I'm asked to determine what $\varphi{(p^k)}$ is for an arbitrary prime $p$. By definition, $\varphi{(p^k)}=p^k\left(1\frac1{p}\right)=p^k\left(\frac{p1}{p}\right)=p^{k1}(p1)$. But I thought that since the Totient function was multiplicative that $\varphi{(p^k)}=\varphi{(p)}^k=(p1)^k$. 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?
Asked
Active
Viewed 157 times
1

6Multiplicative 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 03 '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 03 '13 at 12:23
1 Answers
4
The term multiplicative means (in this context) that when we have $\rm{gcd}(m,n) = 1$ we get $\varphi(mn) = \varphi(m)\varphi(n)$.
Tobias Kildetoft
 20,063
 1
 59
 88

1For the general case, see http://math.stackexchange.com/questions/114841/proofofaformulainvolvingeulerstotientfunction/114847. – lhf Nov 03 '13 at 12:32

See also http://math.stackexchange.com/questions/521233/whatfactorhastobeappliedtophiabproptophiaphibfornoncoprime. – lhf Nov 05 '13 at 15:28