A000290}. Geometrical language for these numbers is justified in Fig.~\ref{polyfigure}. \begin{figure}[hbtp] \centering \includegraphics[width=0.7\textwidth]{polygonal} \caption{Geometrical Representation of Polygonal Numbers} \label{polyfigure} \end{figure} Let $f^{(m,n)}$ be the set of numbers that are simultaneously representable as both $m$-gonal and $n$-gonal numbers, where $m,n\in \mathbb N,$ and let $f_k^{(m,n)},$ be the $k$-th such number. In other words, there exists integers $r$ and $s$ such that $f_k^{(m,n)}=f_r^{m}\cap f_s^n. $ Note that $m$ and $n$ are symmetric and henceforth we assume, without loss of generality, that $n>m$. Much of the historical background of polygonal numbers can be found in books \cite{dickson, gelfond} and the relationship of these numbers was studied in various articles \cite{ando, eggan, lucas}. In particular, the equal values of three simultaneous polygonal numbers and some Pell equations arising from these numbers were studied \cite{krausz}. In addition, the proof for the infinitude of $f^{(m,n)}$ can be found in the literature \cite{chu, chu1, lucas}, i.e., for $(m-2)(n-2)$ not a perfect square, except when $(m,n)=(3,6),\left| f^{(m,n)}\right| =\infty$, where $\left| \quad \right|$ represents cardinality. The first part of this paper focuses on the case where $\left| f^{(m,n)}\right|$ is finite and estimates the cardinality of such integers, employing the techniques from hyperbolic geometry and the theories from Pell-Diophantine equations, i.e., \begin{equation} \label{pell} x^2-Dy^2=\Omega \end{equation} with $D=(m-2)(n-2)$ and $\Omega =(m-2)^2(n-4)^2-D(m-4)^2$, arriving at $$ 1\leq \left| f^{(m,n)}\right| \leq \lfloor \frac{d(\Omega )}2\rfloor -1 $$ where $\Omega = (m-2)^2(n-4)^2 - (m-2)(n-2)(m-4)^2$, $d(\Omega)$ signifies the number of divisors of $\Omega$, and $\lfloor\quad\rfloor$ is the greatest integer function. The second part of this paper begins with a preliminary discussion on the theory of second-degree Diophantine equations in $\mathbb Q$($\sqrt{D})$. We define our arsenal in the theory and examine the fundamental class of solutions of (\ref{pell}) as Chu recently did \cite{chu1}. More importantly, we improve the lower bound in the solution bounds given by Nagell \cite{nagell} regarding the solutions of a generalized Pell's equation. The improved bounds lead to the systematic determination of solutions under two special circumstances, i.e., (a) $f^{(m,m+2)}$ and (b) $f^{(m,m+1)}$, where $m\geq 5$. The result is a formula for directly ascertaining $f^{(m,n)}_k$, where $k=1, 2, 3, \cdots$ and $n=m+1$ or $n=m+2$. We establish a few corollaries to illustrate the application of this algorithm. This systematic approach can be readily extended to other algebraic cases. \end{section} \begin{section}{When $\left| f^{(m,n)} \right|$ is Finite} Chu \cite{chu} and Lucas \cite{lucas} considered the case when {\bf $\left| f^{(m,n)}\right| =\infty$}. That is, for {\bf $\left| f^{(m,n)}\right|$} to be otherwise finite, we must have $(m-2)(n-2)$ be a perfect square, and $(m,n)\neq (3,6)$. Using (\ref{polygonal-equation}) and the fact for every $k$, $\exists r, s\in \mathbb N$, such that $f_k^{(m,n)}=f_r^{m}\cap f_s^n$, we obtain the equation \begin{equation} \label{m-n-equation} f^{(m,n)}_k = f^m_r \cap f^n_s = \frac{(m-2)r^2 - (m-4)r}{2} = \frac{(n-2)s^2 - (n-4)s}{2} \end{equation} or, with a little bit of rearrangement, \begin{equation} \label{s-equation} s = \frac{(n-4) \pm\sqrt{(n-4)^2 + 