Added files

1 files changed, 1297 insertions, 0 deletions

diff --git a/src/Lecture5/notebook/7-SageAlgebra.tex b/src/Lecture5/notebook/7-SageAlgebra.tex
new file mode 100644
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--- /dev/null
+++ b/src/Lecture5/notebook/7-SageAlgebra.tex
@@ -0,0 +1,1297 @@
+\documentclass[11pt]{article}
+
+    \usepackage[breakable]{tcolorbox}
+    \usepackage{parskip} % Stop auto-indenting (to mimic markdown behaviour)
+    
+    \usepackage{iftex}
+    \ifPDFTeX
+        \usepackage[T1]{fontenc}
+        \usepackage{mathpazo}
+    \else
+        \usepackage{fontspec}
+    \fi
+
+    % Basic figure setup, for now with no caption control since it's done
+    % automatically by Pandoc (which extracts ![](path) syntax from Markdown).
+    \usepackage{graphicx}
+    % Maintain compatibility with old templates. Remove in nbconvert 6.0
+    \let\Oldincludegraphics\includegraphics
+    % Ensure that by default, figures have no caption (until we provide a
+    % proper Figure object with a Caption API and a way to capture that
+    % in the conversion process - todo).
+    \usepackage{caption}
+    \DeclareCaptionFormat{nocaption}{}
+    \captionsetup{format=nocaption,aboveskip=0pt,belowskip=0pt}
+
+    \usepackage[Export]{adjustbox} % Used to constrain images to a maximum size
+    \adjustboxset{max size={0.9\linewidth}{0.9\paperheight}}
+    \usepackage{float}
+    \floatplacement{figure}{H} % forces figures to be placed at the correct location
+    \usepackage{xcolor} % Allow colors to be defined
+    \usepackage{enumerate} % Needed for markdown enumerations to work
+    \usepackage{geometry} % Used to adjust the document margins
+    \usepackage{amsmath} % Equations
+    \usepackage{amssymb} % Equations
+    \usepackage{textcomp} % defines textquotesingle
+    % Hack from http://tex.stackexchange.com/a/47451/13684:
+    \AtBeginDocument{%
+        \def\PYZsq{\textquotesingle}% Upright quotes in Pygmentized code
+    }
+    \usepackage{upquote} % Upright quotes for verbatim code
+    \usepackage{eurosym} % defines \euro
+    \usepackage[mathletters]{ucs} % Extended unicode (utf-8) support
+    \usepackage{fancyvrb} % verbatim replacement that allows latex
+    \usepackage{grffile} % extends the file name processing of package graphics 
+                         % to support a larger range
+    \makeatletter % fix for grffile with XeLaTeX
+    \def\Gread@@xetex#1{%
+      \IfFileExists{"\Gin@base".bb}%
+      {\Gread@eps{\Gin@base.bb}}%
+      {\Gread@@xetex@aux#1}%
+    }
+    \makeatother
+
+    % The hyperref package gives us a pdf with properly built
+    % internal navigation ('pdf bookmarks' for the table of contents,
+    % internal cross-reference links, web links for URLs, etc.)
+    \usepackage{hyperref}
+    % The default LaTeX title has an obnoxious amount of whitespace. By default,
+    % titling removes some of it. It also provides customization options.
+    \usepackage{titling}
+    \usepackage{longtable} % longtable support required by pandoc >1.10
+    \usepackage{booktabs}  % table support for pandoc > 1.12.2
+    \usepackage[inline]{enumitem} % IRkernel/repr support (it uses the enumerate* environment)
+    \usepackage[normalem]{ulem} % ulem is needed to support strikethroughs (\sout)
+                                % normalem makes italics be italics, not underlines
+    \usepackage{mathrsfs}
+    
+
+    
+    % Colors for the hyperref package
+    \definecolor{urlcolor}{rgb}{0,.145,.698}
+    \definecolor{linkcolor}{rgb}{.71,0.21,0.01}
+    \definecolor{citecolor}{rgb}{.12,.54,.11}
+
+    % ANSI colors
+    \definecolor{ansi-black}{HTML}{3E424D}
+    \definecolor{ansi-black-intense}{HTML}{282C36}
+    \definecolor{ansi-red}{HTML}{E75C58}
+    \definecolor{ansi-red-intense}{HTML}{B22B31}
+    \definecolor{ansi-green}{HTML}{00A250}
+    \definecolor{ansi-green-intense}{HTML}{007427}
+    \definecolor{ansi-yellow}{HTML}{DDB62B}
+    \definecolor{ansi-yellow-intense}{HTML}{B27D12}
+    \definecolor{ansi-blue}{HTML}{208FFB}
+    \definecolor{ansi-blue-intense}{HTML}{0065CA}
+    \definecolor{ansi-magenta}{HTML}{D160C4}
+    \definecolor{ansi-magenta-intense}{HTML}{A03196}
+    \definecolor{ansi-cyan}{HTML}{60C6C8}
+    \definecolor{ansi-cyan-intense}{HTML}{258F8F}
+    \definecolor{ansi-white}{HTML}{C5C1B4}
+    \definecolor{ansi-white-intense}{HTML}{A1A6B2}
+    \definecolor{ansi-default-inverse-fg}{HTML}{FFFFFF}
+    \definecolor{ansi-default-inverse-bg}{HTML}{000000}
+
+    % commands and environments needed by pandoc snippets
+    % extracted from the output of `pandoc -s`
+    \providecommand{\tightlist}{%
+      \setlength{\itemsep}{0pt}\setlength{\parskip}{0pt}}
+    \DefineVerbatimEnvironment{Highlighting}{Verbatim}{commandchars=\\\{\}}
+    % Add ',fontsize=\small' for more characters per line
+    \newenvironment{Shaded}{}{}
+    \newcommand{\KeywordTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{\textbf{{#1}}}}
+    \newcommand{\DataTypeTok}[1]{\textcolor[rgb]{0.56,0.13,0.00}{{#1}}}
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+    \newcommand{\CommentTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textit{{#1}}}}
+    \newcommand{\OtherTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{{#1}}}
+    \newcommand{\AlertTok}[1]{\textcolor[rgb]{1.00,0.00,0.00}{\textbf{{#1}}}}
+    \newcommand{\FunctionTok}[1]{\textcolor[rgb]{0.02,0.16,0.49}{{#1}}}
+    \newcommand{\RegionMarkerTok}[1]{{#1}}
+    \newcommand{\ErrorTok}[1]{\textcolor[rgb]{1.00,0.00,0.00}{\textbf{{#1}}}}
+    \newcommand{\NormalTok}[1]{{#1}}
+    
+    % Additional commands for more recent versions of Pandoc
+    \newcommand{\ConstantTok}[1]{\textcolor[rgb]{0.53,0.00,0.00}{{#1}}}
+    \newcommand{\SpecialCharTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}}
+    \newcommand{\VerbatimStringTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}}
+    \newcommand{\SpecialStringTok}[1]{\textcolor[rgb]{0.73,0.40,0.53}{{#1}}}
+    \newcommand{\ImportTok}[1]{{#1}}
+    \newcommand{\DocumentationTok}[1]{\textcolor[rgb]{0.73,0.13,0.13}{\textit{{#1}}}}
+    \newcommand{\AnnotationTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textbf{\textit{{#1}}}}}
+    \newcommand{\CommentVarTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textbf{\textit{{#1}}}}}
+    \newcommand{\VariableTok}[1]{\textcolor[rgb]{0.10,0.09,0.49}{{#1}}}
+    \newcommand{\ControlFlowTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{\textbf{{#1}}}}
+    \newcommand{\OperatorTok}[1]{\textcolor[rgb]{0.40,0.40,0.40}{{#1}}}
+    \newcommand{\BuiltInTok}[1]{{#1}}
+    \newcommand{\ExtensionTok}[1]{{#1}}
+    \newcommand{\PreprocessorTok}[1]{\textcolor[rgb]{0.74,0.48,0.00}{{#1}}}
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+    \newcommand{\WarningTok}[1]{\textcolor[rgb]{0.38,0.63,0.69}{\textbf{\textit{{#1}}}}}
+    
+    
+    % Define a nice break command that doesn't care if a line doesn't already
+    % exist.
