Exercise 1.32

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

For $\alpha = a + bi \in \Z[i]$ we defined $\lambda(\alpha) =
a^2 + b^2$. From the properties of $\lambda$ deduce the identity $(a^2+ b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
For all complex numbers $\alpha,\beta$, $\vert \alpha \beta\vert = \vert \alpha\vert \vert \beta\vert$, so
$$\lambda (\alpha \beta) = \lambda(\alpha) \lambda(\beta).$$
If $\alpha = a+bi \in \Z[i), \beta = c + d i \in \Z[i]$, then $\alpha \beta = (ac - bd) + (ad +bc)i$, thus
$$(a^2+ b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2.$$
\end{proof}

Exercise 1.32

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Exercise 1.32

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