Prove that there are infinitely many primes congruent to 3mod4 using euclid's proof for infinitely many prime number.
I guess I don't really know where to start because I don't understand euclid's proof for infinitely many primes. I guess I am kind of confused about the part after "thus it must be divisible by at least one of our finitely many primes.." I understand that part, but I don't understand how proving the p by pn leaves a remainder of 1 shows that p is not divisible by any of the primes. 
If the number $p$ were divisible by one of the $p_j$, then the remainder should be zero. However if you divide $p$ by any of the $p_j$,
$$p = \left( p_j \cdot \prod_{i\neq j} p_i \right) + 1$$
hence $p \equiv 1$ mod $p_j$
Make sense?
We know there is at least one prime $q \cong 3$ mod $4$; namely $3$ itself. Suppose now that there were only a finite number of such primes, call them $q_j$. What can you say about $q_1q_2\cdots q_n$? Or $(q_1q_2\cdots q_n)^2$