Proof that there are infinitely many primes congruent to 3 modulo 4 I'm having difficult proving this.
As a hint the exercise to prove first, that if $a\lneqq \pm 1$ satisfies $a \equiv 3 \pmod4$, then exist $p$ prime, $p \equiv 3 \pmod 4$ such $p\mid4$. But I'm not really getting for what purpose can this be used.
 A: 
Lemma. If $a\equiv 3 \pmod 4$ then there exists a prime $p$ such that $p\mid a$ and $p\equiv 3 \pmod 4$.

Proof. Clearly, all primes dividing $a$ are odd. Suppose all of them would be $\equiv 1 \pmod 4$. Then their product would also be $a\equiv 1\pmod4$, which is a contradiction. $\hspace{3cm}\square$

There exist infinitely many primes $p$ such that $p\equiv 3\pmod 4$.

Suppose that $p_1,\dots,p_n$ would be all such primes. (In particular, we have $p_1=3$.) Consider $a=4p_2\cdots p_n+3$. (Or you can take $a=4p_2\cdots p_n-1$.) Show that $p_i\nmid a$ for $i=1,\dots,n$. (The case $3\nmid a$ is solved differently than the other primes - this is the reason for omitting $p_1$ in the definition of $a$.) Then use the above lemma to get a contradiction. 
A: Use Euclid's proof showing that there are infinitely many primes, i.e., find an Euclidean polynomial you can use for your arithmetic progression $l \mod k$. Since $l^2\equiv 1 \mod k$ such an Euclidean polynomial exists - see http://www.mast.queensu.ca/~murty/murty-thain2.pdf how to do it (in particular, on page one, the case $4n+3$ is given, see [5]). For $8n+1$ see Infinitely many primes of the form $8n+1$.
A: Show that there are infinitely many primes that are congruent to 3 mod 4. (Hint: Use that $4\mid(p_1p_2\cdots p_r + 3)$.
Solution: Suppose there are finitely many primes p congruent to 3 mod 4 and denote them by (noting that 3 is one of them) $3, p_1, p_2, p_3,\dotsc, p_r$.
Now consider $A = 4p_1p_2\cdots p_r+5$. If $A$ is prime, then we are done since it is clearly congruent to 3 mod 4 and it is not one of the $p_i$ (as it is greater than all of them). If not, then $A$ has a prime factorization:
$$A = q_1q_2\cdots q_s.$$
Now observe that any prime greater than 3 has to be congruent to 1 or 3 mod 4. This is because, if it were congruent to 0, 2 or 3 mod 4, it would be even, and if it were congruent to 3 mod 4, then it would have to be 3 (only numbers congruent to 3 mod 6 are multiples of 3). Thus each of the $q_i$ has to be congruent to 1 or 3 modulo 4.
However, not all the $q_i$ can be congruent to 1 mod 4. If they were, then their product, which is $A$, could also have to be congruent to 1 mod 4, and this is not possible by construction (note that it would
be possible if we had included 3 as one of the factors in the product $4p_1p_2\cdots p_r$).
Thus there exists at least one $q_i$ which is congruent to 3 mod 4. However, this cannot be one of the
primes $3, p_1, p_2, p_3,\dotsc,p_r$. If it were, then $q_i$ would divide $A$ (since it is in the prime facorization of $A$) and
it would divide $4p_1p_2\cdots p_r$ (since it is one of the primes in that product), and it thus has to divide their difference, which is 3. Thus $q_i$ has to be 3, but that’s a contradiction since 3 does not divide $4p_1p_2\cdots p_r$.
A: Assume from the sake of contradiction that there exist only finitely many primes. 
Denote them as $P = \{p_1, p_2,\cdots, p_k\}$
Let $A = 4\cdot p_1\cdot p_2\cdots p_k - 1$
Then, we know that $A\equiv -1 \equiv 3 \pmod{4}$. 
Since $\exists p\in P$ such that $p\mid A$
We know $A = k\cdot p_i$, where $k\in\mathbb{Z}\wedge p_i\in P$
Hence, $A = p_i\cdot (4\cdot p_1\cdot p_2\cdots p_{i-1}\cdot p_{i+1}\cdots p_k) - 1 = k\cdot p_i$
This indicates that $p_i\mid 1$
However, by property of primes, $p_i$ should be greater than 1. 
Hence, a contradiction. 
A: If there are only finitely many primes $\equiv 3 \pmod 4$, take the product of them and denote that product by $a$. Now look at $2a + 1$, and try to deduce a contradiction.
A: Does this work?
This is a modification of Euclid’s proof. Let $p_1,p_2,…,p_k$ be the only 3 mod 4 primes. Consider two cases :
(I) $k$ is odd
Then notice $p_1p_2…p_k+4$ is a 3 mod 4 number, and in particular not divisible by any of the $p_i$s we have. Then it must be a product of some 1 mod 4 primes for it to be not a 3 mod 4 prime itself. But product of 1 mod 4 primes is always 1 mod 4, so not possible.
(II) $k$ is even
This is slightly trickier. Consider $p_1p_2…p_{k-1}+2^x$ for some $x \geq 2$. Again, this is 3 mod 4. I claim we can choose $x$ so that this is not divisible by $p_k$. Suppose not, then $2^{a+1} \equiv{2^{a}} \pmod{p_k}$,but theb 2 is congruent 1 mod p so 1 is divisible by p_k, which is absurd.
