\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}}} \newcommand{\Zp}[1]{\mathbb{Z}[\sqrt{#1}]} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf On the Extension of the Diophantine \\ \vskip .1in Pair $\{1,3\}$ in $\Zp{d}$ } \\ { \vskip 1cm \large Zrinka Franu\v{s}i\'{c} \\ Department of Mathematics\\ University of Zagreb\\ Bijeni\v{c}ka cesta 30\\ 10000 Zagreb \\ Croatia\\ \href{mailto:fran@math.hr}{\tt fran@math.hr} } \end{center} \vskip .2in \begin{abstract} In this paper, we consider Diophantine triples of the form $\{1,3,c\}$ in the ring $\Zp{d}$. We prove that the Diophantine pair $\{1,3\}$ cannot be extended to the Diophantine quintuple in $\Zp{d}$ with $d<0$ and $d\not=-2$. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} \newenvironment{example}{\begin{examp}\normalfont\quad}{\end{examp}} \newtheorem{rema}[theorem]{Remark} \newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}} \newtheorem{note}[theorem]{Note} \newenvironment{Note}{\begin{note}\normalfont\quad}{\end{note}} \newcommand{\cd}{{$\cdot$}} \newcommand{\F}{{\mathbb F}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\N}{{\mathbb N}} \newcommand{\R}{{\mathbb R}} \newcommand{\Q}{{\mathbb Q}} \newcommand{\C}{{\mathbb C}} \newcommand{\lra}{\longrightarrow} \newcommand{\lr}{\rightarrow} \newcommand{\sha}{{\amalg \hspace{-1.40mm}\amalg}} \section{Introduction} A {\it Diophantine $m$-tuple} in a commutative ring $R$ with the unit $1$ is a set of $m$ distinct non-zero elements with the property that the product of each two distinct elements increased by $1$ is a perfect square in $R$. These sets are studied in many different rings: the ring of integers $\Z$, the ring of rationals $\Q$, the ring of Gaussian integers $\Z[i]$ (\cite{duje-glasnik,fran-glasnik}), the ring of integers of quadratic fields $\Zp{d}$ and $\Z[(1+\sqrt{d})/2]$ (\cite{fran-ram,fran-hun}), polynomial rings (\cite{duje-fuch,duje-jura}). The most famous Diophantine $m$-tuples and historical examples are quadruples $\{\frac{1}{16},\frac{33}{16},\frac{17}{4},\frac{105}{16}\}$ (found by Diophant) and $\{1,3,8,120\}$ (found by Fermat). More information about these sets can be found on Dujella's web page on Diophantine $m$-tuples \cite{duje-web}. In 1969, Baker and Davenport \cite{bak-dav} showed that the Diophantine triple $\{1,3,8\}$ can be extended uniquely to the Diophantine quadruple $\{1,3,8,120\}$ \seqnum{A030063} . Hence, the triple $\{1,3,8\}$ cannot be extended to a Diophantine quintuple. Jones \cite{jones} proved that if the set $\{1,3,c\}$ is a Diophantine triple then $c$ must be of the form \begin{equation*} c_k=\frac{1}{6}((2+\sqrt{3})(7+4\sqrt{3})^k+(2-\sqrt{3})(7-4\sqrt{3})^k-4), \end{equation*} for some $k\in\N$ (\seqnum{A045899}). It can be easily verified that sets $\{1,3,c_{k},c_{k-1}\}$ and $\{1,3,c_{k},c_{k+1}\}$ are Diophantine quadruples. Indeed, $c_kc_{k+1}+1=t_k^2$, where $(t_k)$ is \seqnum{A051048}. Moreover, the Diophantine triple $\{1,3,c_{k}\}$ can be extended to a quadruple only by the elements $c_{k-1}$ and $c_{k+1}$. This assertion is proved in 1998. by Dujella and Peth\" o \cite{duje-pet}. A direct consequence of their assertion is the fact that the pair $\{1,3\}$ cannot be extended to a Diophantine quintuple. A natural question that may arise is can we extend the pair $\{1,3\}$ to a quintuple in a larger ring. In this paper, we have chosen the ring $\Zp{d}=\{a+b\sqrt{d}:a,b\in\Z\}$ where $d\in\Z$ and $d$ is not a perfect square. So, we assume that $\{1,3,a+b\sqrt{d}\}$ is a Diophantine triple in $\Zp{d}$ and according to a definition it means that there exist integers $\xi_1,\eta_1,\xi_2,\eta_2$ such that \begin{eqnarray} \label{e.00} a+b\sqrt{d}+1 &=& (\xi_1+\eta_1\sqrt{d})^2\\ \label{e.01} 3a+3b\sqrt{d}+1 &=& (\xi_2+\eta_2\sqrt{d})^2. \end{eqnarray} Although, it seams that we have much more 'freedom' in larger rings, the result for some rings stays the same. \begin{theorem}\label{tmain} Let $d$ be a negative integer and $d\not =-2$. The Diophantine pair $\{1,3\}$ cannot be extended to a Diophantine quintuple in the ring $\Zp{d}$. \end{theorem} In certain rings $\Zp{d}$ for positive $d$, Theorem \ref{tmain} is not valid. For example $\{1,3,8,120,1680\}$ is the Diophantine quintuple in $\Zp{8\cdot 1680+1}$. In last two sections, we describe the set of all Diophantine triples $\{1,3,c\}$ in $\Zp{-2}$ and in $\Zp{d}$ for some positive integer $d$, where $c$ is an integer and such that $\{1,3,c\}$ is not a Diophantine triple in $\Z$. The existence of such triples is related to solvability of certain Pellian equations. \section{Main result} \begin{proposition}\label{p1} Let $d$ be a negative integer. If $\{1,3,c\}$ is a Diophantine triple in the ring $\Zp{d}$, then $c$ is an integer. \end{proposition} \begin{proof} Let $c=a+b\sqrt{d}$. There exist $\xi_1,\eta_1,\xi_2,\eta_2\in\Z$ such that \begin{equation}\label{e.1} a+1=\xi_1^2+\eta_1^2d,\ b=2\xi_1\eta_1,\ 3a+1=\xi_2^2+\eta_2^2d,\ 3b=2\xi_2\eta_2. \end{equation} Evidently, $b$ must be an even number. By squaring and subtracting corresponding relations in (\ref{e.1}), we get \begin{equation}\label{e.2} (a+1)^2-db^2=x^2,\ (3a+1)^2-d(3b)^2=y^2, \end{equation} where $x=\xi_1^2-\eta_1^2d$ and $y=\xi_2^2-\eta_2^2d$. By eliminating the quantities $b$ and $d$ from (\ref{e.2}), we obtain \begin{equation}\label{e.4} (3x)^2-y^2=(3a+3)^2-(3a+1)^2=4(3a+2), \end{equation} It is clear, from (\ref{e.4}), that $3x-y$ and $3x+y$ must be even. So, we have \begin{equation*}\label{e.5} 3x-y=2^{k}c_1,\ 3x+y=2^{l}c_2, \end{equation*} where $k,l$ are positive integers and $c_1,c_2$ are odd. Thus, we have $$ (3x-y)(3x+y)=2^{k+l}n, $$ where and $c_1c_2=n$. Furthermore, it follows that $$ y=2^{l-1}c_2-2^{k-1}c_1. $$ The expression $(3a+1)^2-y^2$ can be given in terms of $c_1$ and $c_2$: \begin{eqnarray*} (3a+1)^2-y^2&=&(2^{k+l-2}n-1)^2-(2^{l-1}c_2-2^{k-1}c_1)^2\\ &=&4^{k+l-2}n^2-2^{k+l-1}n+1-4^{l-1}c_2^2+ 