+    \def\br{\hspace*{\fill} \\* }
+    % Math Jax compatibility definitions
+    \def\gt{>}
+    \def\lt{<}
+    \let\Oldtex\TeX
+    \let\Oldlatex\LaTeX
+    \renewcommand{\TeX}{\textrm{\Oldtex}}
+    \renewcommand{\LaTeX}{\textrm{\Oldlatex}}
+    % Document parameters
+    % Document title
+    \title{Algebra and Cryptography with SageMath}
+    \date{2021-04-23}
+    \author{Sebastiano Tronto - \texttt{sebastiano.tronto@uni.lu}}
+    
+    
+    
+    
+    
+% Pygments definitions
+\makeatletter
+\def\PY@reset{\let\PY@it=\relax \let\PY@bf=\relax%
+    \let\PY@ul=\relax \let\PY@tc=\relax%
+    \let\PY@bc=\relax \let\PY@ff=\relax}
+\def\PY@tok#1{\csname PY@tok@#1\endcsname}
+\def\PY@toks#1+{\ifx\relax#1\empty\else%
+    \PY@tok{#1}\expandafter\PY@toks\fi}
+\def\PY@do#1{\PY@bc{\PY@tc{\PY@ul{%
+    \PY@it{\PY@bf{\PY@ff{#1}}}}}}}
+\def\PY#1#2{\PY@reset\PY@toks#1+\relax+\PY@do{#2}}
+
+\expandafter\def\csname PY@tok@w\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}}
+\expandafter\def\csname PY@tok@c\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}}
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+        \def\FancyVerbFormatLine ##1{\hsize\linewidth
+            \vtop{\raggedright\hyphenpenalty\z@\exhyphenpenalty\z@
+                \doublehyphendemerits\z@\finalhyphendemerits\z@
+                \strut ##1\strut}%
+        }%
+        % If the linebreak is at a space, the latter will be displayed as visible
+        % space at end of first line, and a continuation symbol starts next line.
+        % Stretch/shrink are however usually zero for typewriter font.
+        \def\FV@Space {%
+            \nobreak\hskip\z@ plus\fontdimen3\font minus\fontdimen4\font
+            \discretionary{\copy\Wrappedvisiblespacebox}{\Wrappedafterbreak}
+            {\kern\fontdimen2\font}%
+        }%
+        
+        % Allow breaks at special characters using \PYG... macros.
+        \Wrappedbreaksatspecials
+        % Breaks at punctuation characters . , ; ? ! and / need catcode=\active         
+        \OriginalVerbatim[#1,codes*=\Wrappedbreaksatpunct]%
+    }
+    \makeatother
+
+    % Exact colors from NB
+    \definecolor{incolor}{HTML}{303F9F}
+    \definecolor{outcolor}{HTML}{D84315}
+    \definecolor{cellborder}{HTML}{CFCFCF}
+    \definecolor{cellbackground}{HTML}{F7F7F7}
+    
+    % prompt
+    \makeatletter
+    \newcommand{\boxspacing}{\kern\kvtcb@left@rule\kern\kvtcb@boxsep}
+    \makeatother
+    \newcommand{\prompt}[4]{
+        \ttfamily\llap{{\color{#2}[#3]:\hspace{3pt}#4}}\vspace{-\baselineskip}
+    }
+    
+
+    
+    % Prevent overflowing lines due to hard-to-break entities
+    \sloppy 
+    % Setup hyperref package
+    \hypersetup{
+      breaklinks=true,  % so long urls are correctly broken across lines
+      colorlinks=true,
+      urlcolor=urlcolor,
+      linkcolor=linkcolor,
+      citecolor=citecolor,
+      }
+    % Slightly bigger margins than the latex defaults
+    
+    \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
+    
+    
+
+\begin{document}
+    
+    \maketitle
+    
+    
+
+    
+    This lecture's notes are in a different format: the presentations for
+the \(\LaTeX\) part were made with \(\LaTeX\), so this one is made with
+Sage, or rather with the \href{https://jupyter.org/}{Jupyter Notebook}.
+
+\hypertarget{the-jupyter-notebook}{%
+\section{The Jupyter Notebook}\label{the-jupyter-notebook}}
+
+\textbf{Reference:} {[}\href{https://jupyter.org/documentation}{1}{]}
+
+The Jupyter Notebook is one of the default interfaces for SageMath,
+along with the command line interface. You can access it via web
+browser, but it is running locally on your device (notice the strange
+url: \texttt{http://localhost:8888/notebooks...}).
+
+You can create a new notebook by clicking on
+\texttt{New\ \textgreater{}\ SageMath\ 9.2}. You can also create a
+Python 3 notebook to write Python code.
+
+Jupyter saves and reads files in the \texttt{.ipynb} format. If you
+download the file for this lecture you can open it and follow the
+examples interactively.
+
+\hypertarget{cells}{%
+\subsection{Cells}\label{cells}}
+
+The notebook contains one or more \emph{interactive cells} that you can
+run, like this one below:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{2}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{c+c1}{\PYZsh{} Exercise: modify this cell to use the print() command}
+\PY{l+m+mi}{2}\PY{o}{+}\PY{l+m+mi}{2}
+\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{5}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{2}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+2/5
+\end{Verbatim}
+\end{tcolorbox}
+        
+    If you are reading this from Jupyter rather than from the pdf file, you
+can edit the cell above and run it again. You can also add more cells by
+selecting \texttt{Insert} from the menu bar.
+
+Notice that only the last statement produces an output. You can force
+anything to be written as output with the \texttt{print()} command,
+which works like in Python. As an exercise, try to modify the cell above
+to provide more output!
+
+    \hypertarget{markdown}{%
+\subsection{Markdown}\label{markdown}}
+
+\href{https://en.wikipedia.org/wiki/Markdown}{Markdown} is a simple
+markup language - think of LaTeX or html, but much simpler. You can add
+text to your notebook with Markdown cells by selecting
+\texttt{Cell\ \textgreater{}\ Cell\ Type\ \textgreater{}\ Markdown}.
