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\begin{center}
\vskip 1cm{\LARGE\bf
Rational Distance Sets on $x \, y = 1$
}
\vskip 1cm
\large
Allan J. MacLeod\\
Statistics, O.R. and Mathematics Group\\
University of the West of Scotland\\
High St. \\
Paisley\\
Scotland PA1 2BE\\
\href{mailto:allan.macleod@uws.ac.uk}{\tt allan.macleod@uws.ac.uk}
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\begin{abstract}
We consider the problem of finding sets of points
on $x y=1$, where all of the inter-point distances are rational.
The search for such points has links to both congruent and
concordant numbers.
\end{abstract}
\section{Introduction}
Consider a curve in two dimensions, given in the form $f(x,y)=0$.
Let $P = \{P_1,P_2,\ldots,P_n\}$ be a set of $n$ distinct points.
How many of the $n(n-1)/2$ inter-point distances can we make
rational? This problem is obviously related to the discussion in
section D20 of Guy's book \cite{guy}.
If the curve is a straight line then there are clearly an
infinite number of points, so we usually restrict the points in
$P$ to have no more than two on any line.
For the circle $f(x,y)=x^2+y^2-1$, let $\mu$ and $\theta$ be two
distinct angles with $s=\tan \mu$ and $t=\tan \theta$ both
rational. Let $P_1=(\cos 4\mu , \sin 4\mu)$ and $P_2=(\cos
4\theta , \sin 4\theta)$, then the distance from $P_1$ to $P_2$ is
\begin{displaymath}
\frac{4|s-t||1+s t|}{(1+s^2)(1+t^2)}
\end{displaymath}
which is clearly rational. Thus there are an infinite number of
distinct points on the unit circle with rational distances from
each other. This normally forces a second restriction on $P$ - no
more than $3$ points on any circle.
Very recently, Solymosi and de Zeeuw \cite{sz} proved that lines and
circles are the only two-dimensional curves that give an infinite
number of rational distances.
Even before this, Campbell \cite{camp} discussed the problem for the
basic parabola $y=x^2$. Based on an early preprint on this
investigation, the present author found one of the first examples of a
set of $4$ acceptable points on the parabola with all $6$ inter-point
distances rational. Choudhry \cite{chou} also discussed this parabola,
and showed how to find $5$ points with $9$ distances rational.
The current work discusses this problem for the rectangular
hyperbola
\begin{equation}
x \, y = 1 .
\end{equation}
After this paper was initially submitted, Goins and Mugo \cite{gm} released a
preprint which is clearly extremely related to this discussion, but from a different perspective.
They consider the general conic section
\begin{displaymath}
axy+bx+cy+d=0
\end{displaymath}
with $a \ne 0$ and $ad \ne bc$. They show that, if the congruent number elliptic curve
\begin{displaymath}
Y^2=X^3-D^2X \hspace{2cm} D=\frac{ad-bc}{2a^2}
\end{displaymath}
has rank greater than $0$, then there are an infinite number of sets of $4$ points on the conic, with
the usual conditions on lines and circles.
For the rectangular hyperbola $xy=1$ we have $D=-1/2$, but it is straight-forward that the elliptic curve
$Y^2=X^3-X/4$ is equivalent to $V^2=U^3-4U$ which asks if $2$ is a congruent number. It is an old result
that this is not the case, see Chapter XVI of Dickson's History \cite{dick}.
So the Goins-Mugo result does not apply in this special case.
\section{Basic Results}
Let $P=(p,1/p)$ and $Q=(q,1/q)$ be on the hyperbola with $p \ne
q$. Then the distance is given by
\begin{equation}
|PQ|^2= (p-q)^2+(1/p-1/q)^2 = \frac{(p-q)^2}{p^2q^2}(1+p^2q^2)
\end{equation}
Clearly, we can go a long way to making this rational if we
restrict $p$ and $q$ to being rational themselves, and we assume
from now on that all points are themselves rational. Thus $|PQ|$
will be rational if $1+p^2q^2=u^2$ for $u \in \mathbb{Q}$.
Thus, we can assume, without loss of generality, that
\begin{equation}
p \, q= \frac{a^2-1}{2a} \equiv \phi(a)
\end{equation}
where $a \in \mathbb{Q}$.
