Exercise 2.19

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}}

Prove that $\phi(nm)\phi((n, m)) = (n, m)\phi(n)\phi(m)$.

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}
With the  notations of Ex. 2.18,
\begin{align*}
\phi(nm) &= \prod_{i=1}^r p_i^{\alpha_i + \beta_i} \left( 1- \frac{1}{p_i}ight) \prod_{i=1}^s q_i^{\lambda_i} \left( 1- \frac{1}{q_i}ight) \prod_{i=1}^t s_i^{\mu_i} \left( 1- \frac{1}{s_i}ight)\
\phi(n\wedge m) &= \prod_{i=1}^r p_i^{\gamma_i} \left( 1- \frac{1}{p_i}ight)
\end{align*}
so
\begin{align*}
(n\wedge m) \phi(n)\phi(m) &= \prod_{i=1}^r p_i^{\gamma_i} \prod_{i=1}^r \left [ p_i^{\alpha_i + \beta_i} \left( 1- \frac{1}{p_i}ight)^2 ight] \prod_{i=1}^s q_i^{\lambda_i} \left( 1- \frac{1}{q_i}ight) \prod_{i=1}^t s_i^{\mu_i} \left( 1- \frac{1}{s_i}ight)\
&=  \prod_{i=1}^r \left [ p_i^{\alpha_i + \beta_i + \gamma_i} \left( 1- \frac{1}{p_i}ight)^2 ight] \prod_{i=1}^s q_i^{\lambda_i} \left( 1- \frac{1}{q_i}ight) \prod_{i=1}^t s_i^{\mu_i} \left( 1- \frac{1}{s_i}ight)\
&=\phi(nm)\phi(n\wedge m).
\end{align*}
Conclusion: 
$$(n\wedge m) \phi(n)\phi(m) = \phi(nm)\phi(n\wedge m).$$
\end{proof}

Exercise 2.19

Answers

Comments

Exercise 2.19

Answers

Comments

Add answer