Sharedwww / talks / 2011-gross-stein-cubic / rational.texOpen in CoCalc
Author: William A. Stein
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\documentclass[handout]{beamer}
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\usepackage{amssymb,amsmath, amscd}
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\usepackage{times, verbatim}
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\usepackage{graphicx}
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\usepackage{graphics}
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\DeclareFontEncoding{OT2}{}{} % to enable usage of cyrillic fonts
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\newcommand{\textcyr}[1]{%
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\selectfont #1}}
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\newcommand{\Sha}{{\mbox{\textcyr{Sh}}}}
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\DeclareMathOperator{\vol}{vol}
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\DeclareMathOperator{\ord}{ord}
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\DeclareMathOperator{\tor}{tor}
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\DeclareMathOperator{\rank}{rank}
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\DeclareMathOperator{\Reg}{Reg}
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\newcommand{\Q}{\mathbb{Q}}
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\title{Solving Cubic Equations}
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\author{Benedict Gross and William Stein}
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\date{January, 2012}
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\begin{document}
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\maketitle
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\begin{frame}[fragile]
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\frametitle{Algebraic equations}
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\begin{center}
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{\includegraphics[height=.4\textheight]{triangle3.pdf}}
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\pause
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\medskip
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{\includegraphics[height=.2\textheight]{pythag.png}}
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\end{center}
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Pythagoras (600 BCE) \pause ~~~~ Baudh\=ayana (800 BCE)
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\end{frame}
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\begin{frame}
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\frametitle{Differential equations}
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$$F'(T) = F(T)\qquad\qquad dF/dT = F \pause \qquad\qquad F(0) = 1$$
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\pause
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$F(T) = exp(T) \pause = 1 + T + T^2/2 + T^3/6 + T^4/24 + T^5/120 + ...$
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\pause
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\includegraphics[width=\textwidth]{population.png}
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\end{frame}
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\begin{frame}[fragile]
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\frametitle{Pythagorean triples}
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\medskip
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$a^2 + b^2 = c^2$ has solutions $(3,4,5), (5,12,13), (7,24,25),\ldots$
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\medskip
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\pause
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There are more solutions on a Babylonian tablet (1800 BCE):
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\medskip
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\pause
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\begin{tabular}{lr}
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\raisebox{-.45\height}{\includegraphics[width=0.7\textwidth]{plimpton.png}}&
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\footnotesize
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$\begin{array}{|c|}\hline
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\vspace{-2ex}\\
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( 3, 4, 5 )\\
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( 5, 12, 13 )\\
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( 7, 24, 25 )\\
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( 9, 40, 41 )\\
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( 11, 60, 61 )\\
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( 13, 84, 85 )\\
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( 15, 8, 17 )\\
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( 21, 20, 29 )\\
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( 33, 56, 65 )\\
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( 35, 12, 37 )\\
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( 39, 80, 89 )\\
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( 45, 28, 53 )\\
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( 55, 48, 73 )\\
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( 63, 16, 65 )\\
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( 65, 72, 97 )\\
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\hline
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\end{array}$
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\end{tabular}
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\end{frame}
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\begin{frame}
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\frametitle{The general solution of $a^2+b^2=c^2$}
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$x = a/c$ and $y = b/c$ satisfy the equation $x^2 + y^2 = 1$
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\pause
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{\includegraphics[width=.5\textwidth]{param3.pdf}}
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\pause
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$$ t = \frac{y}{1+x} \pause \qquad\qquad x = \frac{1-t^2}{1+t^2}\qquad y = \frac{2t}{1+t^2}$$
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\end{frame}
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\begin{frame}
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Write $ t = p/q$. Then
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$$x = \frac{q^2 - p^2}{q^2 + p^2}\qquad\qquad y = \frac{2qp}{q^2 + p^2}$$
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\pause
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\medskip
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$$a = q^2 - p^2 \qquad\qquad b = 2qp \qquad\qquad c = q^2 + p^2$$
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\pause
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$$t = 1/2 \longrightarrow (a,b,c) = (3,4,5)$$
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$$t = 2/3 \longrightarrow (a,b,c) = (5,12,13)$$
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$$t = 3/4 \longrightarrow (a,b,c) = (7,24,25)$$
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\end{frame}
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\begin{frame}
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\frametitle{Cubic equations}
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After linear and quadratic equations come cubic equations, like
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$$x^3 + y^3 = 1 \qquad\qquad y^2 + y = x^3 - x$$
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\pause
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Here there may be either a finite or an infinite number of rational solutions.
