# Can someone help with a proof involving Euler's phi function?

• September 27th 2010, 03:58 PM
auitto
Can someone help with a proof involving Euler's phi function?
Using the formula for $\varphi(n),$ compute $\varphi(27),\quad \varphi(81),$ and $\varphi(p^\alpha),$ where p is a prime number. Give a proof that the formula for $\varphi(n)$ is valid when $n=p^\alpha,$ where p is a prime number.

Here is what I have so far:

$
27=3^3 ==> \varphi(27)=27\left(1-\frac{1}{3}\right)=18
$

$
81=3^4 ==> \varphi(81)=81\left(1-\frac{1}{3}\right)=54
$

$
\varphi(p^\alpha)=p^\alpha\left(1-\frac{1}{p}\right)=p^\alpha-p^{\alpha-1}=p^{\alpha-1}(p-1).
$

I am trying to find a way to go about proving what it wants me to prove, but I can not see any methods that will work. Does anyone have any suggestions?
• September 27th 2010, 10:25 PM
Traveller
Hint: $\phi (n)$ refers to the number of positive integers not greater than n, that are relatively prime to n. When n = $p^{\alpha}$ where p is a prime, the only numbers less than or equal to $p^{\alpha}$ that are NOT relatively prime to it are the ones that are divisible by p.