Exercise 2.3.28 (Condition for $\phi(mn) = \phi(m)$)

Proof. Using Problem 27,

where $P$ is the product of the primes common to $m$ and $n$. Therefore

If $\varphi (n)>1$, then $P>\varphi (P)$. This is impossible by definition of $\varphi$, so $\varphi (n)=1$, which is equivalent to $n=1$ or $n=2$. Here $n>1$, thus $n=2$. Therefore $P=\varphi (P)$, so $P=1$.

If $m$ is even, then $m$ and $2m$ have the common prime factor $2$, so $P>1$, in contradiction with $P=1$. Therefore $m$ is odd.

If $\varphi (m)=\varphi (\mathit{mn})$ and $n>1$, then $n=2$ and $m$ is odd. □