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\title{Elem. N$^\mathrm o$ Thy II, examination II solutions}
%\author{David Pierce}
\date{\today}
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\address{Mathematics Dept\\
Middle East Technical University\\
Ankara 06531, Turkey}

\email{dpierce@metu.edu.tr}
\urladdr{http://www.math.metu.edu.tr/~dpierce/}
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\newcommand{\Z}{\stnd{Z}}         % integers
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\begin{document}
  \maketitle\thispagestyle{empty}

  \begin{solution}
    By the algorithm for finding continued fractions, when
    $x=\oldsqrt{a^2+1}$:
    \begin{equation*}
a_0=a,\quad\xi_0=\oldsqrt{a^2+1}-a,\quad
      \frac1{\xi_0}=\frac{\oldsqrt{a^2+1}+a}{a^2+1+a^2}=\oldsqrt{a^2+1}+a,
    \end{equation*}
so $a_1=2a$ and $\xi_1=\oldsqrt{a^2+1}-a=\xi_0$.  Therefore
$x=[a;\overline{2a}]$. 
  \end{solution}

  \begin{remark}
    Alternatively, one may let
    \begin{equation*}
      x=[a;\overline{2a}]=[a;a+a,\overline{2a}]=[a;a+[a,\overline{2a}]]
=[a;a+x]=a+\frac1{a+x}=
      \frac{a^2+ax+1}{a+x}, 
    \end{equation*}
so that $ax+x^2=a^2+ax+1$ and therefore $x^2=a^2+1$.  Since $a>0$, we
have $[a;\overline{2a}]>0$ and therefore
$[a;\overline{2a}]=x=\oldsqrt{a^2+1}$. 
  \end{remark}

  \begin{solution}
    \begin{enumerate}
      \item
Since $\sqrt 5$ is a root of $x^2-5$, whose leading coefficient is $1$,
we can conclude $\ord=\lat[1,\sqrt 5]=\mLambda$.
\item
We know that the units of $\roi$ (when $K=\Q(\sqrt 5)$) are
$\pm\gr^n$.  Of these, those that are greater than $1$ form the list
\begin{equation*}
  \gr,1+\gr,1+2\gr,2+3\gr,3+5\gr,5+8\gr,\dots
\end{equation*}
(and in general $\gr^n=\Fib{n-1}+\Fib n\gr$).  But since $2\gr=1+\sqrt
5$, we have $\ord=\lat[1,2\gr]$; also, $\norm{\gr}=-1$.  The first
power of $\gr$ greater than $1$ that belongs to $\ord$ and has norm $1$
is therefore $\gr^6$.  Hence the elements of $\ord$ of norm $1$ are
$\pm\gr^{6n}$, where $n\in\Z$.
    \end{enumerate}
  \end{solution}

  \begin{solution}
We want to solve    
    \begin{equation*}
    19=x^2+2xy+4y^2=(x+y)^2+3y^2.
    \end{equation*}
Hence $y^2\leq19/3<9$, so $\size y<3$.  When $y=\pm2$, the equation
becomes $(x\pm2)^2=7$, which has no solution.  When $y=\pm1$, we get
$(x\pm 1)^2=16$, so $(x\pm1)\in\{4,-4\}$.  When $y=0$, there is no
solution.  So the solutions of the original equation are $(3,1)$,
$(-5,1)$, $(5,-1)$, $(-3,-1)$. 
  \end{solution}

