TITLE: Prove there are infinitely many primes in $\mathbb{Z}[i]$
QUESTION [9 upvotes]: I saw a proof online there are infinitely many primes in $\mathbb{Z}$.  The Euler product let's us factor the harmonic series:
$$ \prod  \left( 1 - \frac{1}{p} \right) = \sum \frac{1}{n}$$
I wonder if this extends to $\mathbb{Z}[i]$.  Joan Baez Week 216 of This Week's Finds defines the zeta function of number field as the sum over non-zero ideals basically: 
$$ \zeta_{\mathbb{Z}[i]}(s) = \sum \frac{1}{(m^2 + n^2)^s} = \prod_{p \in \mathbb{Z}[i]} \left( 1 - \frac{1}{|p|^s} \right) $$
Here $2 = (1+i)(1-i)$ means that $2$ is ramified in $\mathbb{Z}[i]$.  When does this converge?
$$ \sum_{m, n \geq 1} \frac{1}{(m^2 + n^2)^s} \approx \frac{\pi}{4}\int \frac{ dr }{r^{2s-1}} < \infty$$
This zeta function converges for $s > 1$ since we are adding more numbers. So for $s = 1$
$$ \zeta_{\mathbb{Z}[i]}(1) = \sum \frac{1}{m^2 + n^2} = \prod_{p \in \mathbb{Z}[i]} \left( 1 - \frac{1}{|p|} \right) = \infty $$
We should also check that $\mathbb{Z}[i]$ has unique factorization, because it has a Euclidean algorithm

Did we just prove $\mathbb{Z}[i]$ has infinitely many primes?
If that doesn't work, we show that $\zeta(2)$ is irrational.  In fact $\zeta_{\mathbb{Z}}(2) = \frac{\pi^2}{6}$ but what about $\zeta_{\mathbb{Z}[i]}(2) $ ? 

Different ways to prove there are infinitely many primes?
GCD of gaussian integers

REPLY [2 votes]: HINT: 
Modify slightly the argument of   Euclid by taking $q= 4 p_1\cdot \ldots \cdot p_n-1$ to show that $q$ has a prime factor $\equiv -1 \mod 4$ which is also different from all the $p_i$'s. Therefore, there exist infinitely many prime numbers of form $4 k-1$. These will also be prime elements in $\mathbb{Z}[i]$ ( see Gaussian integers)