Let $Q'=(-q,-1/q)$ which is also on the curve, and
\begin{displaymath}
|PQ'|=\frac{(p+q)^2}{p^2q^2}(1+p^2q^2)
\end{displaymath}
so if $|PQ|$ rational so is $|PQ'|$. Thus if we have a point with
negative coordinates, we can always replace it with one with
positive ones. We thus assume, from now on, that all points are
in the first quadrant.
It should be noted that we do not lose anything by this, since we
can never have $Q$ and $Q'$ in $P$, as this would require $1+q^4$
to be a square, which is impossible as shown by Fermat.
As stated in the introduction, the basic assumption in rational
distance sets is that we look for no more than $2$ points on a
single line or more than $3$ points on a single circle. The hyperbola
meets $y=ax+b$ where
\begin{displaymath}
ax^2+bx-1=0
\end{displaymath}
so that any line can only meet the hyperbola at a
maximum of $2$ points, so the first condition is automatically
satisfied.
For a circle, assume $R=(r,1/r)$ with $r \ne p,q$, then the unique
circle through $P,Q,R$ meets the hyperbola at a fourth point with
x-coordinate $1/(pqr)$.
\section{Three Points}
Suppose we have points $P,Q,R$, then, if the $3$ inter-point
distances are rational, we require
\begin{equation}
p \, q =\frac{a^2-1}{2a} \hspace{1.5cm} p \, r=\frac{b^2-1}{2b}
\hspace{1.5cm} q \, r = \frac{c^2-1}{2c}
\end{equation}
with $a,b,c \in \mathbb{Q}$.
Thus
\begin{equation}
\frac{a^2-1}{2ap} \; \; \frac{b^2-1}{2b p}=\frac{c^2-1}{2c}
\end{equation}
so that
\begin{displaymath}
p^2=\frac{(a^2-1)(b^2-1)c}{2ab(c^2-1)}
\end{displaymath}
which implies that there must exist a rational $e$ with
\begin{equation}
e^2 = 2 a b c (a^2-1)(b^2-1)(c^2-1)
\end{equation}
Define $a=A/N$ and $b=B/N$, with $A,B,N \in \mathbb{Z}$, so that
\begin{displaymath}
2ab(a^2-1)(b^2-1)=\frac{2AB(A^2-N^2)(B^2-N^2)}{N^6}
\end{displaymath}
and, if we define $f=eN^3$ and $2AB(A^2-N^2)(B^2-N^2)=KM^2$,
where $K$ is a square-free integer, we get
\begin{displaymath}
f^2=K M^2(c^3-c)
\end{displaymath}
Now, define $y=Kf/M$ and $x=Kc$, giving
\begin{equation}
E_K: y^2=x^3-K^2x
\end{equation}
This is an elliptic curve, and, in fact, is the elliptic curve
related to the congruent number problem. This forms the basis for
the book by Koblitz \cite{kob}.
The curve $E_K$ has the obvious rational points $(0,0) , (K,0) ,
(-K,0)$, which give $c=0,1,-1$ respectively, which cause problems
with the identity $q r=2c/(c^2-1)$. Thus to have a practical
solution we need another rational point, which means that the
rank of $E_K$ must be greater than $0$.
The famous result of Tunnell \cite{tunn} shows that this will be the
case if $K$ is $\equiv 5 , 6 , 7 \bmod 8$, and possibly when $K$
is $\equiv 1 , 2 , 3 \bmod 8$.
For example, when $a=3/2, b=4/3$, we have
$2AB(A^2-N^2)(B^2-N^2)=181440=35 \times 72^2$, so that $K=35,
M=72$. The curve $E_{35}$ has rank $0$ and so there are no trios
of points in this case.
Alternatively, when $a=4/3, b=2/1$, we have
$2AB(A^2-N^2)(B^2-N^2)=9072=7 \times 36^2$, giving $K=7,M=36$.
The curve $E_7$ has rank $1$ with, for example, the point
$(25,120)$. This gives $c=25/7$, and $p q=7/24$ and $p r =3/4$
so that
\begin{displaymath}
q r = \frac{21}{96p^2} = \frac{288}{175}
\end{displaymath}
leading to $p=35/96$, $q=4/5$ and $r=72/35$.
We now analyse what happens when we add torsion points to points of infinite order. As stated, the curve has
$3$ finite torsion points $(0,0), (K,0), (-K,0)$. Let $(g,h)$ be a point of infinite order giving a value of $c$, which we call $c_1$,
and leading to the $3$
points on the hyperbola $(p_1,1/p_1)$, $(q_1,1/q_1)$ and $(r_1,1/r_1)$.
