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\glsnoundefn{quad_irr}{quadratic
irrational}{quadratic irrationals}
We now study the \glspl{cfe} of \emph{quadratic
irrationals} $\theta \in \RR \setminus \QQ$: i.e.
irrational $\theta$ such that $\theta$ satisfies
an equation
\[ a\theta^2 + b\theta + c = 0 \qquad a, b, c \in
\ZZ .\]
(or equivalently $\theta$ of the form $r +
s\sqrt{d}$, $r, s \in \QQ$, $s \neq 0$, $d \in
\NN$ not a square).

\begin{example*}
  $d = 6$, $\theta = \sqrt{6}$. $2 < \sqrt{6} < 3$
  so $\acf_0 = \left\lfloor \sqrt{6} \right\rfloor =
  2$, $\theta_1 = \frac{1}{\sqrt{6} - 2} =
  \frac{\sqrt{6} + 2}{2} = \frac{\sqrt{6} - 2}{2}
  + 2$. Hence $\acf_1 = \left\lfloor \theta_1
  \right\rfloor = 2$, $\theta_2 =
  \frac{2}{\sqrt{6} - 2} = \sqrt{6} + 2 =
  (\sqrt{6} - 2) + 4$, so $\acf_2 = \left\lfloor
  \theta_2 \right\rfloor = 4$, $\theta_3 =
  \frac{1}{\sqrt{6} - 2} = \theta_1$. So
  \begin{align*}
    \theta
    &= \contf[\acf_0, \theta_1] \\
    &= \contf[\acf_0, \acf_1, \theta_2] \\
    &= \contf[\acf_0, \acf_1, \acf_2, \acf_1, \acf_2, \theta_1]
    \\
    &= \contf[2, 2, 4, 2, 4, 2, 4, \ldots] \\
    &= \contf[2, \ol{2, 4}]
  \end{align*}
  \glssymboldefn{recurring_cfe}{$[1, \overline{2,
  3}]$}{$[1, \overline{2, 3}]$}
  (overline means repeat this pattern
  indefinitely).
\end{example*}

\begin{flashcard}[essentially-periodic-defn]
\begin{definition*}[Essentially periodic]
  \glsadjdefn{ess_pdic}{essentially
  periodic}{\gls{cfe}}
  \glsadjdefn{ply_pdic}{purely
  periodic}{\gls{cfe}}
  \cloze{Let $\theta \in \RR \setminus \QQ$ have
  \gls{cfe} $\contf[\acf_0, \acf_1, \acf_2,
  \ldots]$. Then the \gls{cfe} of $\theta$ is
  \emph{essentially periodic} of period $k$ if it has the
  form $\contf[\acf_0, \acf_1, \ldots, \acf_{m -
  1}, \rcring{\acf_m, \ldots, \acf_{m + k -
  1}}]$. It is \emph{purely periodic} if we can
  take $m = 0$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \Gls{cfe} of $\sqrt{6}$ is \gls{ess_pdic};
  \gls{cfe} of $\frac{1}{\sqrt{6} - 2}$ is
  \gls{ply_pdic}.
\end{example*}

\begin{flashcard}[Lagrange-cfe-thm]
\begin{theorem}[Lagrange]
  % Theorem 5.8
  If $\theta \in \RR \setminus \QQ$, \cloze{then
  \gls{cfe} of $\theta$ is \gls{ess_pdic} $\iff$
  $\theta$ is a \gls{quad_irr}.}
\end{theorem}

\begin{proof}
  \cloze{
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] If $\theta =
      \contf[\rcring{\acf_0, \ldots, \acf_{k -
      1}}]$ is \gls{ply_pdic}, then
      \[ \theta = \contf[\acf_0, \ldots, \acf_{k -
      1}, \theta] \implies \theta = \frac{p_{k -
      1}\theta + p_{k - 2} \theta}{q_{k - 1}\theta +
      q_{k - 2}} \]
      Rearrange: get a quadratic equation
      satisfied by $\theta$.

