# Thread: Euler Phi primes with just definition of Phi

1. ## Euler Phi primes with just definition of Phi

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?

2. I think they are expecting you to calculate the value from the definition of $\displaystyle \phi(n)$.

How many numbers below 2p are coprime to 2?
How many numbers below 2p are coprime to p?

3. Originally Posted by SimonM
I think they are expecting you to calculate the value from the definition of $\displaystyle \phi(n)$.

How many numbers below 2p are coprime to 2?
How many numbers below 2p are coprime to p?
Hmm,
I get something like
when p != 2
How many numbers below 2p are coprime to 2? $\displaystyle 2p-2$
How many numbers below 2p are coprime to p? $\displaystyle (1/2) 2p$
with overlapping = $\displaystyle p-1$, but I don't know how to show this other than example values of p?