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\pause
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\begin{center}
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\includegraphics[width=.4\textwidth]{fermat.jpg}
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\end{center}
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\end{frame}
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\begin{frame}
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\frametitle{The graph}
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$$y^2 + y = x^3 - x$$
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\begin{center}
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\only<1>{\includegraphics[width=.6\textwidth]{secant1.pdf}}%
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\only<2>{\includegraphics[width=.6\textwidth]{secant2.pdf}}%
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\only<3>{\includegraphics[width=.6\textwidth]{secant3.pdf}}
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\end{center}
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\end{frame}
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\begin{frame}
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\frametitle{The limit of a secant line is a tangent}
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$$y^2 + y = x^3 - x$$
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\begin{center}
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\includegraphics[width=.6\textwidth]{tangent.pdf}
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\end{center}
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\end{frame}
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\begin{frame}[fragile]
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\frametitle{Large solutions}
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If the number of solutions is infinite, they quickly become large.
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$$y^2 + y = x^3 - x$$
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\vspace{-10ex}
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\scriptsize
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\begin{verbatim}
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(0, 0)
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(1, 0)
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(-1, -1)
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(2, -3)
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(1/4, -5/8)
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(6, 14)
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(-5/9, 8/27)
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(21/25, -69/125)
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(-20/49, -435/343)
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(161/16, -2065/64)
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(116/529, -3612/12167)
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(1357/841, 28888/24389)
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(-3741/3481, -43355/205379)
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(18526/16641, -2616119/2146689)
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(8385/98596, -28076979/30959144)
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(480106/4225, 332513754/274625)
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(-239785/2337841, 331948240/3574558889)
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(12551561/13608721, -8280062505/50202571769)
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(-59997896/67387681, -641260644409/553185473329)
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(683916417/264517696, -18784454671297/4302115807744)
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(1849037896/6941055969, -318128427505160/578280195945297)
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(51678803961/12925188721, 10663732503571536/1469451780501769)
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(-270896443865/384768368209, 66316334575107447/238670664494938073)
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\end{verbatim}
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% (4881674119706/5677664356225, -8938035295591025771/13528653463047586625)
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% (-16683000076735/61935294530404, -588310630753491921045/487424450554237378792)
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% (997454379905326/49020596163841, -31636113722016288336230/343216282443844010111)
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% (2786836257692691/16063784753682169, -435912379274109872312968/2035972062206737347698803)
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% (213822353304561757/158432514799144041, 41974401721854929811774227/63061816101171948456692661)
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% (-3148929681285740316/2846153597907293521, -2181616293371330311419201915/4801616835579099275862827431)
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% (79799551268268089761/62586636021357187216, -754388827236735824355996347601/495133617181351428873673516736)
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\end{frame}
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\begin{frame}
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\frametitle{Even the simplest solution can be large}
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\begin{center}
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$\displaystyle{}y^2 + y = x^3 - 5115523309x - 140826120488927$
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\end{center}
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Numerator of $x$-coordinate of smallest solution (5454 digits):
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\begin{center}
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\includegraphics[width=\textwidth, height=.3\textheight]{numer.png}
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\end{center}
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Denominator:
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\begin{center}
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\includegraphics[width=\textwidth, height=.3\textheight]{denom.png}
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\end{center}
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\end{frame}
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\begin{frame}
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\frametitle{The rank}
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The rank of $E$ is essentially the number of independent solutions.