4r(n-2)[(m-2)r-(m-4)]} }{2(n-2)} \end{equation} Since $s\in \mathbb N$, the discriminant of the last equation is a square integer. In other words, $\exists q \in \mathbb N$, such that $$ (n-4)^2 + 4r(n-2)[(m-2)r - (m-4)] = q^2 $$ To rearrange (\ref{s-equation}) into a Pell equation of the form in (\ref{pell}), we multiply the above equation pairwise by $(m-2)^2$, arriving at \begin{equation} \label{var-representation} \left\{ \begin{array}{l} x = (m-2)q \\ y = 2r(m-2) - (m-4) \\ D = (m-2)(n-2) \\ \Omega = (m-2)^2(n-4)^2 - (m-4)^2(m-2)(n-2) \end{array} \right. \end{equation} Since $s \in \mathbb N$, we can rewrite (\ref{s-equation}) into $s=\frac{(n-4)\pm q}{2(n-2)}$, or $q=2s(n-2)\mp(n-4)$, whereas from (\ref{var-representation}), $r=\frac{(m-4)+y}{2(m-2)}$. Note that \begin{equation} \label{s-equation2} s = \left\{ \begin{array}{l} \frac{(n-4)-q}{2(n-2)}, \quad\text{when} \quad s=0\\ \frac{(n-4)+q}{2(n-2)}, \quad\text{when} \quad s\geq1\\ \end{array} \right. \end{equation} Since $x=(m-2)q = 2s(m-2)(n-2)\mp (m-2)(n-4)$, we can transform $s$ in terms of $x$ into \begin{equation} s = \frac{x\pm (m-2)(n-4)}{2(m-2)(n-2)}. \end{equation} Thus, for every $r, s \in \mathbb N$, we must have \begin{equation} \label{r-general-equation} r = \frac{(m-4)+y}{2(m-2)}, \quad\text{when} \quad r = 0, 1, 2,\cdots \end{equation} and \begin{equation} \label{s-general-equation} s = \left\{ \begin{array}{l} \frac{(n-4)-q}{2(n-2)} = \frac{-x+(m-2)(n-4)}{2(m-2)(n-2)}, \quad\text{when} \quad s=0\\ \frac{(n-4)+q}{2(n-2)} = \frac{x+(m-2)(n-4)}{2(m-2)(n-2)}, \quad\text{when} \quad s\geq1\\ \end{array} \right. \end{equation} In other words, \begin{equation} \label{x-equation} x \equiv \left\{ \begin{array}{l} +(m-2)(n-4)[\bmod \quad2(m-2)(n-2)], \quad\text{if} \quad s=0.\\ -(m-2)(n-4)[\bmod \quad2(m-2)(n-2)], \quad\text{if} \quad s\geq1. \end{array} \right. \end{equation} and \begin{equation} \label{y-equation} y \equiv -(m-4)[\bmod \quad2(m-2)], \quad r\geq0. \end{equation} We note that (\ref{m-n-equation}) has solutions when either $(r_0, s_0) = (0,0)$ or $(r_1, s_1) = (1,1)$, or, in other words \begin{equation} \label{r-s-equation} (r,s) = \left\{ \begin{array}{l} (0,0), {\begin{array}{l}x_0 = + (m-2)(n-4)\\y_0=-(m-4)\end{array}}\\ (1,1), {\begin{array}{l}x_1 = n(m-2)\\y_1=m\end{array}} \end{array} \right. \end{equation} We now proceed to estimate bounds on the cardinality of $f^{(m,n)}$. \begin{theorem} There are only finitely many numbers $f_k^{(m,n)}$ for $(m-2)(n-2)$ a perfect square except when $(m,n)=(3,6)$. Lower and upper bounds for $\left| f^{(m,n)}\right|$ are given by $1\leq \left| f^{(m,n)}\right| \leq \lfloor \frac{d(\Omega )}2\rfloor -1,$ where $d(\Omega)$ signifies the number of divisors of $\Omega$. \end{theorem} \begin{pf} We rewrite (\ref{pell}) into $x^2-g^2y^2=\Omega$, where $D=(m-2)(n-2)=g^2$ and $g\in \mathbb Z$ with the requirements set forth in (\ref{var-representation}): \begin{equation} \label{revise} x^2-\gamma ^2=\Omega \end{equation} where $\gamma =gy=\pm \sqrt{x^2-\Omega }$. Eq.~(\ref{revise}) can also be rewritten as $\frac{x^2}\Omega -\frac{\gamma ^2}\Omega =1$, the equation of a hyperbola with asymptotes $$ \gamma =\pm \frac{\sqrt{\Omega }}{\sqrt{\Omega}}x=\pm x $$ and vertices $$ (\pm \sqrt{\Omega },0). $$ The geometric language for the above is justified in Fig.