\documentclass{article} \voffset=-0.05\textheight \textheight=1.1\textheight \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \title{Math 480 (Spring 2007): Homework 6} \author{\bf Due: Monday, May 7} \date{} \include{macros} \begin{document} \maketitle \noindent{\bf There are 5 problems.} Each problem is worth 6 points and parts of multipart problems are worth equal amounts. You may work with other people and use a computer, unless otherwise stated. Acknowledge those who help you.\\ \begin{enumerate} \item Calculate the following Legendre symbols by hand (you may use the quadratic reciprocity law): $\kr{4}{10007}, \kr{3}{37}, \kr{5}{11}, \kr{5!}{7}$. \item \label{ex:powering} Let $G$ be an abelian group and let $n$ be a positive integer. \begin{enumerate} \item Prove that the map $\vphi:G\to G$ given by $\vphi(x) = x^n$ is a group homomorphism. \item Prove that the subset $H$ of $G$ of squares of elements of $G$ is a subgroup. \end{enumerate} \item Give an example of an abelian group $G$ and two distinct subgroups $H$ and $K$ both of index $2$. Note that $G$ will not be cyclic. \item\label{ex:rec14}(*) Prove that for any $n\in\Z$ the integer $n^2+n+1$ does not have any divisors of the form $6k-1$. (Hint: First reduce to the case that $6k-1$ is prime, by using that if $p$ and $q$ are primes not of the form $6k-1$, then neither is their product. If $p=6k-1$ divides $n^2+n+1$, it divides $4n^2+4n+4 = (2n+1)^2+3$, so $-3$ is a quadratic residue modulo~$p$. Now use quadratic reciprocity to show that $-3$ is not a quadratic residue modulo~$p$.) \item For each of the following equations, either find all integer solutions with $0\leq x