So I was given a pretty easy problem

Calculate $\displaystyle \phi(2p)$ where $\displaystyle p$ is prime.

Which I can solve with what I have read online:

* Given $\displaystyle \phi(ab)$ if $\displaystyle (a,b)=1$, then $\displaystyle \phi(a) \phi(b)$

Then it's pretty easy:

$\displaystyle \phi(2p)$ and since$\displaystyle (2,p)=1$, then$\displaystyle \phi(2) \phi(p) = 1*(p-1) = p-1$

But the problem is, in class we haven't proved * ,so we can't use it. Pretty much all we know about$\displaystyle \phi(m)$ is that it is the number of positive integers wich are less than m and relatively prime to m. AND $\displaystyle \phi(p) = p-1 $.

I'm just wondering how I would go about doing this?

Thanks

EDIT:

After some thought and more reading I came to this:

CASE 1 : if p != 2

$\displaystyle \phi(2p) = \phi(2^1p^1)=$

$\displaystyle 2p(1-1/2)(1-1/p) = p-1$

Case 2:$\displaystyle p = 2$

$\displaystyle \phi(2*2) = \phi(4) =\phi(2^2)=$

$\displaystyle 4(1-1/2) = 2$

Am I correct in my reasoning?