Exercise 8.11

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}}

Use Gauss' theorem to find the number of solutions to $x^3+y^3 = 1$ in $\F_p$ for $p=13,19,37$, and $97$.

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}
$\bullet$ $p = 13$.

$4 	imes 13 = 52 =(-5)^2+27	imes 1^2$, where $-5 \equiv 1 \pmod 3$, so $A = -5$.

If $p=13$, $N(x^3+y^3=1) = p-2+A = 13 - 2  - 5 = 6$ : the solutions are only the trivial solutions.

\medskip

$\bullet$ $p=19$.

$4 	imes 19 = 76 = 7^2+27	imes 1^2$, where $7\equiv1 \pmod 3$, so $A = 7$.

If $p = 19$, $N(x^3+y^3 = 1) = 19 -2 + 7 = 24$.

\medskip

$\bullet$ $p=37$.

$4	imes 37 = 148 = (-11)^2 + 27 	imes 1^2$, where $-11 \equiv 1 \pmod 3$, so $A = -11$.

If $p=37$, $N(x^3+y^3 = 1) = 37 - 2 -11 = 24$.

\medskip

$\bullet$ $p=97$.

$4 	imes 97 = 388 = 19^2 + 27 	imes 1^2$, where $19 \equiv 1 \pmod 3$, so $A = 19$.

If $p=97$, $N(x^3+y^3=1) = 97 - 2 + 19 = 114$.

(These results were verified on pari/gp).)
\end{proof}

Answers