Forming $(g,h)+(0,0)$ gives $x=-K^2/g$, which gives a new value of $c$ given by $c_2=-1/c_1$.
Thus $c_2/(c_2^2-1)=c_1/(c_1^2-1)$ and so $p_2=p_1$, giving
the same set of three points.
Forming $(g,h)+(K,0)$ gives $x=K(g+K)/(g-K)$ so $c_3=(c_1+1)/(c_1-1)$ and $p_3=p_1(c_1^2-1)/2c_1$. Thus $q_3=1/r_1$
and $r_3=1/q_1$, and we have a different set of three points $(1/r_1,r_1)$, $(1/q_1,q_1)$ and $(p_3,1/p_3)$. We can invert each
of the x-coordinates to give the set $(q_1,1/q_1)$, $(r_1,1/r_1)$ and $(1/p_3,p_3)$. Note, however, that
\begin{displaymath}
p_1q_1r_1\frac{1}{p_3}=p_1 \, \frac{(a^2-1)}{2ap_1} \, \frac{(b^2-1)}{2bp_1} \, \frac{2c_1}{(c_1^2-1)p_1} = 1
\end{displaymath}
so the four points actually lie on a circle.
Similarly, if we add $(g,h)+(-K,0)$, we get the same situation.
Finally, in this section, note that, if $p,q,r$ give $3$ points with rational distances, the point $1/(pqr)$ (on
the circle through the three points) is at a rational distance from the other points.
\section{Four Points}
Suppose we wish a fourth point $S=(s,1/s)$ to have a rational distance to each of $P, Q, R$. We assume $pqrs \ne 1$,
so that the four points do not lie on a circle.
Then we require
\begin{equation}
1+p^2s^2=\Box \hspace{1.5cm} 1+q^2s^2=\Box \hspace{1.5cm} 1+r^2s^2=\Box
\end{equation}
Consider the first two conditions, noting that $p,q,r$ are known quantities. We have
\begin{equation}
p\,s=\frac{f^2-1}{2f} \hspace{2cm} q\,s=\frac{g^2-1}{2g}
\end{equation}
with $f,g \in \mathbb{Q}$, giving
\begin{displaymath}
p\,f(g^2-1)-q(f^2-1)g=0
\end{displaymath}
This quadratic must have rational roots, so the discriminant must be a rational square, giving
\begin{displaymath}
h^2=q^2(f^2-1)^2+4p^2f^2
\end{displaymath}
with $h \in \mathbb{Q}$, and so, defining $Y=qh, X=qf$, we have the quartic
\begin{equation}
Y^2=X^4+(4p^2-2q^2)X^2+q^4
\end{equation}
This quartic has an obvious rational point $(0,q^2)$, so is birationally equivalent to the elliptic curve
\begin{equation}
V^2=U(U+p^2)(U+q^2)
\end{equation}
with $f=V/(q(U+p^2))$.
Since $p,q \in \mathbb{Q}$, express $p=J/I, \, q=L/I, \, U=Z/I^2, \, V=W/I^3$
with $I,J,L \in \mathbb{Z}$, so that
\begin{equation}
W^2=Z(Z+J^2)(Z+L^2) \hspace{1cm} , \hspace{1cm} f=\frac{W}{L(Z+J^2)}
\end{equation}
This is the elliptic curve for a specific subset of Euler's concordant forms,
namely $x^2+J^2y^2=\Box \, , \, x^2+L^2y^2=\Box$, as discussed by Ono \cite{ono}.
The curve has
\begin{enumerate}
\item[(a)]$3$ points of order $2$, at $(0,0),(-J^2,0),(-L^2,0)$,
\item[(b)] $4$ points of order $4$ at $(JL,\pm JL(J+L))$ and $(-JL,\pm JL(J-L))$,
\item[(c)] the point at infinity.
\end{enumerate}
It is even possible to have points of order $8$,
so the torsion subgroup is usually $\mathbb{Z}_2 \times \mathbb{Z}_4$ but could be $\mathbb{Z}_2 \times \mathbb{Z}_8$.
None of these torsion points lead to a non-trivial value of $s$.
Note that $r$ is already a solution to equation $(9)$, which gives a point on this curve. From equation $(4)$, let $X_1=q\,b$ and
$(X_1,Y_1)$ be the solution of $(10)$ then
\begin{displaymath}
U=\frac{X_1^2+Y_1-q^2}{2}
\end{displaymath}
gives a point on $(10)$, meaning that the curves all have rank at least $1$.
