Exercise 15.4.8

Proof. (a) By Exercise 1, $|\mathbb{Z}[i]∕2(1+i)\mathbb{Z}[i]|=N(2(1+i))=8$. We want to find a complete system of representatives of the Gaussian integers modulo $2(1+i)$.

Note first that $2i\equiv -2(\mathit{mod}2(1+i))$, and $4=(1-i)[2(1+i)]\equiv 0(\mathit{mod}2(1+i))$.

Let $z=a+\mathit{ib}$ be any Gaussian integer. The Euclidean division in $\mathbb{Z}$ gives $b=2q+r$ where $r\in \{0,1\}$. Then

where $c\in \mathbb{Z},r\in \{0,1\}$. If we write $c=4q^{\prime }+s,0\le s<4$, then

Two distinct Gaussian integers among the set $\{s+\mathit{ri}\mid s\in \{0,1,2,3\},r\in \{0,1\}\}$ are not congruent modulo $2(1+i)$, otherwise $|\mathbb{Z}[i]∕2(1+i)\mathbb{Z}[i]|<8$. Thus

since $-3\equiv 1$ and $2+i=-i+(2+2i)\equiv -i(\mathit{mod}2(1+i))$.

A coset $[\alpha ]$ , where $\alpha \in \mathbb{Z}[i]$, is invertible in $\mathbb{Z}[i]∕2(1+i)\mathbb{Z}[i]$ if and only if $\alpha$ and $2(1+i)$ are relatively prime. Since $2(1+i)=-i{(1+i)}^3$, where $-i$ is a unit, and $1+i$ is prime. Thus $\alpha$ and $2(1+i)$ are relatively prime if and only if $\alpha$ and $1+i$ are relatively prime. By Section A about Gaussian integers, this is equivalent to $\alpha$ is odd. Therefore

Since $\beta$ is odd, $\beta$ and $1+i$ are relatively prime. By the preceding reasoning, $\beta \in {\left(\mathbb{Z}[i]∕2(1+i)\mathbb{Z}[i]\right)}^{\text{∗}}$, therefore

Thus $\beta =i^{𝜀},𝜀\in \{0,1,2,3\}$. (b) By (15.44), when $\beta$ is odd,

Then

We know from Section (15.1) that $\phi \left(\frac{\varpi }{2}\right)=1$. Moreover Theorem 15.4.4 shows that $Q_{\beta }(u)=u^d{P}_{\beta }(1∕u)$. Taking $u=1$, we obtain $P(1)=Q(1)$, therefore

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