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\title{Math 480 (Spring 2007): Homework 6}
\author{\bf Due: Monday, May 7}
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\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 $0\leq x <p$ or prove that no solutions exist:
\begin{enumerate}
\item $x^2 + 2x + 3 \con 0 \pmod{7}$, where $p=7$.
\item $x^2 - x + 7 \con 0 \pmod{11}$, where $p=11$.
\item $x^2 + x + 1 \con 0 \pmod{2}$, where $p=2$.
\item $x^2 - 3 \con 0 \pmod{389}$, where $p=389$.
\item $x^2 + x + 1 \con 0 \pmod{3}$, where $p=3$.
\item $2x^2 + 3x - 2 \con 0 \pmod{5}$, where $p=5$.

\end{enumerate}

\end{enumerate}

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