2^{k+l-1}c_1c_2-4^{k-1}c_1^2\\ &=& 4^{k+l-2}n^2+1-4^{l-1}c_2^2-4^{k-1}c_1^2\\ &=& (4^{k-1}c_1^2-1)(4^{l-1}c_2^2-1). \end{eqnarray*} On the other hand, $(3a+1)^2-y^2=d(3b)^2$ (by (\ref{e.2})). Since $|c_i|\geq 1$ and $k,l\geq 1$, we obtain that $d<0$ implies $b=0$. \end{proof} According to the previous proposition, it may have sense to study Diophantine triples of the form $\{1,3,c\}$ in $\Zp{d}$ such that $c$ is an integer. The triple $\{1,3,c\}$, $c\in\Z$, will be called a \emph{proper} triple in $\Zp{d}$ if it is a Diophantine triple $\Zp{d}$, but not a Diophantine triple in $\Z$. For instance, $\{1,3,161\}$ is a proper triple in $\Zp{2}$, $\{1,3,-3\}$ is a proper triple in $\Zp{-2}$, but $\{1,3,120\}$ is not a proper triple in $\Zp{d}$ ($d\not =1$). Let us, now, assume that $\{1,3,c\}$ is proper triple in $\Zp{d}$. So, one of the following cases may occur \begin{equation} \label{sl.1} \begin{array}{rcl} c+1 &=& \xi^2\\ 3c+1 &=& d\eta^2, \end{array} \end{equation} \begin{equation} \label{sl.2} \begin{array}{rcl} c+1 &=& d\eta^2\\ 3c+1 &=& \xi^2, \end{array} \end{equation} \begin{equation} \label{sl.3} \begin{array}{rcl} c+1 &=& d\eta_1^2\\ 3c+1 &=& d\eta_2^2. \end{array} \end{equation} Following proposition gives a connection between the extensibility of the Diophantine pair $\{1,3\}$ and solvability of certain Pellian equations. \begin{proposition}\label{p2} The Diophantine pair $\{1,3\}$ can be extended to a proper Diophantine triple $\{1,3,c\}\subset\Z$ in $\Zp{d}$ if and only if one of the following equations \begin{equation} \label{e.11} 3x^2-dy^2=2 \end{equation} and \begin{equation} \label{e.21} x^2-3dy^2=-2 \end{equation} is solvable in $\Z$. \end{proposition} \begin{proof} \begin{tabular}{|c|} \hline $\Rightarrow$\\ \hline \end{tabular} If $\{1,3,c\}$ is a proper triple and (\ref{sl.1}) is fulfilled, then by eliminating $c$, we get $3\xi^2-d\eta^2=2$. So, the equation (\ref{e.11}) is solvable in $\Z$. Analogously, by eliminating $c$ from (\ref{sl.2}), we obtain that the equation (\ref{e.21}) is solvable in $\Z$. Finally, from (\ref{sl.3}) we have \begin{equation*} d(\eta_2^2-3\eta_1^2)=-2, \end{equation*} and the only possibilities are: \begin{description} \item[(a)] $d=-1$ and $\eta_2^2-3\eta_1^2=2$, which is impossible (because the $x^2-3y^2=2$ is not solvable in $\Z$) , \item[(b)] $d=2$ and $\eta_2^2-3\eta_1^2=-1$, which is also impossible for the same reason, \item[(c)] $d=-2$ and $\eta_2^2-3\eta_1^2=1$, which is possible because the Pell's equation $x^2-3y^2=1$ has infinitely many solution. Note that the equation (\ref{e.11}) is also solvable if $d=-2$, it has a unique solution $(0,1)$. \end{description} \begin{tabular}{|c|} \hline $\Leftarrow$\\ \hline \end{tabular} Assume that $(\xi,\eta)\in\Z^2$ is a solution of the equation (\ref{e.11}). Then the set $\{1,3,\xi^2-1\}$ represents a proper triple in $\Zp{d}$. Indeed, $3(\xi^2-1)+1=d\eta^2=(n\sqrt{d})^2$. Similarly, if $(\xi,\eta)\in\Z^2$ is a solution