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Method to find solutions to Pell's equation $X^2 -
dY^2 = 1$:

Compute the \gls{cfe} of $\sqrt{d}$ as
\[ \sqrt{d} = \contf[\acf_0, \rcring{\acf_1,
\ldots, \acf_n}] .\]
Look at $\frac{p_{n - 1}}{q_{n - 1}}$, the $(n -
1)$-th convergent of $\sqrt{d}$. If $n$ is even,
then $p_{n - 1}^2 - dq_{n - 1}^2 = 1$. If $n$ is
odd, then $p_{n - 1}^2 - dq_{n - 1}^2 = -1$, but
$p_{2n - 1}$, $q_{2n - 1}$ will give a solution.

\begin{example*}
  $d = 6$, $\sqrt{d} = \contf[2, \rcring{2, 4}]$.
  $n = 2$, $\frac{p_1}{q_1} = 2 + \half =
  \frac{5}{2}$. $5^2 - 6 \cdot 2^2 = 25 - 24 = 1$.

  $d = 17$, $4 < \sqrt{17} < 5$, $\acf_0 = 4$,
  $\theta_1 = \frac{1}{\sqrt{17} - 4} =
  \frac{\sqrt{17} + 4}{17 - 16} = (\sqrt{17} - 4)
  + 8$. Then $\acf_1 = 8$, $\theta_2 =
  \frac{1}{\sqrt{17} - 4} = \theta_1$, so
  $\sqrt{17} = \contf[4, \rcring{8}]$. So $n = 1$,
  $\frac{p_0}{q_0} = \frac{4}{1}$, $4^2
  - 17 \cdot 1^2 = -1$. $\frac{p_1}{q_1} = 4 +
  \frac{1}{8} = \frac{33}{8}$. Then $33^2 - 17
  \cdot 8^2 = 1$.
\end{example*}

\begin{remark*}
  The solutions $(x, y) \in \ZZ^2$ to $x^2 - dy^2
  = \pm 1$ correspond to \emph{units} in the ring
  of integers in $\QQ(\sqrt{d})$ ($\to$
  \courseref[Number Fields]{NF}), via the formula
  \[ (x, y) \leftrightarrow x + \sqrt{d} y .\]
  You can show that the solutions $(x, y)$ to $x^2
  - dy^2 = \pm 1$ are precisely the pairs $\pm
  (p_{kn - 1}, q_{kn - 1})$, where $k \ge 0$, and $n$
  is minimal such that $\sqrt{d} = \contf[\acf_0,
  \rcring{\acf_1, \ldots, \acf_n}]$ (if $k = 0$,
  then $(p_{-1}, q_{-1}) = (1, 0)$ gives the trivial
  solution).
\end{remark*}

\newpage
\section{Primality testing and factorisation}

Want to find processes to:
\begin{itemize}
  \item Test whether a given integer $N \in \NN$
    is \gls{prime},
  \item If $N$ is not \gls{prime}, find a
    non-trivial factor.
\end{itemize}
Hope to do these in polynomial-time.

Can test primality in polynomial-time. Don't know
how to factorise in polynomial-time, but there are
algorithms that are much faster than trial
division.

We'll usually assume $N > 1$ and that $N$ is odd.
(Can always divide by powres of $2$ if $N$ is
even).

Begin by looking at necessary conditions for $N$
to be \gls{prime}. For example:
\begin{example*}
  If $N$ is \gls{prime}, $a \in \ZZ$,
  $\gcdbrack(a, N) = 1$, then $a^{N - 1} \equiv
  1 \pmod{N}$ (\nameref{fermats_little_thm}).

