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QUESTION

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\item[(i)]
Prove that there are infinitely many primes congruent to 3 mod 8.

[Hint: Suppose $p_1,p_2\ldots ,p_n$ are the only such primes.
Consider $N=(p_1p_2\ldots p_n)^2+2$, and use the result of
question 5.]

\item[(ii)]
Prove that there are infinitely many primes congruent to 1 mod 6.

[Hint: Suppose $p_1,p_2\ldots,p_n$ are the only such primes.
Consider $(2p_1p_2\ldots p_n)^2+2$ and use the result of question
5.]

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ANSWER

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\item[(i)]
Following the hint, suppose $p_1,p_2\ldots p_n$ are the only
primes $\equiv3$ mod 8. (The list is non-empty- $p_1=3,p_2=11$,
etc.) Set $N=(p_1p_2\ldots p_n)^2+2$. Now each $p_I$ is odd, so
$N$ is odd, and hence all prime divisors of $N$ ore odd. Let $p$
be a prime divisor of $N$. Then $N\equiv0$ mod $p$, so that
$(p_1p_2\ldots p_n)^2\equiv-2$ mod $p$. This shows that $-2$ is a
square mod $p$, so that $\left(\frac{-2}{p}\right)=1$. Thus we may
use question 5(i) to deduce that $p\equiv 1$ or 3 mod 8.

Now identify $(8k+1)(8l+1)=8(8kl+k+l)+1$ shows that if every prime
dividing $N$ is $\equiv1$ mod 8, then $N$ is also $\equiv1$ mod 8.
But $N=(p_1p_2\ldots p_n)^2+2=p_1^2p_2^2\ldots p_n^2+2$, and as
each $p_i\equiv3$ mod 8, each $p_I^2\equiv9\equiv1$ mod 8. Thus
$p_1^2p_2^2\ldots p_n^2\equiv1$ mod 8, so $N\equiv1+2\equiv3$ mod
8. Thus $N$ has at least one prime divisor $p\equiv3$ mod 8. As
$p_1,p_2,\ldots ,p_n$ are the only primes $\equiv3$ mod 8, $p=p_i$
for some $i$. Thus $p|p_1p_2\ldots p_n$. But $p|N$. Hence
$p|N-(p_1p_2\ldots p_n)^2=2$. But we have already remarked that
every prime divisor of $N$ is odd, so $p\neq2$. This contradiction
shows that our original assumption was wrong, so there are
infinitely many primes congruent to 3 mod 8.
\item[(ii)]
Suppose $p_1,p_2,\ldots,p_n$ are the only primes$\equiv1$ mod 6.
(The list is non-empty, e.g. $p_1=7$) Set $N=(2p_1p_2\ldots
p_n)^2+3$. We note that as $3\not|p_i$ for each $i,\ 3\not|N$.
Also $N$ is odd, so each prime divisor of $N$ is odd.

Let $p$ be a prime divisor of $N$. Then $N\equiv0$ mod $p$, so
$(2p_1p_2\ldots p_n)^2\equiv-3$ mod $p$, and so
$\left(\frac{-3}{p}\right)=1$, Thus by question 5(ii), $p\equiv1$
mod 6. But by assumption, $p_1,p_2\ldots p_n$ are the only primes
$\equiv1$ mod 6. Thus $p=p_i$ for some $i$, and hence
$p|(2p_1p_2\ldots p_n)$. But $p|N$, so $p|N-(2p_1p_2\ldots
p_n)^2=3$. But we have already seen $3\not|N$, so $p\neq3$. This
contradiction shows that our original assumption was wrong, and so
there are infinitely many primes $\not\equiv1$ mod 6.

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