  \begin{remark}
    Solving an equation means not only finding solutions, but showing
    that there are no other solutions.  This is done here by noting
    that there are only 5 possibilities for~$y$.  Alternatively, one
    may rewrite the equation as
    \begin{equation*}
19=(x+y+y\sqrt{{-3}})(x+y-y\sqrt{{-3}})=\norm{x+2\omega y},
    \end{equation*}
where we work in $\Q(\sqrt{{-3}})$.  Since $(x,y)$ is a solution if
and only if $(\size x,\size y)$ is a solution, we can obtain all
solutions from Figure~\ref{fig:1}.
\begin{figure}[ht]
  \begin{pspicture}(-5,-0.5)(5,4.4)
%\psgrid
\psarc(0,0){4.359}{-5}{185}
\psset{dotsize=2pt 3}
    \psdots
(0,0)(1,0)(2,0)(3,0)(4,0)(5,0)
(0,1.732)(1,1.732)(2,1.732)(3,1.732)(4,1.732)
(0,3.464)(1,3.464)(2,3.464)(3,3.464)%(4,3.464)
(-1,0)(-2,0)(-3,0)(-4,0)(-5,0)
(-1,1.732)(-2,1.732)(-3,1.732)(-4,1.732)
(-1,3.464)(-2,3.464)(-3,3.464)%(-4,3.464)
\psset{dotstyle=o}
\psdots
(0.5,4.330)(1.5,4.330)(2.5,4.330)%(3.5,4.330)
(0.5,2.598)(1.5,2.598)(2.5,2.598)(3.5,2.598)
(0.5,0.866)(1.5,0.866)(2.5,0.866)(3.5,0.866)(4.5,0.866) 
(-0.5,4.330)(-1.5,4.330)(-2.5,4.330)%(-3.5,4.330)
(-0.5,2.598)(-1.5,2.598)(-2.5,2.598)(-3.5,2.598)
(-0.5,0.866)(-1.5,0.866)(-2.5,0.866)(-3.5,0.866)(-4.5,0.866) 
\uput[ur](1,1.732){$2\omega$}
\uput[ur](4,1.732){$3+2\omega$}
\uput[ul](-4,1.732){$-5+2\omega$}
\uput[d](0,0){$0$}
\uput[d](1,0){$1$}
  \end{pspicture}
\caption{Solutions of $\norm{\xi}=7$ from
  $\lat[1,2\omega]$ in $\Q(\sqrt{{-3}})$}\label{fig:1} 
\end{figure}
  \end{remark}

  \begin{solution}
    \begin{enumerate}
      \item
In $\Q(\sqrt{21})$, we have $(5+\sqrt{21})/2=2+\omega$,
and
 $\norm{(5+\sqrt{21})/2}=1$.  Hence $(5+\sqrt{21})/2$ is a unit of
  $\roi$, so its powers are also units of $\roi$.  Let
$\alpha\in\roi$.  Since
\begin{equation*}
\roi=\lat[1,\omega]=\xlat[1,\frac{1+\sqrt{21}}2],
\end{equation*}
we have $2\alpha\in\lat[2,1+\sqrt{21}]\included\lat[1,\sqrt{21}]$.  This
proves 
\begin{equation*}
a_n+b_n\sqrt{21}=2\Bigl(\frac{5+\sqrt{21}}2\Bigr)^n\in\lat[1,\sqrt{21}], 
\end{equation*}
so
$a_n,b_n\in\Z$. 
\item
Since $\norm{a_n+b_n\sqrt{21}}=4$, the pairs $(\pm a_n,\pm b_n)$ are
solutions of $x^2-21y^2=4$.
\item
Suppose $(a,b)$ is an arbitrary solution of $x^2-21y^2=4$.  Then
$a\equiv b\pmod 2$, so $2\divides a-b$.  Hence
\begin{equation*}
  \frac{a+b\sqrt{21}}2
  =\frac{a-b+b+b\sqrt{21}}2
  =\frac{a-b}2+b\frac{1+\sqrt{21}}2
 =\frac{a-b}2+b\omega
\in\lat[1,\omega],
\end{equation*}
so $(a+b\sqrt{21})/2$ is an element of $\roi$ of norm $1$.  But there
is $\epsilon$ or $r+s\omega$ in $\roi$ of norm $1$ such that $r,s>0$
and every element of $\roi$ of norm $1$ is $\pm\epsilon^n$ for some
$n$.  But $(5+\sqrt{21})/2=2+\omega$ and has norm $1$, so it must be
$\epsilon$.  Hence $(a,b)=(\pm a_n,\pm b_n)$ for some $n$.
    \end{enumerate}
  \end{solution}