      If $\theta = \contf[\acf_0, \ldots, \acf_{m
      - 1}, \rcring{\acf_m, \ldots, \acf_{m + k -
      1}}]$ is \gls{ess_pdic}, then $\theta =
      \contf[\acf_0, \ldots, \acf_{m - 1},
      \beta]$, where $\beta$ has a \gls{ply_pdic}
      \gls{cfe}, so $\beta = r + s\sqrt{d}$. Now:
      \[ \theta = \frac{p_{m - 1} \beta + p_{m -
      2}}{q_{m - 1}\beta + q_{m - 2}} \]
      Rearrange to see $\theta$ has the form $r' +
      s'\sqrt{d}$, hence $\theta$ is a
      \gls{quad_irr}.
    \item[$\Leftarrow$] Now suppose $\theta$ is a
      \gls{quad_irr}, with \gls{cfe} $\contf[\acf_0,
      \acf_1, \acf_2, \ldots]$.a We know $\theta$
      satisfies an equation $a\theta^2 + b\theta +
      c = 0$, $a, b, c \in \ZZ$. We define
      \[ f(x, y) = ax^2 + bxy + cy^2 ,\]
      a \gls{bqf}, with $f(0, 1) = 0$. For $n \ge
      1$, define
      \[ f_n(x, y) = f \left(
      \begin{pmatrix}
        x & y
      \end{pmatrix}
      \begin{pmatrix}
        p_n & q_n \\
        p_{n - 1} & q_{n - 1}
      \end{pmatrix}
      \right)
      = f(p_n x + p_{n - 1} y, q_nx + q_{n - 1} y)
      .\]
      Claim: As $n$ varies, the sequence $f_n(x,
      y)$ takes on finitely many distinct
      \glspl{bqf}.
      This implies the Theorem: For
      all $n \ge 1$,
      \[ \theta = \contf[\acf_0, \ldots, \acf_n,
      \theta_{n + 1}] = \frac{\theta_{n + 1} p_n +
      p_{n - 1}}{\theta_{n + 1} q_n + q_{n - 1}} .\]
      So
      \begin{align*}
        f_n(\theta_{n + 1}, 1)
        &= f(p_n \theta_{n + 1} + p_{n - 1},
        q_n\theta_{n + 1} + q_{n - 1}) \\
        &= (q_n\theta_{n - 1} + q_{n - 1})^2
        f \left( \frac{p_n \theta_{n + 2} + p_{n -
        2}}{q_n \theta_{n + 2} + q_{n - 2}},
        1\right) \\
        &= (q_n \theta_{n + 1} + q_{n - 1})^2
        f(\theta, 1) \\
        &= 0
      \end{align*}
      Claim shows that as $n$ varies, $\theta_{n +
      1}$ can take on only finitely many distinct
      values. Hence, there exist $n, k \ge 1$ such
      that $\theta_n = \theta_{n + k}$, hence
      \gls{cfe} of $\theta$ is \gls{ess_pdic}.

      Now we prove the claim: write $f_n(x, y) =
      A_n x^2 + B_n xy + C_n y^2$.
      \begin{align*}
        A_n
        &= f_n(1, 0) = f(p_n, q_n) \\
        C_n
        &= f_n(0, 1) = f(p_{n - 1}, q_{n - 1}) = A_{n - 1}
      \end{align*}
      So
      \[ \disc f_n = B_n^2 - 4A_n C_n = \disc f
      \cdot \det
      \begin{pmatrix}
        p_n & q_n \\
        p_{n - 1} & q_{n - 1}
      \end{pmatrix}^2
      = \disc f \]
      (as $p_n q_{n - 1} - p_{n - 1} q_n = (-1)^{n
      - 1}$). To show that claim, it's enough to
      show $|A_n|$ is bounded as $n$ varies. Let's
      write $\theta'$ for the other root of $a x^2
      + bx + c = 0$, so $f(x, 1) = a(x - \theta)(x
      - \theta')$. Then
      \[ |A_n| = |f(p_n, q_n)| = q_n^2 \left| f
      \left( \frac{p_n}{q_n}, 1 \right) \right| =
      q_n^2 |a| \left| \frac{p_n}{q_n} - \theta
      \right| \left| \frac{p_n}{q_n} - \theta'
      \right| .\]
      We know $\left| \theta - \frac{p_n}{q_n}
      \right| \le \frac{1}{q_n q_{n + 1}}$ (proved
      last time). So
      \[ |A_n| \le \frac{q_n^2}{q_n q_{n + 1}} |a|
      \left| \frac{p_n}{q_n} - \theta' \right| \le
      |a| \left| \frac{p_n}{q_n} - \theta' \right|
      .\]
      We know $\left| \frac{p_n}{q_n} - \theta'
      \right| \to \left| \theta - \theta' \right|$
      as $n \to \infty$. Therefore $|a| \left|
      \frac{p_n}{q_n} - \theta' \right|$ is
      bounded as $n$ varies. \qedhere
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[galois-ply-pdic-thm]
\begin{theorem}[Galois]
  % Theorem 5.9
  \label{galois_thm}
  \cloze{Let $\theta = r + s\sqrt{d}$ be a
  \gls{quad_irr}. Let $\theta' = r - s\sqrt{d}$
  (``the other root of the quadratic''). Then the
  \gls{cfe} of $\theta$ is \gls{ply_pdic} $\iff$
  $\theta > 1$, $\theta' \in (-1, 0)$.

