Exercise 2.3.27 (Comparison between $\phi(mn)$ and $\phi(m)\phi(n)$)

Question

If $P$ denotes the product of primes common to $m$ and $n$, prove that $\phi(mn) = P \phi(m)\phi(n)/\phi(P)$. Hence if $(m,n) >1$, prove $\phi(mn)> \phi(m)\phi(n)$.

Richard Ganaye · Accepted Answer

\begin{proof}
\begin{enumerate}
\item[a)] If $p_1,p_2,\ldots,p_k$ are the prime numbers common to $m,n$, we write the decompositions of $m,n$ in prime numbers under the form
\begin{align*}
m &= p_1^{\alpha_1} \cdots p_k^{\beta_k}q_1^{\gamma_1} \cdots q_l^{\gamma_l},\
n &=p_1^{\beta_1} \cdots p_k^{\beta_k}r_1^{\delta_1} \cdots r_s^{\delta_s}.
\end{align*}
so
$$nm = \prod_{i=1}^k p_i^{\alpha_i + \beta_i}  \prod_{j=1}^l q_j^{\beta_j} \prod_{k=1}^r r_k^{\beta_k}.
$$
Here
$$P = p_1p_2\cdots p_k$$
($P = 1$ if $k = 0$, i.e. if the product $\prod_{i=1}^k p_i$ is empty).

Then
\begin{align*}
\phi(mn) &= \prod_{i=1}^k p_i^{\alpha_i + \beta_i - 1} (p_i-1) \prod_{j=1}^l q_j^{\beta_j -1}(q_j - 1) \prod_{k=1}^r r_k^{\beta_k - 1}(r_k - 1)\
\phi(m)\phi(n) &= \prod_{i=1}^k p_i^{\alpha_i  - 1} (p_i-1) \prod_{j=1}^l q_j^{\beta_j -1}(q_j - 1) \prod_{i=1}^k p_i^{\beta_i  - 1} (p_i-1)\prod_{k=1}^r r_k^{\beta_k - 1}(r_k - 1)\
&= \prod_{i=1}^k p_i^{\alpha_i  + \beta_i - 2} (p_i-1)^2 \prod_{j=1}^l q_j^{\beta_j -1}(q_j - 1) \prod_{k=1}^r r_k^{\beta_k - 1}(r_k - 1)\
\end{align*}
Therefore
\begin{align*}
\frac{\phi(nm)}{\phi(n)\phi(m)} =& \frac{\prod_{i=1}^k p_i}{\prod_{i=1}^k (p_i-1)}\
&= \frac{P}{\phi(P)}.
\end{align*}
So
$$\phi(mn) = \frac{P}{\phi(P)}\,  \phi(m)\phi(n).$$

\item[b)] If $m\wedge n >1$, then the product $P$ is not empty, thus $P>1$, and then $\phi(P) < P$. Therefore
$$\phi(mn) = \frac{P}{\phi(P)}\,  \phi(m)\phi(n) > \phi(m)\phi(n).$$

\end{enumerate}
\end{proof}

Exercise 2.3.27 (Comparison between $\phi(mn)$ and $\phi(m)\phi(n)$)

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Exercise 2.3.27 (Comparison between $\phi(mn)$ and $\phi(m)\phi(n)$)

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