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QUESTION
\begin{description}
\item[(i)]
Prove that any prime $p>3$ is either of the form $6k+1$ or of the
form $6k+5$ for some integer $k$.
\item[(ii)]
Prove that the product of any two integers of the form $6k+1$ is
of that same form.
\item[(iii)]
Adapt the proof of Theorem 2.7 to prove that there are infinitely
many primes of the form $6k+5$.
\end{description}
ANSWER
\begin{description}
\item[(i)]
By the division algorithm, any integer can be written in one of
the forms $6k,6k+1,6k+2,6k+3,6k+4,6k+5$. Of these, $6k,6k+2$ and
$6k+4$ are even, and $6k+3$ is divisible by 3. Thus none of these
can represent a prime $>3$. Thus $p$ must be of the form $6k+1$ or
$6k+5$.
\item[(ii)]
$(6k+1)(6l+1)=36kl+6k+6l+1=6(6kl+k+l)+1$.
Thus the product of two integers of the form $6n+1$ is again of
the form $6n+1$
\item[(iii)]
Suppose there are only finitely many primes of the form $6k+5$.
Let them be $p_1,\ldots p_n$, and consider $N=6(p_1\ldots p_n)-1$
Now $N=6((p_i\ldots p_n)-1)+5$, so $N$ is of the form $6k+5$. Thus
neither 2 nor 3 divides $N$, so by (i) the prime divisors of $N$
are either of the form $6k+1$ or $6k+5$. Suppose every prime
dividing $N$ is of the form $6k+1$. Then, by repeated use of (ii),
$N$ would also be of that form- but we have seen that this is not
the case.
Thus $N$ has a prime divisor, $p$ say, of form $6k+5$. By
assumption, $p_1,\ldots,p_n$ are the only such primes, so $p=p_i$
for some $i$, and in particular $p|p_1\ldots p_n$. But $p|N$, so
by theorem 1.3(4), $p|6(p_1\ldots p_n)-N$, i.e. $p|1$- but this
contradicts $p$ prime, as all primes are $>1$.
This contradiction shows our original assumption was wrong, so
there are infinitely many primes of the form $6k+5$.
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