Claim: Show that any Gaussian integer is a unit or can be written as a finite product of primes.
Proof attempt: Let $x=a+ib$ be a Gaussian Integer with $a,b\in \mathbb{Z}.$ If $x=0$ then I am not sure how to argue. But here is the argument for $N(x)\geq 1.$
Let $P(k)$ be the proposition that any Gaussian integer $x$ with $N(x)=k$ is either a unit or can be written as a finite product of primes, where $k\geq 1.$
Base case: If $N(x)=1$ then $x$ is a unit and so $P(1)$ holds.
Induction Step: Assume that $P(n)$ holds for some $n\geq 1.$ We will show that $P(n+1)$ holds using $P(n).$ Let $x$ be a Gaussian Integer with $N(x)=n+1$ then $x$ is not a unit. If $x$ is a Gaussian prime then $x=x*1$ otherwise assume that $x$ is not a Gaussian prime. Then since it is not a unit it must the case that there exists $y,z\in \mathbb{Z}[i]$ such that $x=y*z$ with $y$ and $z$ both not being units. Then $N(y)N(z)=N(x)=n+1.$ Thus $N(y)\leq n$ and $N(z)\leq n.$ And so by the induction step we have that $y$ and $z$ can be written as a finite product of primes.
Here $N(a+ib)=a^2+b^2$ where $a,b\in \mathbb{Z}.$
I have two questions: How do I make the proof work when the norm $N(x)=0$. And is this proof correct?
Every non-zero element can be written as a finite product of irreducible elements, in rings of integers like the Gaussian integers:
$x\in\mathcal{O}_K$ can be written as product of irreducible elements
The induction is explained there, too.
Furthermore, $\mathbb{Z}[i]$ is a PID so that any irreducible element is prime.