$n\cdot \phi(n)=m\cdot \phi(m)$

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?

Solution 1:

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 $m=1$.