We now investigate the effect of adding torsion points to a point on the curve. Suppose $(Z_0,W_0)$ is a point
on the elliptic curve so that $W_0^2=Z_0(Z_0+J^2)(Z_0+L^2)$, and let $f_0=W_0/(L(Z_0+J^2))$, from which we find a value for $s$.
Adding $(0,0)$ gives $Z_1=J^2L^2/Z_0$ and $W_1=-J^2L^2W_0/Z_0^2$ leading to
\begin{displaymath}
f_1=-\frac{LW_0}{Z_0(Z_0+L^2)}=-\frac{L(Z_0+J^2)}{W_0}=-\frac{1}{f_0}
\end{displaymath}
Adding $(-J^2,0)$ gives
\begin{displaymath}
Z_2=\frac{-J^2(Z_0+L^2)}{Z_0+J^2} \hspace{1cm} , \hspace{1cm} W_2=\frac{W_0J^2(L^2-J^2)}{(Z_0+J^2)^2}
\end{displaymath}
which just gives $f_2=-f_0$.
Adding $(-L^2,0)$ gives
\begin{displaymath}
Z_3=\frac{-L^2(Z_0+J^2)}{Z_0+L^2} \hspace{1cm} , \hspace{1cm} W_3=\frac{W_0L^2(J^2-L^2)}{(Z_0+L^2)^2}
\end{displaymath}
which just gives $f_3=1/f_0$.
These $3$ new values of $f$ just lead to the same values of $s$.
If, however, we add the points of order $4$ we do get a different point.
To understand the effect of adding the points of order $4$, it is simpler to use the elliptic curve
given in equation $(10)$. Let $(U_0,V_0)$ be a non-torsion rational point on the curve, which gives
\begin{equation}
s_0=\frac{U_0^2-p^2q^2}{2pqV_0}
\end{equation}
Now, consider adding the point of order $4$ given by $(pq,pq(p+q))$. Using a symbolic algebra package
gives
\begin{displaymath}
s_4=\frac{2(U_0^2+U_0(p^2+q^2)-V_0(p+q)+p^2q^2)(U_0(p+q)-V_0)}{-(U_0-pq)(U_0+pq)(U_0^2+U_0(2p^2+2pq+2q^2)-2V_0(p+q)+p^2q^2)}
\end{displaymath}
which does not seem too helpful.
Using $V_0^2=U_0^3+(p^2+q^2)U_0^2+p^2q^2U_0$, however, allows us to simplify this to
\begin{equation}
s_4=\frac{2V_0}{U_0^2-p^2q^2}=\frac{1}{pqs_0}
\end{equation}
which is just the fourth point on the hyperbola and the circle through $p,q,s$. The other points of order $4$
also give this value, which we disallow.
\section{Numerical Results}
To find size $3$ and $4$ rational distance sets, we need to determine points of infinite order on
both types of elliptic curves. Determining the rank of an elliptic curve and a full set of generators is
a non-trivial process. We avoid this problem, and use both a simple search procedure and a very simple
descent procedure to determine independent points of infinite order moderately quickly.
All the programming was done using the software package Pari.
For the search, we determine the minimal form of the curve and search this for points with
integer coordinates up to a specified size. Clearly, we must restrict the search on the infinite
component, and we often must also do this on the closed finite component. Thus it
is possible that we might miss a possible generator.
For the descent, we note that the curves we must search are of the general form
\begin{equation}
y^2=x^3+Gx^2+Hx
\end{equation}
A rational point on such curves has form $(du^2/v^2,duw/v^3)$ where $d$ is squarefree and
$\gcd(d,v)=1$. Substituting gives
\begin{displaymath}
dw^2=d^2u^4+dGu^2v^2+Hv^4
\end{displaymath}
so that $d|H$, allowing us to find possible values of $d$ easily.
We have
\begin{displaymath}
4dw^2=4d^2u^4+4dGu^2v^2+4Hv^4=(2du^2+Gv^2)^2-(G^2-4H)v^4
\end{displaymath}
and we find that, in both equations $(7)$ and $(12)$, the corresponding values of $G$ and $H$
give $G^2-4H=\Box=C^2$ say.
