Exercise 5.24

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}}

If $p \equiv 1 \pmod 4$, show that $p$ is a sum of two squares, i.e. $p = a^2 + b^2$ with $a, b \in \Z$.(Hint : $p = \alpha \beta$, with $ \alpha$ and $\beta$ being non units in $\Z[i]$. Remember that $\Z[i]$ has unique factorisation.)

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} $\Z[i]$ is a principal ideal domain, thus $p$ is prime in $ \Z[i]$ iff $p$ is irreducible in $ \Z[i]$. If $p \equiv 1\pmod 4$, $p$ is not a prime by Ex.5.23, so it is not irreducible. Therefore $p = \alpha \beta, \ \alpha,\beta \in \mathbb{Z}[i]$, where $\alpha, \beta$ are not units, so that $N(\alpha)>1, N(\beta)>1$ (where $N(a+bi) = a^2 + b^2$ is the complex norm). $N(p) = p^2= N(u) N(v), 1 < N(u) < p^2$ Thus $N(u) = p$, that is $p = a^2+b^2$, where $u = a+bi$. Conclusion : if $p$ is prime in $\mathbb{N}$, $p \equiv 1\pmod 4$, then $p = a^2+b^2,\ a,b \in \Z$, $p$ is a sum of two squares. \end{proof}

Exercise 5.24

Answers

Comments

Exercise 5.24

Answers

Comments

Add answer