Question: Are there infinitely many primes of the form 4n+3 and 4n-1?
My attempt: Suppose the contrary that there exists finitely many primes of the form 4n+3, say k+1 of them: 3,$p_1,p_2,....,p_k$
Consider $N$ = 4$p_1p_2p_3...p_k$+3 $N$ cannot be a prime of this form. So suppose that $N$=$q_1...q_r$ where $q_i∈P$
Claim: At least one of the $q_i$'s is of the form 4n+3:
Proof for my claim: $N$ is odd => $q_1,...,q_r$ are odd => $q_i$ ≡ 1 (mod 4) or $q_i# ≡ 3 (mod 4)
If all $q_1,...q_r$ are of the form 4n+1, then (4n+1)(4m+1)=16nm+4n+4m+1 = 4() +1
Therefore, $N=q_1...q_r$ = 4m+1. But $N=4p_1..p_k$+3 i.e. $N$≡3 (mod 4), N is congruent to 1 mod 4 which is a contradiction.
Therefore, at least one of $q_i$ ≡ 3 (mod 4). Suppose $q_j$ ≡ 3 (mod 4)
=> $q_j=p_i$ for some 1$\leq$i $\leq$ k or $q_j$ =3
If $q_j=p_i≠3$ then $q_j$ | $N$ = 4$p_1...p_k$ + 3 => $q_j$=3 Contradiction!
If $q_j$=3 (≠$p_i$ , 1$\leq$i $\leq$ k) then $q_j$ | $N$ = 4$p_1...p_k$ + 3 => $q_j=p_t$ for some 1$\leq$i $\leq$ k Contradiction!
In fact, there must be also infinitely many primes of the form 4n+1 (according to my search), but the above method does not work for its proof. I could not understand why it does not work. Could you please show me?