\documentstyle{article}
\begin{document}
\def\QQ{{\bf Q}}
\def\ZZ{{\bf Z}}
\def\N{{\cal N}}
\def\Tr{{\cal T}}
\def\a{\alpha}
\def\b{\beta}
\def\g{\gamma}
\def\d{\delta}
\def\eps{\epsilon}
\def\th{\theta}

\title{No more undulating squares}
\author{David Moews\\
\small Center for Communications Research\\[-0.8ex] 
\small 4320 Westerra Court\\[-0.8ex] 
\small San Diego, CA 92121\\[-0.8ex]
\small USA\\[-0.8ex]
\small\tt dmoews@ccrwest.org}
\date{October 1998}

\maketitle

A natural number is {\em undulating} if its decimal expansion is of
the form $ab\dots{}abab$ or $bab\dots{}abab$ for some digits $a$ and
$b$.  We will prove that the only undulating squares are 0, 1, 4, 9,
16, 25, 36, 49, 64, 81, 121, 484, 676, and 69696.

Examining the undulating numbers with 8 digits or fewer finds only
the squares above. If we have an undulating square with 9 or more
digits, however, $babababab$ will be a square mod $10^9$, which
can happen only if $(a,b)\in\{(0,0),(2,1),(2,9),(6,1),(6,9),(8,4)\}$.  If there
is an undulating square with $(a,b)=(8,4)$, there will be one of equal length
with $(a,b)=(2,1)$, so it will do to disprove the existence of
undulating squares with 9 or more digits, $a$ 2 or 6, and $b$ 1 or 9.

For an undulating square $y^2$ with an even number, $2m>0$, of digits,
we have $$99y^2=(10a+b)(100^m-1).$$  We can write $10a+b=qr$, where
$q$ is 1 or 3 and $r\in\{7,29,61,23\}$ is prime; then $r|y$, and if we
write $y=rz$ we have $$100^m-(99r/q)z^2=1,$$ or, if we set
$v=10^m$, $w=(3/q)z$, and $x=11qr$, $$v^2-xw^2=1.$$  If we let
$\a$ be $\sqrt{x}$,
$K$ be the quadratic extension $\QQ(\a)$, and $\N$ be the norm
from $K$ to $\QQ$, this will be the
norm equation $\N(v+w\a)=1$.  Since for
our $K$'s the norm of the fundamental unit is always 1,
the solution
to this will be $v+w\a=\pm f^N$, where $f$ is the fundamental
unit of $K$, as shown below:

\bigskip

\begin{center}
\begin{tabular}{cccccc}
$(a,b)$ & $q$ & $r$ & $\a$ & $f$ & $\N(f)$ \\
  \\
$(2,1)$ & 3 & 7 & $\sqrt{3\cdot7\cdot11}$ & $76+5\a$ & 1\\
  \\
$(2,9)$ & 1 & 29 & $\sqrt{11\cdot29}$ & $12901780+722361\a$ & 1 \\
  \\
$(6,1)$ & 1 & 61 & $\sqrt{11\cdot61}$ & $58620+2263\a$ & 1\\
  \\
$(6,9)$ & 3 & 23 & $\sqrt{3\cdot11\cdot23}$ & $551+20\a$ & 1
\end{tabular}
\end{center}

\bigskip

Let $R=\ZZ[\a]$.  
For $(a,b)=(2,1)$ and $(a,b)=(6,9)$, we observe that $f=1$
in $R/5R$, so all powers of $f$ are 1 in $R/5R$ and
$v$ must be $\pm 1$ modulo 5, contradicting
$v=10^m$.  For $(a,b)=(6,1)$, $f=\a$ in $R/3R$, so $f^2=-1$
and the powers of $f$ are 1, $\a$, $-1$, $-\a$, 1, \dots.
Recalling that $3|w$, we see that $N$ must be even; 
but since $f=-2\a$ in $R/5R$, the powers of
$f$ in $R/5R$ are 1, $-2\a$, $-1$, $2\a$, 1, \dots, so
$v$ is $\pm 1$ modulo 5, contradicting $v=10^m$.
Finally, if $(a,b)=(2,9)$, $f=\a$ in $R/5R$, with powers 
1, $\a$, $-1$, $-\a$, 1, \dots, so $N$
must be odd; but $f=4+\a$ in $R/8R$, making
the powers of $f$ 1, $4+\a$, $-1$, $4-\a$, 1, \dots,
so $v$ is congruent to 4 modulo 8, contradicting
$v=10^m$ (unless $m=2$, which we have already
ruled out.)

