[Number Theory] Proving special cases of Dirichlet's Theorem

Let's prove that there are infinitely many primes $p$ such that $p \equiv 3~ (mod~4)$.

Let $p_1, p_2, ... p_k$ be a finite list of primes congruent to 3 mod 4, and construct a new integer

$4 \times (p_1, p_2, ... p_k) -1$

that is 3 mod 4, and not divisible by any of the $p_i$'s, $( 0<i \leq k)$.

Note that $4 \times (p_1, p_2, ... p_k) -1$ is an odd integer, so it can only have prime factors that are 1 or 3 mod 4. Let's suppose that all its prime factors were 1 and 4. Then

$1\cdot 1 \cdot 1 \cdots 1=1 ~mod~ 4,$

so $4 \times (p_1, p_2, ... p_k) -1$ would also be 1 mod 4, but this is a contradiction, because our integer is 3 mod 4 by construction. So there must be at least one prime factor that is 3 mod 4. But since $4 \times (p_1, p_2, ... p_k) -1$ is not divisible by any $p_i$, which are primes congruent to 3 mod 4, we have thus found a new prime $p_{k+1} = 3~mod~ 4$ that is not in our original list $p_1, p_2, ... p_k$.

$\therefore$ there must be an infinitude of primes that are congruent to $3 ~mod ~4$.

Another special case is that there exists infinitely many primes $p$ such that $p \equiv 1~(mod~4)$. This case can be shown to be true in the same way, by constructing an integer $(p_1p_2 \cdots p_t)^2 +1$, and following the same procedure above.

In fact, Dirichlet's theorem states that for $(k,m)=1$, any coprime integers $m$ and $k$, there exists infinitely many primes $p$ such that $p \equiv m ~(mod ~k)$. His own proof, consisting of Dirichlet L-functions, is often considered to have contributed to the origins of Analytic Number Theory.