# Euler Totient clarification

I'm asked to determine what $\varphi{(p^k)}$ is for an arbitrary prime $p$. By definition, $\varphi{(p^k)}=p^k\left(1-\frac1{p}\right)=p^k\left(\frac{p-1}{p}\right)=p^{k-1}(p-1)$. But I thought that since the Totient function was multiplicative that $\varphi{(p^k)}=\varphi{(p)}^k=(p-1)^k$. I've checked small values and the first formula is correct. But why if the Totient function is a multiplicative function is the second formula not correct?

• Multiplicative here means that whenever $n$ and $m$ and coprime, we have $\varphi(mn) = \varphi(m)\varphi(n)$. But it only holds when we have that extra assumption. – Tobias Kildetoft Nov 3 '13 at 12:19
• Okay. Thank you! It was about a year ago that I learned that and didn't remember the coprimality condition. – Iceman Nov 3 '13 at 12:23

## 1 Answer

The term multiplicative means (in this context) that when we have $\rm{gcd}(m,n) = 1$ we get $\varphi(mn) = \varphi(m)\varphi(n)$.