+
+You can also include some LaTeX code in Markdown cells, with dollar
+signs \$ or align environments:
+
+\begin{align*}
+\frac{(x+y)^2}{x+1} = \frac{x^2+y^2}{x+1}
+\end{align*}
+
+When you are done writing a Markdown cell, you can run it to see the
+well-formatted text. To edit the text again, double-click on the cell.
+Try doing it now to fix the formula above!
+
+    \hypertarget{symbolic-expressions}{%
+\section{Symbolic expressions}\label{symbolic-expressions}}
+
+\textbf{Reference:}
+{[}\href{https://doc.sagemath.org/html/en/reference/calculus/sage/symbolic/expression.html}{2}{]}
+
+Now, let's get started with Sage. One thing you might want to do is
+manipulating symbolic expressions, like the following:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{3}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{f} \PY{o}{=}  \PY{n}{x}\PY{o}{\PYZca{}}\PY{l+m+mi}{2} \PY{o}{+} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{x} \PY{o}{\PYZhy{}} \PY{l+m+mi}{5} \PY{o}{==} \PY{l+m+mi}{0}
+\PY{n}{solve}\PY{p}{(}\PY{n}{f}\PY{p}{,}\PY{n}{x}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{3}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+[x == -sqrt(6) - 1, x == sqrt(6) - 1]
+\end{Verbatim}
+\end{tcolorbox}
+        
+    Notice that the single \texttt{=} is part of an assignment, as in
+Python: we are \emph{assigning} to the variable \texttt{f} the value
+\texttt{x\^{}2\ +\ 2*x\ -\ 5\ \textgreater{}=\ 0}, which in this case is
+an equation, so it contains the symbol \texttt{==}. Keep in mind the
+difference between the two!
+
+\textbf{Exercise:} change the code above to solve the corresponding
+inequality \(x^2+2x-5\geq 0\).
+
+    \hypertarget{mathematical-variables}{%
+\subsection{Mathematical variables}\label{mathematical-variables}}
+
+Last time we saw what \emph{variables} are in Python, and that they are
+a little bit different from the \emph{Mathematical variables} that you
+use in Mathematics. In Sage, both concepts are present, but they are
+still distinct. For example in the cell above \texttt{f} is a variable
+in the sense of computer science, while \texttt{x} is a Mathematical
+variable.
+
+If you want to use Mathematical variables other than \texttt{x}, you
+first need to \emph{declare} them with the \texttt{var()} command:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{14}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{var}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{y}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
+\PY{n}{solve}\PY{p}{(}\PY{n}{y}\PY{o}{\PYZca{}}\PY{l+m+mi}{2} \PY{o}{+} \PY{p}{(}\PY{n}{x}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{y} \PY{o}{\PYZhy{}} \PY{l+m+mi}{2} \PY{o}{==} \PY{l+m+mi}{0}\PY{p}{,} \PY{n}{y}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{14}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+[y == -1/2*x - 1/2*sqrt(x\^{}2 + 2*x + 9) - 1/2, y == -1/2*x + 1/2*sqrt(x\^{}2 + 2*x +
+9) - 1/2]
+\end{Verbatim}
+\end{tcolorbox}
+        
+    Try removing the first line in the cell above and see what error you
+get!
+
+Here is another example:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{16}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{var}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{a}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{b}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
+\PY{n}{f} \PY{o}{=} \PY{n}{x}\PY{o}{\PYZca{}}\PY{l+m+mi}{2}\PY{o}{+}\PY{n}{a}\PY{o}{*}\PY{n}{x}\PY{o}{+}\PY{n}{b}
+\PY{n}{solve}\PY{p}{(}\PY{n}{f}\PY{p}{,}\PY{n}{x}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{16}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+[x == -1/2*a - 1/2*sqrt(a\^{}2 - 4*b), x == -1/2*a + 1/2*sqrt(a\^{}2 - 4*b)]
+\end{Verbatim}
+\end{tcolorbox}
+        
+    Some common constants are
+\href{https://doc.sagemath.org/html/en/reference/calculus/sage/symbolic/expression.html}{already
+defined} in Sage:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{17}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{e}\PY{o}{\PYZca{}}\PY{p}{(}\PY{n}{pi}\PY{o}{*}\PY{n}{I}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{17}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+-1
+\end{Verbatim}
+\end{tcolorbox}
+        
+    We will study symbolic expressions more in detail next time, in the
+context of calculus/analysis.
+
+    \hypertarget{basic-rings-and-fields}{%
+\section{Basic rings and fields}\label{basic-rings-and-fields}}
+
+\textbf{References:}
+{[}\href{https://doc.sagemath.org/html/en/reference/rings_standard/index.html}{3}{]}
+{[}\href{https://doc.sagemath.org/html/en/reference/rings_numerical/index.html}{4}{]}
+{[}\href{https://doc.sagemath.org/html/en/reference/finite_rings/index.html}{5}{]}
+
+As you should know, a \emph{field} is a Mathematical structure with two
+operations, addition and multiplication, which respect certain rules
+(distributivity, associativity, commutativity\ldots). Some examples of
+fields are the Rational numbers \(\mathbb Q\), the Real numbers
+\(\mathbb R\) and the Complex numbers \(\mathbb C\), but there are many
+more. As you should also know, a \emph{(commutative) ring} is like a
+field, except not all elements different from \(0\) need have a
+multiplicative inverse. For example the integers
+\(\mathbb Z = \{ \dots, -1, 0, 1, 2, \dots\}\) are a ring, but not a
+field.
+
+These structures are already implemented in Sage. Some of the most
+common are listed in the following table:
+
+\begin{longtable}[]{@{}rcl@{}}
+\toprule
+Mathematical object & Math symbol & Sage name \\
+\midrule
+\endhead
+Integers & \(\mathbb Z\) & \texttt{ZZ} \\
+Rational numbers & \(\mathbb Q\) & \texttt{QQ} \\
+Real numbers & \(\mathbb R\) & \texttt{RR} \\
+Complex numbers & \(\mathbb C\) & \texttt{CC} \\
+Integers modulo \(n\) & \(\mathbb Z/n\mathbb Z\) &
+\texttt{Integers(n)} \\
+Finite fields & \(\mathbb F_p\) & GF(p) \\
+\(\dots\) & \(\dots\) & \(\dots\) \\
+\bottomrule
+\end{longtable}
+
+    If you write a number or an expression, Sage will figure out where it
+``lives'', choosing the most restrictive interpretation possible. For
+example \texttt{3} will be interpreted to be an integer, even if it is
+also a rational number, a real number and a complex number.