of (\ref{e.21}), then we get a proper triple $\{1,3,d\eta^2-1\}$. \end{proof} In order to prove Theorem \ref{tmain}, we need a special case of Theorem 8 from \cite{jones}. \begin{lemma}[Theorem 8, \cite{jones}]\label{l1} If $\{1,3,c\}$ is a Diophantine triple in $\Z$, then $c=c_k$ for some integer $k\geq 2$ where $$ c_k=14c_{k-1}-c_{k-2}+6,\, c_1=8,\, c_0=0. $$ \end{lemma} By solving the recursion from Lemma \ref{l1} we get \begin{equation}\label{ck} c_k=\frac{1}{6}((2+\sqrt{3})(7+4\sqrt{3})^k+(2-\sqrt{3})(7-4\sqrt{3})^k-4),\ k\geq 0 \end{equation} \begin{proposition}\label{p3} Let $d$ be a negative integer and $d\not =-2$. If $\{1,3,c\}$ is a Diophantine triple in $\Zp{d}$, then $c=c_k$ for some positive integer $k$, where $c_k$ is given by (\ref{ck}). \end{proposition} \begin{proof} According to Proposition \ref{p1}, $c$ is an integer. Hence, $\{1,3,c\}$ is a triple in $\Z$ or a proper triple in $\Zp{d}$. Since the equations (\ref{e.11}) and (\ref{e.21}) are not solvable in $\Z$ for $d<0$ and $d\not =-2$, Proposition \ref{p2} implies that $\{1,3,c\}$ is not a proper triple in $\Zp{d}$. Hence, it is a Diophantine triple in $\Z$. Finally, from Lemma \ref{l1} we get that $c=c_k$ for some $k$. \end{proof} \begin{lemma}[Theorem 1, \cite{duje-pet}]\label{l2} If $\{1,3,c_k,d\}$ is a Diophantine quadruple in $\Z$ where $c_k$ is given by (\ref{ck}) and $k$ is a positive integer, then $d\in\{c_{k-1},c_{k+1}\}$. \end{lemma} At this point we can conclude that Theorem \ref{tmain} is proved. (It follows directly from Proposition \ref{p3} and Lemma \ref{l2}.)\\ Since the extension of the pair $\{1,3\}$ is closely related to the solvability of the Pellian equation $x^2-3y^2=-2$, our result can be understood as the following statement. \begin{corollary} Let $d$ be a negative integer and $d\not =-2$. The equation \begin{equation}\label{e.30} x^2-3y^2=-2 \end{equation} in $\Zp{d}$ has only real solutions. \end{corollary} \begin{proof} Suppose that $(\xi_1+\eta_1\sqrt{d},\xi_2+\eta_2\sqrt{d})$ is a solution of the equation (\ref{e.30}). For $$a+b\sqrt{d}=(\xi_2+\eta_2\sqrt{d})^2-1,$$ we obtain that $\{1,3,a+b\sqrt{d}\}$ is a Diophantine triple u $\Zp{d}$. Indeed, $$3(a+b\sqrt{d})+1=3(\xi_2+\eta_2\sqrt{d})^2-2=(\xi_1+\eta_1\sqrt{d})^2.$$ According to Proposition \ref{p3}, there exists some $k$ such that $a+b\sqrt{d}=c_k$. Immediately, we conclude that $\eta_1=\eta_2=0$. \end{proof} \section{The case $d=-2$} In this section, we will take a closer look at the Diophantine triples $\{1,3,c\}$ in $\Zp{-2}$. In fact, we are able to describe all Diophantine triples in $\Zp{-2}$. \begin{proposition}\label{p4} If $\{1,3,c\}$ is a Diophantine triple in $\Zp{-2}$, then $c\in\{-1,c_k,d_k\}$ for some $k\in\N$, where $c_k$ is given by (\ref{ck}) and \begin{equation}\label{dk} d_k=-\frac{1}{6}\left((7+4\sqrt{3})^k+(7-4\sqrt{3})^k+4\right) \end{equation} \end{proposition} \begin{proof} By Proposition \ref{p1}, we have that $c$ is an integer. Since