If $(a+bi)$ is a unit in $\mathbb{Z} [i]$, then $N(a+bi) = 1$?

Question

How does one prove that, if some $a+bi \in \mathbb{Z}[i]$ is a unit, then the norm of $a+bi$, $N(a+bi) = 1$ without simply checking each unit? I've tried a few different things (applying the Well-Ordering Principle considering only the norms, trying to use the fact that $(a+bi)$ has a multiplicative inverse and distributing out, etc) but I haven't gotten anywhere. Even a hint would be great!

Answer 1

Suppose $\alpha=a+bi$ is a unit. Then there exists $\beta \in \Bbb{Z}[i]$ so that $\alpha \beta=1$. Then $$N(\alpha)N(\beta)=N(\alpha\beta)=N(1)=1.$$ So, $N(\alpha)=N(\beta)\in \{\pm 1\}$. Now, $N(\alpha)=\alpha\overline{\alpha}=(a+bi)(a-bi)=a^2+b^2\ge 0$. So, $N(\alpha)=1$.

Answer 2

Hint: $N(\alpha\beta)=N(\alpha)N(\beta)$ and the fact that the only units in $\mathbb Z$ are $\pm 1$.

Answer 3

Since $a+bi$ is a unit, then $(a+bi)|1$, so $ N(a+bi)|N(1)$ and $N(a+bi)|1$, so $0<N(a+bi) = 1$