  For example, if $N = 15$, $a = 2$, then
  $\gcdbrack(a, N) = 1$, but
  \[ a^{N - 1} = 2^{14} = (2^4)^3 2^2 \equiv 4
  \not\equiv 1 \pmod{15} \]
\end{example*}

\begin{remark*}[Binary exponentiation]
  Suppose $a, x, N \in \NN$. Then we can compute
  $a^x \mod N$ in polynomial-time. Write
  \[ x = \sum_{i = 0}^k b_i 2^i, \qquad b_i \in
  \{0, 1\} .\]
  Compute $a, a^2, a^4 = (a^2)^2, \ldots, a^{2^k}
  = (a^{2^{k - 1}})^2$. Then
  \[ a^x = \prod_{i = 0}^k (a^{2^i})^{b_i} .\]
\end{remark*}

\begin{example*}
  $N = 91$, $a = 3$. Then $3^{90} = 3^{N - 1}
  \equiv 1 \pmod{91}$. However, $N = 7 \times 13$
  is composite.
\end{example*}

\begin{flashcard}[fermat-pseudoprime]
\begin{definition}[Fermat pseudoprime]
  % Definition 6.1
  \glsnoundefn{ferm_pp}{Fermat pseudoprime}{Fermat
  pseudoprimes}
  \cloze{Let $N \in \NN$ be an odd composite integer, $b
  \in \ZZ$, $\gcdbrack(b, N) = 1$. We say $N$ is a
  \emph{Fermat pseudoprime to the base $b$} if
  $b^{N - 1} \equiv 1 \pmod{N}$.}
\end{definition}
\end{flashcard}

\begin{remark*}
  For fixed $N$, the condition of $N$ being a
  \gls{ferm_pp} to the base $b$ only depends on
  $b \mod N$. So it makes sense for $b \in (\ZZ /
  N\ZZ)^\times$.
\end{remark*}

\begin{proposition}
  % Proposition 6.2
  \label{ferm_pp_base_subgroup_prop}
  Let $N \in \NN$ be odd, composite. Then
  \begin{enumerate}[(1)]
    \item $\{b \in (\ZZ / N\ZZ)^\times \st
      \text{$N$ is a \gls{ferm_pp} to the base
      $b$}\}$ is a subgroup of $(\ZZ /
      N\ZZ)^\times$.
    \item If $\exists ~b_0 \in (\ZZ / N\ZZ)^\times$
      such that $N$ is not a \gls{ferm_pp} to the
      base $b_0$ then the same is true for at
      least half of all $b \in (\ZZ /
      N\ZZ)^\times$.
  \end{enumerate}
\end{proposition}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Call this set $H$. We need to show $1
      \in H$, and $H$ closed under multiplication
      (since $(\ZZ / N\ZZ)^\times$ is finite).
      $1^{N - 1} \equiv 1 \pmod{N}$, so $1 \in H$.

      If $b_1, b_2 \in H$, then $b_1^{N - 1}
      \equiv 1 \equiv b_2^{N - 1} \pmod{N}$. So
      $(b_1 b_2)^{N - 1} \equiv b_1^{N - 1} b_2^{N
      - 1} \equiv 1 \pmod{N}$. So $b_1 b_2 \in H$.
    \item $b_0$ exists implies $H \neq (\ZZ /
      N\ZZ)^\times$. We need to show $\#((\ZZ /
      N\ZZ)^\times \setminus H) \ge \frac{\#(\ZZ /
      N\ZZ)^\times}{2}$. We know $\#(\ZZ /
      N\ZZ)^\times = \#H \cdot [(\ZZ /
      N\ZZ)^\times : H] \ge 2\#H$. \qedhere
  \end{enumerate}
\end{proof}

Idea for primality test: choose $b \in (\ZZ /
N\ZZ)^\times$ at random and testing whether $N$ is
a \gls{ferm_pp} to the base $b$.

\begin{flashcard}[carmichael-defn]
\begin{definition}[Carmichael number]
  % Definition 6.3
  \glsnoundefn{carm_num}{Carmichael
  number}{Carmichael numbers}
  \cloze{Let $N \in \NN$ be odd and composite. We say $N$
  is a \emph{Carmichael number} if it's a
  \gls{ferm_pp} to every base $b \in (\ZZ /
  N\ZZ)^\times$.}
\end{definition}
\end{flashcard}
\vspace{-1em}

There exist infinitely many \glspl{carm_num}.