the equation (\ref{e.11}) is solvable in $\Z$, Proposition \ref{p2} implies that $\{1,3,c\}$ can be a triple in $\Z$ or a proper triple in $\Zp{-2}$. If $\{1,3,c\}$ is a triple in $\Z$, then $c=c_k$ from Lemma \ref{l1}. If $\{1,3,c\}$ is a proper triple in $\Zp{-2}$, then one of the cases (\ref{sl.1}) and (\ref{sl.3}) may occur. The case (\ref{sl.1}) implies that $c=-1$, because $(0,1)$ is the unique solution of (\ref{e.11}). In the case of (\ref{sl.3}), $c=(-2x^2-1)/3$ where $x$ is a solution of the Pell's equation \begin{equation}\label{e.50} x^2-3y^2=1. \end{equation} All solutions of (\ref{e.50}) are $$ x_{k+2}=4x_{k+1}-x_{k},\, x_1=2,\, x_0=1, $$ for $k\geq 0$. By solving the recursion we obtain that $$x_k=\frac{1}{2} ((2-\sqrt{3})^k+(2+\sqrt{3} )^k),$$ and $d_k=(-2x_k^2-1)/3$. \end{proof} Note that for $k=0$, we have $d_0=-1$. Also, it is interesting that $(-d_k)$ is \seqnum{A011922}. \begin{rema} It is known that $\{1,3,c_k,c_{k+1}\}$ is a Diophantine quadruple in $\Z$ for $k\geq 1$ (Lemma \ref{l2}). The same can be proven for the set $\{1,3,d_k,d_{k+1}\}$ in $\Zp{-2}$ for $k\geq 0$. We have \begin{eqnarray*} 36(d_kd_{k+1}+1) &=&66+ (7-4 \sqrt{3})^{2 k+1}+(7+4 \sqrt{3})^{2 k+1}+\\ && 16((7-4\sqrt{3})^k (2-\sqrt{3})+(2+\sqrt{3}) (7+4\sqrt{3})^k)\\ &=&64+16((2+\sqrt{3})^{2k+1}+(2-\sqrt{3})^{2k+1})+\\ && ((2+\sqrt{3})^{2k+1}+(2-\sqrt{3})^{2k+1})^2\\ &=& 4(x_{2k+1}^2+8x_{2k+1}+16)=4(x_{2k+1}+4)^2, \end{eqnarray*} where $x_{2k+1}$ is a solution of $x^2-3y^2=1$. It can be easily shown that $x_{2k+1}\equiv 2\pmod 3$ (because $x_{n+2}=4x_{n+1}-x_n$, $x_0=1$, $x_1=2$), so $$ \sqrt{d_kd_{k+1}+1}=\frac{1}{3}(x_{2k+1}+4) $$ is an integer. \end{rema} Further, the set $\{1,3,c_k,d_l\}$ is not a Diophantine quadruple for $k\geq 1$ and $l\geq 0$. Indeed, it can be easily seen that $c_kd_l+1$ is a negative odd number (from Lemma \ref{l1} it follows that $c_k$ is even, and from (\ref{dk}) that $d_l<0$). Hence, $\sqrt{c_kd_l+1}\not\in\Zp{-2}$. \section{Proper triples in $\Zp{d}$ for some $d>0$} According to Proposition \ref{p2}, the set $\{1,3,c\}\subset\Z$ is a proper Diophantine triple in $\Zp{d}$ if and only if one of the equations (\ref{e.11}), (\ref{e.21}) is solvable in $\Z$. \begin{examp}\label{exa1} Determine all proper Diophantine triples $\{1,3,c\}$ in $\Zp{10}$. \end{examp} Note that $d=10$ is the smallest positive integer, $d\not =1$, such that the equation (\ref{e.11}) is solvable in $\Z$. If $(x_n,y_n)$ is a solution of (\ref{e.11}), then $\{1,3,x_n^2-1\}$ is a proper triple in $\Zp{10}$. All solutions of (\ref{e.11}) are $$ x_{n+2}=22x_{n+1}-x_n,\, x_1=42, x_0=2,$$ for $n\geq0$. Hence, $$ x_n=\frac{1}{6}((6+\sqrt{30})(11+2\sqrt{30})^n+(6-\sqrt{30})(11-2\sqrt{30})^n),\,n\geq0.