Starting to do exercises on Elliptic Curves

.github/workflows/blank.yml

@@ -24,6 +24,7 @@ jobs:
             Misc/Recurrence/Recurrence.tex
             Misc/pde/pde.tex
             Misc/Ufd/Ufd.tex
+            Misc/Elliptic/Elliptic.tex
             Papers/GK/GK.tex
       - name: Compile Chinese LaTeX documents
         uses: xu-cheng/latex-action@v2
@@ -52,3 +53,4 @@ jobs:
             Ufd.pdf
             GK.pdf
             DifferentialForm.pdf
+            Elliptic.pdf

MIT/Solutions/MIT.tex

@@ -7,7 +7,7 @@
 \usepackage{mathrsfs}
 
 \newcommand{\emptymacro}{}    % Used to test against an empty macro
-% \newcommand{\linearalgebra}{} 
+\newcommand{\linearalgebra}{} 
 
 \title{MIT Math Problems}
 \author{Cecilia}

Misc/Elliptic/Elliptic.tex

@@ -0,0 +1,150 @@
+\documentclass{article}
+\usepackage{amsmath}
+\usepackage[legalpaper, margin=0.5in]{geometry}
+\usepackage[final]{pdfpages}
+
+\title{Elliptic Curve Exercises}
+\author{Cecilia Chan}
+
+\begin{document}
+\maketitle
+
+\includepdf{Misc/Elliptic/MATH-5020-hw1-1.pdf}
+
+\section*{Problem 1}
+Consider:
+\begin{eqnarray*}
+  a_0 + \sum\limits_{i = 1}^{n} a_i \left(\frac{s}{t}\right)^i &=& 0                                     \\
+            a_0 t^n  + \sum\limits_{i = 1}^{n} a_i s^i t^{n-i} &=& 0                                     \\
+                                                   a_0 t^n &=& - \sum\limits_{i = 1}^{n} a_i s^i t^{n-i}
+\end{eqnarray*}
+Note that every term of the right hand size is a multiple of $ s $, but $ \gcd(s^n, t) = \gcd(s, t) = 1 $, so $ a_0 $ is a multiple of $ s $.
+
+Consider:
+\begin{eqnarray*}
+  \sum\limits_{i = 0}^{n-1} a_i \left(\frac{s}{t}\right)^i + a_n \left(\frac{s}{t}\right)^n &=& 0 \\
+  \sum\limits_{i = 0}^{n-1}  a_i s^i t^{n-i} + a_n s^n &=& 0 \\
+  a_n s^n &=& -\sum\limits_{i = 0}^{n-1} a_i s^i t^{n-i}
+\end{eqnarray*}
+Note that every term on the right hand side is a multiple of $ t $, but $ \gcd(s^n, t) = \gcd(s, t) = 1 $, so $ a_n $ is a multiple of $ t $.
+
+\section*{Problem 2}
+$ \Rightarrow $:
+
+Suppose $ ax + by = c $ has integral solution $ (x, y) $, then $ \frac{a}{\gcd(a, b)} x + \frac{b}{\gcd(a, b)} y = \frac{c}{\gcd(a, b)} $. In this case, left hand side is an integer, so is the right hand side, and therefore $ c $ is a multiple of $ \gcd(a, b) $.
+
+$ \Leftarrow $:
+
+Suppose $ c $ is a multiple of $ \gcd(a, b) $, then $ c = k \gcd(a, b) $ for some integer $ k $. By Bezout's identity, there exists $ x_0, y_0 $ such that $ ax_0 + by_0 = \gcd(a, b) $. Then $ a(kx_0) + b(ky_0) = c $, and $ (kx_0, ky_0) $ is an integral solution to $ ax + by = c $.
+
+\section*{Problem 3}
+Let the number of rooster, hen and chicken be $ r, h, c $ respectively. Then we have:
+
+\begin{eqnarray*}
+  r + h + c &=& 100 \\
+  5r + 3h + \frac{1}{3}c &=& 100
+\end{eqnarray*}
+
+To avoid fraction, we can have:
+\begin{eqnarray*}
+  r + h + 3d &=& 100 \\
+  5r + 3h + d &=& 100
+\end{eqnarray*}
+
+Or in matrix form 
+
+\begin{eqnarray*}
+  A &=& \left(
+  \begin{array}{ccc}
+    1 & 1 & 3 \\
+    5 & 3 & 1 
+  \end{array}\right) \\
+  b &=& \left(\begin{array}{c}
+    r \\
+    h \\
+    d
+  \end{array}\right) \\
+  b &=& \left(\begin{array}{c}
+    100 \\
+    100
+  \end{array}\right) \\
+  Ax &=& b
+\end{eqnarray*}
+
+Smith normal form allow us to write $ A = P^{-1} S Q^{-1} $ where $ P, Q $ are unimodular matrices, and $ S $ is a diagonal matrix.
+
+\begin{eqnarray*}
+  P &=& \left(
+  \begin{array}{cc}
+    1 & 0 \\
+    5 & -1
+  \end{array}\right) \\
+  Q &=& \left(
+  \begin{array}{ccc}
+    1 & -1 & 4 \\
+    0 & 1 & -7 \\
+    0 & 0 & 1
+  \end{array}\right) \\
+  S &=& \left(
+  \begin{array}{ccc}
+    1 & 0 & 0 \\
+    0 & 2 & 0 \\
+  \end{array}\right) \\
+\end{eqnarray*}
+
+Now we can easily solve the system as follows:
+\begin{eqnarray*}
+  Ax &=& b \\
+  P^{-1} S Q^{-1} x &=& b \\
+  S Q^{-1} x &=& P b \\
+  \left(
+  \begin{array}{ccc}
+    1 & 0 & 0 \\
+    0 & 2 & 0 \\
+  \end{array}\right) Q^{-1}x &=& \left(\begin{array}{c}
+               100 \\
+               400 
+              \end{array}\right) \\
+              Q^{-1}x &=& \left(\begin{array}{c}
+               100 \\
+               200 \\
+               t
+              \end{array}\right) \\
+  x &=& Q \left(\begin{array}{c}
+               100 \\
+               200 \\
+               t
+              \end{array}\right) \\
+  x &=& \left(
+    \begin{array}{ccc}
+      1 & -1 & 4 \\
+      0 &  1 & -7 \\
+      0 & 0 & 1 
+    \end{array}\right) \left(\begin{array}{c}
+      100 \\
+      200 \\
+      t
+     \end{array}\right) \\
+  \left(\begin{array}{c}
+    r \\
+    h \\
+    d
+  \end{array}\right) &=& \left(\begin{array}{c}
+    4t - 100 \\
+    -7t + 200 \\
+    t
+  \end{array}\right)
+\end{eqnarray*}
+
+All of these are number of animals, so it must be the case that they are non-negative. Therefore we have:
+\begin{eqnarray*}
+   4t - 100 &\geq& 0 \\
+          t &\geq& 25 \\
+  -7t + 200 &\geq& 0 \\
+          t &\leq& \frac{200}{7} \\
+          t &\leq& 28 \\
+\end{eqnarray*}
+
+So the answer is obvious, $ t $ can only be $ 25, 26, 27, 28 $, and the corresponding number of rooster, hen and chicken are $ 0, 25, 75 $, $ 4, 18, 78 $, $ 8, 11, 81 $, $ 12, 4, 84 $ respectively.
+
+\end{document}