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zin!t7Jb#b`a6`mA0MymmlVKAX65`7Z3!|{O27+t@{Ptq{8FJ@|#9(R&XP^Yi2@TC#ve*BS+YE%9FoQr3LGNK3y2_&tc}AYX=}p}9Q-qeG!_7v z%2=iG0ex3lA7Kb^h)g>U&Lm`Q7{a0X5<@YzEItgyS#%pg&}|%qW&@pNF=H5=0}La+ z(wD`NjSUP(II{l5a1N6;2T4$8tQ1!mNz&KBMoHOygt2k9Y`owo%9NcK2PT@0;LtDw z>S<>Q4&q*xE1bn-%ElkgK~Y*e4o=9lZ1RnnJLPAQrs; zgT&KWC hn?c`ivH&tOQxZpTsWctjGaM4(5K0CH*7IzX{twqfCy)RD literal 0 HcmV?d00001 diff --git a/UE/ue3.tex b/UE/ue3.tex new file mode 100644 index 0000000..70a8ae2 --- /dev/null +++ b/UE/ue3.tex @@ -0,0 +1,279 @@ +\documentclass[a4paper,10pt,fleqn]{article} +\usepackage[utf8x]{inputenc} +\usepackage{amsmath,amssymb,ulsy} +\usepackage{fullpage} +\usepackage{txfonts} +\usepackage[ngerman]{babel} +\usepackage{fixltx2e} %Deutschsprach Bugs +%\usepackage[T1]{fontenc} +%\usepackage{lmodern} +\usepackage{amsthm} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{emaxima} +%\usepackage{ngerman} + +\pagestyle{fancy} +\chead{3. Übung ZtuA} +\rhead{Mi, 9. Mai 2012} + +\def\P{\mathbb{P}} +\def\N{\mathbb{N}} +\def\R{\mathbb{R}} +\def\Z{\mathbb{Z}} +\def\oder{\vee} +\def\und{\wedge} + +\def\kgV{\text{kgV}} +\def\ggT{\text{ggT}} +\def\sgn{\text{sgn}} +%opening +\title{$3$. Übung ZtuA} +\author{} + +\begin{document} + +%\section*{$3$. Übung} +\subsection*{$13$. Aufgabe} +{\texttt{Man beweise, dass es je unendlich viele Primzahlen der Form a) 4k+3 und b) 4k+1 gibt. (Hinweis: Man verwende dazu jeweils eine geeignete Variante des klassischen Beweises von Euklid über die Unendlichkeit der Menge der Primzahlen, wobei speziell für den Beweisteil b) der erste Ergänzungssatz benötigt wird.)} \newline + \begin{enumerate} + \item[(a)] Angenommen, die Menge aller Primzahlen der Form $4k+3$, d.h. $p_{1}, \ldots, p_{n}$ sind alle. $7 \equiv 3 \mod 4$, daher ist diese Menge nichtleer. Definiere + \begin{equation} + m:=4p_{1} \ldots p_{n} - 1 \equiv 3 \mod 4 + \end{equation} +insbesondere ist $m$ ungerade. Nun gilt +\begin{equation} + \forall i: p_{i} < 2 p_{i} < 3p_{i} - 1 < 4p_{1} \ldots p_{n} - 1 +\end{equation} +Nach dem Fundamentalsatz der Zahlentheorie hat $m$ mindestens einen Primteiler $p$. Dieses $p$ kann nicht von der Form $4k+1$ sein, da sonst der Rest $-1$ bleiben würde. Daher hat $m$ nur Primteiler der Form $4k+1$, woraus folgt, dass $m \equiv 1 \mod 4$ ist, was ein Widerspruch zur Konstruktion von $m$ ist. \newline +(Anders: $m \equiv 3 \mod 4$, d.h. m kann nicht nur Primfaktoren der Form $4k+1$ haben, sei $p \equiv 3 \mod 4 \land p \mid m \Rightarrow p \mid 1$. WS!) +\item[(b)] Angenommen, es gibt nur endlich viele Primzahlen der Form $4k+1$, diese seien $p_{1}, \ldots, p_{r}$. $5 \equiv 1 \mod 4$, daher $r\geq 1$. Mit + \begin{equation} + \alpha \equiv 1 \mod 4 \land \beta \equiv 1 \mod 4 \Rightarrow \alpha \beta \equiv 1 \mod 4, + \end{equation} +erhält man für $n$: +\begin{equation} +n:=\left(2p_{1} \cdots p_{r} \right)^{2} +1 = 4 \left( p_{1} \cdots p_{r} \right)^{2} + 1 \Rightarrow n \equiv 1 \mod 4 +\end{equation} +Sei $p \in \P \land p \mid n$ (nach dem Fundamentalsatz der Zahlentheorie): +\begin{equation} +\forall i \in \lbrace 1 , \ldots, r \rbrace: p \neq p_{i} \textsl{ (da Rest 1 bleibt) } +\end{equation} + +Insbesondere folgt daraus, dass $p \equiv 3 \mod 4$. Man erhält