If every prime that divides $n$ also divides $m$, show that $\phi(mn)=n\phi(m)$ and $\phi(mn)=m\phi(n)$ 
If every prime that divides $n$ also divides $m$, show that $\phi(mn)=n\phi(m)$ and $\phi(mn)=m\phi(n)$.

My attempt.
As every prime that divides $n$ also divides $m$, this implies that $(m,n)=d$ where $d$ is a prime. I cannot go further.
 A: If we write $m = p_1^{a_1}p_2^{a_2}\dotsm p_k^{a_k}p_{k+1}^{a_{k+1}}p_{k+2}^{a_{k+2}}\dotsm p_{k+\ell}^{a_{k+\ell}}$, then $n = p_1^{b_1}p_2^{b_2}\dotsm p_k^{b_k}$: each rpime that divides $n$ also divides $m$, but $m$ may have more factors (in $p_{k+1}^{a_{k+1}}p_{k+2}^{a_{k+2}}\dotsm p_{k+\ell}^{a_{k+\ell}}$). We know that
$$\phi(k) = k\prod_{p \mid\ k} \left(1 - \frac{1}{p}\right)$$
for prime $p$ (see the Wolfram MathWorld article). So we can write
$$\phi(mn) = mn\left(1-\frac{1}{p_1}\right)\left(1- \frac{1}{p_2}\right)\dotsm\left(1- \frac{1}{p_k}\right)\left(1- \frac{1}{p_{k+1}}\right)\left(1- \frac{1}{p_{k+2}}\right)\dotsm \left(1- \frac{1}{p_{k+\ell}}\right)$$
But then, note that
$$\phi(m) = m\left(1-\frac{1}{p_1}\right)\left(1- \frac{1}{p_2}\right)\dotsm\left(1- \frac{1}{p_k}\right)\left(1- \frac{1}{p_{k+1}}\right)\left(1- \frac{1}{p_{k+2}}\right)\dotsm \left(1- \frac{1}{p_{k+\ell}}\right)$$
and thus we get:
$$\phi(mn) = n\phi(m)$$
A: Since $\phi$ is a multiplicative function, it suffices to check for $n=p^j$, and $m =p^k$ for some fixed prime $p$.
$\phi(p^k) = p^k-p^{k-1}$, and 
$\phi(nm)=\phi(p^{k+j}) =p^{k+j}-p^{k+j-1} = p^j(p^k-p^{k-1}) = n\phi(m)$.
This argument is symmetric in $n$ and $m$, so $\phi(mn) = m\phi(n)$ as well. Multiplicativity gives that this is true in general.
