\documentclass[11pt]{article}

    \usepackage[breakable]{tcolorbox}
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    \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:
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                         % to support a larger range
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      {\Gread@eps{\Gin@base.bb}}%
      {\Gread@@xetex@aux#1}%
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    \makeatother

    % The hyperref package gives us a pdf with properly built
    % internal navigation ('pdf bookmarks' for the table of contents,
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    \usepackage{hyperref}
    % The default LaTeX title has an obnoxious amount of whitespace. By default,
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    \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}
    

    
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    % Define a nice break command that doesn't care if a line doesn't already
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    \def\lt{<}
    \let\Oldtex\TeX
    \let\Oldlatex\LaTeX
    \renewcommand{\TeX}{\textrm{\Oldtex}}
    \renewcommand{\LaTeX}{\textrm{\Oldlatex}}
    % Document parameters
    % Document title
    \title{Mathematical Software - Homework 3}
    \date{Deadline: Sunday, May 9th}
    
    
    
    
    
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\def\PYZbs{\char`\\}
\def\PYZus{\char`\_}
\def\PYZob{\char`\{}
\def\PYZcb{\char`\}}
\def\PYZca{\char`\^}
\def\PYZam{\char`\&}
\def\PYZlt{\char`\<}
\def\PYZgt{\char`\>}
\def\PYZsh{\char`\#}
\def\PYZpc{\char`\%}
\def\PYZdl{\char`\$}
\def\PYZhy{\char`\-}
\def\PYZsq{\char`\'}
\def\PYZdq{\char`\"}
\def\PYZti{\char`\~}
% for compatibility with earlier versions
\def\PYZat{@}
\def\PYZlb{[}
\def\PYZrb{]}
\makeatother


    % For linebreaks inside Verbatim environment from package fancyvrb. 
    \makeatletter
        \newbox\Wrappedcontinuationbox 
        \newbox\Wrappedvisiblespacebox 
        \newcommand*\Wrappedvisiblespace {\textcolor{red}{\textvisiblespace}} 
        \newcommand*\Wrappedcontinuationsymbol {\textcolor{red}{\llap{\tiny$\m@th\hookrightarrow$}}} 
        \newcommand*\Wrappedcontinuationindent {3ex } 
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        % Take advantage of the already applied Pygments mark-up to insert 
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        %        {, <, #, %, $, ' and ": go to next line. 
        %        _, }, ^, &, >, - and ~: stay at end of broken line. 
        % Use of \textquotesingle for straight quote. 
        \newcommand*\Wrappedbreaksatspecials {% 
            \def\PYGZus{\discretionary{\char`\_}{\Wrappedafterbreak}{\char`\_}}% 
            \def\PYGZob{\discretionary{}{\Wrappedafterbreak\char`\{}{\char`\{}}% 
            \def\PYGZcb{\discretionary{\char`\}}{\Wrappedafterbreak}{\char`\}}}% 
            \def\PYGZca{\discretionary{\char`\^}{\Wrappedafterbreak}{\char`\^}}% 
            \def\PYGZam{\discretionary{\char`\&}{\Wrappedafterbreak}{\char`\&}}% 
            \def\PYGZlt{\discretionary{}{\Wrappedafterbreak\char`\<}{\char`\<}}% 
            \def\PYGZgt{\discretionary{\char`\>}{\Wrappedafterbreak}{\char`\>}}% 
            \def\PYGZsh{\discretionary{}{\Wrappedafterbreak\char`\#}{\char`\#}}% 
            \def\PYGZpc{\discretionary{}{\Wrappedafterbreak\char`\%}{\char`\%}}% 
            \def\PYGZdl{\discretionary{}{\Wrappedafterbreak\char`\$}{\char`\$}}% 
            \def\PYGZhy{\discretionary{\char`\-}{\Wrappedafterbreak}{\char`\-}}% 
            \def\PYGZsq{\discretionary{}{\Wrappedafterbreak\textquotesingle}{\textquotesingle}}% 
            \def\PYGZdq{\discretionary{}{\Wrappedafterbreak\char`\"}{\char`\"}}% 
            \def\PYGZti{\discretionary{\char`\~}{\Wrappedafterbreak}{\char`\~}}% 
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        % Some characters . , ; ? ! / are not pygmentized. 
        % This macro makes them "active" and they will insert potential linebreaks 
        \newcommand*\Wrappedbreaksatpunct {% 
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            \lccode`\~`\,\lowercase{\def~}{\discretionary{\hbox{\char`\,}}{\Wrappedafterbreak}{\hbox{\char`\,}}}% 
            \lccode`\~`\;\lowercase{\def~}{\discretionary{\hbox{\char`\;}}{\Wrappedafterbreak}{\hbox{\char`\;}}}% 
            \lccode`\~`\:\lowercase{\def~}{\discretionary{\hbox{\char`\:}}{\Wrappedafterbreak}{\hbox{\char`\:}}}% 
            \lccode`\~`\?\lowercase{\def~}{\discretionary{\hbox{\char`\?}}{\Wrappedafterbreak}{\hbox{\char`\?}}}% 
            \lccode`\~`\!\lowercase{\def~}{\discretionary{\hbox{\char`\!}}{\Wrappedafterbreak}{\hbox{\char`\!}}}% 
            \lccode`\~`\/\lowercase{\def~}{\discretionary{\hbox{\char`\/}}{\Wrappedafterbreak}{\hbox{\char`\/}}}% 
            \catcode`\.\active
            \catcode`\,\active 
            \catcode`\;\active
            \catcode`\:\active
            \catcode`\?\active
            \catcode`\!\active
            \catcode`\/\active 
            \lccode`\~`\~ 	
        }
    \makeatother

