Exercise 15.4.7

Proof. We want to prove that, given $\beta \in \mathbb{Z}[i]$, there are polynomials $P_{\beta }(u),Q_{\beta }(u)\in \mathbb{Z}[i][u]$ such that $Q_{\beta }(0)=1$ and

and

To prove this theorem by induction for all Gaussian integers $\beta =n+\mathit{mi},n\ge 0,m\ge 0$, it is sufficient to verify that it is true for $0,1,i,i+1$, and assuming that it is true for $\beta -1,\beta$ , prove that it is true for $\beta +1$, and also assuming that it is is true for $\beta -i,\beta$, prove that is is true for $\beta +i$.

If $\beta =0$, $\phi (0z)=0$, thus $P_0(u)=0,Q_0(u)=1$ satisfy the Theorem.

If $\beta =1$, $\phi (1z)=\phi (z)$, thus $P_1(u)=1,Q_1(u)=1$ are suitable.

If $\beta =i$, $\beta$ is odd, and

where $P_i(u)=i,Q_i(u)=1$.

If $\beta =1+i$, $\beta$ is even, and by (15.36) (see Exercise 1),

where $P_{1+i}(u)=1+i,Q_{1+i}(u)=1-u$.

In these four cases, $Q_{\beta }(0)=1$.

Now suppose that the property holds for $\beta -1$ and $\beta$, $\beta \in \mathbb{Z}[i]$. The third line of (15.42) (see Exercise 6) gives,

$\text{∙}$ If $\beta$ is odd, then $\beta -1$ is even, thus

so that, with the same computing as in the odd case of Exercise 15.2.7(a),

Writing $a=\phi (z),p_{\beta }=P_{\beta }({\phi }^4(z)),q_{\beta }=Q_{\beta }({\phi }^4(z))$ and $a^{\prime }={\phi }^{\prime }(z)$, this gives

that is

where

Moreover $Q_{\beta +1}(0)=Q_{\beta -1}(0)(Q_{\beta }^2(0)+0\times P_{\beta }^2(0))=1$.

$\text{∙}$ If $\beta$ is even, $\beta -1$ is odd, thus

so that, with the same computing as in the even case of Exercise 15.2.7 (a),

With the same notations as in first case, using ${\phi }^{\prime }{(x)}^2=1-{\phi }^4(x)$,

that is

where

This gives also $Q_{n+1+i}(0)=1$.

Now consider $\phi ((\beta +i)z)$, where $\beta \in \mathbb{Z}[i]$.

$\text{∙}$ If $\beta$ is even, $\beta -i$ is odd, thus

Then, using the fourth line of (15.42),

As usual, writing $a=\phi (z),a^{\prime }={\phi }^{\prime }(z),p_{\beta }=P_{\beta }({\phi }^4(z)),q_{\beta }=Q_{\mathit{n\beta }}({\phi }^4(z))$, and using ${\phi {}^{\prime }}^2(z)=1-{\phi }^4(z)$, we obtain

so that

where

This implies $Q_{\beta +i}(0)=1$.

$\text{∙}$ If $\beta$ is odd, $\beta -i$ is even, thus

Then

With the same notations,

so that

where

We obtain $Q_{\beta +i}(0)=1$. The induction is done. □