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\pause
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\medskip
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\begin{itemize}
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\item rank $(E) = 0$ means there are finitely many solutions.
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\medskip
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\pause
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\item rank $(E) > 0$ means there are infinitely many solutions.
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\medskip
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\pause
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\item The curve $E(a)$ with equation
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$$y(y+1) = x(x-1)(x+a)$$
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has rank $= 0,1,2,3,4$ for $a=0,1,2,4,16$. % (N = 11, 37, 389, 5077, ?).
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\end{itemize}
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%\begin{center}
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%\includegraphics[height=.6\textheight]{pics/Ea}
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%\end{center}
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% Records:
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%
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% 0 0
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% 1 1
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% 2 2
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% 3 4
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% 4 16
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% 5 79
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% 6 298
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%
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\end{frame}
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\begin{frame}
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\frametitle{The rank is finite}
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\begin{center}
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\includegraphics[height=.45\textheight]{mordell}
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\hspace{3em}\includegraphics[height=.45\textheight]{weil}
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\end{center}
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\pause
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\medskip
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Can it be arbitrarily large?
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\end{frame}
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\begin{frame}
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\frametitle{The current record is rank($E$) = $28$}
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\vspace{1ex}
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{\tiny
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$
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y^2 + xy + y = x^3 - x^2 -
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20067762415575526585033208209338542750930230312178956502x +$\vspace{-1.1ex}\\
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\hfill$344816117950305564670329856903907203748559443593191803612660082962919394\
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48732243429$}
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\vspace{-2ex}
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\begin{center}
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\includegraphics[height=0.7\textheight]{egens.png}
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\includegraphics[width=.2\textwidth]{elkies.png}
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\end{center}
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\end{frame}
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\begin{frame}
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Bryan Birch and Peter Swinnerton-Dyer made a prediction for the
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rank, based on the average number of solutions at prime numbers
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$p$.
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\begin{center}
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\includegraphics[height=.8\textheight]{weddingphoto.png}
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\end{center}
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\end{frame}
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\begin{frame}
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\frametitle{Primes}
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\medskip
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A prime $p$ is a number greater than $1$ that is not divisible by any smaller number.
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\pause
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\medskip
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2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
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71, 73, 79, 83, 89, 97, 101, 103, 107, 109, $\ldots$
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\medskip
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\pause
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There are infinitely many primes. The largest explicit prime known is $2^{43112609} - 1$ with 12,978,189 digits.
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\pause
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\medskip
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\begin{center}
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\includegraphics[width=.8\textwidth]{primepi.pdf}
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\end{center}
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\end{frame}
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\begin{frame}[fragile]
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\frametitle{The Prime Number $2^{9689}-1=$}
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{\tiny
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\begin{verbatim}
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478220278805461202952839298660005909741497172402236500851334510991837895094266297027892768
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611270789458682472098152425631930658505267683408748083442943326479742589324762368833102163
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320895484735480579994334130982598901374380618710958104314868081377832153049671560156328262
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441404039814320762203627219040859079053720347525610556407157926386787524098557335652265610
388
854212857732105787905232886503535587361567936365588992571157442015383209175242284304691881
389