    \let\OriginalVerbatim=\Verbatim
    \makeatletter
    \renewcommand{\Verbatim}[1][1]{%
        %\parskip\z@skip
        \sbox\Wrappedcontinuationbox {\Wrappedcontinuationsymbol}%
        \sbox\Wrappedvisiblespacebox {\FV@SetupFont\Wrappedvisiblespace}%
        \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
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    \definecolor{cellborder}{HTML}{CFCFCF}
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    % prompt
    \makeatletter
    \newcommand{\boxspacing}{\kern\kvtcb@left@rule\kern\kvtcb@boxsep}
    \makeatother
    \newcommand{\prompt}[4]{
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\begin{document}
    
    \maketitle
    
    

    \begin{center}
    \emph{For this exercise you should have received this text in .ipynb
format. Complete the exercises by modifying this file, and submit the
modified version}
	\end{center}

	\vspace{1cm}
    \textbf{Exercise 1}

Use SageMath to solve the following problems:

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  Find the roots of the following polynomial over \(\mathbb Q\):
  \begin{align*}
  p = 4 x^{7} + 4 x^{6} + 3 x^{5} - 13 x^{4} - 13 x^{3} - 9 x^{2} + 3 x + 3 \in \mathbb Q[x]
  \end{align*}
\item
  Find the roots of the same polynomial \(p\) over \(\mathbb R\) and
  over \(\mathbb C\).
\item
  Find the determinant, the trace and the characteristic polynomial of
  the following matrix: \begin{align*}
  A=\left(\begin{array}{rrrr}
  -1 & 1 & -1 & 0 \\
  1 & \frac{1}{2} & 1 & 0 \\
  \frac{1}{2} & -\frac{1}{2} & -2 & 1 \\
  0 & 0 & 1 & 1
  \end{array}\right)
  \end{align*}
\item
  Find a solution to the linear system \(A\mathbf x =\mathbf v\), where
  \(A\) is the matrix above and \(\mathbf v=(1, 2, 3, 4)\).
\end{enumerate}

Write your code in the cell below.

    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
\prompt{In}{incolor}{ }{\boxspacing}
\begin{Verbatim}[commandchars=\\\{\}]

\end{Verbatim}
\end{tcolorbox}

	\vspace{1cm}
    \textbf{Exercise 2}

After exchanging messages with the RSA protocol seen in class, Alice and
Bob decide to meet and play their favorite game: flip a coin. They like
this game very much because it does not take long to set it up and they
have exactly the same chances of winning.

Unfortunately, due to the COVID-19 pandemic they cannot meet in person,
and despite being good friends they don't trust each other enough to
play this game via Webex call. Luckily, Alice is an expert in
cryptography and she knows how to play this game using the Chinese
remainder theorem.