also die folgende Kongruenz: +\begin{equation} +\left( 2p_{1} \cdots p_{r} \right)^{2} \equiv -1 \mod p, +\end{equation} +es ist also $-1$ quadratischer Rest $\mod p$. Dies steht nun im Widersrpuch zum 1. Ergänzungssatz. + \end{enumerate} + +\newpage +\subsection*{$14$. Aufgabe} +{\texttt{(``Briefmarkenproblem'') Unter der Annahme, dass man von zwei Briefmarkensorten mit den Werten $a$ und $b$, wobei $a,b>0$ ganz und teilerfremd vorausgesetzt werden, beliebig viele Briefmarken zur Verfügung hat, zeige man, dass ab einer gewissen Schranke $s$ jede ganzzahlige Frankierung damit möglich ist. Was ist die kleinste derartige Schranke? (Hinweis: Man verwende zunächst den Chinesischen Restsatz, um zu zeigen, dass die Menge $S=\lbrace ax+by \mid 0 \leq x < b, 0 \leq y < a \rbrace$ ein volles Restsystem $\mod ab$ ist und betrachte dann die Partition $S=S_{0} \cup S_{1}$ von $S$, wobei $S_{0} = \lbrace x \in S \mid x ab$ folgt, dass $S_{0} \neq \emptyset, S_{1} \neq \emptyset$. +Sei nun $n \in \N \implies \exists k \in \N: n = k \cdot (ab) + r \land 0 \leq r < ab$. Ist $k \geq 1$, so kann man n sicher mit Hilfe des vollen Restsystems und der Abschätzung \eqref{abschS} darstellen. Die größte nicht darstellbare Zahl $\mod ab$ ist $2ab-a-b-1 \Rightarrow s = (a-1)(b-1)$. + +\newpage +\subsection*{$15$. Aufgabe} +{\texttt{Man zeige: Ist $p \in \P$ der Form $4k+3 \Rightarrow x^{2} \equiv -1 \mod p$ ist sicher nicht lösbar, ist $p$ der Form $4k+1$, so ist $x_{0} := \left( \frac{p-1}{2} \right)! \mod p$ eine Lösung. (Hinweis: Für den ersten Teil Primitivwurzel $\mod p$, und über Potenzen von g argumierentieren. Für den zweiten Teil zeige zunächst $x_{0}^{2} \equiv (p-1)! \mod p$ und zeige dann $(p-1)! \equiv -1 \mod p$). }} \newline +Sei $p \in \P$ und $p \equiv 3 \mod 4$. Nach dem Satz von Gauß existiert eine Primitivwurzel $ g \mod p$. +Weiters hat das Polynom $x^{2} - 1 = 0$ genau zwei verschiedene Lösungen in $\Z_{p}$, nämlich $\pm 1$. Nun gilt +\begin{equation} +g^{\left( \frac{p-1}{2} \right)^{2}} \equiv 1 \mod n \Rightarrow g^{\left( \frac{p-1}{2} \right)} = \pm 1 \mod p +\end{equation} +Da $g$ nach Voraussetzung die Ordnung $p-1$ hat, folgt +\begin{equation} +g^{\left( \frac{p-1}{2} \right) } \neq 1 \implies g^{\left( \frac{p-1}{2} \right)} \equiv -1 \mod p +\end{equation} +Aus +\begin{subequations} +\begin{align} +p \equiv 3 \mod 4 \Rightarrow p-1 \equiv 2 \mod 4 \\ +\Rightarrow \exists j \in \N: p-1 = 4j+2 \\ +\Rightarrow \frac{p-1}{2} = \underbrace{2j+1}_{\textsl{ungerade}} +\end{align} +\end{subequations} +Nach Satz $3.2$ kann daher $-1$ kein quadratischer Rest modulo $p$ sein. +\paragraph{} +Sei nun $p \in \P \land p \equiv 1 \mod 4$. Es gilt +\begin{subequations} +\begin{align} +p-1 \equiv -1 \mod p \\ +p-2 \equiv -2 \mod p \\ +\vdots \\ +p-\frac{p-1}{2} \equiv \frac{p+1}{2} \\ +\Rightarrow \left( \frac{p-1}{2} \right) ! \equiv (p-1)\cdot (p-2) \cdots (\frac{p+1}{2}) +\end{align} +\end{subequations} +Weiters gilt +\begin{equation} +a \in \Z_{p}^{*}: \exists! a^{-1} \in \Z_{p}^{*} : a \cdot a^{-1} \equiv 1 \mod p +\end{equation} +Die Zahlen $1$ und $p-1$ sind klarerweise selbstinvers. Die Zahlen $2,\cdots, p-2$ gilt $\vert \lbrace 2,3,\cdots, p-2 \rbrace \vert = p-3$, $p-3$ ist gerade, daher finden sich immer zwei, welche zueinander invers sind. Daher folgt +\begin{subequations} +\begin{align} +1 \cdot 2 \cdot 3 \cdots p-2 \equiv 1 \mod p \\ +1 \cdot 2 \cdot 3 \cdots p-2 \cdot p-1 \equiv p-1 \equiv -1 \mod p +\end{align} +\end{subequations} + +\newpage +\subsection*{$16$. Aufgabe} +{\texttt{Man berechne die Legendresymbole (700/769) und (1215/1381) zuerst ohne und dann mit Verwendung von Jacobisymbolen.