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\title{Number-theory exercises, V}
\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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As usual, $p$ and $q$ are primes.

  \begin{ex}
    The number $32\dsp 970\dsp 563$ is the product of two primes.  Find them.
  \end{ex}

  \begin{ex}
    Factorize $1\dsp 003\dsp 207$ (the product of two primes) knowing
    \begin{equation*}
      1\dsp 835^2\equiv598^2\pmod{1\dsp 003\dsp 207}.
    \end{equation*}
  \end{ex}

  \begin{ex}
    Compute $16{200}$ \emph{modulo} $19$.
  \end{ex}

  \begin{ex}
    If $p\neq q$, and $\gcd(a,pq)=1$, and
    $n=\lcm(p-1,q-1)$, show
    \begin{equation*}
    a^n\equiv1\pmod{pq}.
    \end{equation*}
  \end{ex}

  \begin{ex}
    Show $a^{13}\equiv a\pmod{70}$.
  \end{ex}

  \begin{ex}
    Assuming $\gcd(n,p)=1$, and $0\leq n<p$, solve the congruence
    \begin{equation*}
    a^nx\equiv b\pmod p.
    \end{equation*}
  \end{ex}

  \begin{ex}
    Solve $2^{14}x\equiv 3\pmod{23}$.
  \end{ex}

  \begin{ex}
    Show $\displaystyle\sum_{k=1}^{p-1}k^p\equiv0\pmod p$.
  \end{ex}

  \begin{ex}
    We can write the congruence $2^p\equiv2\pmod p$ as
    \begin{equation*}
      2^p-1\equiv 1\pmod p.
    \end{equation*}
Show that, if $n\divides 2^p-1$, then $n\equiv 1\pmod p$.
(\emph{Suggestion:}  Do this first if $n$ is a prime $q$.  Then
$2^{q-1}\equiv1\pmod q$.  If $q\not\equiv1\pmod p$, then $\gcd(p,q-1)=1$,
so $pa+(q-1)b=1$ for some $a$ and $b$.  Now look at
$2^{pa}\cdot2^{(q-1)b}$ \emph{modulo} $n$.)
  \end{ex}

  \begin{ex}
    Let $F_n=2^{2^n}+1$.  (Then $F_0,\dots,F_4$ are primes.)  Show 
    \begin{equation*}
      2^{F_n}\equiv2\pmod{F_n}.
    \end{equation*}
  \end{ex}



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