The game plays out as follows:

\begin{itemize}
\item[(A1)] Alice picks two large prime numbers \(p\) and \(q\), she computes
\(n=pq\) and sends \(n\) to Bob, keeping \(p\) and \(q\) secret.

\item[(B1)] Bob picks a random number \(a\) with \(1<a<n\) and \(\gcd(a,n)=1\),
computes \(b=a^2\mod n\) and sends \(b\) to Alice, keeping \(a\) secret.

\item[(A2)] Alice computes two numbers \(x\) and \(y\) such that
\(x^2\equiv b\pmod p\) and \(y^2\equiv b\pmod q\) and she uses the
Chinese remainder theorem to compute a number \(z\) such that
\(z\equiv x\pmod p\) and \(z\equiv y\pmod q\), so that
\(z^2\equiv b\pmod n\). Then she sends \(z\) to Bob.

Since \(n\) is the product of two primes, there are \(4\) possible
square roots of \(b\) modulo \(n\), corresponding to the solutions of
the four systems of congruences (one for each possible combination of
\(\pm\)) \begin{align*}\begin{cases}
z\equiv \pm x\pmod p\\
z \equiv \pm y\pmod q
\end{cases}\end{align*}

One of those solutions is \(a\) and another is \(-a\), and Bob knows
them. Alice is picking one of the \(4\) possible roots at random (she
chooses between \(x\) and \(-x\) and between \(y\) and \(-y\)), so she
has \(50\%\) chance of picking one that Bob already knows. This
corresponds to Alice flipping a coin, and she wins if she picks
\(\pm a\):

\item[(B2)] If \(z\equiv\pm a\pmod n\), Bob declares to have lost. Otherwise,
Bob claims to have won, and as proof he produces one prime factor of
\(n\) by computing \(g=\gcd(n,a+z)\). \emph{(One can prove that in this
situation \(g\) is always one of the two prime factors of \(n\).)}

Since factoring a number without extra information is very hard, Alice
will be convinced that she must have given Bob one of the square roots
that he did not know, so she admits the loss.

\end{itemize}

Now to the actual exercise:

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  Write the code for the functions A1, B1 and B2 as indicated in the
  cell below. The function A2 is already written.
\item
  Modify the functions B1, A2 and B2 to check that the opponent is not
  cheating. More precisely:

\begin{itemize}
\tightlist
\item
  In B1, Bob should check that \(n\) is not a prime power. \emph{(This
  is the only way Alice can try to cheat: if she sends Bob a number
  \(n\) that is the product of more than two primes, than she has less
  than \(50\%\) chance of winning!)}
\item
  In A2, Alice should check that \(b\) is a square modulo \(n\).
\item
  In B2, Bob should check that \(z^2\equiv a^2\pmod n\).
\end{itemize}

In case cheating is detected, a message should be printed saying that
the person is cheating.
\end{enumerate}

\newpage

    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
\prompt{In}{incolor}{1}{\boxspacing}
\begin{Verbatim}[commandchars=\\\{\}]
\PY{c+c1}{\PYZsh{} Alice needs this to compute the square roots}
\PY{k+kn}{from} \PY{n+nn}{sage}\PY{n+nn}{.}\PY{n+nn}{rings}\PY{n+nn}{.}\PY{n+nn}{finite\PYZus{}rings}\PY{n+nn}{.}\PY{n+nn}{integer\PYZus{}mod} \PY{k}{import} \PY{n}{square\PYZus{}root\PYZus{}mod\PYZus{}prime}

\PY{k}{def} \PY{n+nf}{A1}\PY{p}{(}\PY{p}{)}\PY{p}{:}
    \PY{c+c1}{\PYZsh{} This function must return two distinct primes and their product.}
    
\PY{k}{def} \PY{n+nf}{B1}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{:}
    \PY{c+c1}{\PYZsh{} This function must return a random integer a}
    \PY{c+c1}{\PYZsh{} with 1\PYZlt{}a\PYZlt{}n and gcd(a,n)=1.}

