Exercise 3.23

Proof. If $a,b,c$ are in $\mathbb{Z}[i]$ we say that $a\equiv b(\mathit{mod}c)$ if there exists $q\in \mathbb{Z}[i]$ such that $a-b=\mathit{qc}$.

As $i\equiv -1(\mathit{mod}1+i),a+\mathit{bi}\equiv a-b(mod1+i)$.

$(1-i)(1+i)=2$, so $2\equiv 0(\mathit{mod}1+i)$.

If $a-b$ is even, $a-b=2k,k\in \mathbb{Z}\subset \mathbb{Z}[i]$, so $a-b\equiv 0(\mathit{mod}1+i)$.

If $a-b$ is odd, $a-b=2k+1,k\in \mathbb{Z}$, so $a-b\equiv 1(\mathit{mod}1+i)$.

Conclusion : for all $z\in \mathbb{Z}[i]$, $z\equiv 0,1(\mathit{mod}1+i)$. □