Exercise 2.3.26 (Condition for $\phi(nm) = n \phi(m)$)

Question

Show that $\phi(nm) = n \phi(m)$ if every prime that divides $n$ also divides $m$.

Richard Ganaye · Accepted Answer

\begin{proof} We write the decomposition of $n$ and $m$ in prime numbers under the form
\begin{align*}
n &= p_1^{\alpha_1} \cdots p_k^{\alpha_k},\qquad \qquad \qquad (\alpha_i \geq 1),\
m &=p_1^{\beta_1} \cdots p_k^{\beta_k}q_1^{\gamma_1} \cdots q_l^{\gamma_l}, \qquad (\beta_i \geq 1, \gamma_j \geq 1).
\end{align*}
Then 
$$nm = \prod_{i=1}^k p_i^{\alpha_i + \beta_i} \prod_{j=1}^l q_j^{\gamma_j},$$
and
\begin{align*}
\phi(m) &= \prod_{i=1}^k p_i^{\beta_i -1}(p_i-1) \prod_{j=1}^l q_j^{\gamma_j - 1} (q_j-1)\
\phi(nm) &= \prod_{i=1}^k p_i^{\alpha_i + \beta_i -1}(p_i-1)  \prod_{j=1}^l q_j^{\gamma_j - 1} (q_j-1)\
&= \prod_{i=1}^k p_i^{\alpha_i} \prod_{i=1}^k p_i^{ \beta_i -1}(p_i-1)  \prod_{j=1}^l q_j^{\gamma_j - 1} (q_j-1)\
&= n \phi(m).
\end{align*}
So $\phi(nm) = n \phi(m)$ if every prime that divides $n$ also divides $m$.
\end{proof}

Exercise 2.3.26 (Condition for $\phi(nm) = n \phi(m)$)

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Exercise 2.3.26 (Condition for $\phi(nm) = n \phi(m)$)

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