~\ref{asymtote}. It shows that all the solutions are symmetric over the real axis. This permits us to be concerned only with positive $x$ and $\gamma$ in the first quadrant. \begin{figure} \centering\includegraphics[width=0.7\textwidth]{asymptot} \caption{An Equation of Hyperbola: $x^2 - \gamma^2 = \Omega$}\label{asymtote} \end{figure} We denote $d$ as the vertical distance from the asymptote $\gamma =x$ to the curve $\gamma =\sqrt{x^2-\Omega }$. Thus \begin{equation} \label{distance}d=x-\sqrt{x^2-\Omega }. \end{equation} The trivial solution when $x=y=0$ is eliminated. As seen from Fig.~\ref{asymtote}, $\sqrt{\Omega }$ cannot be an integer otherwise $\gamma =y=0$. Hence $$ \max (d)=\lfloor \sqrt{\Omega }\rfloor +1-\sqrt{(\lfloor \sqrt{\Omega}\rfloor +1)^2-\Omega }, \qquad \min (d)=1. $$ where $\sqrt{\Omega }\in \mathbb R$. Therefore \begin{equation} \label{ineq}1\leq d\leq \lfloor \sqrt{\Omega }\rfloor +1-\sqrt{(\lfloor \sqrt{ \Omega}\rfloor+1)^2-\Omega }=\Phi . \end{equation} From (\ref{distance}) we see that $x\in \mathbb N$ hence $\sqrt{x^2-\Omega }\in \mathbb N$, leading to integral solutions $(x,\gamma ).$ Solving for $x$ in (\ref{distance}), we get \begin{equation} \label{xvalue}x=\frac{\Omega +d^2}{2d}. \end{equation} Similarly $$ y=\sqrt{\frac{x^2-\Omega }{g^2}}=\frac{\Omega -d^2}{2dg}. $$ It immediately follows from (\ref{xvalue}) that $x\in \mathbb N$, $2d\left| \Omega +d^2,\right. $ and $\Omega \equiv -d^2$ (mod $2d$), hence $ d\left| \Omega .\right.$ On the other hand, it is apparent that $\Omega =(m-2)^2(n-4)^2-(m-4)^2(m-2)(n-2)$ is an even number, therefore (\ref{xvalue}) implies that $d$ is an even number as well. Hence, the cardinality of $d$ becomes \begin{equation}\label{ineq2} 1\leq \left| d\right| \leq \sum_{d\left| \Omega \right. , 2\leq d\leq \Phi}1. \end{equation} Since $(x,y)=(\frac{\Omega +d^2}{2d},\frac{\Omega -d^2}{2dg})$ for each $d$ satisfying (\ref{ineq2}), $$ 1\leq \left| (x,y)\right| \leq \sum_{d\left| \Omega \right. , 2\leq d\leq \Phi}1, $$ which implies \begin{equation}\label{number} 1\leq \left| f^{(m,n)}\right| \leq \sum_{d\left| \Omega \right. , 2\leq d\leq \Phi}1. \end{equation} To get sharper bounds, we need two lemmas --- each of whose proofs is self-evident. \begin{lemma}\label{lemma1} $\lim \limits_{\Omega \rightarrow \infty }\Phi =\sqrt{\Omega }.$ \end{lemma} \begin{lemma}\label{lemma2} There are exactly $\lfloor \frac{d(\Omega )}2\rfloor -1$ $d$'s such that $2\leq d\leq \sqrt{\Omega }$ with $d\left| \Omega \right. .$ \end{lemma} By both Lemmas~\ref{lemma1} and \ref{lemma2}, it is apparent that $$ \sum_{d\left| \Omega \right., 2\leq d\leq \Phi}1=\lfloor \frac{d(\Omega )}2\rfloor -1. $$ Therefore an analogy to (\ref{number}) is \begin{equation} \label{final-bound} 1\leq \left| f^{(m,n)}\right| \leq \lfloor \frac{d(\Omega )}2\rfloor -1. \end{equation} We thereby complete our proof. \end{pf} \begin{example} In the case of $f^{(3,11)}$, $\Omega =40$ and $1\leq \left| f^{(3,11)}\right| \leq 3.$ \end{example} \begin{example} In the case of $f^{(27,102)}$, $\Omega =4680000$ and $1\leq \left| f^{(27,102)}\right| \leq 209.$ \end{example} \end{section} \begin{section}{Preliminaries - Diophantine Equations of the Second Degree} We define a common language used in describing the properties of the equation (\ref{pell}) expressed in the quadratic field $\mathbb Q$($\sqrt{D }).