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

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?

• Yes, there is. The linked Luca & Munagi paper has it as part of Theorem 2 on page 3. – Lisa Nov 17 '15 at 23:13
• @Lisa I don't understand. The only related fact I can find in that paper is that $a(n,n) = n \phi(n)$. Is $a(n,n)$ easily shown to never take the same value twice? – A.P. Nov 18 '15 at 0:07
• $\phi(n)/n$ is injective as well. – lhf Nov 19 '15 at 12:23

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 $m=1$.

• You might consider submitting this proof to the OEIS sequence mentioned in the question! – Charles Nov 18 '15 at 0:45
• I feel very stupid, but if $p<q$, I see that $q$ cannot divide $n$. But why can't it divide $\phi(n)$? – Jason DeVito May 19 '18 at 14:29
• @JasonDeVito The prime factors of $\phi(n)$ cannot exceed the largest prime factor of $n$ – Peter May 19 '18 at 18:11