For
an undulating square $y^2$ with an odd number, $2m+1$, of digits,
we have $$99y^2=10a(100^m-1)+b(100^{m+1}-1)=(10a+100b)100^m-(10a+b).$$
Multiplying by 99 and writing $v=99y$ and $z=3\cdot 10^m$, we
get $$v^2-11(10a+100b)z^2=-99(10a+b);$$ let $4^L$ be the
maximal power of 4 dividing $10a+100b$, $w$ be $2^L z$,
$t$ be $-99(10a+b)$,
and $x$ be $11(10a+100b)/4^L$; then $$v^2-xw^2=t,$$
so letting $\a=\sqrt{x}$, $K=\QQ(\a)$ and $\N$ be the norm
from $K$ to $\QQ$ as before, we get the norm equation $\N(v+w\a)=t$.
$K$'s fundamental unit, as it turns out, still has norm 1, so the
solution to this will be $v+w\a=\pm cf^N$, where $f$
is the fundamental unit, and $c$ is some integral 
element of $K$ with norm $t$.
To find $c$ we may
factor $-99(10a+b)$ as a product of prime ideals in $K$
and then attempt to split this product into two conjugate
principal ideals, in the usual way.
Note that the ring of integers $R$ in $K$
is $\ZZ[\a]$ except when $(a,b)=(6,9)$, when it is
$\ZZ[\th]$ for $\th={1\over2}(1+\a)$ a root of $X^2-X-41=0$.
We show $f$ and the possible values for $c$ below.

\bigskip

\begin{center}
\begin{tabular}{ccccccc}
$(a,b)$ & $2^L$ & $\a$ & $t$ & Possible $c$'s & $f$ & $\N(f)$ \\
  \\
$(2,1)$ & 2 & $\sqrt{2\cdot3\cdot5\cdot11}$ & $-3^3\cdot7\cdot11$ & 
$99\pm6\a$ & $109+6\a$ & 1 \\
  \\
$(2,9)$ & 2 & $\sqrt{2\cdot5\cdot11\cdot23}$ & $-3^2\cdot11\cdot29$ &
$18007\pm 358\a$ & $1259389+25038\a$ & 1 \\
&&&& $297\pm 6\a$ &&\\
&&&& $803\pm 16\a$ && \\
  \\
$(6,1)$ & 4 & $\sqrt{2\cdot5\cdot11}$ & $-3^2\cdot11\cdot61$ & 
$99\pm 12\a$ & $21+2\a$ & 1 \\
  \\
$(6,9)$ & 8 & $\sqrt{3\cdot5\cdot11}$ & $-3^3\cdot11\cdot23$ &
$15\th+42$ & $6+\th$ & 1 \\
&&&& $-15\th+57$ && \\
\end{tabular}
\end{center}

\bigskip

For $(a,b)=(6,9)$, we must impose the constraint that $v$
and $w$ are integral, i.e., that when we look in $R/2R$, the coefficient
of $\th$ in $c f^N$ vanishes.  However, $c$ equals $\th$ or $1+\th$
in $R/2R$, depending on which value of $c$ we pick, and $f^N$
will equal 1, $\th$, or $1+\th$, according to whether $N$ is congruent
to 0, 1, or 2 modulo 3; so $N\equiv-1$ modulo 3 if $c=15\th+42$ and
$N\equiv1$ modulo 3 if $c=-15\th+57$, meaning that we can replace
$f$ by
$f^3=1079+84\a$ and $c$ by $-(-15\th+57)f=297+24\a$
or $-(15\th+42)f^{-1}=297-24\a$.

For $(a,b)=(6,1)$, recall that $v=99y$, so $9|v$.  We can write
$(v+w\a)/3=\pm (33\pm 4\a)f^N$.  In $R/3R$, we have $33\pm 4\a=\pm\a$,
and $f=-\a$ has $f^2=-1$; therefore, $N$ must be even and we can
replace $f$ by $f^2=881+84\a$.
For $(a,b)=(2,9)$, similarly, $3|v$, so look at $c f^N$ in $R/3R$;
we have $f=1$ in $R/3R$; but our three pairs of values for $c$ equal 
$1\pm\a$, 0,
and $-1\pm\a$, respectively, in $R/3R$, so we must take the second
pair of values for $c$, $c=297\pm 6\a$.  Observe that, in all cases, we have
forced $c$ to be the value of $c$ coming from a 1-digit undulating
square, i.e., $c=99\sqrt{b}\pm 2^L\cdot3\a.$  The new values for
$c$ and $f$ are shown below.