142740066213555930351685370397681268638575037622778794958058208183126172570100349820651232
390
987267723348951095346937568303703837399969677158578890563911552261340549570718452415821920
391
822376644205901459333065700972215396237685342377048613857808977562130116781129916640736174
392
660669780818675796691467124607371290420058840892318638773788767529288695379706698096740605
393
353012285353903696549022478492464900795489867850331465554647550450168618735486696437455261
394
412064078294962245202778896213860266593314768769632208950427879162465151931232783175655377
395
937719452467339581928148666857638401959072017941334958297031939388438881049454604034208753
396
656362833215207318161430072176937142623851754052084521466531330118355196259184955893849902
397
534878037671647707393063443684008446825593744345169031599934913766463896897261419901530490
398
654781905622717122494707073971630095377574344130792050186353223446654564569577433188504497
399
825014866346737213039209989485214519099823287877248665051301081676990289251871925006694721
400
570653621624869624056925686555429622155221156042777866254593699880107018616260147647429345
401
983018365127336346273267588306070141035925482914977433929717368076561095959991130918978823
402
835013163567266143596921823997719693387439540399662367558052821120713639637085805605116078
403
177098545257698803233381293927275210194462952749031383555198519709592888523641530178921867
404
514101454120309619127093436903952209828031766894206132557234964363840305648734929088422378
405
629288747223121903238528103409182430661894774072726552428489330447486145494207679904173944
406
716583828167141043583120679050191452732628737033997470720601688256282740427017032260672798
407
034347932642573009183981307771932245539476396060658821432660315614149074055769805516626304
408
444758375671151649018119344223685942415184379538933576543212994405485534515585927342456182
409
514681371472060628778102124092370802149229834963517952727030296297015692768651163505008040
410
728267425236264469571076976886613730278931360967438271901738550848466337347612084356798306
411
505955807293511063754424080735066708298723377976887493898358452309563899612061631863439196
412
711208646438464947096323007272920091258614726799976249670985276950353573392441620265772074
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124868359220282898331114083392330243391779797699031142584361935093675448381119440881276338
414
808420445180491245438388418080094527562666805762895476338464130510775377324708249580453335
415
571748196502507081973046642282610569751056428979895118219288597635222905389894873761464213
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9910911535864505818992696826225754111
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\end{verbatim}
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}
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\end{frame}
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\begin{frame}
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\frametitle{Primality testing}
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Determining that $n > 1$ is a prime can be done quickly.
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\pause
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\medskip
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\begin{center}
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"PRIMES is in P"
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\end{center}
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\medskip
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AKS: Manindra Agrawal, Neeraj Kayal, and Nitin Saxena (2002)
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\pause
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\medskip
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If $n$ fails the primality test, it is more difficult to factor it.
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\pause
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\medskip
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123018668453011775513049495838496272077285356959533
445
479219732245215172640050726365751874520219978646938
446
995647494277406384592519255732630345373154826850791
447
702612214291346167042921431160222124047927473779408
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0665351419597459856902143413 $=$ RSA-$768$ $=$
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\pause
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\medskip
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453
334780716989568987860441698482126908177047949837137
454
685689124313889828837938780022876147116525317430877
455
37814467999489\\
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$\times$
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367460436667995904282446337996279526322791581643430
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876426760322838157396665112792333734171433968102700
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92798736308917
460
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\end{frame}
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\begin{frame}
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What do we mean by a solution of the cubic equation at the prime number $p$?
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\pause
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$$y^2 + y = x^3 - x$$
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$(x,y) \equiv (3,1)$ is a solution at $p = 11$
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\pause
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\medskip
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There are finitely many solutions $A(p)$ at each prime $p$.
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\pause
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\begin{center}
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\includegraphics[width=.47\textwidth]{A23.pdf}
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\hfill
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\includegraphics[width=.47\textwidth]{A71.pdf}
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\end{center}
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\end{frame}
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\begin{frame}
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It is common to write
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$$A(p) = p + 1 - a(p)$$
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\pause
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We define the $L$-function of $E$ by the infinite product
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$$L(E,s) = \prod_p (1 - a(p)p^{-s} + p^{1-2s})^{-1} = \sum a(n)n^{-s}$$
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\pause
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This definition only works in the region $s > 3/2$, where the infinite product converges.
