Nonunit element in $\mathbb{Z}[i]$

Question

If $a+bi$ is not a unit of $\mathbb{Z}[i]$ prove that $a^2+b^2>1$.

Definition: Let $R$ be commutative ring with unit element. An element $u\in R$ we call unit if it's inverse $u^{-1}$ also lies in $R$.

Proof: Suppose by contradiction that $a^2+b^2\leq 1$ then $a^2+b^2\in \{0,1\}$.

If $a^2+b^2=1$ then we have four cases $a+bi\in \{\pm i, \pm 1\}$. But each of them is unit of $\mathbb{Z}[i]$. In this case we get contradiction.

If $a^2+b^2=0$ then $a+bi=0$. Here we have no contradiction, since by definition $0$ is NOT unit.

Can anyone explain this to me, please?

Answer 1

If $a+ib$ is unit so there exists $c+id\in \mathbb{Z}[i]$ such that $(a+ib)(c+id)=1$ hence $(a^2+b^2)(c^2+d^2)=1$ then $a^2+b^2=1$. Since $a,b\in \mathbb{Z}$, $(a,b)=(\pm 1,0)$ or $(a,b)=(0,\pm 1)$ thus $a+ib\in \{\pm 1,\pm i\}$. Conversely, it is clear that $\pm 1, \pm i$ are units therefore the units of $\mathbb{Z}[i]$ are $\pm 1, \pm i$.

Answer 2

In general, for a number field $K$ we have that $|Nm(\alpha)|>1$ for every (nonzero) non-unit $\alpha\in \mathcal{O}_K$, see this duplicate. Now $Nm(\alpha)=a^2+b^2$ for $K=\Bbb{Q}[i]$, and $\mathcal{O}_K=\Bbb{Z}[i]$.