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{\bf Review for Final Exam Math 431}
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\medskip
\noindent
{\bf Example}\;{\it There exists infiitely many primes of
the form $2^rk+1, r \in {\bf N}$.}

\medskip
\noindent
{\it Proof.}\;The case $r = 1$ means that there are
infinitely many odd primes which is certainly true. Suppose
there are finitely many primes of the form $4k+1$ then we
we may label them all as $\{p_1, ..., p_n\}$. Let $N =
(2p_1 ... p_n)^2 + 1$ and suppose that $N$ is not a prime
then there is a prime factor $p$, i.e., $N = 0$ (mod $p$). It
is clear that
$p$ is odd and $p \ne p_i$ for all $i = 1, ..., n$; indeed
we have
\begin{align*}
(2p_1 ... p_n)^2 = - 1\;\; ({\rm mod}\; p).
\tag{1}\end{align*}
Squaring yields
$
(2p_1 ... p_n)^4 = 1\;\; ({\rm mod}\; p).
$
Let $d = (4, p-1)$ be the greatest common divisor then $d =
1, 2$ or $4$. By the Corollary $(2p_1 ... p_n)^d = 1\;\;
({\rm mod}\; p)$ hence, in view of (1), $d \ne 1$ and $d \ne
2$. Thus $4$ divides $p - 1$ and so $p$ must be one of the
$p_i$. This absurdity means that there must be infinitely
many primes of the form $4k+1$.



\medskip
\noindent
{\bf Problem 1.}\;Complete the proof of the preceding
Example.

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\noindent
{\bf Problem 2.}\;The Fermats numbers are of the form $F_n =
2^{2^n} + 1$. it is easily checked that $F_n$ is prime for
$n = 1, 2, 3$ and $4$. Use Fermat's Theorem and its Corollary
to show that
$F_5$ is not a prime. (Hint: First show that if $p$ is a
prime factor of $2^{2^n} + 1$ then $2^{2^{n+1}} = 1$ (mod
$p$). Next show that the GCD of $2^{n+1}$ and $p - 1$ must
be $2^{n+1}$. This means that $p$ must be of the form
$2^{n+1} k + 1$. Set $n = 5$ and use brute force to find a
prime factor of $F_5$.)


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\noindent
{\bf Problem 3.}\;Find $n$ so that $3^n + 2^n$ is divisible
by $7$.


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\noindent
{\bf Problem 4.}\; If $m = p_1 p_2$ where $p_1$ and $p_2$
are primes. Show that
$\phi(1) + \phi(p_1) +
\phi(p_2) + \phi(p_1p_2) = m$.


\medskip
\noindent
{\bf Problem 5.}\; Extend the preceding problem to the case
where $m$ is the product of 3 primes.


\medskip
\noindent
{\bf Problem 6.}\; Find another Carmichael number.


\medskip
\noindent
{\bf Problem 7.} If $p - 1 = 6k$ how many solutions does the
equation $x^{6k} = 1$ (respectively $x^{3k} = 1, x^{3k} =
-1$) (mod $p$) have?

\medskip
\noindent
{\bf Problem 8.} Find all positive integer solutions of the equation $x^2
- 3y^2 = 1$.




 
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