$ The following definitions are well-known and have appeared in various literature \cite{chu, chu1, dickson, nagell}. Both Definitions~\ref{definitionb} and \ref{definitionc} can be proven by finite induction. \begin{definition}\label{definitiona} Let $D $ and $\Omega$ be two integers. If $x=u$ and $y=v$ are integers which satisfy (\ref{pell}), we say, for simplicity, that the ordered pair $$ (u,v)=u+v\sqrt{D } $$ is a solution to (\ref{pell}). In general, we consider all of the solutions $(x,y)$ of \begin{equation} \label{aaaaa}x^2-Dy^2=1,\qquad \Omega=1 \end{equation} with $x,y>0$. Among these there is a least integer ordered pair $(x^{\prime },y^{\prime })$ , such that $x^{\prime }$ and $y^{\prime }$ have their least positive values. The ordered pair $(x^{\prime },y^{\prime }),$ or $x^{\prime }+y^{\prime } \sqrt{D },$ is called the\ fundamental solution of (\ref{aaaaa}). \end{definition} \begin{definition}\label{definitionb} If $(x^{\prime },y^{\prime })$ is the fundamental solution of (\ref{aaaaa}), then we may obtain all of the solutions, $(x^{*},y^{*})$, of (\ref{aaaaa}) through $$ (x^{*},y^{*})=(x^{\prime },y^{\prime })^n, $$ where $n=1,2,3,\ldots, etc.$ \end{definition} \begin{ex-def} The fundamental solution of $x^2-5y^2=1$ is $(9,4),$ so all of the solutions of this equation are given by $(9,4)^n$, where $n = 1,2,3,\ldots, etc.$ \end{ex-def} \begin{definition}\label{definitionc} There can be multiple classes of solutions to (\ref{pell}). If $(u_1,v_1)$ is the fundamental solution that occurs in the class $K$ of (\ref{pell}), then we may obtain all of the solutions $(u^{\prime},v^{\prime})$ belonging to the class $K$ through $$ (u^{\prime },v^{\prime })=(u_1,v_1)(x^{*},y^{*}) = (u_1,v_1)(x^{\prime },y^{\prime })^n, $$ where $(x^{*},y^{*})$ and $(x^{\prime },y^{\prime })^n$ are as defined in Definition~\ref{definitionb}. \end{definition} \begin{ex-def} There are only two classes of solutions to the equation $u^2-5v^2=4$. $(3,\pm 1)$ are the least integer ordered pairs satisfying the equation. Therefore each of $(3,\pm 1)$ is the fundamental solution of the said equation in its own respective class. If we take $(3,1)$, and the fundamental solution (9,4) of $x^2-5y^2=1$, then all of the solutions belonging to that class of (\ref{pell}) are given by (3,1)(9,4)$^n$, where $n=1,2,3, \ldots, etc,$ whereas all of the solutions belonging to the other class of (\ref{pell}) are given by (3,-1)(9,4)$^n$, where $n=1,2,3, \ldots, etc.$ \end{ex-def} \begin{definition}\label{definitiond} $K^{\prime}$=$(u,v)(x,y)^n$ and $\overline{K}^{\prime \prime }$=$(u,-v)(x,y)^n$ are the solutions of (\ref{pell}) in two distinct classes, $K$ and $\overline{K},$ respectively, where $n=1,2,3, \ldots, etc,$ and $(x,y)$ is the fundamental solution of (\ref{aaaaa}) and $% (u,v)$ of (\ref{pell}). In general, if the solutions $K^{\prime }$ and $\overline{K} ^{\prime \prime }$ coincide with each other, or $K^{\prime }$ = $\overline{K}^{\prime \prime }$ for $n=1,2,3, \ldots, etc,$ we call either $K$ or $\overline{K}$ an ambiguous class. In the case when either $u$ or $v=0$, the class that contains this $(u,v)$ is an ambiguous class. \end{definition} Next we need to introduce an important lemma, which is the generalization of the bounds presented in Nagell's book \cite{nagell}. \begin{lemma}\label{lemmae} The fundamental solutions, $(u,v)$, of all of the classes of (\ref{pell}) satisfy the bounds $$ \begin{array}{c} 0\leq v\leq y^{\prime} \sqrt{\frac \Omega{2(x^{\prime }+1)}}, \\ \sqrt{\Omega}\leq \left| u\right| \leq \sqrt{ \frac{(x^{\prime }+1)\Omega}2} \end{array} $$ where $(x^{\prime },y^{\prime })$ is the fundamental solution of (\ref{aaaaa}) and $u = x$ and $v = y$ are integers which satisfy (\ref{pell}). \end{lemma} \begin{pf} The lower bound on $u,$ given as $0$ by Nagell, can be improved to $\sqrt{\Omega}$ since $v\geq 0$. \end{pf} \end{section} \begin{section}{Class Number Arguments} We begin this section by introducing the idea of mapping and defining the $proper$ solutions that satisfy a set of rules involving the number of the classes of solutions to the general Pell equation, (\ref{pell}). \begin{definition} Let $A$ be a set of the integral solutions $(u,v)$ of $u^2-Dv^2=\Omega$ and $B$ be a set of the integer solutions $(r,s)$, where $r, s \in \mathbb N$, such that we can map $A$ into $B$. In other words, we denote $\theta $ as which assigns to each $(u,v)$ of $A$ an $(r,s)$ of $B$, i.e., $\theta :A\rightarrow B.$ \end{definition} \noindent Note that both $r$ and $s$ are as defined in the aforesaid relationship, $i.e$, {\bf $f^{(m,n)}_k=f_r^m\cap f_s^n$}. \begin{definition}\label{definitiong} $(u,v)$ of (\ref{pell}) is a {\it proper solution} if: \begin{description} \item[i.] $(u,v)$ satisfies the bounds in Lemma~\ref{lemmae} or, in other words, $(u,v)$ is a fundamental solution pair of a solution class of (\ref{pell}), and \item[ii.] $(u,v)$ produces a valid mapping, i.e., $\theta :A\rightarrow B.$ \end{description} \end{definition} \noindent For example, we know from Definition~\ref{definitionc}, $(3,1)$ is a fundamental solution which constitutes one class of solutions to the equation $x^2-5y^2=4,$ hence satisfying the bounds in Lemma~\ref{lemmae}. We term $(3,1)$ a {\it proper solution} because $(3,1)(9,4)^n$ also produces a valid mapping $\theta :A\rightarrow B$ for some $n\in \mathbb N$ and $(r, s)$ of $B$. \begin{theorem}\label{thm2} In most cases, there are 4, and perhaps more $proper$ solutions associated with $f^{(m,n)},$ a set of numbers that are simultaneously representable as both $m$-gonal and $n$-gonal numbers, for $m,n\in \mathbb N.$ \end{theorem} \begin{pf} From (\ref{r-s-equation}) we know that $(x_o,\pm y_o)=[(n-4)(m-2),$ $% \pm (m-4)]$ and $(x_1,\pm y_1)=$ $[n(m-2),$ $\pm m]$ are the solutions to (\ref{pell}). In general, if $(a,b)$ and $(a\prime ,b\prime )$ belong to the same class, then it is easy to check that $\Omega \left| a^{\prime \prime },b^{\prime \prime },\right. $ where $(a^{\prime \prime },b^{\prime \prime })=(aa^{\prime }\pm bb^{\prime }D,ab^{\prime }\pm ba^{\prime }).$ It is evident that if $K$ is the class consisting of the solutions $% (x_i,y_i),i=1,2,3,\ldots,$ and $\overline{K},$ or the conjugate of $% K,(x_i,-y_i),$ $i=1,2,3,\ldots,$ will constitute another class. Since the case $% (-x,-y)$ collapses onto class $K$, the same for the case $(-x,y)$, which collapses onto $\overline{K}.$ Therefore under necessary and sufficient conditions, $[(n-4)(m-2),$ $\pm (m-4)]$ and $[n(m-2),$ $\pm m]$ represent four different classes of solutions, though they are not necessarily {\it fundamental}. We can thus have at least 4 $proper$ solutions, i.e., $[(n-4)(m-2),$ $\pm (m-4)]$ or $[n(m-2),$ $\pm m].$ \end{pf} Next, we speak of an ambiguous class. We check certain conditions that will put both of the aforesaid $(x_o,y_o)$ and $(x_1,y_1)$ in the same solution class. As $\Omega$ in Lemma~\ref{lemmae} becomes large, we can have a considerable but finite number of fundamental solutions belonging to different classes that satisfy the bounds. We will see that the number of $proper$ solutions limits to, in some cases, no more than four. \begin{theorem}\label{thm3} There are at least two $proper$ solutions when \begin{description} \item[i.] $n=m+1$ and $n=m+2$ for $m\geq 5;$ \item[ii.] when $n-m>2,$ $(3,7),(3,8)$\ and $(3,10)$\ are the only cases. \end{description} \noindent Each of {\em i} and {\em ii} represents the necessary and sufficient conditions such that $(x_o,y_o)$ and $(x_1,y_1)$ both belong to the same solutions class, i.e., both are ambiguous. \end{theorem} \begin{pf} From Theorem~\ref{thm2}, $(x_o,\pm y_o)=$ $[(m-2)(n-4),$ $\pm (m-4)]$ and $(x_1,\pm y_1)=$ $[n(m-2),$ $\pm m]$ can represent at most 4 classes of solutions. We can write $(x_o,\pm y_o)(x^{\prime },y^{\prime })=(x_1,\pm y_1)$ as $$ \left\{ \begin{array}{ll} x_1 & =(m-2)(n-4)x^{\prime }\pm (m-4)(m-2)(n-2)y^{\prime }=n(m-2) \\ y_1 & =(m-2)(n-4)y^{\prime }\pm (m-4)x^{\prime }=\pm m \end{array} \right. $$ Multiplying $x_1$ by $\frac{(m-4)}{(m-2)}$ and $y_1$ by $(n-4)$ we get $$ \left\{ \begin{array}{ll} \frac{(m-4)}{(m-2)}x_1= & (m-4)(n-4)x^{\prime }\pm (m-4)^2(n-2)y^{\prime }=n(m-4) \\ (n-4)y_1= & (m-2)(n-4)^2y^{\prime }\pm (m-4)(n-4)x^{\prime }=\pm m(n-4) \end{array} \right. $$ Note that the signs above are member-wise linked with each other, so by subtracting $\frac{(m-4)}{(m-2)}x_1$ from $(n-4)y_1,$ we can only have $$ y^{\prime }[(m-2)(n-4)^2-(m-4)^2(n-2)]=\left\{ \begin{array}{l} 4(n-m) \\ -2(mn-2m-2n) \end{array} ,\right. $$ whereas $$ y^{\prime }[(m-2)(n-4)^2+(m-4)^2(n-2)]=\left\{ \begin{array}{l} 4(n-m) \\ -2(mn-2m-2n) \end{array} \right. $$ is impossible. Further simplified, $$ y^{\prime }(n-m)(mn-2m-2n)=\left\{ \begin{array}{l} 4(n-m) \\ -2(mn-2m-2n) \end{array} \right. $$ and $$ y^{\prime }=\left\{ \begin{array}{l} \frac 4{mn-2m-2n}\\ \frac {-2}{n-m} \end{array} \right. . $$ Knowing $n>m$ and $y^{\prime }\in \mathbb Z,$ we easily deduce that when $% y^{\prime }=\frac {-2}{n-m},$ $n=m+1$ and $n=m+2$ are the two satisfying cases. In the case when $y^{\prime }=\frac 4{mn-2m-2n},$ $$ mn-2m-2n=\left\{ \begin{array}{l} \pm 1 \\ \pm 2 \\ \pm 4 \end{array} \right. $$ giving us $$ n=\left\{ \begin{array}{l} \frac{2m\pm 1}{m-2} \\ \frac{2m\pm 2}{m-2} \\ \frac{2m\pm 4}{m-2} \end{array} \right. . $$ A simple calculation yields the only 3 satisfying cases when $n-m > 2$: $% (m,n)=(3,7),$ $(3,8),$ and $(3,10).$ We prove and therefore single out the cases when $(x_o,\pm y_o)$ and $(x_1,\pm y_1)$ belong to the same class, or when $(x_1,\pm y_1)$ is generated from $(x_o,\pm y_o).$ We need to ascertain whether $(x_o,\pm y_o)$ satisfies the bounds in Lemma~\ref{lemmae}. If they do then $(x_o,\pm y_o)$ are the two $proper$ solutions, otherwise we must assume there exists at least two other $proper$ solutions that will each generate $(x_o,\pm y_o).$ However, we must note that $x_o,y_o\neq 0,$ i.e., $[(m-2)(n-4),$ $\pm (m-4)]% \not =(0,0),$ otherwise according to Definition~\ref{definitiond}, we will have one less class and the remaining one is ambiguous. The case when this does not occur is $m\geq 5.$ \end{pf} We restrict our attention to the two aforesaid special cases, focusing our interest on $m\geq 5$: (a) $f^{(m,m+2)}$ and (b) $f^{(m,m+1)}.$ We prove that $(x_o,\pm y_o)$ indeed satisfy the bounds and therefore are the $proper$ solutions. The following lemma is useful and readily obtained: \begin{lemma}\label{specialcase} We designate $(x^{\prime },y^{\prime })$ as the fundamental solution of $x^2-Dy^2=1.$ Therefore, when $n=m+2,$ $% (x^{\prime },y^{\prime })=(m-1,1),$ whereas $n=m+1,$ $(x^{\prime },y^{\prime })=(2m-3,2).