+
+    \hypertarget{parents-and-coercion}{%
+\subsection{Parents and coercion}\label{parents-and-coercion}}
+
+\textbf{Reference:}
+{[}\href{https://doc.sagemath.org/html/en/tutorial/tour_coercion.html}{6}{]}
+
+You can check where an object ``lives'' with the \texttt{parent()}
+command. It works more or less like the Python command \texttt{type()},
+but it gives a more Mathematically inclined answer. Check the reference
+link {[}6{]} above if you want more details.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{18}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{c+c1}{\PYZsh{}Edit this cell to find out the type of other objects that we used}
+\PY{n}{parent}\PY{p}{(}\PY{l+m+mi}{3}\PY{o}{/}\PY{l+m+mi}{5}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{18}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+Rational Field
+\end{Verbatim}
+\end{tcolorbox}
+        
+    Sometimes Sage does not give you the best possible interpretation, so
+you can force something to be interpreted as living in a smaller ring as
+follows:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{4}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{minus\PYZus{}one} \PY{o}{=} \PY{n}{e}\PY{o}{\PYZca{}}\PY{p}{(}\PY{n}{pi}\PY{o}{*}\PY{n}{I}\PY{p}{)}
+\PY{n}{minus\PYZus{}one\PYZus{}coerced} \PY{o}{=} \PY{n}{ZZ}\PY{p}{(}\PY{n}{e}\PY{o}{\PYZca{}}\PY{p}{(}\PY{n}{pi}\PY{o}{*}\PY{n}{I}\PY{p}{)}\PY{p}{)}  \PY{c+c1}{\PYZsh{} coercion}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{parent}\PY{p}{(}\PY{n}{minus\PYZus{}one}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{parent}\PY{p}{(}\PY{n}{minus\PYZus{}one\PYZus{}coerced}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+Symbolic Ring
+Integer Ring
+    \end{Verbatim}
+
+    \textbf{Remark.} Notice that there is a fundamental difference between
+the rings \texttt{RR} and \texttt{CC} and all the others in the table
+above: the real and complex numbers are \emph{approximated}.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{1}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{QQ}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{RR}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+3
+3.00000000000000
+    \end{Verbatim}
+
+    You can also choose the precision of this approximation using the
+alternative name \texttt{RealField}.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{4}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{RR}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{RealField}\PY{p}{(}\PY{n}{prec}\PY{o}{=}\PY{l+m+mi}{1000}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+Real Field with 53 bits of precision
+Real Field with 1000 bits of precision
+    \end{Verbatim}
+
+    \hypertarget{polynomial-rings}{%
+\section{Polynomial rings}\label{polynomial-rings}}
+
+\textbf{Reference:}
+{[}\href{https://doc.sagemath.org/html/en/reference/polynomial_rings/index.html}{7}{]}
+
+If you want to work with polynomials over a certain ring it is better to
+use this specific construction, rather than the symbolic expressions
+introduced above.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{5}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{polring}\PY{o}{.}\PY{o}{\PYZlt{}}\PY{n}{x}\PY{p}{,}\PY{n}{y}\PY{p}{,}\PY{n}{z}\PY{o}{\PYZgt{}} \PY{o}{=} \PY{n}{RR}\PY{p}{[}\PY{p}{]}  \PY{c+c1}{\PYZsh{} Alternative: polring.\PYZlt{}x,y,z\PYZgt{} = PolynomialRing(RR)}
+\PY{n}{polring}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{5}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+Multivariate Polynomial Ring in x, y, z over Real Field with 53 bits of
+precision
+\end{Verbatim}
+\end{tcolorbox}
+        
+    You can use as many variables as you like, and you can replace
+\texttt{RR} with any ring. In the example above \texttt{polring} is just
+the name of the variable (in the computer science sense) associated with
+this polynomial ring.
+
+\hypertarget{operations-on-polynomials}{%
+\subsection{Operations on polynomials}\label{operations-on-polynomials}}
+
+The usual Mathematical operations are available on polynomial rings,
+including Euclidean division \texttt{//} and remainder \texttt{\%}.
+There is also the single-slash division \texttt{/}, but the result may
+not be a polynomial anymore.
+
+\textbf{Exercise:} use the \texttt{parent()} command to find out what
+the quotient of two polynomials is.
+
+\textbf{Question:} what happens if you remove the first line in the cell
+below? What if we used the variable \texttt{y} instead of \texttt{x}?
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{6}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{polring}\PY{o}{.}\PY{o}{\PYZlt{}}\PY{n}{x}\PY{o}{\PYZgt{}} \PY{o}{=} \PY{n}{QQ}\PY{p}{[}\PY{p}{]}
+\PY{n}{p} \PY{o}{=} \PY{n}{x}\PY{o}{\PYZca{}}\PY{l+m+mi}{2} \PY{o}{+} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{x} \PY{o}{\PYZhy{}} \PY{l+m+mi}{3}   \PY{c+c1}{\PYZsh{} Don\PYZsq{}t forget * for multiplication!}
+\PY{n}{q} \PY{o}{=} \PY{n}{p} \PY{o}{/}\PY{o}{/} \PY{p}{(}\PY{n}{x}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{)}
+\PY{n}{r} \PY{o}{=} \PY{n}{p} \PY{o}{\PYZpc{}}  \PY{p}{(}\PY{n}{x}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{)}
+\PY{n}{f} \PY{o}{=} \PY{n}{p} \PY{o}{/}  \PY{p}{(}\PY{n}{x}\PY{o}{+}\PY{l+m+mi}{1}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{q}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{r}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{f}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+x + 1
+-4
+(x\^{}2 + 2*x - 3)/(x + 1)
+    \end{Verbatim}
+
+    You can do more complex operations. Try out \texttt{roots()} and
+\texttt{factor} in the cell below.
+
+\textbf{Remark.} Notice how the result can change substantially if you
+change the base ring.