}} \newline +Zuest ist Folgendes zu überprüfen, um von Legendre- bzw Jacobisymbolen sprechen zu können: +\begin{maxima} +primep(769); +primep(1381); +\maximaoutput* +\m \mathbf{true} \\ +\m \mathbf{true} \\ +\end{maxima} +Mit Legendresymbol: +\begin{subequations} +\begin{align} +\left( \frac{700}{769} \right)_{L} = \left( \frac{2^{2} 5^{2} 7}{769} \right)_{L} = \left( \frac{7}{769} \right)_{L} = \\ += \left( \frac{769}{7} \right)_{L} = \left( \frac{6}{7} \right)_{L} = \left( \frac{-1}{7} \right)_{L} = -1 +\end{align} +\end{subequations} +Mit Jacobisymbol: +\begin{subequations} +\begin{align} +\left( \frac{700}{769} \right)_{J} = \left( \frac{2^{2} 175}{769} \right)_{J} = \left( \frac{175}{769} \right)_{J} = \left( \frac{769}{175} \right)_{J} = \left( \frac{69}{175} \right)_{J} = \left( \frac{175}{69} \right)_{J} = \left( \frac{37}{69} \right)_{J} = \\ +=\left( \frac{69}{37} \right)_{J} = \left( \frac{32}{37} \right)_{J} = \left( \left( \frac{2}{37} \right)_{J} \right)^{5} = \left( \frac{2}{37} \right)_{J} = -1 +\end{align} +\end{subequations} +Mit Legendresymbol +\begin{subequations} +\begin{align} +\left( \frac{1215}{1381} \right)_{L} = \left( \frac{3^{5} 5}{1381} \right)_{L} = \left( \frac{3}{1381} \right)_{L}^{5} \cdot \left( \frac{5}{1381} \right)_{L} = \left( \frac{3}{1381} \right)_{L} \cdot \left( \frac{5}{1381} \right)_{L} = \\ += \left( \frac{1381}{3} \right)_{L} \cdot \left( \frac{1381}{5} \right)_{L} = \left( \frac{1}{3} \right)_{L} \cdot \left( \frac{1}{5} \right)_{L} = 1 +\end{align} +\end{subequations} +Mit Jacobisymbol: +\begin{subequations} +\begin{align} +\left( \frac{1215}{1381} \right)_{J} = \left( \frac{1381}{1215} \right)_{J} = \left( \frac{166}{1215} \right)_{J} = \left( \frac{2 \cdot 83}{1215} \right)_{J} = \left( \frac{2}{1215} \right)_{J} \cdot \left( \frac{83}{1215} \right)_{J} = \\ +\stackrel{1215 \equiv -1 \mod 8} = \left( \frac{83}{1215} \right)_{J} = \left( \frac{1215}{83} \right)_{J} = \left( \frac{53}{83} \right)_{J} = \left( \frac{83}{53} \right)_{J} = \left( \frac{30}{53} \right)_{J} = \left( \frac{2}{53} \right)_{J} \cdot \left( \frac{15}{53} \right)_{J} = \\ += - \left( \frac{53}{15} \right)_{J} = - \left( \frac{8}{15} \right)_{J} = - \left( \left( \frac{2}{15} \right)_{J} \right)^{4} = -1 +\end{align} +\end{subequations} + +\newpage +\subsection*{$17$. Aufgabe} +{\texttt{Man bestimme alle ungeraden Primzahlen $p$, für welche $10$ quadratischer Rest ist.