\begin{flashcard}[euler-pseudoprime-defn]
\begin{definition}[Euler pseudoprime]
  % Definition 6.4
  \glsnoundefn{eul_pp}{Euler pseudoprime}{Euler
  pseudoprimes}
  \cloze{Let $N \in \NN$ be odd and composite. Let $b
  \in \ZZ$ with $\gcdbrack(b, N) = 1$. Then we
  say that $N$ is an \emph{Euler pseudoprime to
  the base $b$} if
  \[ b^{\frac{N - 1}{2}} \equiv \jacobi{b}{N}
  \pmod{N} .\]}
\end{definition}
\end{flashcard}
\vspace{-1em}

Recall: If $p$ is an odd \gls{prime},
$\gcdbrack(b, p) = 1$, then $b^{\frac{p - 1}{2}}
\equiv \legendre{b}{p} \pmod{p}$
(\nameref{eulers_criterion}).

\begin{remark*}
  If $N $is an \gls{eul_pp} to the base $b$, then
  it's a \gls{ferm_pp} to the base $b$. This
  definition makes sense for $b \in (\ZZ /
  N\ZZ)^\times$, and it's again the case that
  \[ \{b \in (\ZZ / N\ZZ)^\times \st \text{$N$ is
  an \gls{eul_pp} to the base $b$}\} \]
  is a subgroup of $(\ZZ / N\ZZ)^\times$.
\end{remark*}

\begin{hiddenflashcard}[fakeproof-carmichael-numbers]
\begin{theorem}
  Let $N \in \NN$ be odd, composite. Then there
  exists $b \in (\ZZ / N\ZZ)^\times$ such that $N$
  is not a \gls{ferm_pp} to the base $b$.
\end{theorem}

\begin{proof}
  \cloze{This is false. The numbers for which this
  fails are called \glspl{carm_num}.}
  \vspace{15em}

  \phantom{}
\end{proof}
\end{hiddenflashcard}

\begin{flashcard}[non-existence-of-absolute-euler-pseudoprimes-thm]
\begin{theorem}
  % Theorem 6.5
  \label{nexists_absolute_eul_pp}
  Let $N \in \NN$ be odd, composite. Then there
  exists $b \in (\ZZ / N\ZZ)^\times$ such that $N$
  is not an \gls{eul_pp} to the base $b$.
\end{theorem}

\begin{proof}
  \cloze{First assume $N$ is squarefree, $N = pN_0$, $p$
  \gls{prime}, $N_0 \ge 3$, $p \nmid N_0$. Since
  $p$ is odd, there exists $u \in \ZZ$ such that
  $\legendre{u}{p} = -1$. Choose $b \in \ZZ$ such
  that $b \equiv u \pmod{p}$, $b \equiv 1
  \pmod{N_0}$ (using \nameref{CRT}). Then
  \[ \jacobi{b}{N} = \jacobi{b}{p} \jacobi{b}{N_0}
  = \legendre{u}{p} \jacobi{1}{N_0} = -1 .\]
  We know
  \[ b^{\frac{N - 1}{2}}
  \equiv 1^{\frac{N - 1}{2}}
  \equiv 1 \not\equiv -1 \pmod{N_0} .\]
  So $b^{\frac{N - 1}{2}} \not\equiv \jacobi{b}{N}
  \pmod{N}$. So $b$ works.

  Next suppose $N$ is not squarefree, and choose
  $p$ \gls{prime} such that $p^2 \mid N$. Choose
  $b \in \ZZ$ such that $b \equiv 1 + p
  \pmod{p^2}$, $\gcdbrack(b, N) = 1$
  (\nameref{CRT}). Then
  \[ b^{N - 1} \equiv (1 + p)^{N - 1} \equiv 1 +
  (N - 1)p \equiv 1 - p \not\equiv 1 \pmod{p^2} .\]
  So $b^{N - 1} \not\equiv 1 \pmod{N}$, so $N$ is
  not a \gls{ferm_pp} to the base $b$, so
  certainly not an \gls{eul_pp} to the base $b$.}
\end{proof}
\end{flashcard}