$ \end{lemma} \begin{theorem}\label{thm4} In the case of $f^{(m,m+2)},$ when $m\geq 4,$ two proper solutions of (\ref{pell}) are $$ [(\lfloor \sqrt{\frac \Omega {2m}}\rfloor +1)^2,\pm (\lfloor \sqrt{\frac \Omega {2m}}\rfloor -1)] $$ \end{theorem} \begin{pf} From Theorem~\ref{thm2}, $(x_o,\pm y_o)=[(m-2)^2,\pm (m-4)] $ by substitution. We also have \begin{equation} \label{JoeSu}\left\{ \begin{array}{l} D=m(m-2) \\ \Omega =(m-2)^4-m(m-2)(m-4)^2=2m^3-8m^2+16 \end{array} \right. \end{equation} Applying Lemma~\ref{lemmae}, we get \begin{equation} \label{JoeSu0} \begin{array}{c} 0\leq v\leq \sqrt{\frac \Omega {2m}} \\ \sqrt{\Omega }\leq \left| u\right| \leq \sqrt{ \frac{m\Omega }2} \end{array} \end{equation} Considering the upper bound for $v$, we yield $\sqrt{(m-3)^2}<\sqrt{\frac \Omega {2m}}=\sqrt{m^2-4m+\frac 8m}<\sqrt{(m-2)^2},$ which means $\lfloor \sqrt{\frac \Omega {2m}}\rfloor =m-3.$ It is inferred from (\ref{JoeSu0}) that $$ 0\leq \lfloor\sqrt{\frac \Omega {2m}}\rfloor -1=m-4<\sqrt{\frac \Omega {2m}}% ,\qquad m\geq 4. $$ Because $\lfloor \sqrt{\frac \Omega {2m}}\rfloor -1=m-4$ satisfies the bounds, we know that it is a fundamental solution of a particular class. Since the mapping $\theta :(x_o,y_o)\rightarrow (r_o,s_o)$ is valid, we know that $(x_o,y_o)$ and its conjugate, $(x_o,-y_o),$ are the 2 proper solutions. In other words, $y_o=\lfloor\sqrt{\frac \Omega {2m}}% \rfloor -1.$ By substitution we get $(x_o,\pm y_o)=[(m-2)^2,\pm (m-4)]=[(\lfloor\sqrt{\frac \Omega {2m}}\rfloor +1)^2,\pm (\lfloor\sqrt{% \frac \Omega {2m}}\rfloor -1)].$ \end{pf} In a similar fashion, we also get two $proper$ solutions in the case of $f^{(m,m+1)}$. For brevity we omit the proof herein: \begin{theorem}\label{thm5} In the case of $f^{(m,m+1)}$, when m$\geq $4, two proper solutions of (\ref{pell}) are $$ [(\lfloor \sqrt{\frac \Omega {m-1}}\rfloor \lfloor \sqrt{\frac \Omega {m-1}}% \rfloor +\lfloor\sqrt{\frac \Omega {m-1}}\rfloor ),\pm (\lfloor\sqrt{\frac \Omega {m-1}}\rfloor -1)] $$ \end{theorem} We will go on to prove that\ $(x_o,\pm y_o)$ from Theorem~\ref{thm5} are indeed the {\it only} two proper solutions: \begin{theorem}\label{thm6} There are only two proper solutions, i.e., $(x_o,\pm y_o),$ in the case $n=m+1,$ as long as $m\geq 5$. \end{theorem} \begin{pf} By Lemma~\ref{specialcase}, we take $(x^{\prime },y^{\prime })=(2m-3,2) $ as the fundamental solution of $x^2-Dy^2=1.$ From Theorem~\ref{thm3}, we know that for $n=m+1,$ $$ (x_o,y_o)(x^{\prime },y^{\prime })=(x_1,y_1) $$ or $$ [\underbrace{(m-2)(m-3)}_{x_o},\underbrace{(m-4) }_{y_o}](\underbrace{2m-3}_{x^{\prime }}, \underbrace{2}_{y^{\prime }})=[\underbrace{(m+1)(m-2)}_{x_1}, \underbrace{m}_{y_1}] $$ For brevity, we say in $\mathbb Q[\sqrt{D}],$ \begin{equation} \label{super}(x_1,y_1)\equiv (x_ox^{\prime }+y_oy^{\prime }D,x^{\prime }y_o+x_oy^{\prime })\bmod \left\{ \begin{array}{l} 2(n-2) \\ 2(m-2) \end{array} .\right. \end{equation} hence by way of both (\ref{x-equation}) and (\ref{y-equation}) $$ (x_1,y_1)\equiv [-(m-2)(n-4),-(m-4)]\bmod \left\{ \begin{array}{l} 2(n-2) \\ 2(m-2) \end{array} .\right. $$ Since $y^{\prime }=2$ and $D=(m-2)(n-2),$ we deduce (\ref{super}) to \begin{equation} \label{k}(x_1,y_1)\equiv (x_ox^{\prime },x^{\prime }y_o+x_oy^{\prime })\bmod \left\{ \begin{array}{l} 2(n-2) \\ 2(m-2) \end{array} .