Thus, define $Z=2w$, $X=2du^2-Gv^2$ and $Y=Cv^2$, leading to $X^2=Y^2+dZ^2$. This has
a simple parametric solution $Y=p^2-dq^2$, $Z=2pq$ and $X=p^2+dq^2$. This gives
\begin{equation}
\frac{u^2}{v^2}=\frac{(C-G)p^2+d(C+G)q^2}{2d(p^2-dq^2)}
\end{equation}
We generate small-size pairs $(p,q)$, with $\gcd(p,q)=1$, compute the right-hand-side of
this ratio and test whether it is square. We find several non-integer point of infinite order using this approach,
which are usually generators, but might not be.
Combining the above two approaches, we find $k$ independent points of infinite order
$G_1,\ldots,G_k$ and then compute
\begin{equation}
a_1G_1+\ldots+a_kG_k+T
\end{equation}
with $T$ a torsion-point and $|a_i|$ less than some preset limit.
Using this allows us to find, after a reasonable amount of computation, the following table of
size $4$ rational distance sets on $x\,y=1$.
\begin{center}TABLE 1 \\Size $4$ rational distance sets\\[0.5em]
\begin{tabular}{rrrr}
p \hspace{0.5cm}& q \hspace{0.3cm} & r \hspace{0.3cm} & s \hspace{0.3cm} \\
77/340& 15/56& 96/55& 77/18\\
14/65& 60/77& 462/325& 26/9\\
11/60& 117/418& 5681/17490& 80/39\\
27/80&320/231&319/168&55/19\\
287/1380&595/1254&715/238&99/10\\
3/11& 1275/2288& 319/416& 52/45\\
7/78& 145/264& 25/28& 864/235\\
437/702& 63/46& 8/5& 697/210\\
9/34&9675/16898&208/81&3689/390\\
207/434& 221/300& 160/161& 91/24\\
21/110& 424/1071& 39/62& 24/5\\
2/9&876/1771&4125/5168&884/495\\
715/2352&1548/16575&34/33&28/17
\end{tabular}
\end{center}
The method of Choudhry, for finding larger sets, is based on the fact that $(s,s^2)$ and $(t,t^2)$ have a rational distance if
$s+t=\phi(k)$ for $k \in \mathbb{Q}$. For the hyperbola, $(s,1/s)$ and $(t,1/t)$ have a rational distance
if $s\,t=\phi(k)$. So we move from an additive approach to a multiplicative one. This makes finding
relations such as those given by Choudhry much more difficult.
\section{Acknowledgement}
The author would like to express sincere thanks to the anonymous referee and the Editor for several very helpful comments
on both submissions of this paper.
\begin{thebibliography}{20}
\bibitem{camp} G. Campbell, Points on $y=x^2$ at rational distance,
{\it Math. Comp.}, {\bf 73} (2004), 2093--2108.
\bibitem{chou} A. Choudhry, Points at rational distances on a parabola,
{\it Rocky Mt. J. Math.}, {\bf 36} (2006), 413--424.
\bibitem{dick} L. E. Dickson, {\it History of the Theory of Numbers
Vol.\ 2: Diophantine Analysis}, Chelsea, 1971.
\bibitem{gm} E. H. Goins and K. Mugo, Points on hyperbolas at rational
distance, preprint, \url{http://arxiv.org/abs/1108.0690}.
\bibitem{guy} R. K. Guy, {\it Unsolved Problems in Number Theory}, 3rd
edition, Springer, 2004.
\bibitem{kob} N. Koblitz, {\it Introduction to Elliptic Curves and
Modular Forms}, Springer, 1984.
\bibitem{ono} K. Ono, Euler's concordant forms, {\it Acta Arith.}, {\bf
65} (1996), 101--123.
\bibitem{st} J. H. Silverman and J. Tate, {\it Rational Points on
Elliptic Curves}, Springer, 1992.
\bibitem{sz} J. Solymosi and F. de Zeeuw, On a question of Erd\"{o}s and Ulam, {\it Disc. \& Comp. Geom.}, {\bf 43} (2010), 393--401.
\bibitem{tunn} J. B. Tunnell, A classical Diophantine problem and
modular forms of weight 3/2, {\it Invent. Math.} {\bf 72} (1983),
323--334.
\end{thebibliography}
\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11D25; Secondary 11Y50, 11G05.
\noindent \emph{Keywords: } hyperbola, rational distance set, elliptic
curve, congruent number, concordant form.
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received April 21 2011;
revised versions received October 10 2011; January 10 2012.
Published in {\it Journal of Integer Sequences}, January 14 2012.
\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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