Show that $n\equiv 3 \pmod 4$ has a prime factor $p\equiv 3\pmod 4$ I have to prove two things, but I don't know where to start. Can anyone offer some guidance?
$1.$ Let $n\in\mathbb{N}$. Show that if $n$ is congruent to $3 \pmod 4$, then $n$ has a prime factor which is congruent to $3 \pmod 4$.
$2.$ Show that there is infinitely  many prime numbers congruent to $3 \pmod 4$.
UPDATE:
I have trouble understanding the questions. Could someone provide a walkthrough to the proof of both 1) and 2)?
 A: 1) $n\equiv 3\pmod{4}$, therefore all the prime factors of $n$ are odd. For contradiction, assume all of the prime factors are equivalent to $1$ mod $4$. Then prove $n$ is also equivalent to $1$ mod $4$, which is a contradiction (note $(4k+1)(4m+1)=4(km+k+m)+1$).
2) Suppose for contradiction there are only finitely many primes of the form $4k+3$, and call the set of all of them $\{p_1,p_2,\ldots,p_n\}$. But then $$\left(p_1p_2\cdots p_n\right)^2+2\equiv 3\pmod{4}$$
by (1) has a prime factor $p_{n+1}$ equivalent to $3$ mod $4$. By definition of $\{p_1,p_2,\ldots,p_n\}$, we get $p_{n+1}\in\{p_1,p_2,\ldots, p_n\}$, thus $$p_{n+1}\mid \left(p_1p_2\cdots p_n\right)^2+2\implies p_{n+1}\mid 2,$$ contradiction.
A: Hint: For 1, prove the contrapositive: if all prime factors of an odd number $n$ are  congruent to 1 modulo 4, then so is $n$. Then extend this to even numbers.
A: For the second exercise a prime number $p>3$ can be written in the form $6n\pm 1$ therefore  a number prime can be written as $$p=6n+1$$. You can control the congruence module $4$. Indeed $$p=6n+1\equiv 2+1=3 \pmod 4$$
Edit: the thing works if $$n=4k+1$$
