TITLE: $n\cdot \phi(n)=m\cdot \phi(m)$
QUESTION [11 upvotes]: If I am not mistaken here in OEIS says that  
$n\cdot \phi(n)=m\cdot \phi(m)$ is possible only if $n=m$.  
$\phi(n)$  denotes Euler's totient function.
Is there a proof of this fact?

REPLY [9 votes]: Suppose $n$ is the smallest number such that a number $m\ne n$ exists with
$$n\ \phi(n)=m\ \phi(m)$$
Let $p$ be the largest prime factor of $n$ and $q$ be the largest prime factor 
of $m$. Then, $p<q$ is impossible because $n\ \phi(n)$ would not be divisivle
by $q$. $p>q$ is impossible because $m\ \phi(m)$ would not be divisible by $p$.
So, we have $p=q$. 
The valuation of $p$ in $n\ \phi(n)$ is uniquely determined
by the valuation of $p$ in $n$, so the valuations of $p$ in $n\ \phi(n)$ and
$m\ \phi(m)$ must coincide.
Therefore we have $\frac{n}{p}\phi(\frac{n}{p})=\frac{m}{p}\phi(\frac{m}{p})$
contradicting the assumption that $n$ is the smallest number, such that
an $m$ exists with $n\ \phi(n)=m\ \phi(m)$.
Hence $n=1$, but $1=m\ \phi(m)$ has the only solution $m=1$.