  In this
  case, if $\theta = \contf[\rcring{\acf_0,
  \ldots, \acf_n}]$, then $-\frac{1}{\theta'} =
  \contf[\rcring{\acf_n, \ldots, \acf_0}]$.}
\end{theorem}
\end{flashcard}

\begin{proof}
  Omitted.
\end{proof}

\textbf{Application:} $\theta = \sqrt{d}$, $d \in
\ZZ$ a non-square. Then $\acf_0 = \left\lfloor
\sqrt{d} \right\rfloor$, $\theta_1 =
\frac{1}{\sqrt{d} - \acf_0} > 1$.
\[ \theta_1' = \frac{1}{-(\sqrt{d} + \acf_0)} \in
(-1, 0)\]
Hence $\theta_1$ satisfies hypothesis of
\cref{galois_thm}. Hence
\[ \sqrt{d} = \contf[\acf_0, \rcring{\acf_1,
\ldots, \acf_n}] ,\]
for some $n \ge 1$.

\begin{flashcard}[pells-has-sol-thm]
\begin{theorem}
  % Theorem 5.10
  Let $d \in \NN$ be a \cloze{non-square}. Then the
  equation $X^2 - dY^2 = 1$ has a solution with
  $X, Y \in \NN$.
\end{theorem}

\begin{proof}
  \cloze{Let $\sqrt{d} = \contf[\acf_0, \rcring{\acf_1,
  \ldots, \acf_n}] = \contf[\acf_0, \theta_1]$,
  $\theta_1 = \contf[\rcring{\acf_1, \ldots,
  \acf_n}]$ (using the application of
  \cref{galois_thm} above). Then
  \[ \sqrt{d} = \contf[\acf_0, \acf_1, \ldots,
  \acf_n, \rcring{\acf_1, \ldots, \acf_n}] =
  \contf[\acf_0, \ldots, \acf_n, \theta_1] =
  \frac{p_n \theta_1 + p_{n - 1}}{q_n \theta_1 +
  q_{n - 1}} \]
  where $\theta_1 = \frac{1}{\sqrt{d} - \acf_0}$.
  Hence
  \begin{align*}
    \sqrt{d}
    &= \frac{p_n + p_{n - 1} (\sqrt{d} - \acf_0)}{q_n
    + q_{n - 1}(\sqrt{d} - \acf_0)} \\
    \implies dq_{n - 1} + (q_n - \acf_0 q_{n - 1})
    \sqrt{d}
    &= (p_n - p_{n - 1} \acf_0) + p_{n - 1}
    \sqrt{d}
  \end{align*}
  Equate $\sqrt{d}$ and rational parts: $dq_{n -
  1} = p_n - p_{n - 1} \acf_0$, $p_{n - 1} = q_n -
  \acf_0 q_{n - 1}$.
  \[ p_{n - 1}^2 - dq_{n - 1}^2 = p_{n - 1} (q_n -
  \acf_0 q_{n - 1}) - p_n q_{n - 1} + p_{n - 1}
  q_{n - 1} \acf_0 = p_{n - 1} q_n - q_n q_{n - 1} =
  (-1)^n .\]
  If $n$ is even, then $p_{n - 1}^2 - dq_{n - 1}^2
  = 1$, and we've found a solution. If $n$ is odd,
  we run the same argument using
  \[ \sqrt{d} = \contf[\acf_0, \rcring{\acf_1,
  \ldots, \acf_n, \acf_1, \ldots, \acf_n}].
  \qedhere \]}
\end{proof}
\end{flashcard}