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\pause
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\begin{center}
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\includegraphics[width=.85\textwidth]{lplot32.pdf}
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\end{center}
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\end{frame}
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\begin{frame}
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If we formally set $s=1$ in the product, we get
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$$\prod_p(1 - a(p)p^{-1} + p^{-1})^{-1} = \prod_p p/A(p)$$
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\pause
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\medskip
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If $A(p)$ is large on average compared with $p$, this will approach
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0. The larger $A(p)$ is on average, the faster it will tend to 0.
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\pause
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\medskip
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%\begin{center}
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%{\scriptsize
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%\begin{tabular}{|r|r|r|r|}\hline
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%$p$& $A(p)$ & $\frac{p}{A(p)}$ & $\prod \frac{p}{A_p}$\\\hline
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%23 & 33 & 0.697 & 0.697\\
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%29 & 40 & 0.725 & 0.505\\
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%31 & 40 & 0.775 & 0.392\\
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%37 & 49 & 0.755 & 0.296\\
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%41 & 52 & 0.788 & 0.233\\
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%43 & 56 & 0.768 & 0.179\\
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%47 & 60 & 0.783 & 0.140\\
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%53 & 63 & 0.841 & 0.118\\\hline
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% 59 & 72 & 0.819 & 0.097\\
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% 61 & 77 & 0.792 & 0.077\\
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% 67 & 84 & 0.798 & 0.061\\\hline
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%\end{tabular}}
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%\end{center}
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\includegraphics[width=\textwidth]{Approd.pdf}
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\end{frame}
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\begin{frame}
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\frametitle{The conjecture of Birch and Swinnerton-Dyer}
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\begin{enumerate}
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\item The function $L(E,s)$ has a natural (analytic) continuation to a neighborhood of $s = 1$.
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\pause
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\medskip
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\item The order of vanishing of $L(E,s)$ at $s =1$ is equal to the rank of $E$.
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\pause
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\medskip
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\item The leading term in the Taylor expansion of $L(E,s)$ at $s=1$ is given by certain arithmetic invariants of $E$.
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\end {enumerate}
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\medskip
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$$L(E,s) = c(E)(s-1)^{\rank(E)} + \dots $$
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\end{frame}
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\begin{frame}
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The most mysterious arithmetic invariant was studied by John Tate and Igor Shafarevich, who conjectured that it is finite. Tate called this invariant $\Sha$.
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\pause
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\begin{center}
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\includegraphics[height=.45\textheight]{tate.png}
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\hspace{3em}\includegraphics[height=.45\textheight]{shafarevich.png}
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\end{center}
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\
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%\begin{center}
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%\includegraphics[width=.3\textwidth]{tate.png}
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%\end{center}
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\end{frame}
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%\pause
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%\begin{center}
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%\includegraphics[width=.35\textwidth]{shafarevich.png}
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%\end{center}
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\begin{frame}
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\frametitle{The Birch and Swinnerton-Dyer Conjecture}
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\Large
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\vspace{-1em}
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$$\begin{align*}
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L(E,s) &= c(E) (s-1)^{\rank(E)}+ \cdots \\
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c(E) &=
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\frac{\Omega_E \cdot \Reg_E \cdot \#\Sha_E \cdot \prod c_p}{\#E(\Q)_{\tor}^2}
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\end{align*}
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$$
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\large
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Each quantity on the right measures the size of
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an abelian group attached to $E$.
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\vfill
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\begin{center}
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\includegraphics[width=.6\textwidth]{bsd}
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\end{center}
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\end{frame}
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\begin{frame}
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\frametitle{Natural (analytic) continuation}
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\mbox{}\vspace{-5em}
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The infinite sum $\sum_{n=0}^{\infty}x^n$ converges when $-1 < x < 1$.
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\pause
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\begin{center}
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\includegraphics[width=.7\textwidth]{continue}
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\vspace{-10em}
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$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\qquad\qquad\qquad$
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\end{center}
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\end{frame}
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\begin{frame}
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The natural (analytic) continuation of $L(E,s) = \sum a(n)n^{-s}$ was obtained by Andrew Wiles and Richard Taylor (1995).\pause~ They proved that the function defined by the infinite series
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$$F(\tau) = \sum a(n) e^{2\pi i n \tau}$$
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is a modular form.