\PY{k}{def} \PY{n+nf}{A2}\PY{p}{(}\PY{n}{b}\PY{p}{,} \PY{n}{p}\PY{p}{,} \PY{n}{q}\PY{p}{)}\PY{p}{:}
    \PY{n}{x} \PY{o}{=} \PY{n}{ZZ}\PY{p}{(}\PY{n}{square\PYZus{}root\PYZus{}mod\PYZus{}prime}\PY{p}{(}\PY{n}{Integers}\PY{p}{(}\PY{n}{p}\PY{p}{)}\PY{p}{(}\PY{n}{b}\PY{p}{)}\PY{p}{,} \PY{n}{p}\PY{p}{)}\PY{p}{)}
    \PY{n}{y} \PY{o}{=} \PY{n}{ZZ}\PY{p}{(}\PY{n}{square\PYZus{}root\PYZus{}mod\PYZus{}prime}\PY{p}{(}\PY{n}{Integers}\PY{p}{(}\PY{n}{q}\PY{p}{)}\PY{p}{(}\PY{n}{b}\PY{p}{)}\PY{p}{,} \PY{n}{q}\PY{p}{)}\PY{p}{)}
    \PY{k}{return} \PY{n}{crt}\PY{p}{(}\PY{n}{x}\PY{p}{,} \PY{n}{y}\PY{p}{,} \PY{n}{p}\PY{p}{,} \PY{n}{q}\PY{p}{)}

\PY{k}{def} \PY{n+nf}{B2}\PY{p}{(}\PY{n}{a}\PY{p}{,} \PY{n}{z}\PY{p}{,} \PY{n}{n}\PY{p}{)}\PY{p}{:}
    \PY{c+c1}{\PYZsh{} This function must print out one of two messages:}
    \PY{c+c1}{\PYZsh{} \PYZdq{}Bob has lost\PYZdq{} if z is congruent to a or \PYZhy{}a modulo n.}
    \PY{c+c1}{\PYZsh{} \PYZdq{}Bob has won, proof: \PYZdq{} followed by a prime factor of n otherwise.}
    \PY{c+c1}{\PYZsh{} In this case the prime must be calculated as explained above.}


\PY{c+c1}{\PYZsh{} This is how the game plays out:}
\PY{n}{p}\PY{p}{,} \PY{n}{q}\PY{p}{,} \PY{n}{n} \PY{o}{=} \PY{n}{A1}\PY{p}{(}\PY{p}{)}
\PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Alice picked n =}\PY{l+s+s2}{\PYZdq{}}\PY{p}{,} \PY{n}{n}\PY{p}{)}
\PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{[[ Alice}\PY{l+s+s2}{\PYZsq{}}\PY{l+s+s2}{s secret:}\PY{l+s+s2}{\PYZdq{}}\PY{p}{,} \PY{n}{p}\PY{p}{,} \PY{n}{q}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{]]}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
\PY{n}{a} \PY{o}{=} \PY{n}{B1}\PY{p}{(}\PY{n}{n}\PY{p}{)}
\PY{n}{b} \PY{o}{=} \PY{n}{a}\PY{o}{\PYZca{}}\PY{l+m+mi}{2} \PY{o}{\PYZpc{}} \PY{n}{n}
\PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Bob picked b =}\PY{l+s+s2}{\PYZdq{}}\PY{p}{,} \PY{n}{b}\PY{p}{)}
\PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{[[ Bob}\PY{l+s+s2}{\PYZsq{}}\PY{l+s+s2}{s secret:}\PY{l+s+s2}{\PYZdq{}}\PY{p}{,} \PY{n}{a}\PY{p}{,} \PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{]]}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
\PY{n}{z} \PY{o}{=} \PY{n}{A2}\PY{p}{(}\PY{n}{b}\PY{p}{,} \PY{n}{p}\PY{p}{,} \PY{n}{q}\PY{p}{)}
\PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Alice picked z =}\PY{l+s+s2}{\PYZdq{}}\PY{p}{,} \PY{n}{z}\PY{p}{)}
\PY{n}{B2}\PY{p}{(}\PY{n}{a}\PY{p}{,} \PY{n}{z}\PY{p}{,} \PY{n}{n}\PY{p}{)}
\end{Verbatim}
\end{tcolorbox}

    \vspace{1cm}
    \textbf{Grading}

This homework assignment is worth \(20\%\) of your final grade. Exercise
1 is worth 4 points (one for each part) and Exercise 2 is worth 12
points (8 points for part (a) and 4 points for part (b)), for a total of
\textbf{16 points}.

    
    
\end{document}