Does $\varphi(mn) = \varphi(m)\varphi(n)$ imply that $\gcd(m,n) = 1$?

Question

I know that Euler's totient function is multiplicative, in other words $\varphi(mn) = \varphi(m)\varphi(n)$ whenever $\gcd(m,n) = 1$. This is not true in general, for example $\varphi(2 \cdot 2) \neq \varphi(2)\varphi(2)$. What about the converse? Does $\varphi(mn) = \varphi(m)\varphi(n)$ imply that $m$ and $n$ are coprime? I have a feeling that this is true. With a computer search I found no small counterexamples. I was able prove a few special cases too, for example $\varphi(p^ap^b) \neq \varphi(p^a)\varphi(p^b)$ whenever $a, b > 1$. No idea how to prove this in general.

Answer 1

Yes. In general, $\varphi(mn)=\varphi(m)\varphi(n)\frac{d}{\varphi(d)}$ where $d=\gcd(m,n)$. When $d\neq 1$, this is certainly more than $\varphi(m)\varphi(n)$.

Answer 2

If $\{p_i\}$ are primes dividing $m$, and $\{q_i\}$ are primes dividing $n$, then

$\phi(m)\cdot \phi(n)=m(1-\frac{1}{p_1})\dots(1-\frac{1}{p_k})\cdot n(1-\frac{1}{q_1})\dots(1-\frac{1}{q_l}) =$

$ =m n (1-\frac{1}{p_1})\dots(1-\frac{1}{p_k})(1-\frac{1}{q_1})\dots(1-\frac{1}{q_l})$

which is equal to $\phi(mn)$ iff $\{p_i\}$ and $\{q_i\}$ are disjoint.