}} \newline +Aus $10 = 2 \cdot 5$ erhält man aus Satz 3.2, (4): +\begin{equation}\label{starkeMultLegendre} +\left( \frac{10}{p} \right) = \left( \frac{2}{p} \right) \cdot \left( \frac{5}{p} \right) +\end{equation} +Daher ist $10$ genau dann quadratischer Rest, wenn beide Faktoren auf der rechten Seite von \eqref{starkeMultLegendre} gleich $1$, oder wenn beide gleich $-1$ sind. \newline +Betrachte den Fall $p = 5, p=2$ getrennt: $\left( \frac{10}{5} \right) = \left( \frac{10}{2} \right) = 0$. \newline +Sei $p \in \P \setminus \lbrace 2,5 \rbrace$: +\begin{itemize} +\item Seien beide Faktoren gleich $1$. Dann folgt aus Satz 3.5, dem 2. Ergänzungssatz, dass $p \equiv \pm 1 \mod 8$ gilt. Da $5 \equiv 1 \mod 4$, erhält man aus dem Quadratischen Reziprozitätsgesetz: +\begin{equation} +\left( \frac{5}{p} \right) = \left( \frac{p}{5} \right) +\end{equation} +Eine ungerade Primzahl $p \neq 5$ ist kongruent zu $1,2,3,4 \mod p$. Es gilt +\begin{equation} +\left( \frac{1}{5} \right) = 1, \left( \frac{2}{5} \right) = -1, \left( \frac{3}{5} \right) = -1, \left( \frac{4}{5} \right) = 1 +\end{equation} +Daher muss notwendigerweise gelten: $p \equiv \pm 1 \mod 5$. +Zusammen erhält man also, dass aus $p \equiv \pm 1 \mod 8 \land p \equiv \pm 1 \mod 5$ folgt, dass $\left( \frac{10}{p} \right) = 1$. Man erhält mit dem Chinesischen Restsatz folgendes System: +\begin{subequations} +\begin{cases} p \equiv \pm 1 \mod 40 \\ p \equiv \pm 9 \mod 40 \end{cases} +\end{subequations} +\item Seien beide Faktoren gleich $-1$. Daher ist $p \equiv \pm 3 \mod 8 \land \left( p \equiv \pm 2 \mod 5 \right)$. Man erhält daher mit dem Chinesischen Restsatz: +\begin{subequations} +\begin{cases} p \equiv \pm 3 \mod 40 \\ p \equiv \pm 13 \mod 40 \end{cases} +\end{subequations} +\end{itemize} +Weiters beachte man $\varphi(40)=\varphi(5 \cdot 8 )=4 \cdot 4 = 16$. +\newpage +\subsection*{$18$. Aufgabe} +{\texttt{Man zeige: Ist $p$ eine Primzahl, sodass auch $q=2p+1$ prim ist, so teilt $q$ entweder $2^{p}-1$ oder $2^{p}+1$ und zwar in Abhängigkeit davon, ob $2$ quadratischer Rest $\mod q$ ist oder nicht. (Für welche Mersenn'sche Zahlen $2^{p}-1$ mit $p<100$ sieht man so sofort, dass sie zusammengesetzt sind?).}} \newline +\begin{enumerate} +\item Sei $\left( \frac{2}{q} \right) = 1$, d.h. sei $2$ quadratischer Rest $\mod q$. Daher + \begin{equation} + \left( \frac{2}{q} \right) = 1 \implies \exists x \in \Z_{q}: x^{2} \equiv 2 \mod q + \end{equation} +Setzt man diese Tatsache ein, erhält man +\begin{equation} + 2^{p}-1=\left( x^{2} \right)^{p} - 1 = x^{2p} -1 +\end{equation} +Aus dem kleinen Fermat erhält man nun direkt +\begin{equation} + x^{(2p+1)-1} = x^{2p} \equiv 1 \mod 2p+1 \Rightarrow x^{2p}-1 \equiv 0 \mod q \Rightarrow q \mid 2^{p}-1 +\end{equation} +\item Sei $\left( \frac{2}{q} \right) = -1$. Aus dem Euler'schen Kriterium erhält man nun sofort unter Beachtung von $\frac{q-1}{2} = p$, dass + \begin{equation} + \underbrace{2^{\frac{q-1}{2}}}_{\equiv -1 \mod q } +1 \equiv -1 + 1 \equiv 0 \mod q \Rightarrow q \mid 2^{p}+1 + \end{equation} +\end{enumerate} +\begin{maxima} +for p:3 thru 97 step 1 do if primep(p) and primep(2*p+1) and power_mod(2,(p-1)/2,p) = 1 then ldisplay(p); +for i in [23,41,89] do ldisplay(primep(2^i-1)); +618970019642690137449562111-341550071728321; +\maximaoutput* +\t9. p=23 \\ +\t10. p=41 \\ +\t11. p=89 \\ +\m \mathbf{done} \\ +\t12. \mathrm{primep}\left(8388607\right)=\mathbf{false} \\ +\t13. \mathrm{primep}\left(2199023255551\right)=\mathbf{false} \\ +\t14. \mathrm{primep}\left(618970019642690137449562111\right)=\mathbf{true} \\ +\m \mathbf{done} \\ +\m 618970019642348587377833790 \\ +\end{maxima} +\end{document} -- 2.47.3