I am almost done with my proof that there are infinitely many primes of the form $4k+3$, but have one remaining concern.
My chosen $N$, the subject of my contradictions, is defined as $4(p_1p_2\cdots p_n-1)+3$ where $p_1p_2\cdots p_n$ are the finite primes of the form $4k+3$ (assumed finite earlier for contradiction.)
My proof hinges on the fact that none of the $p_i$'s can divide $N$ without remainder. I claim that this is by "construction of $N$," but lack the more rigorous proof. Any tips?
The usual proof goes along the following lines: assume that $3=p_1,7=p_2,p_3,\ldots,p_m$ are some distinct primes of the form $4k+3$. The huge number $$ N = -1+4\prod_{j=1}^{m}p_j $$ is $\equiv -1\pmod{4}$, hence it must have a prime divisor $\equiv -1\pmod{4}$. However, neither $p_1,p_2,\ldots,p_{m-1}$ or $p_m$ are divisors of $N$, since they all divide $N+1$, hence there must be an extra prime number of the form $4k+3$.
A similar approach is to construct a sequence of integers $\{a_n\}_{n\geq 1}$ with the properties that $a_n\equiv -1\pmod{4}$ for every $n\geq 1$, and for every $m>n\geq 1$ $$ \gcd(a_n,a_m)=1 $$ holds. It follows that there must be at least one different prime of the form $4k+3$ for each element of the sequence. A sequence fulfilling such constraints is, for instance, $$ a_1=3,\qquad a_n = -1+4\prod_{k=1}^{n-1}a_k.$$