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{\Large METU MATH 365, EXAM 2}\hspace{\stretch{1}} 
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Thursday, December 16, 2010, at 17:40

Instructor: David Pierce

\vspace{2ex}

\parbox{13cm}{\emph{Instructions:} 
  There are 4 numbered problems (with a total of 8 parts) on 4 pages.
Please work carefully.
It should be
obvious how to read your solutions.  }


\begin{problem}
  Exactly one of $1458$ and $1536$ has a primitive root.  Which one,
  and why?  Find a primitive root of the number that has one.
\end{problem}

\newpage



\begin{problem}
Remembering that $p$ is always prime,
  define the arithmetic function $\upomega$ by
  \begin{equation*}
 \upomega(n)=\sum_{p\divides n}1.
  \end{equation*}
  \begin{enumerate}
  \item
  Define $\upmu$ (the M\"obius function), preferably using $\upomega$.
  \item 
Prove that, if $m$ and $n$ are co-prime, then
$\upomega(mn)=\upomega(m)+\upomega(n)$. 
\item
Prove that
\begin{equation*}
  \sum_{d\divides n}\uptau(d)\cdot\upmu(d)=(-1)^{\upomega(n)}.
\end{equation*}
\item
Find a simple description of the function $f$ given by
\begin{equation*}
f(n)=\sum_{d\divides n}\upomega(d)\cdot\upmu\Bigl(\frac nd\Bigr).
\end{equation*}
  \end{enumerate}
\end{problem}

\newpage


\begin{problem}
  Find the least positive $x$ such that
  \begin{equation*}
    11^{5117}x\equiv5\pmod{600}.
  \end{equation*}
\end{problem}

\newpage

\begin{problem}
\mbox{}
  \begin{enumerate}
  \item
Since $2$ is a primitive root of $29$, the function $x\mapsto\log_2x$
from $\Zmodu[29]$ to $\Zmod[28]$ is defined.  Considering this as a
function from the set $\{-14,\dots,-1,1,\dots14\}$ to $\{-14,\dots,14\}$,
fill out a table like the one below.  
\begin{center}
\makebox[0pt][c]
{$
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  \begin{array}{|r||*{14}{p{5mm}|}}\hline
        m & \hfill$1$ & \hfill$2$ & \hfill$3$ & \hfill$4$ & \hfill$5$
        & \hfill$6$ & \hfill$7$ & \hfill$8$ & \hfill$9$ & \hfill$10$ &
        \hfill$11$ & \hfill$12$ & \hfill$13$ & \hfill$14$\\\hline\hline
\log_2  m & & & & & & & & & &  &  &  &  &  \\\hline
\log_2(-m)& & & & & & & & & &  &  &  &  &  \\\hline
  \end{array}
$
}
\end{center}
\item
With respect to the modulus $29$, exactly
one of the two congruences
\begin{align*}
  x^{400}&\equiv13,&x^{400}&\equiv-13
\end{align*}
has a solution.  Find all of its solutions (\emph{modulo} $29$), and
explain why the other congruence has no solutions.
  \end{enumerate}
\end{problem}




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