Exercise 15.5.8

Proof.

By Theorem 15.3.2, the zeros of $\phi$ occur at $z=(m+\mathit{in})\varpi$ for $m,n\in \mathbb{Z}$. If $\beta =1$ then $\phi (\varpi )=0$, so we assume that $n\ge 2$.

If $N(\beta )=1$, then $\beta \in \{1,i,-1,-i\}$, thus $\frac{\varpi }{\beta }\in \{\varpi ,-\mathit{i\varpi },-\varpi ,\mathit{i\varpi }\}$ so that

We must assume $N(\beta )\ge 2$ to obtain the conclusion.

If we assume that $\phi \left(\frac{\varpi }{\beta }\right)=0$ with $N(\beta )\ge 2$, the same Theorem 15.3.2 shows that

Then $1=(m+\mathit{in})\beta$, where $\beta \in \mathbb{Z}[i]$ . This shows that $\beta$ is invertible in $\mathbb{Z}[i]$, and $1=N(1)=N(m+\mathit{in})N(\beta )$ shows that $N(\beta )\mid 1$, where $N(\beta )>0$, thus $N(\beta )=1$, which contradicts our hypothesis $N(\beta )\ge 2$.

We have proved

This remains true when $\beta$ is even. □