Prove using Gaussian primes that there are infinitely many primes numbers in the arithmetic progression $1, 5, 9, 13, 17, 21, ...$
Hint: If not, multiply them together and add $i$; the result has a Gaussian prime factor $q$ of type $(S)$. Demonstrate that $q\bar{q}$ is a "new" prime.
I know that $(S)$ means split which is when $q$ is a prime in ordinary integers and is not in Gaussian and $p= q\bar{q}$, where $\bar{q}$ is the complex conjugate. I also know that despite Chebyshev's bias, a prime number asymptotically has an equal chance of being split or inert. I also know that 13 is split. Overall, I am very confused on how to prove this with Gaussian integers so any help is appreciated.
As suggested in the hint, suppose we have a finite list of prime numbers $p_k$ with $p_k\equiv 1\mod{4}$. Then the number $N=\prod p_k + i$ can be written as a product of Gaussian primes. We know that Gaussian primes are equal $a+bi$ where $a^2+b^2$ is a prime number not equivalent to $3\mod{4}$, or equal to a prime number equivalent to $3\mod{4}$. Note that the norm of $N$ is equivalent to $2\mod{4}$. From this follows that $N$ is a multiple of $1+i$, but not a multiple of $(1+i)^2$.
Since $N/(1+i)$ is not a totally real or a totally imaginary number, it follows that there is at least one Gaussian prime factor $a+bi$ where $a^2+b^2$ is a prime number not equivalent to $3\mod{4}$. Since 1+i is not a factor of $N/(1+i)$, it follows that $a^2+b^2\equiv 1\mod{4} $, which gives the “new“ prime.