+
+\textbf{Remark.}
+\href{https://doc.sagemath.org/html/en/reference/structure/sage/structure/factorization.html}{Factorizations}
+are a particular object in Sage. They are kinda like a list, but not
+really. You can get a list of pairs (factor, power) with
+\texttt{list(factor(f))}.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{7}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{polring\PYZus{}onevar}\PY{o}{.}\PY{o}{\PYZlt{}}\PY{n}{t}\PY{o}{\PYZgt{}} \PY{o}{=} \PY{n}{QQ}\PY{p}{[}\PY{p}{]}
+
+\PY{n}{f} \PY{o}{=} \PY{n}{t}\PY{o}{\PYZca{}}\PY{l+m+mi}{5} \PY{o}{+} \PY{n}{t}\PY{o}{\PYZca{}}\PY{l+m+mi}{4} \PY{o}{\PYZhy{}} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{t}\PY{o}{\PYZca{}}\PY{l+m+mi}{3} \PY{o}{\PYZhy{}} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{t}\PY{o}{\PYZca{}}\PY{l+m+mi}{2} \PY{o}{\PYZhy{}} \PY{l+m+mi}{3}\PY{o}{*}\PY{n}{t} \PY{o}{\PYZhy{}} \PY{l+m+mi}{3}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{factor}\PY{p}{(}\PY{n}{f}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{f}\PY{o}{.}\PY{n}{roots}\PY{p}{(}\PY{p}{)}\PY{p}{)}  \PY{c+c1}{\PYZsh{} Result: list of pairs (root,multiplicity)}
+
+\PY{n}{polring\PYZus{}manyvar}\PY{o}{.}\PY{o}{\PYZlt{}}\PY{n}{x}\PY{p}{,}\PY{n}{y}\PY{p}{,}\PY{n}{z}\PY{o}{\PYZgt{}} \PY{o}{=} \PY{n}{QQ}\PY{p}{[}\PY{p}{]}
+\PY{n}{factor}\PY{p}{(}\PY{n}{x}\PY{o}{*}\PY{n}{y}\PY{o}{+}\PY{n}{x}\PY{p}{)}
+
+\PY{c+c1}{\PYZsh{} The following line gives an error, because the polynomial}
+\PY{c+c1}{\PYZsh{} is understood to possibly have many variables:}
+\PY{c+c1}{\PYZsh{}(x\PYZca{}2\PYZhy{}1).roots()}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+(t + 1) * (t\^{}2 - 3) * (t\^{}2 + 1)
+[(-1, 1)]
+    \end{Verbatim}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{7}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+(y + 1) * x
+\end{Verbatim}
+\end{tcolorbox}
+        
+    \hypertarget{matrices-and-vectors}{%
+\section{Matrices and vectors}\label{matrices-and-vectors}}
+
+\textbf{References:}
+{[}\href{https://doc.sagemath.org/html/en/reference/matrices/index.html}{8}{]},
+but in particular the subections
+{[}\href{https://doc.sagemath.org/html/en/reference/matrices/sage/matrix/docs.html}{9}{]}
+and
+{[}\href{https://doc.sagemath.org/html/en/reference/matrices/sage/matrix/matrix2.html}{10}{]}
+
+In Sage you can easily manipulate matrices and vectors
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{77}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{A} \PY{o}{=} \PY{n}{matrix}\PY{p}{(}\PY{p}{[}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{4}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{22}\PY{o}{/}\PY{l+m+mi}{7}\PY{p}{]}\PY{p}{]}\PY{p}{)}
+\PY{n}{B} \PY{o}{=} \PY{n}{matrix}\PY{p}{(}\PY{p}{[}\PY{p}{[}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{7}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{]}\PY{p}{)}
+\PY{n}{v} \PY{o}{=} \PY{n}{vector}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{)}
+
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{A}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}  \PY{c+c1}{\PYZsh{} \PYZbs{}n just means \PYZdq{}newline\PYZdq{}}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{B}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{B}\PY{o}{*}\PY{n}{v}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{A}\PY{o}{\PYZca{}}\PY{l+m+mi}{2} \PY{o}{+} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{B} \PY{o}{\PYZhy{}} \PY{n}{A}\PY{o}{*}\PY{n}{B}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+
+\PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Rank of A =}\PY{l+s+s2}{\PYZdq{}}\PY{p}{,} \PY{n}{rank}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} \PY{c+c1}{\PYZsh{} You can also use A.rank()}
+\PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Rank of B =}\PY{l+s+s2}{\PYZdq{}}\PY{p}{,} \PY{n}{rank}\PY{p}{(}\PY{n}{B}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+[   1    2    3]
+[   0    0    1]
+[   4   -3 22/7]
+
+[1/2   0   0]
+[  7   0   0]
+[  1   1   1]
+
+(3/2, 21, 6)
+
+[  -7/2    -10   80/7]
+[    17     -4   15/7]
+[ 241/7  -18/7 869/49]
+
+Rank of A = 3
+Rank of B = 2
+    \end{Verbatim}
+
+    \textbf{Exercise:} in the cell above, compute the determinant, inverse
+and characteristic polynomial of the matrix \texttt{A}. \emph{Hint: look
+at the reference {[}10{]} above (the functions are listed in alphabetic
+order).}
+
+As for polynomials, you can specify where a matrix or a vector lives
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{57}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{M} \PY{o}{=} \PY{n}{matrix}\PY{p}{(}\PY{n}{CC}\PY{p}{,} \PY{p}{[}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{]}\PY{p}{)}
+\PY{n}{parent}\PY{p}{(}\PY{n}{M}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{57}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+Full MatrixSpace of 2 by 2 dense matrices over Complex Field with 53 bits of
+precision
+\end{Verbatim}
+\end{tcolorbox}
+        
+    You can also solve linear systems and compute eigenvalues and
+eigenvectors of a matrix
+
+\textbf{Warning.} In linear algebra there are distinct concepts of
+\emph{left} and \emph{right} eigenvalues (and eigenvector). The one you
+know is probably that of \textbf{right} eigen-\{value,vector\}, that is
+an element \(\lambda\) of the base field and a non-zero vector
+\(\mathbf v\) with \(A\mathbf v=\lambda\mathbf v\). The other concept
+corresponds to the equality \(\mathbf v^TA=\lambda \mathbf v\).
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{60}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{A} \PY{o}{=} \PY{n}{Matrix}\PY{p}{(}\PY{n}{RR}\PY{p}{,} \PY{p}{[}\PY{p}{[}\PY{n}{sqrt}\PY{p}{(}\PY{l+m+mi}{59}\PY{p}{)}\PY{p}{,}\PY{l+m+mi}{32}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}\PY{p}{]}\PY{p}{)}
+\PY{n}{v} \PY{o}{=} \PY{n}{vector}\PY{p}{(}\PY{n}{RR}\PY{p}{,} \PY{p}{[}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{)}
+\PY{n}{A}\PY{o}{.}\PY{n}{solve\PYZus{}right}\PY{p}{(}\PY{n}{v}\PY{p}{)} \PY{c+c1}{\PYZsh{} Solve Ax=v. Alternative: A \PYZbs{} v}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{60}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+(0.289916349448506, 0.0241596957873755)
+\end{Verbatim}
+\end{tcolorbox}
+        
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{64}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{A} \PY{o}{=} \PY{n}{Matrix}\PY{p}{(}\PY{n}{QQ}\PY{p}{,} \PY{p}{[}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{]}\PY{p}{]}\PY{p}{)}
+\PY{n}{A}\PY{o}{.}\PY{n}{eigenspaces\PYZus{}right}\PY{p}{(}\PY{p}{)} \PY{c+c1}{\PYZsh{} Also: A.eigenvalues(), A.eigenvectors\PYZus{}right()}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{64}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+[
+(-0.3722813232690144?, Vector space of degree 2 and dimension 1 over Algebraic
+Field
+User basis matrix:
+[                   1 -0.6861406616345072?]),
+(5.372281323269015?, Vector space of degree 2 and dimension 1 over Algebraic
+Field
+User basis matrix:
+[                 1 2.186140661634508?])