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\medskip
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\pause
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\begin{center}
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\hfill
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\includegraphics[height=.45\textwidth]{wiles.png}
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\hfill
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\includegraphics[height=.45\textwidth]{taylor.png}
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\hfill
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\end{center}
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\end{frame}
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\begin{frame}
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Combining a limit formula I proved with Don Zagier (1983) with work of Victor Kolyvagin (1986) we can now show the following.
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\medskip
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\pause
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If $L(E,1) \neq 0$ the rank is zero, so there are finitely many solutions.
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\medskip
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\pause
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If $L(E,1) = 0$ and $L'(E,1) \neq 0$ the rank is one, so there are infinitely many solutions.
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\medskip
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\pause
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In both cases, we can also show that $\Sha$ is finite.
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\pause
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\begin{center}
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\includegraphics[height=.4\textwidth]{india.png}
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\hfill
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\includegraphics[height=.4\textwidth]{zagier.png}
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\hfill
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\includegraphics[height=.4\textwidth]{kolyvagin.png}
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\hfill
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\end{center}
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\end{frame}
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\begin{frame}
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When the order of $L(E,s)$ at $s = 1$ is greater than one we cannot prove anything in general\ldots
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\pause
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\medskip
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But the computer has been a great guide.
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\pause
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\medskip
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Here is a summary of the evidence for the simplest rank 2 curve
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$$y(y+1) = x(x-1)(x+2)$$
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\pause
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\begin{itemize}
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\item the order of vanishing is equal to 2
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\item most primes up to 50,000 do not divide the order of $\Sha$
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% no ordinary prime up to 48,859 divides the order of group appearing in leading term $L"(E,1)$(excluding possibly $p=16231$).
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\end{itemize}
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\pause
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\begin{center}
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\includegraphics[width=.5\textwidth]{wstein.png}
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\end{center}
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\end{frame}
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\begin{frame}
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\frametitle{The average rank}
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Manjul Bhargava has recently made progress on the study of the average rank, for ALL cubic
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curves with rational coefficients.
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\pause
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\begin{center}
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\includegraphics[height=.7\textheight]{manjul.png}
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\end{center}
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\end{frame}
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\begin{frame}
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\frametitle{Enumerating the curves}
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\medskip
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\begin{itemize}
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\item Every such curve has a unique equation of the form $y^2 = x^3 + Ax + B$ where $A$ and $B$ are integers (not divisible by $p^4$ and $p^6$, for any prime $p$).
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\pause
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\medskip
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\item Define the height $H(E)$ as the maximum of the positive integers $|A|^3$ and $|B|^2$.
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\pause
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\medskip
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\item For any positive real number $X$, there are only finitely many curves with $H(E) \leq X$.
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\pause
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\medskip
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\item Call this number $N(X)$. It grows at the same rate as $(X)^{1/2}(X)^{1/3} = X^{5/6}$.
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\end{itemize}
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\end{frame}
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\begin{frame}
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\begin{itemize}
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\item Define the average rank by the limit as $X \rightarrow \infty$ of $$\frac{1}{N(X)} \sum_{H(E)\leq X} rank(E)$$
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\pause
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\medskip
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\item We suspect that this limit exists, and is equal to $1/2$.
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\pause
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\medskip
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\item In fact, we think that on average half the curves have rank zero and half have rank one.
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\pause
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\medskip
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\item Bhargava and Shankar have shown why there is an upper bound on the limit, and have obtained a specific upper bound which is less than $1$.
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\end{itemize}
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\end{frame}
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\begin{frame}
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\frametitle{Thank you}
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\begin{center}
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\includegraphics[height=.9\textheight]{stein_invert}
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\end{center}
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\end{frame}
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\end{document}