Suppose we know that $(m,n)=2$. Show that this implies that $\phi{(mn)}=2\phi{(m)}\phi{(n)}$.
My attempt:
So let $m=p_1^{r_1}p_2^{r_2}...p_k^{r_k}, n=p_1^{s_1}p_2^{s_2}...p_k^{s_k}$. Then
$(m,n)=p_1^{\min{(r_1,s_1)}}p_2^{\min{(r_2,s_2)}}...p_k^{\min{(r_k,s_k)}}=2$, So $p_1=2$ and $\min{(r_1,s_1)}=1$ and for all other $\min{(r_i,s_i)}=0$.
This means that $m$ and $n$ differ in their prime decomposition for all primes $p_j$ except $p_1=2$.
How to finish? I'm almost there I feel like and I just need the next argument...hints would be appreciated...