+]
+\end{Verbatim}
+\end{tcolorbox}
+        
+    We can also extract a specific submatrix by selecting only some rows and
+columns, with a syntax similar to that of Python's lists. Check out more
+examples in the reference {[}9{]} above, and try them in the cell below.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{94}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{A} \PY{o}{=} \PY{n}{MatrixSpace}\PY{p}{(}\PY{n}{ZZ}\PY{p}{,} \PY{l+m+mi}{7}\PY{p}{)}\PY{o}{.}\PY{n}{random\PYZus{}element}\PY{p}{(}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{A}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{A}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{:}\PY{l+m+mi}{5}\PY{p}{]}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)} \PY{c+c1}{\PYZsh{} Rows from 1 to 3, columns from 2 to 5}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{A}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{:}\PY{p}{]}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}    \PY{c+c1}{\PYZsh{} First row, all columns}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{A}\PY{p}{[}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{5}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{:}\PY{l+m+mi}{5}\PY{p}{]}\PY{p}{)}   \PY{c+c1}{\PYZsh{} Rows 0, 5 and 2 (in this order) and columns 0 to 5}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+[-14   2   0  -1   1  -2  -1]
+[  0  -8   0   9  -2  11   1]
+[  0   3   1  -1   1   1 221]
+[ -1   2   1 -25 -10   4   0]
+[ -3   0   0   2  16  -1  -2]
+[  1  -3   3 -41   1   0   0]
+[ -2   1   0   0  -6   2  12]
+
+[ 0  9 -2]
+[ 1 -1  1]
+
+[-14   2   0  -1   1  -2  -1]
+
+[-14   2   0  -1   1]
+[  1  -3   3 -41   1]
+[  0   3   1  -1   1]
+    \end{Verbatim}
+
+    \textbf{Exercise:} write a sage function that computes the determinant
+of an \(n\times n\) matrix \(A=(a_{ij})\) using Laplace's rule by the
+first row, that is \begin{align*}
+    \operatorname{det}A = \sum_{j=1}^n (-1)^ja_{0j}M_{0j}
+\end{align*} where \(M_{0j}\) is the determinant of the
+\((n-1)\times(n-1)\) matrix obtained by removing the \(0\)-th row and
+the \(j\)-th column from \(A\).
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{91}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{k}{def} \PY{n+nf}{my\PYZus{}det}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{:}
+    \PY{k}{if} \PY{o+ow}{not} \PY{n}{A}\PY{o}{.}\PY{n}{is\PYZus{}square}\PY{p}{(}\PY{p}{)}\PY{p}{:}
+        \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Error: matrix is not square}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+    
+    \PY{n}{n} \PY{o}{=} \PY{n}{A}\PY{o}{.}\PY{n}{nrows}\PY{p}{(}\PY{p}{)}  \PY{c+c1}{\PYZsh{} size of the matrix}
+    
+    \PY{c+c1}{\PYZsh{} Continue from here!}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \hypertarget{number-theory}{%
+\section{Number Theory}\label{number-theory}}
+
+\textbf{Reference:}
+{[}\href{https://doc.sagemath.org/html/en/reference/rings_standard/sage/rings/integer.html}{11}{]}
+
+Sage includes a large library of functions for computing with the
+integers, see the link above.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{8}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{n} \PY{o}{=} \PY{l+m+mi}{123456789}
+\PY{n}{m} \PY{o}{=} \PY{l+m+mi}{987654321}
+\PY{n}{p} \PY{o}{=} \PY{l+m+mi}{3607}
+
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{factor}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{is\PYZus{}prime}\PY{p}{(}\PY{n}{p}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{p}\PY{o}{.}\PY{n}{divides}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{euler\PYZus{}phi}\PY{p}{(}\PY{n}{m}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{gcd}\PY{p}{(}\PY{n}{n}\PY{p}{,} \PY{n}{m}\PY{p}{)}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{lcm}\PY{p}{(}\PY{n}{n}\PY{p}{,} \PY{n}{m}\PY{p}{)}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+3\^{}2 * 3607 * 3803
+True
+True
+619703040
+9
+13548070123626141
+    \end{Verbatim}
+
+    \hypertarget{primes}{%
+\subsection{Primes}\label{primes}}
+
+\textbf{Reference:}
+{[}\href{https://doc.sagemath.org/html/en/reference/sets/sage/sets/primes.html}{12}{]}
+
+The set of prime numbers is called \texttt{Primes()}. It is like an
+infinite list: for example you can get the one-millionth prime number or
+you can use this list to create other lists. You can also check what the
+first prime number larger than a given number is.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{9}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{PP} \PY{o}{=} \PY{n}{Primes}\PY{p}{(}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{PP}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{PP}\PY{p}{[}\PY{l+m+mi}{10}\PY{p}{]}\PY{p}{,} \PY{n}{PP}\PY{p}{[}\PY{l+m+mi}{10}\PY{o}{\PYZca{}}\PY{l+m+mi}{6}\PY{p}{]}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{PP}\PY{o}{.}\PY{n}{next}\PY{p}{(}\PY{l+m+mi}{44}\PY{p}{)}\PY{p}{)}
+
+\PY{n}{First\PYZus{}Thousand\PYZus{}Primes} \PY{o}{=} \PY{n}{PP}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{:}\PY{l+m+mi}{1000}\PY{p}{]}
+\PY{n+nb}{print}\PY{p}{(}\PY{p}{[}\PY{n}{p} \PY{k}{for} \PY{n}{p} \PY{o+ow}{in} \PY{n}{First\PYZus{}Thousand\PYZus{}Primes} \PY{k}{if} \PY{n}{p} \PY{o}{\PYZlt{}} \PY{l+m+mi}{100} \PY{o+ow}{and} \PY{n}{p} \PY{o}{\PYZgt{}} \PY{l+m+mi}{75}\PY{p}{]}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+Set of all prime numbers: 2, 3, 5, 7, {\ldots}
+31 15485867
+47
+[79, 83, 89, 97]
+    \end{Verbatim}
+
+    \hypertarget{the-chinese-remainder-theorem-crt}{%
+\subsection{The Chinese remainder theorem
+(CRT)}\label{the-chinese-remainder-theorem-crt}}
+
+We say that two integers \(a\) and \(b\) are \emph{congruent} modulo
+another integer \(n>0\) if they have the same remainder when divided by
+\(n\). We denote this by \(a\equiv b\pmod n\), or in Python/Sage syntax
+\texttt{a\ \%\ n\ ==\ b\ \%\ n}.
+
+The Chinese remainder theorem states that if \(a,b\in\mathbb Z\) and
+\(n,m\in \mathbb Z_{>0}\) are such that \(\gcd(n,m)=1\) then the system
+of congruences
+
+\begin{align*}
+\begin{cases}
+    x \equiv a \pmod n\\
+    x \equiv b \pmod m
+\end{cases}
+\end{align*}
+
+has exactly one solution modulo \(mn\). This means that there is one and
+only one number \(x\) with \(0\leq x<mn\) such that \(x\equiv a\pmod n\)
+and \(x\equiv b\pmod m\).
