There are only finitely many elements $z \in \mathbb{Z}[i]$ with $N(z) \leq K$

Question

We know $\mathbb{Z}[i]$ is a Euclidean domain, so the question is given any integer $K > 0$, show that $N(z) \leq K$ for only finitely many $z \in \mathbb{Z}[i]$. Once I have this, I think I see how I can use the division algorithm to prove that the quoteint ring $\mathbb{Z}[i]/I$ is finite for any nonzero ideal $I$ of $\mathbb{Z}[i]$, because the Gaussian integers are a P.I.D.

Answer 1

For $z=a+bi$, the equation $N(z)\leq K$ determines the circle $a^2+b^2\leq K$, or $|z|^2\leq K$. This is a ball of radius $\sqrt{K}$ centered on the origin on the complex plane; in particular, it is bounded.

On the other hand, $\mathbb{Z}[i]\subset \mathbb{C}$ is discrete, and hence its intersection with any bounded set must be finite.