Euler function $\phi(mn) = \phi(n)$ Characterize all pairs of integers $m$ and $n$ for which $\phi(mn) = \phi(n)$
I know that if $\gcd(m,n) = 1$, then $\phi(mn) = \phi(m)\phi(n)$
How can I use the fact to prove $\phi(mn) = \phi(n)$?
 A: It easy to show that for all pairs $(m,n)$ the following identity holds:
$$
\varphi(mn)=\varphi(m)\varphi(n)\frac{d}{\varphi(d)},
$$
where $d=\gcd(m,n)$.
So, it's enough to describe all pairs $(m,d)$, where $d\mid m$ and $\varphi(d)=d\varphi(m)$. Since $\varphi(d)\leqslant d$ and $1\leqslant \varphi(m)$ for all numbers $m$ and $d$, we conclude that $\varphi(m)=1$ and $\varphi(d)=d$.
Therefore, $m=1$ or $m=2$ and $d=1$. So, the answer is all pairs of the forms $(1,k)$ and $(2, 2k+1)$.
A: Let $n=p_1^{\alpha_1}\dots p_r^{\alpha _r}$. Then $\varphi(n)=n\prod\limits_{i=1}^r\frac{p_i-1}{p}$.
From here we can see $\varphi(m)\cdot \varphi(n)\leq \varphi(mn)$ with equality only when the numbers are relatively prime. (if $p|n$ and $p|m$ then the factor $\frac{p-1}{p}$ only appears once in $\varphi(nm)$ but twice in $\varphi(n)\varphi(m)$.
therefore we have: $\varphi(n)\varphi(m)\leq \varphi(nm)=\varphi(n)$ and so $\varphi(m)=1$ and $(n,m)=1$. Of course $\varphi(m)=1\iff m=1$ or $m=2$.
And hence the only pairs are: $(2k,2k),(2k+1,2k+1),(2k+1,4k+2)$