\bigskip

\begin{center}
\begin{tabular}{ccccccc}
$(a,b)$ & $c$ & $f$ & $\N(f)$ \\
  \\
$(2,1)$ & $99\pm6\a$ & $109+6\a$ & 1 \\
  \\
$(2,9)$ & $297\pm 6\a$ & $1259389+25038\a$ & 1 \\
  \\
$(6,1)$ & $99\pm 12\a$ & $881+84\a$ & 1 \\
  \\
$(6,9)$ & $297\pm24\a$ & $1079+84\a$ & 1 \\
\end{tabular}
\end{center}

\bigskip

Let the conjugate of $\b$ in $K$ be $\bar\b$.
Shifting to our revised values of $f$ and $c$, we have the equation
\begin{equation}
\label{e1}
\pm(cf^N-\bar c{\bar f}^N)=2w\a=2^{L+1}\cdot3\cdot10^m\a.
\end{equation}
Observe that, for each $(a,b)$, our two values of $c$ are conjugates
of each other.  Since $\N(f)=1$ implies that $\bar f=f^{-1}$,
$cf^N-\bar c{\bar f}^N=-(\bar cf^{-N}-c{\bar f}^{-N})$, so we
may negate the left-hand side of (\ref{e1}) if necessary
to make $c=99\sqrt{b}+2^L\cdot3\a$.  We will then have $c>0$,
so since $\N(c)=t<0$, $\bar c<0$; similarly, $f>0$,
and since $\N(f)=1$, $\bar f>0$.
The parenthesis in (\ref{e1}) is then positive, and so is the
right-hand side, so we may take the positive sign in (\ref{e1}):
$$c f^N-\bar c{\bar f}^N=2^{L+1}\cdot3\cdot10^m\a=
(c-\bar c)10^m.$$
If $N=0$, then $m=0$, so assume that $N>0$.  We can then 
rearrange terms to get
$$
1-{10^m(c-\bar c)\over cf^N}={\bar c\over c}
({\bar f\over f})^N,
$$
so
$$ |1-{10^m(c-\bar c)\over cf^N}| \le |{\bar c\over c}| f^{-2N}.$$
For $|\eps|<{1\over 2}$, we have $|\log (1+\eps)|\le2|\eps|$;
but $|\bar c/c|<1$ and $f>100$, so $|\bar c/c|f^{-2N}<{1\over 2}$
and
$$
|\log{10^m(c-\bar c)\over cf^N}|\le 2|{\bar c\over c}|
f^{-2N},$$
i.e.,
\begin{equation}
\label{e2}
|\log (1-{\bar c\over c})+m\log 10-N\log f|\le2|{\bar c\over c}|
f^{-2N}.
\end{equation}
If $N<0$, negate $N$; since $\bar f=f^{-1}$, we get
$$c{\bar f}^N-\bar c f^N=(c-\bar c)10^m;$$
we can then proceed as above to get
$$ |1-{10^m(\bar c-c)\over \bar cf^N}| \le |{c\over \bar c}| f^{-2N}.$$
We have 
\begin{eqnarray*}
-{c\over\bar c}\le 1-{c\over\bar c}&\le& (1-{c\over\bar c})
(1-{\bar c\over c}) \\
&=& \N({{c-\bar c}\over c}) \\
&=& \N({2^{L+1}\cdot3\a\over c}) \\
&=& {-4^{L+1}\cdot9x\over t} \\
&=& {4(10a+100b)\over 10a+b} \\
&\le& 400,
\end{eqnarray*}
but $f>100$, so $|c/\bar c|f^{-2N}<{1\over 2}$ and, in the same way
as before, we get
\begin{equation}
\label{e3}
|\log (1-{c\over\bar c})+m\log 10-N\log f|\le2|{c\over\bar c}| f^{-2N}.
\end{equation}

