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\title{Number-theory exercises, VI}
\author{David Pierce}
\date{\today}

\address{Mathematics Dept\\
Middle East Tech.\ Univ.\\
Ankara 06531, Turkey}

\email{dpierce@metu.edu.tr}
\urladdr{http://www.math.metu.edu.tr/~dpierce/}


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\begin{document}
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The variables $n$, $k$, and $d$ range over the positive integers.
  \begin{ex}
Assuming $p$ is an \emph{odd} prime:
    \begin{enumerate}
      \item
$(p-1)!\equiv p-1\pmod{1+2+\dotsb+(p-1)}$;
\item
$1\cdot3\dotsm(p-2)\equiv(-1)^{(p-1)/2}\cdot(p-1)\cdot(p-3)\dotsm2\pmod
  p$;
\item
$1\cdot3\dotsm(p-2)\equiv(-1)^{(p-1)/2}\cdot2\cdot4\dotsm(p-1)\pmod
  p$;
\item
$1^2\cdot 3^2\dotsm(p-2)^2\equiv(-1)^{(p+1)/2}\pmod p$. 
    \end{enumerate}
  \end{ex}

  \begin{ex}
$\tau(n)\leq2\sqrt n$.
  \end{ex}

  \begin{ex}
$\tau(n)$ is odd if and only if $n$ is square.
  \end{ex}

  \begin{ex}
    Assuming $n$ is odd: $\sigma(n)$ is odd if and only if $n$ is
    square. 
  \end{ex}

  \begin{ex}
    $\displaystyle\sum_{d\divides n}\frac 1d=\frac{\sigma(n)}n$.
  \end{ex}

  \begin{ex}
    $\{n\colon\tau(n)=k\}$ is infinite (when $k>1$), but
    $\{n\colon\sigma(n)=k\}$ is finite.
  \end{ex}

  \begin{ex}
Let $m\in\Z$.
    The number-theoretic function $n\mapsto n^m$ is multiplicative.
  \end{ex}

  \begin{ex}
    Let $\omega(n)$ be the number of \emph{distinct} prime divisors of
    $n$, and let $m$ be a non-zero integer.  Then $n\mapsto
    m^{\omega(n)}$ is multiplicative. 
  \end{ex}

  \begin{ex}
    Let $\Lambda(n)=
    \begin{cases}
      \log p,&\text{ if $n=p^m$ for some positive $m$};\\
     0,&\text{ otherwise.}
    \end{cases}$
     \begin{enumerate}
       \item
$\log n=\displaystyle\sum_{d\divides n}\Lambda(d)$.
\item
$\Lambda(n)=\displaystyle\sum_{d\divides n}\mu\Bigl(\frac nd\Bigr)\log
  d$.
\item
$\Lambda(n)=-\displaystyle\sum_{d\divides n}\mu(d)\log d$.
     \end{enumerate}
  \end{ex}

  \begin{ex}
    Suppose $n=p_1{}^{k(1)}\dotsm p_r{}^{k(r)}$, where the $p_i$ are
    distinct. 
    \begin{enumerate}
      \item
If $f$ is multiplicative and non-zero, then $\displaystyle\sum_{d\divides
  n}\mu(d)\cdot f(d)=\prod_{i=1}^r(1-f(p_i))$;
\item
$\displaystyle\sum_{d\divides n}\mu(d)\cdot\tau(d)=(-1)^r$.
    \end{enumerate}
  \end{ex}

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