I decided to take the product of each term in the sequence and add I to the result. My main issue is that I do not know how to show that this resulting number must have a Gaussian prime factor q which has a Split primagon. If I can show this, then my idea was to show that q*q' is a prime integer within the sequence where q' is the conjugate of q.  Showing the product has a Gaussian prime factor isn't difficult, but I'm not sure how to how it's split, nor am I sure why only the split case allows for q*q' to produce a prime within the sequence.
Question: Are there infinitely many primes of the form $4n+3$ and $4n+1$?  My attempt: Suppose the contrary that there exist finitely many primes of the form $4n+3$, say $k+1$ of them: $3,p_1,p_2,....,p_k$  Consider $N = 4p_1p_2p_3...p_k+3$, $N$ 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 $\Rightarrow q_1,...,q_r$ are odd $\Rightarrow q_i \equiv 1\ (\text{mod }4)$ or $q_i ≡ 3\ (\text{mod }4)$  If all $q_1,...q_r$ are of the form $4n+1$, then $(4n+1)(4m+1)=16nm+4n+4m+1 = 4(\cdots) +1$  Therefore, $N=q_1...q_r = 4m+1$. But $N=4p_1..p_k+3$, i.e. $N≡3\ (\text{mod }4)$, $N$ is congruent to $1\ \text{mod }4$ which is a contradiction.  Therefore, at least one of $q_i \equiv 3\ (\text{mod }4)$. Suppose $q_j\equiv 3\ (\text{mod }4)$  $\Rightarrow$ $q_j=p_i$ for some $1\leq i \leq k$ or $q_j =3$  If $q_j=p_i≠3$ then $q_j$ | $N = 4p_1...p_k + 3 \Rightarrow q_j=3$ Contradiction!  If $q_j=3$ ($\neq p_i$, $1\leq i \leq k$) then $q_j | N = 4 p_1...p_k + 3 \Rightarrow q_j=p_t$ 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?  Regards
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