For an algebraic number $\d$, let the {\em minimal polynomial} of $\d$ be
the unique polynomial in $\ZZ[X]$ with $\d$ as a root that has minimal
degree, unit content, and positive leading coefficient, and let the
{\em absolute logarithmic Weil height} \cite{bcs} of $\d$ be
$$h_0(\d)={1\over d}(\log A+\sum_{1\le i\le d} \log\max(1,|\d^{(i)}|)),$$
where $d$ is the degree of the minimal polynomial of $\d$, $A$ is its
leading coefficient, and $\d^{(1)}$, \dots, $\d^{(d)}$ are its roots.
In \cite{bcs} we find the following theorem:

Let $\log \g_1$, \dots, $\log \g_n$ be logarithms of algebraic numbers,
and let $d=[\QQ(\g_1,\ldots,\g_n):\QQ]$.  Define
$$h'(\g_i)=\max(h_0(\g_i),{1\over d}|\log \g_i|,{1\over d}).$$
If $k_1$, \dots, $k_n\in\ZZ$ have $\sum_i k_i\log \g_i\ne 0$,
and if $B\ge\max(e,k_1,\ldots,k_n)$, then
$$
\log |\sum_i k_i\log\g_i|\ge
$$
\begin{equation}
\label{ea}
-18(n+1)!\,n^{n+1}(32d)^{n+2}
\log B\log (2nd) \prod_i h'(\g_i).
\end{equation}

We wish to apply this theorem with $n=3$, $\g_1=10$, $\g_2=f$,
and $\g_3$ equalling $1-\bar c/c=2^{L+1}\cdot3\a/c$, or its conjugate,
$1-c/\bar c=-2^{L+1}\cdot3\a/\bar c$.  In this case, we clearly cannot have
$\sum_i k_i\log\g_i=0$ for $k_i\in\ZZ$ not all zero, 
since then $\g_1^{k_1}\g_2^{k_2}
\g_3^{k_3}=1$, and the sets of primes dividing $\g_1$ and $\g_3$ would
have to be the same, which they are not.
Evidently, $\QQ(\g_1,\g_2,\g_3)=K$,
so $d=2$.  It is obvious that $h_0(10)=\log 10$, and since $10>\sqrt{e}$,
$h'(10)=\log 10$ as well.
Since $f$ is an algebraic integer, it has monic minimal
polynomial, and since $f>1$ and $f\bar f=1$, $0<\bar f<1$.  This
implies that $h_0(f)={1\over 2}\log f$, and since $f>e$, 
$h'(f)={1\over 2}\log f$ as well.

We have already seen that $\N(1-\bar c/c)=4(10a+100b)/(10a+b)$,
and if $\Tr$ is the trace from $K$ to $\QQ$,
$$
\Tr(1-{\bar c\over c})=2-{\bar c\over c}-{c\over\bar c}=
(1-{\bar c\over c})(1-{c\over\bar c})=\N(1-{\bar c\over c}).
$$
Write, as before, $10a+b=qr$, where $q\in\{1,3\}$ and
$r\in\{7,23,29,61\}$.  Then $r$ will be minimal with
$r\Tr(1-\bar c/c)$ and $r\N(1-\bar c/c)$ integral, so $r$
will be the leading coefficient of the minimal polynomial of
$1-\bar c/c$, and
\begin{eqnarray*}
h_0(1-{\bar c\over c})&=&{1\over 2}(\log r+\log (1-{\bar c\over c})
+\log (1-{c\over\bar c})) \\
&=& {1\over 2}\log r\N(1-{\bar c\over c}) \\
&=& {1\over 2}\log {4(10a+100b)\over q},
\end{eqnarray*}
and this is clearly also equal to the height of the conjugate,
$h_0(1-c/\bar c)$.  This height is a sum of nonnegative terms,
one of which is ${1\over 2}\log \g_3$, so it is at least ${1\over 2}\log
\g_3$, and since $4(10a+100b)\ge qe$, it is at least $1\over2$.  Therefore,
$$h'(\g_3)={1\over 2}\log {4(10a+100b)\over q}.$$