  \begin{remark}
    The pair $(a_n/2,b_n/2)$ solves the Pell equation $x^2-21y^2=1$,
    but its entries need not belong to $\Z$.  For example,
    $(a_1/2,b_1/2)=(5/2,1/2)$. 
  \end{remark}

  \begin{solution}
    \begin{enumerate}
      \item
We have
$2=2x^2-3y^2\Iff 4=4x^2-6y^2=\norm{2x+y\sqrt 6}$ in $\Q(\sqrt 6)$.
So let
\begin{equation*}
  K=\Q(\sqrt 6),\quad \alpha=2,\quad \beta=\sqrt 6,\quad m=4.
\end{equation*}
\item\label{item:b}
We want the elements of $\ord$ of norm $1$.  But
$(1/2)\mLambda=\lat[1,\sqrt 6/2]$, and $\sqrt6/2$ is a root of
$2x^2-3$.  Hence
$\ord=\lat[1,2\sqrt6/2]=\lat[1,\sqrt6]=\lat[1,\omega]=\roi$. 
We obtain the units of $\roi$ from the continued-fraction expansion
of $\sqrt 6$:
\begin{align*}
  x&=\sqrt 6,& a_0&=2,& \xi_0&=\sqrt6-2;\\
\frac1{\xi_0}&=\frac{\sqrt6+2}2,&a_1&=2,&\xi_1&=\frac{\sqrt6-2}2;\\
\frac1{\xi_1}&=\sqrt6+2,&a_2&=4,&\xi_2&=\sqrt6-2=\xi_0;
\end{align*}
so $\sqrt6=[2;\overline{2,4}]$.  Since $[2;2]=5/2$, and
$\norm{5+2\omega}=1$, we can conclude that the elements of $\roi$ of
norm $1$ are $\pm(5+2\omega)^n$.  Therefore the desired parallelogram
$\mPi$ can be bounded by the straight lines given by
\begin{equation*}
  2x+y\sqrt6=1;\qquad2x+y\sqrt6=5+2\sqrt6.
\end{equation*}
Also, we are looking for points on the hyperbola
\begin{equation*}
  4=4x^2-6y^2=(2x+y\sqrt6)(2x-y\sqrt6),
\end{equation*}
one of whose asymptotes, given by
\begin{equation*}
  2x-y\sqrt6=0,
\end{equation*}
forms a third side of $\mPi$; the fourth side is given by
\begin{equation*}
  2x-y\sqrt6=4,
\end{equation*}
since this line meets the hyperbola where $2x+y\sqrt6=1$ does. 
\item\label{item:c}
Same as \eqref{item:b}.
    \end{enumerate}
  \end{solution}

  \begin{remark}
    The point in~\eqref{item:b} is that, if $\gamma$ is from
    $\mLambda$ and solves $\norm{\xi}=4$, then the same is true of
    $\delta$ or $2a+b\omega$, where $\delta=
    \pm(5+2\omega)^n\gamma$; and we should be able to pick the sign and
    $n$ so that $(a,b)\in\mPi$.  But we can
    pick $n$ so that
    \begin{equation*}
      1\leq\size{\delta}<\size{5+2\omega}<10.
    \end{equation*}
We also have $4=\delta\delta'$, so
\begin{equation*}
  \size{\delta'}=\frac4{\delta}\leq4.
\end{equation*}
Since
\begin{equation*}
  \delta=\frac{\delta+\delta'}2+\frac{\delta-\delta'}{2\omega}\omega.
\end{equation*}
we conclude
\begin{equation*}
  \size a=\xsize{\frac{\delta+\delta'}4}<4, \quad \size
  b=\xsize{\frac{\delta-\delta'}{2\omega}}<4.
\end{equation*}
Thus, in~\eqref{item:b}, $\mPi$ can be the square with vertices
$(\pm4,4)$ and $(\pm4,-4)$.  But this isn't good enough
for~\eqref{item:c}. 
  \end{remark}

\end{document}