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?