It follows from (\ref{ea}), then, that if $B\ge\max(e,k_1,k_2,k_3)$ and
the $k_i$'s are not all zero, then
$$\log|\sum_i k_i\log\g_i|\ge$$
$$-18\cdot4!\cdot3^4\cdot64^5\cdot\log B\cdot\log 12
\cdot \log 10\cdot {1\over2}\log f\cdot {1\over 2} \log
{4(10a+100b)\over q},$$
and using $f\le f+\bar f=\Tr(f)\le 3\cdot 10^6$ and $4(10a+100b)/q
\le 4000$, we get
\begin{equation}
\label{e4}
|\sum_i k_i\log\g_i|\ge \exp(-10^{17}\log B).
\end{equation}
Since $\log\g_3\ge 0$, and since $|\log \g_3+m\log 10-N\log f|\le 
2\cdot{1\over2}=1$,
we have $m\log 10-N\log f\le 1$, so
$m\le (1+N\log f)/\log 10$, and we can set $B=3+N\log f/\log 10$.
Recalling that $|\bar c/c|\le 1$ and $|c/\bar c|\le 400$, we see that
for (\ref{e2}) or (\ref{e3}) to be compatible with (\ref{e4}), we need
\begin{eqnarray*}
10^{17}\log B &\ge& 2N\log f-\log 800\\
&=& 2(B-3)\log 10-\log 800,
\end{eqnarray*}
and therefore $B<10^{18}$.

We need to verify that (\ref{e2}) and (\ref{e3}) are not satisfied with
$B<10^{18}$.  Let
$\g$ be the 3-tuple $(\g_1,\g_2,\g_3)$.  Let $j_1$, $j_2$, $j_3\in\ZZ^3$
satisfy $\ZZ j_1+\ZZ j_2+\ZZ j_3=\ZZ^3$, and point in the same approximate
direction as $\g$.

If $0\ne l\in\ZZ^3$ has $|l\cdot\g|\le\eps$, then there must be some 
$i$ with $l\cdot j_i\ne 0$, so $|l\cdot j_i|\ge 1$ and
\begin{eqnarray*}
l\cdot j_i&=& l\cdot(j_i-{j_i\cdot \g\over ||\g||^2}\g
+{j_i\cdot\g\over ||\g||^2}\g) \\
&=&l\cdot(j_i-{j_i\cdot\g\over ||\g||^2}\g)
+{(j_i\cdot\g)(l\cdot\g)\over||\g||^2},
\end{eqnarray*}
so
\begin{eqnarray*}
|l\cdot j_i|&\le&
{||l||\over ||\g||^2} \, ||\,||\g||^2 j_i-(j_i\cdot\g)\g ||
+{||j_i||\,||\g||\, |l\cdot\g|\over ||\g||^2} \\
&\le& {||l||\over ||\g||^2} \, ||\, ||\g||^2 j_i-(j_i\cdot\g)\g ||
+{\eps||j_i||\over ||\g||}, 
\end{eqnarray*}
so if 
\begin{equation}
\label{e6}
\eps ||j_i||<{1\over 2}||\g||, \qquad \hbox{$i=1$, 2, 3,}
\end{equation}
and
\begin{equation}
\label{e7}
||l||\, ||\, ||\g||^2 j_i-(j_i\cdot\g)\g||<{1\over 2}||\g||^2,
\qquad \hbox{$i=1$, 2, 3,}
\end{equation}
we will have a contradiction. 

3-tuples of vectors $(j_1,j_2,j_3)$ satisfying (\ref{e6}) and (\ref{e7})
are listed in the Appendix, for all of our four values
of $(a,b)$ and two values of $\g_3$, and for $\eps=10^{-58}$
and $||l||\le \sqrt{3}\cdot 10^{18}$.  It follows that any solution to 
(\ref{e2}) must have $2|\bar c/c|f^{-2N}>10^{-58}$, and any solution
to (\ref{e3}) must have $2|c/\bar c|f^{-2N}>10^{-58}$.  Keeping in mind
that $f>100$, this implies that $N\le 30$, but since 
$f\le 3\cdot10^6\le10^{13/2}$,
we have $m\le 1+N\log f/\log 10\le 1+15\cdot13=196$.  Therefore, any
undulating square other than those already seen must have no more than 
$1+2\cdot196=393$ digits.  A computer search for such undulating numbers
which are squares finds no squares other than those already seen, so there 
are no undulating squares other than those we have mentioned.

\begin{thebibliography}{9}
\bibitem{bcs} A.~Baker and G.~W\"ustholz, Logarithmic forms and
group varieties, {\em J. reine angew. Math.} {\bf 442} (1993), pp.~19--62.
\bibitem{hjls} J.~Hastad, B.~Just, J.~C.~Lagarias, and 
C.~P.~Schnorr, Polynomial time algorithms for finding integer relations
among real numbers, {\em SIAM J. Comput.} {\bf 18} (1989), pp.~859--881.
\end{thebibliography}

\section*{Appendix}

These $j_i$'s were found by the algorithm of Hastad, Just, Lagarias,
and Schnorr \cite{hjls}.  For each choice of $(a,b)$ and $\g_3$, we 
list $j_1$, $j_2$, and $j_3$ in order.