$$ So, all proper triples $\{1,3,c\}$ in $\Zp{10}$ are $\{1,3,d_n\}$ where \begin{equation}\label{dn1} d_n=x_n^2-1=\frac{1}{6}((11+2\sqrt{30})^{2n+1}+(11-2\sqrt{30})^{2n+1}-4),\, n\geq0. \end{equation} Besides these triples, sets $\{1,3,25\pm8 \sqrt{10}\}$, $\{1,3,160\pm44 \sqrt{10}\}$, $\{1,3,355\pm112 \sqrt{10}\}$ are also Diophantine triples in $\Zp{10}$. The third element of these triples is related to the solution in $\Zp{10}$ of the equation $x^2-3y^2=-2$. For instance, $(6+2\sqrt{10},4+\sqrt{10})$ is solution of $x^2-3y^2=-2$ and $25+8 \sqrt{10}=(4+\sqrt{10})^2-1$. More generally, if $X^2-3Y^2=-2$ and $X,Y\in\Zp{10}$ then $\{1,3,Y^2-1\}$ is a Diophantine triple in $\Zp{10}$. \begin{examp}\label{exa2} Determine all proper Diophantine triples $\{1,3,c\}$ in $\Zp{2}$. \end{examp} Since the equation (\ref{e.21}) is solvable in $\Z$, $\{1,3,2y_n^2-1\}$ is a proper triple in $\Zp{2}$ where $(x_n,y_n)$ is a solution of (\ref{e.21}). All solutions of (\ref{e.21}) are $$ y_{n+2}=10y_{n+1}-y_n,\, y_1=9, y_0=1,$$ for $n\geq0$, i.e., $$ y_n=\frac{1}{6} ((5-2 \sqrt{6})^n (3-\sqrt{6})+(3+\sqrt{6}) (5+2 \sqrt{6})^n). $$ So, all proper triples $\{1,3,c\}$ in $\Zp{2}$ are $\{1,3,d_n\}$ where \begin{equation}\label{dn2} d_n=2y_n^2-1=\frac{1}{6} ((5+2 \sqrt{6})^{2n+1}+(5-2 \sqrt{6})^{2n+1}-4). \end{equation} Also, because the equation (\ref{e.21}) is solvable in $\Zp{2}$, there exist triples like $\{1,3,11\pm8\sqrt{2}\}$, $\{1,3,32\pm20\sqrt{2}\}$, $\{1,3,161\pm112\sqrt{2}\}$. Like in Example \ref{exa1}, the third element of these triples is obtained from the solution $(X,Y)\in\Zp{2}^2$ of the equation $x^2-3y^2=-2$, i.e., $\{1,3,Y^2-1\}$ is a triple in $\Zp{2}$.\\ It is interesting to note that the expression (\ref{dn1}) contains $11+2\sqrt{30}$ (i.e. $(11,2)$) which is a fundamental solution of the Pell's equation $x^2-30y^2=1$. Similarly, in (\ref{dn2}), $5+2 \sqrt{6}$ represents a fundamental solution of $x^2-6y^2=1$.\\ In what follows, we will give the complete set of proper Diophantine triples $\{1,3,c\}$ in $\Zp{d}$ for positive $d$ which satisfies certain conditions. This problem is related to the solvability of the equation $x^2-3dy^2=6$. For this equation, we have the following Nagell's result. \begin{lemma}[Theorem 11, \cite{nagell}]\label{l3} Let $D$ be a positive integer which is not a perfect square, and let $C\not =1,-D$ be a square-free integer which divides $2D$. Then if $(x_0,y_0)$ is the least positive (fundamental) solution of the equation \begin{equation}\label{e.100} x^2-Dy^2=C, \end{equation} then $$\frac{1}{|C|}(x_0+y_0\sqrt{D})^2=\frac{x_0^2+Dy_0^2}{|C|}+\frac{2x_0y_0}{|C|}\sqrt{D}$$ is a fundamental solution of the related Pell's equation $x^2-Dy^2=1$. All solutions of (\ref{e.100}) are $$ \frac{x_n+y_n\sqrt{D}}{\sqrt{|C|}}=\left(\frac{x_0+y_0\sqrt{D}}{\sqrt{|C|}}\right)^{n}, $$ where $n$ is a positive odd integer. \end{lemma} \begin{theorem}\label{tm2} Let $d$ be a positive integer such that neither $d$ nor $3d$ are perfect squares and such that one of the equations (\ref{e.11}), (\ref{e.21}) is solvable in $\Z$. If $\{1,3,c\}$ is a proper Diophantine triple in $\Zp{d}$, $d\not =2,10$, then there exists nonnegative integer $n$ such that \begin{equation}\label{dnn} c=d_n=\frac{1}{6} ((\xi+\eta\sqrt{3d})^{2n+1}+(\xi-\eta\sqrt{3d})^{2n+1}-4), \end{equation} where $(\xi,\eta)$ is a fundamental solution of the Pell's equation \begin{equation}\label{e.111} x^2-3dy^2=1. \end{equation} If $d\in\{2,10\}$, then $c=d_n$ for some $n\geq1$. \end{theorem} \begin{proof} First, assume that the equation (\ref{e.11}) is solvable in $\Z$. Hence, $c=x'^2-1$, where $(x',y')$ is a solution of (\ref{e.11}). By Lemma \ref{l3}, we get that all solutions of (\ref{e.11}) are $$ x_n=\frac{1}{6}((\alpha+\beta\sqrt{3d})(\xi+\eta\sqrt{3d})^{n}+ (\alpha-\beta\sqrt{3d})(\xi-\eta\sqrt{3d})^{n}),\, n\geq0, $$ where $(\alpha,\beta)$ is a fundamental solution of the equation \begin{equation}\label{e.110} x^2-3dy^2=6 \end{equation} and $(\xi,\eta)$ is a fundamental solution of the related Pell's equation (\ref{e.111}). (Note that (\ref{e.110}) has exactly one class of solutions, since $6|x'x''-3dy'y''$ i $6|x'y''-y'x''$ for each two solutions $(x',y')$ and $(x'',y'')$ of (\ref{e.110})). Hence, $$d_n=\frac{1}{36}((\alpha+\beta\sqrt{3d})^2(\xi+\eta\sqrt{3d})^{2n}+ (\alpha-\beta\sqrt{3d})^2(\xi-\eta\sqrt{3d})^{2n}+12)-1.$$ According to Lemma \ref{l3}, $\frac{1}{6}(\alpha+\beta\sqrt{3d})^2$ (i.e., $(\frac{1}{6}(\alpha^2+3d\beta^2),\frac{1}{3}\alpha\beta)$ is a fundamental solution of $x^2-3dy^2=1$, and so, we get (\ref{dnn}). If (\ref{e.21}) is solvable in $\Z$, we obtain (\ref{dnn}) in similar manner. Indeed, if $(x',y')$ is a solution of (\ref{e.21}), then $c=dy'^2-1$. All solutions of (\ref{e.21}) are $$ y_n=\frac{1}{2 \sqrt{3d}}((x_0+y_0\sqrt{3d})(\xi+\eta\sqrt{3d})^n+ (-x_0+y_0\sqrt{3d})(\xi-\eta\sqrt{3d})^n),\, n\geq 0,$$ where $(x_0,y_0)$ and $(\xi,\eta)$ are fundamental solutions of (\ref{e.21}) and (\ref{e.111}), respectively. Further, we have $$ d_n=\frac{1}{6}\left(\frac{1}{2}(x_0+y_0\sqrt{3d})^2(\xi+\eta\sqrt{3d})^{2n}+ \frac{1}{2}(x_0-y_0\sqrt{3d})^2(\xi-\eta\sqrt{3d})^{2n}-4\right). $$ From Lemma \ref{l3}, we get that $\frac{1}{2}(x_0\pm y_0\sqrt{3d})^2=\xi\pm\eta\sqrt{3d}$ and again obtain (\ref{dnn}). Finally, we must determine in which cases $d_n\in\{0,1,3\}$, since a Diophantine triple consists of three nonzero distinct elements. Obviously, $d_n>3$ for $n\geq 1$, so we have to see if $d_0\in\{0,1,3\}$. Because $d_0=(\xi-2)/3$, it follows that $\xi\in\{2,5,11\}$ where $\xi$ is a fundamental solution of (\ref{e.111}). If \begin{itemize} \item $\xi=2$, then $5=3d\eta^2$ which is not possible, \item $\xi=5$, then $8=d\eta^2$ which implies that $\eta=1$, $d=8$ or $\eta=2$, $d=2$, \item $\xi=11$, then $40=d\eta^2$ which implies that $\eta=1$, $d=40$ or $\eta=2$, $d=10$. \end{itemize} For $d=8,40$ equations (\ref{e.11}) and (\ref{e.21}) have no solution in $\Z$, but (\ref{e.11}) is solvable for $d=10$ and (\ref{e.21}) is solvable for $d=2$. That is the reason why the element $d_0$ in $\Zp{2}$ and $\Zp{10}$ is omitted. \end{proof} In Table 1 we give a list of $1