\right. \end{equation} We have to prove the non-existence of another proper solution, besides the two from Theorem~\ref{thm5}, i.e., $$ (x_o,\pm y_o)=[(\lfloor \sqrt{\frac \Omega {m-1}}\rfloor \lfloor \sqrt{\frac \Omega {m-1}}\rfloor +\lfloor \sqrt{\frac \Omega {m-1}}\rfloor ),\pm (\lfloor \sqrt{\frac \Omega {m-1}}\rfloor -1)]. $$ We will proceed with this by assuming that there exists a proper solution $% (a,b)$, such that $$ \theta :(a,b)\rightarrow (r,s) $$ is valid, where $02\sqrt{D}+1,$ and therefore $D$ is a quadratic residue of $\Omega $ and hence of its divisors. Second, we examined and singled out the solution classes, or proper solutions, that can be used to generate $f^{(m,n)}$, using the theory of Pell diophantine equations and the improved bounds of \cite{nagell}. In the cases $n=m+1$ and $n=m+2$, there are only two proper solutions. This fact enabled us to quickly generate $f^{(m,m+1)}_k$ and $f^{(m,m+2)}_k$ where $k = 1, 2, 3, . . .$. The approach used in arriving at the two special cases can be readily extended to algebraic cases involving other $m$ and $n$ as well. Pell-Diophantine equations play an important role in this paper, as well in the field of binary quadratic forms \cite{gauss}. For example, in the case when we put $(m,n)=(3,7),$ or $\Omega =4,$ (\ref{pell}) of this kind is heavily used in the theory of biquadratic forms \cite{flath} and the Lucas test for Mersenne primes \cite{shank}. The Pell equation encountered in this paper may have some applications to these fields. \end{section} \begin{thebibliography}{99} \bibitem{ando} S. Ando, On a system of Diophantine equations concerning the polygonal numbers, \emph{Fibonacci Quart.} \textbf{20} (1982), 349--353. \bibitem{chu} W. Chu, Regular polygonal numbers and generalized Pell equations, \emph{Int. Math. Forum.} \textbf{2} (2007), 781--802. \bibitem{chu1} W. Chu and P. Magli, Polygonal numbers and Diophantine equations, \emph{Rend. Accad. Naz. Sci. XL Mem. Mat. Appl.} \textbf{5} (2003), 1-34. \bibitem{dickson} L. E. Dickson, \emph{History of the Theory of Numbers: Volume II}, G.E. Stechert \& Co., 1934. \bibitem{eggan} L. C. Eggan, P. C. Eggan, J. L. Selfridge, Polygonal products of polygonal numbers and the Pell equation, \emph{Fibonacci Quart.} \textbf{20} (1982), 24--28. \bibitem{flath} D. E. Flath, \emph{Introduction to Number Theory}, John Wesley \& Sons, 1989. \bibitem{gauss} C. F. Gauss, \emph{Disquisitiones Arithmeticae}, Yale University, 1966. \bibitem{gelfond} A. O. Gelfond, \emph{The Solution of Equations in Integers}, P.Noordhoff. LTD-Groningen, 1960. \bibitem{krausz} T. Krausz, A note on equal values of polygonal numbers, \emph{Publ. Math. Debrecen.} \textbf{54} (1999), 321--325. \bibitem{lucas} D. S. Lucas, Numbers common to two polygonal sequences, \emph{Fibonacci Quart.} \textbf{11} (1973), 78--84. \bibitem{nagell} T. Nagell, \emph{Introduction to Number Theory: 2nd Edition}, Chelsea, 1981. \bibitem{shank} D. Shank, \emph{Solved and Unsolved Problems in Number Theory: 3rd Edition}, Chelsea, 1985. \bibitem{Sloane} N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Published electronically at \htmladdnormallink{http://www.research.att.com/$\sim $njas/sequences/}{http://www.research.att.com/~njas/sequences/}. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: \ \ Primary 11D09. \noindent \emph{Keywords:} polygonal number, Pell Diophantine equation, quadratic equation, fundamental class, generating function. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000217}, \seqnum{A000290}, \seqnum{A000326}, \seqnum{A000384}, \seqnum{A014979}, \seqnum{A046194}, and \seqnum{A129654}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received July 25 2007; revised version received December 30 2007. Published in {\it Journal of Integer Sequences}, December 30 2007. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}