+
+The procedure to find such a number is not too hard to describe (you
+might see it in an algebra or number theory course), but it can be a bit
+long. Luckily, Sage can do this for you:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{10}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{a} \PY{o}{=} \PY{l+m+mi}{2}
+\PY{n}{b} \PY{o}{=} \PY{o}{\PYZhy{}}\PY{l+m+mi}{1}
+\PY{n}{n} \PY{o}{=} \PY{l+m+mi}{172}
+\PY{n}{m} \PY{o}{=} \PY{l+m+mi}{799}
+
+\PY{k}{if} \PY{n}{gcd}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{n}{m}\PY{p}{)} \PY{o}{!=} \PY{l+m+mi}{1}\PY{p}{:}
+    \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{The numbers are not comprime, I can}\PY{l+s+s2}{\PYZsq{}}\PY{l+s+s2}{t solve this!}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+\PY{k}{else}\PY{p}{:}
+    \PY{n}{x} \PY{o}{=} \PY{n}{crt}\PY{p}{(}\PY{n}{a}\PY{p}{,} \PY{n}{b}\PY{p}{,} \PY{n}{n}\PY{p}{,} \PY{n}{m}\PY{p}{)}
+    \PY{n+nb}{print}\PY{p}{(}\PY{n}{x}\PY{p}{,} \PY{n}{x}\PY{o}{\PYZpc{}}\PY{k}{n}, x\PYZpc{}m)
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+74306 2 798
+    \end{Verbatim}
+
+    \textbf{Exercise.} There is a more general version of the Chinese
+remainder theorem which says that if
+\(a_0, a_1, \dots, a_k\in\mathbb Z\) and
+\(n_0, n_2, \dots, n_k\in\mathbb Z_{>0}\) are such that
+\(\gcd(n_i, n_j)=1\) for \(i\neq j\), then the system of congruences
+
+\begin{align*}
+\begin{cases}
+    x \equiv a_0 \pmod {n_0}\\
+    x \equiv a_1 \pmod {n_1}\\
+    \dots \\
+    x \equiv a_k \pmod {n_k}
+\end{cases}
+\end{align*}
+
+has exactly one solution modulo \(\prod_{i=0}^kn_i\). Use the
+\texttt{crt()} function to find a solution to such a system. *Hint:
+start by running the command \texttt{help(crt)}.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{127}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{c+c1}{\PYZsh{}help(crt)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \hypertarget{cryptography-rsa}{%
+\section{Cryptography: RSA}\label{cryptography-rsa}}
+
+\href{https://en.wikipedia.org/wiki/Cryptography}{Cryptography} is the
+discipline that studies methods to communicate secrets in such a way
+that any unauthorized listener would not be able to understand the
+message.
+
+A simple cryptographic protocol could be changing every letter of your
+text following a fixed scheme (or \emph{cypher}), for example by turning
+every A into a B, every B into a C and so on. However this is not a very
+secure method, for many reasons. One of them is that at some point the
+people who want to communicate need to agree on what method to use, and
+anyone listening to that conversation would be able to decypher every
+subsequent conversation. A public-key cryptographic protocol solves this
+problem.
+
+\hypertarget{public-key-cryptography}{%
+\subsection{Public-key cryptography}\label{public-key-cryptography}}
+
+Public-key cryptographic protocols, such as RSA, work like this: there
+are two keys, a \emph{private} key that is only known to person A
+(traditionally called Alice in every example), and a \emph{public} key
+that does not need to be secret.
+
+The public key is used to \emph{encrypt} the message (that is to
+``lock'' it, or ``hyde'' it), but one needs the private key to
+\emph{decrypt} it. Imagine having two keys for your door, but one can
+only be used to lock it, while the other only to open it.
+
+The message exchange works like this: suppose that person B (Bob) wants
+to send a secret message to Alice. Then Alice secretely generates a
+private and a public key and sends only the public one to Bob. Now Bob
+encrypts the message and sends it to Alice, who can use her private key
+to decrypt it. Even if Eve (short for \emph{eavesdropper}, an
+unauthorized listener) listens to every message exchanged, she won't be
+able to decypher the secret: the private key has never left Alice's
+house!
+
+Notice that such a protocol is \emph{asymmetric}: if Alice wanted to
+send a secret to Bob in reply, Bob would need to generate a pair of keys
+of his own.
+
+Let's see how we can do this in practice, using number theory!
+
+\hypertarget{rsa}{%
+\subsection{RSA}\label{rsa}}
+
+As many other cryptography protocols, RSA is based on a Mathematical
+process that is easy to do in one direction, but very hard to invert. In
+this case the hard process is integer factorization, that is decomposing
+an integer number as a product of primes.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{2}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{p} \PY{o}{=} \PY{l+m+mi}{100003100019100043100057100069}
+\PY{n}{q} \PY{o}{=} \PY{l+m+mi}{100144655312449572059845328443}
+\PY{n}{n} \PY{o}{=} \PY{n}{p}\PY{o}{*}\PY{n}{q}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{is\PYZus{}prime}\PY{p}{(}\PY{n}{p}\PY{p}{)}\PY{p}{,} \PY{n}{is\PYZus{}prime}\PY{p}{(}\PY{n}{q}\PY{p}{)}\PY{p}{,} \PY{n}{is\PYZus{}prime}\PY{p}{(}\PY{n}{p}\PY{o}{*}\PY{n}{q}\PY{p}{)}\PY{p}{)}
+
+\PY{c+c1}{\PYZsh{} Use the command below to see how long it takes}
+\PY{c+c1}{\PYZsh{}timeit(\PYZdq{}factor(n)\PYZdq{}, number=1, repeat=1)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+True True False
+    \end{Verbatim}
+
+    In order to generate the keys, Alice picks a number \(n\) which is the
+product of two large primes \(p\) and \(q\) of more or less the same
+size. Finding such primes is relatively easy compared to factoring the
+number \(n\) she obtained. Then she computes the Euler totient
+\(\varphi(n)=(p-1)(q-1)\) of \(n\), which she can do because she knows
+that \(n=pq\) - it would be impossible otherwise!
+
+Then Alice can compute two integers \((d,e)\) such that
+\(de\equiv 1\pmod{\varphi(n)}\). She will send the numbers \(n\) and
+\(d\) to Bob and keep \(e\) secret. In this case the public key is the
+pair \((n,d)\), while \(e\) is the private key.
+
+Of course, she does all of this using Sage!
