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\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 $d&lt;0$ and $d\not=-2$.
\end{abstract}

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\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 &amp;=&amp; (\xi_1+\eta_1\sqrt{d})^2\\
\label{e.01} 3a+3b\sqrt{d}+1 &amp;=&amp; (\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&amp;=&amp;(2^{k+l-2}n-1)^2-(2^{l-1}c_2-2^{k-1}c_1)^2\\
&amp;=&amp;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\\
&amp;=&amp; 4^{k+l-2}n^2+1-4^{l-1}c_2^2-4^{k-1}c_1^2\\
&amp;=&amp; (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 $d&lt;0$ 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 &amp;=&amp; \xi^2\\  3c+1 &amp;=&amp; d\eta^2, \end{array}
\end{equation}
\begin{equation} \label{sl.2}
\begin{array}{rcl}
c+1 &amp;=&amp; d\eta^2\\  3c+1 &amp;=&amp; \xi^2,
\end{array}
\end{equation}
\begin{equation} \label{sl.3}
\begin{array}{rcl}
c+1 &amp;=&amp; d\eta_1^2\\  3c+1 &amp;=&amp; 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 $d&lt;0$ 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) &amp;=&amp;66+
 (7-4 \sqrt{3})^{2 k+1}+(7+4 \sqrt{3})^{2 k+1}+\\ &amp;&amp;
 16((7-4\sqrt{3})^k (2-\sqrt{3})+(2+\sqrt{3})
 (7+4\sqrt{3})^k)\\
 &amp;=&amp;64+16((2+\sqrt{3})^{2k+1}+(2-\sqrt{3})^{2k+1})+\\ &amp;&amp;
 ((2+\sqrt{3})^{2k+1}+(2-\sqrt{3})^{2k+1})^2\\
&amp;=&amp; 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 $d_l&lt;0$). Hence,
$\sqrt{c_kd_l+1}\not\in\Zp{-2}$.


\section{Proper triples in $\Zp{d}$ for some $d&gt;0$}

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, $d_n&gt;3$ 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 $1&lt;d&lt; 100$ such that the equation
(\ref{e.11}) is solvable. In the last column of Table 1, the
smallest $c$ of a proper triple $\{1,3,c\}$ in $\Zp{d}$ is given.

\begin{table}[!ht]
\label{tab1}
\begin{center}
\begin{tabular}{|c|c|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  $d$ &amp; fund. solution of (\ref{e.11}) &amp; fund. solution of (\ref{e.111}) &amp; $c_{min}$ \\ \hline \hline
  10 &amp; (2,1) &amp; (11,2) &amp; 1763\\ \hline
  46 &amp; (4,1) &amp;  (47,4) &amp; 15 \\ \hline
  58 &amp; (22,5) &amp;  (1451,110) &amp; 483\\ \hline
  73 &amp; (5,1) &amp;  (74,5) &amp; 24 \\ \hline
  94 &amp; (28,5) &amp;  (2351,140) &amp; 783\\
  \hline
\end{tabular}
\end{center}
\caption{\small A list of $1&lt;d&lt; 100$ such that (\ref{e.11}) is
solvable}
\end{table}

If the equation (\ref{e.21}) is solvable, we obtain Table 2.

\begin{table}[!ht]
\label{tab2}
\begin{center}
\begin{tabular}{|c|c|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  $d$ &amp; fund. solution of (\ref{e.11}) &amp; fund. solution of (\ref{e.111}) &amp; $c_{min}$ \\
  \hline \hline
  2 &amp; (2,1) &amp; (5,2) &amp; 161 \\ \hline
  6 &amp; (4,1) &amp; (17,4) &amp; 5\\ \hline
  17 &amp; (7,1) &amp; (50,7) &amp; 16 \\ \hline
  18 &amp; (22,3) &amp; (485,66) &amp; 161 \\ \hline
  22 &amp; (8,1) &amp; (65,8) &amp; 21 \\ \hline
  34 &amp; (10,1) &amp; (101,10) &amp; 33 \\ \hline
  38 &amp; (32,3) &amp; (1025,96) &amp; 341 \\ \hline
  41 &amp; (11,1) &amp; (122,11) &amp; 40 \\ \hline
  54 &amp; (140,11) &amp; (19601,1540) &amp; 6533 \\ \hline
  57 &amp; (13,1) &amp; (170,13) &amp; 56 \\ \hline
  66 &amp; (14,1) &amp; (197,14) &amp; 65 \\ \hline
  82 &amp; (298,19) &amp; (88805,5662) &amp; 29601 \\ \hline
  86 &amp; (16,1) &amp; (257,16) &amp; 85 \\ \hline
  89 &amp; (49,3) &amp; (2402,147) &amp; 800 \\ \hline
  97 &amp; (17,1) &amp; (290,17) &amp; 96 \\
  \hline
\end{tabular}
\end{center}
\caption{\small A list of $1&lt;d&lt; 100$ such that (\ref{e.21}) is
solvable}
\end{table}

\section{Acknowledgements}
The author would like to thank A. Dujella for his helpful  comments.


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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: 11D09,
11R11.

\noindent \emph{Keywords: } Diophantine sets,
 quadratic fields, Pellian equations.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A011922},
\seqnum{A030063},
\seqnum{A045899}, and
\seqnum{A051048}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received August 19 2010;
revised version received August 31 2010; November 17 2010.
Published in {\it Journal of Integer Sequences},
December 7 2010.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in

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