Exercise 15.4.17

Proof. Let $n=2k+1,k\in \mathbb{Z}$ an odd integer.

$\text{∙}$

If $k$ is even, $k=2k^{\prime }$ for some $k^{\prime }\in \mathbb{Z}$. Then $n=4k^{\prime }+1\equiv 1(\mathit{mod}4)$.

Since $4=2(1+i)(1-i)\equiv 0(\mathit{mod}2(1+i))$,

$$n=4k^{\prime }+1\equiv 1={(-1)}^{(n-1)∕2}(\mathit{mod}2(1+i)).$$

$\text{∙}$

If $k$ is odd, $k=2k^{\prime }+1$ for some $k^{\prime }\in \mathbb{Z}$. Then $n=4k^{\prime }+3\equiv -1(\mathit{mod}4)$.

Since $4\equiv 0(\mathit{mod}2(1+i))$,

$$n=4k^{\prime }+3\equiv -1={(-1)}^{(n-1)∕2}(\mathit{mod}2(1+i)).$$

In both cases,

This shows that when $n=\beta$ is an odd integer, since $n=\beta \equiv i^{𝜀}(\mathit{mod}2(1+i))$ by Theorem 15.4.4, we have

This gives, for every odd integer $n\in \mathbb{Z}$,

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