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{105}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{k}{def} \PY{n+nf}{two\PYZus{}large\PYZus{}primes}\PY{p}{(}\PY{p}{)}\PY{p}{:}
+    \PY{n}{p}\PY{p}{,} \PY{n}{q} \PY{o}{=} \PY{l+m+mi}{0}\PY{p}{,} \PY{l+m+mi}{0}
+    \PY{c+c1}{\PYZsh{} We make sure that they are different}
+    \PY{k}{while} \PY{n}{p} \PY{o}{==} \PY{n}{q}\PY{p}{:}
+        \PY{n}{p} \PY{o}{=} \PY{n}{Primes}\PY{p}{(}\PY{p}{)}\PY{p}{[}\PY{n}{randint}\PY{p}{(}\PY{l+m+mi}{10}\PY{o}{\PYZca{}}\PY{l+m+mi}{6}\PY{p}{,} \PY{l+m+mi}{2}\PY{o}{*}\PY{l+m+mi}{10}\PY{o}{\PYZca{}}\PY{l+m+mi}{6}\PY{p}{)}\PY{p}{]}
+        \PY{n}{q} \PY{o}{=} \PY{n}{Primes}\PY{p}{(}\PY{p}{)}\PY{p}{[}\PY{n}{randint}\PY{p}{(}\PY{l+m+mi}{10}\PY{o}{\PYZca{}}\PY{l+m+mi}{6}\PY{p}{,} \PY{l+m+mi}{2}\PY{o}{*}\PY{l+m+mi}{10}\PY{o}{\PYZca{}}\PY{l+m+mi}{6}\PY{p}{)}\PY{p}{]}
+    \PY{k}{return} \PY{n}{p}\PY{p}{,} \PY{n}{q}
+
+\PY{k}{def} \PY{n+nf}{random\PYZus{}unit\PYZus{}mod}\PY{p}{(}\PY{n}{N}\PY{p}{)}\PY{p}{:}
+    \PY{n}{R} \PY{o}{=} \PY{n}{Integers}\PY{p}{(}\PY{n}{N}\PY{p}{)}
+    \PY{n}{d} \PY{o}{=} \PY{n}{R}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{)}
+    \PY{c+c1}{\PYZsh{} We make sure that it is invertible}
+    \PY{k}{while} \PY{o+ow}{not} \PY{n}{d}\PY{o}{.}\PY{n}{is\PYZus{}unit}\PY{p}{(}\PY{p}{)}\PY{p}{:}
+        \PY{n}{d} \PY{o}{=} \PY{n}{R}\PY{o}{.}\PY{n}{random\PYZus{}element}\PY{p}{(}\PY{p}{)}
+    \PY{k}{return} \PY{n}{d}
+
+\PY{k}{def} \PY{n+nf}{Alice\PYZus{}generate\PYZus{}keys}\PY{p}{(}\PY{p}{)}\PY{p}{:}
+    \PY{n}{p}\PY{p}{,} \PY{n}{q} \PY{o}{=} \PY{n}{two\PYZus{}large\PYZus{}primes}\PY{p}{(}\PY{p}{)}
+    \PY{n}{n} \PY{o}{=} \PY{n}{p}\PY{o}{*}\PY{n}{q}
+    \PY{n}{phi\PYZus{}n} \PY{o}{=} \PY{p}{(}\PY{n}{p}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{q}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{)} \PY{c+c1}{\PYZsh{} euler\PYZus{}phi(n) is slow!}
+    
+    \PY{n}{d} \PY{o}{=} \PY{n}{random\PYZus{}unit\PYZus{}mod}\PY{p}{(}\PY{n}{phi\PYZus{}n}\PY{p}{)}
+    \PY{n}{e} \PY{o}{=} \PY{n}{d}\PY{o}{\PYZca{}}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}
+    \PY{k}{return} \PY{n}{n}\PY{p}{,} \PY{n}{d}\PY{p}{,} \PY{n}{e}
+
+\PY{n}{Alice\PYZus{}generate\PYZus{}keys}\PY{p}{(}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{105}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+(419199544978969, 235530823946467, 80799425863927)
+\end{Verbatim}
+\end{tcolorbox}
+        
+    Now, how does Bob encrypt his message? Let's say he wants to send to
+Alice the number \(m\) with \(1<m<n\) (In practice he would like to send
+her some text with emojis, or maybe a voice message; but for computers
+everything is a number, and there are different ways to translate any
+sort of information to a number. He just chooses one of the many
+standard methods that already exist, no cryptography is needed in this
+step. If the message \(m\) is too long, he can split it up in some
+pieces and repeat the process multiple times.)
+
+Now he computes \(m^d\pmod n\) and sends it back to Alice.
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{3}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{k}{def} \PY{n+nf}{Bob\PYZus{}encrypt}\PY{p}{(}\PY{n}{m}\PY{p}{,} \PY{n}{n}\PY{p}{,} \PY{n}{d}\PY{p}{)}\PY{p}{:}
+    \PY{n}{R} \PY{o}{=} \PY{n}{Integers}\PY{p}{(}\PY{n}{n}\PY{p}{)}
+    \PY{k}{return} \PY{n}{R}\PY{p}{(}\PY{n}{m}\PY{p}{)}\PY{o}{\PYZca{}}\PY{n}{d}  \PY{c+c1}{\PYZsh{} Assume that n is large enough}
+    
+\PY{n}{message} \PY{o}{=} \PY{l+m+mi}{42424242}
+\PY{n}{Bob\PYZus{}encrypt}\PY{p}{(}\PY{n}{message}\PY{p}{,} \PY{l+m+mi}{419199544978969}\PY{p}{,} \PY{l+m+mi}{235530823946467}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{3}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+149461597163501
+\end{Verbatim}
+\end{tcolorbox}
+        
+    Since \(de\equiv 1\pmod{\varphi(n)}\), it follows that
+\((m^d)^e\equiv m\pmod n\) (see
+\href{https://en.wikipedia.org/wiki/Euler\%27s_theorem}{Wikipedia:
+Euler's theorem}). So for Alice it is very easy to get back the original
+message:
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{108}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{k}{def} \PY{n+nf}{Alice\PYZus{}decrypt}\PY{p}{(}\PY{n}{m\PYZus{}encrypted}\PY{p}{,} \PY{n}{n}\PY{p}{,} \PY{n}{e}\PY{p}{)}\PY{p}{:}
+    \PY{n}{R} \PY{o}{=} \PY{n}{Integers}\PY{p}{(}\PY{n}{n}\PY{p}{)}
+    \PY{k}{return} \PY{n}{R}\PY{p}{(}\PY{n}{m\PYZus{}encrypted}\PY{p}{)}\PY{o}{\PYZca{}}\PY{n}{e}
+
+\PY{n}{Alice\PYZus{}decrypt}\PY{p}{(}\PY{l+m+mi}{149461597163501}\PY{p}{,} \PY{l+m+mi}{419199544978969}\PY{p}{,} \PY{l+m+mi}{80799425863927}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+            \begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]
+\prompt{Out}{outcolor}{108}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+42424242
+\end{Verbatim}
+\end{tcolorbox}
+        
+    Another assumption on which RSA relies is that even if one knows
+\(M=m^e\) and \(e\), extracting the \(e\)-th root of \(M\) modulo \(n\)
+(and thus obtaining \(m\)) is very hard. Currently the best known way to
+do this is by factorizing \(n\) first, which is considered to be a very
+hard problem. However, there is no proof that faster algorithms can't be
+devised.
+
+Moreover, one day we will overcome the current technological
+difficulties and quantum computers will be available. Quantum computers
+are not just ``more powerful'' than classical hardware, but they work
+based on completely different logical foundations and they make the
+factorization problem much easier to solve: for example
+\href{https://en.wikipedia.org/wiki/Shor\%27s_algorithm}{Shor's
+algorithm} takes advantage of this different logic and can factorize
+numbers quickly, if run on a quantum computer.
+
+To this day the largest number factorized with a quantum computer is
+\(21=3\times 7\). Nonetheless, quantum-safe cryptography protocols
+(i.e.~based on problems that are hard to solve also with quantum
+computers) have already been developed.
+
+
+    % Add a bibliography block to the postdoc
+    
+    
+    
+\end{document}