Exercise 2.2.18

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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{\U}{\mathbb{U}}

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

ewcommand{\im}{\,\mathrm{Im}\,}

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

ewcommand{\Gal}{\mathrm{Gal}}

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

Suppose that complex numbers $\alpha,\beta,\gamma$ satisfy the equations
\begin{align*}
\alpha+\beta+\gamma &= 3,\
\alpha^2+\beta^2+\gamma^2 &= 5,\
\alpha^3+\beta^3+\gamma^3 &=12.
\end{align*}
Show that $\alpha^n+\beta^n+\gamma^n \in \Z$ for all $n\geq 4$. Also compute $\alpha^4+\beta^4+\gamma^4$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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{\U}{\mathbb{U}}

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

ewcommand{\im}{\,\mathrm{Im}\,}

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

ewcommand{\Gal}{\mathrm{Gal}}

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

\begin{proof}
$\alpha,\beta,\gamma$ are the root of $$p = (x-\alpha)(x-\beta)(x-\gamma) = x^3-\sigma_1x^2+\sigma_2x-\sigma_3.$$
(We write $\sigma_i$ in place of $\sigma_i(\alpha,\beta,\gamma)$.)

By Exercise 17, with $n=3$ :
$$
\left\{
\begin{array}{cccl}
 3 &  = & s_1 &=\sigma_1  \
  5&  = &  s_2 &= \sigma_1^2 - 2 \sigma_2 \
  12&=   &   s_3& = \sigma_1^3-3\sigma_1 \sigma_2 + 3 \sigma_3.
\end{array}
ight.
$$
Thus $\sigma_1 = 3, \sigma_2 = \frac{1}{2}(\sigma_1^2-5) = \frac{1}{2}(9-5) = 2$.

$\sigma_3 = \frac{1}{3}(12 - \sigma_1^3+3\sigma_1\sigma_2) = 4 +\sigma_1\sigma_2 -\frac{\sigma_1^3}{3} = 4 + 6 - 9 = 1$.

$\alpha,\beta,\gamma$ are the roots of $p = x^3-3x^2+2x-1$.

If $n \geq 4$, $\alpha^n = 3 \alpha^{n-1} - 2 \alpha^{n-2}+\alpha^{n-3}$, and similar equations for $\beta,\gamma$. Summing these equations, we obtain
\begin{align}
s_n = 3 s_{n-1} -2 s_{n-2}+s_{n-3}. \label{eq2.2.18:1}
\end{align}
(This is a particular case of Newton identities (2.22).)

$s_0 = 3, s_1,s_2,s_3$ are in $\mathbb{Z}$.  If we suppose that $s_k \in \mathbb{Z}$ for all $k, 1\leq k<n$, then \eqref{eq2.2.18:1} show that $s_n \in \mathbb{Z}$, and the induction is done.

$$\forall k \in \mathbb{N}, \ s_n \in \mathbb{Z}.$$

In particular, $s_4 = 3 s_3 - 2 s_2+3 s_1= 3	imes 12- 2 	imes 5 + 3 	imes 3 = 35$.
\end{proof}

Exercise 2.2.18

Answers

Comments

Exercise 2.2.18

Answers

Comments

Add answer