Exercise 3.4

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}}

Show that the equation $3x^2 + 2 = y^2$ has no solution in integers.

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}
If $3x^2+2 = y^2$, then  $\overline{y}^2 = \overline{2}$ in $\Z/3\Z$.

As $\{-1,0,1\}$ is a complete set of residues modulo $3$, the squares in $\Z/3\Z$ are $\overline{0} = \overline{0}^2$ and  $\overline{1} = \overline{1}^2 = (\overline{-1})^2$, so $\overline{2}$ is not a square in $\Z/3\Z$ : $\overline{y}^2 = \overline{2}$ is impossible in $\Z/3\Z$.

Thus $3x^2+2 = y^2$ has no solution in integers.
\end{proof}

Exercise 3.4

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Exercise 3.4

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