How to prove $\phi(mn) > \phi(m)\phi(n)$ if $(m,n) \ne 1$

Question

I need to prove that

$$\phi(mn) > \phi(m)\phi(n)$$

if $m$ and $n$ have a common factor greater than 1.

I have read up on the case where $m$ and $n$ are relatively prime, then $\phi(mn)=\phi(m)\phi(n)$.

Answer 1

Hint: Write the prime factorization of both $m,n$ and observe that $$ \varphi(p^k) = p^{k-1}(p-1) > \varphi(p)^k = (p-1)^{k-1}(p-1).$$

I'm sorry I'm too lazy this moment to work a whole proof. But here is a simple case. If $$ m = pq, n = ph$$ for primes $p,q,h$. Then $$\varphi(m)\varphi(n) = \varphi(p)^2 \varphi(q)\varphi(h)$$ and $$ \varphi(mn) = \varphi(p^2)\varphi(q)\varphi(h) $$ Result follows since $\varphi(p^2) > \varphi(p)^2$ as above.

Answer 2

According this recently asked question, $$ \phi(mn) = \phi(m) \phi(n) \frac{d}{\phi(d)} , $$ where $d = \gcd(m,n)$. Your question follows from the fact that $\varphi(d) < d$ whenever $d>1$.