\bigskip

$(a,b)=(2,1)$:

$\g_3=1-{\bar c}/c:$

{\small
(3948118173343244105682583592769898564801588, 9232466499254518020861284669396465384855728, 80480253437189587064810585227654314740903)

(4158664947885075407681022377053451749292621, 9724819047262851004501080672742246485672783, 84772135552045483109368006388939100324992)

(1470832383030380866640407152053448434003994, 3439464095586508935050252297191881208385991, 29982122558827312529890796407407243841829)
}

\medskip

$\g_3=1-c/{\bar c}:$

{\small
(14656430978118393601985937383773149683367287, 34273292252934317435719663441139385972010462, 19619647882210804351045171509480436465404184)

(7874756644794125017461484425840903581048099, 18414703846434976225456813498058337608568225, 10541444418468879075711143979300529946910579)

(164647407714390634723340734820737212495007, 385019295059498672212302290773264214757090, 220403191534004352998224336986633657016439)
}

\bigskip

$(a,b)=(2,9)$:

$\g_3=1-{\bar c}/c:$

{\small
(222063231539469202116387209192446702907373463994794328383, \hfill\break
1421468912970367647036122762476742177348738777064373804880, \hfill\break
769140016609597630987243664588874711252968957440120464)\hfill\break

(264390766734959308982781228307626219236845815441297382611, \hfill\break
1692415503389386450275636134778713053914201095022061707427, \hfill\break
915746012107398749115991993481158331559709836545941558) \hfill\break

(190788939406782776105226280792627012582819609622732762316, \hfill\break
1221276230311572557698305351313201322409113501512643011490, \hfill\break
660818123770203966944359980963683613686222412714915369) \hfill\break
}

\medskip

$\g_3=1-c/{\bar c}:$

{\small
(7609726769213731904858676172135435105979921, 48711306071005973184800480059700890324973800, 15980321297256252138384260306800990657205236)

(1634424594336045733119800838807173178190775, 10462262191433325590758890335076067275881194, 3432269113694657094646759354977984607672839)

(28374472744252381901596080495840310614652, 181630388102818144431502035750421101979435, 59586001553672598431544987411483933245141)
}

\bigskip

$(a,b)=(6,1)$:

$\g_3=1-{\bar c}/c:$

{\small
(358842098738994946724802666613558555886785, 1164803520941003036762137810649018711113545, 17583640654261146216101581219597913790182)

(8266732593832182328439251001191336944936, 26833861650600952008250402441415185499063, 405078600937903940303079368746695358932)

(101587244231780424318763326496512506500910, 329752780344589895606077149330164237424887, 4977881926076747002868793801217208893025)
}

\medskip

$\g_3=1-c/{\bar c}:$

{\small
(1373888988363038021143329105883018025547243, 4459651575584310223922374777416636137304179, 1335212757740216676754635643863749333092071)

(241202372140900123689637608530813823598180, 782944290305789847984692019915460623302175, 234412304929714369056012762447403094035602)

(3795167286035620301946862038227844615157, 12319134886538933581581919590240254453370, 3688329858521342429307975174564386798308)
}

\bigskip

$(a,b)=(6,9):$

$\g_3=1-{\bar c}/c:$

{\small
(4116736281246703786084754764070761112107677, 13725410144104289757206788143567657980812602, 33028083200032285507591556974390092072476)

(1965748780966300628904098728373993764416980, 6553907371220952512377224157729949875519849, 15770967546275588441309217904538260226626)

(400038325001276561820747842753419733855606, 1333748316358078919416218870860783428163691, 3209459673560258347199843102236447039041)
}

\medskip

$\g_3=1-c/{\bar c}:$

{\small
(195241826410046637693552320988234780575093, 650946274350701367349745450924255654763638, 339224717895823151503590522039512677805407)

(366296257469152206257891128655079841493330, 1221250530649002550702889547560020228766996, 636424821929831278932518317441568656538223)

(38477574409878730099994608407626305606502, 128286208794006656800340159128213307583223, 66853217696774250907713557322046036908953)
}

\end{document}
