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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?

Iceman
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    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 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

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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
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    For the general case, see http://math.stackexchange.com/questions/114841/proof-of-a-formula-involving-eulers-totient-function/114847. – lhf Nov 03 '13 at 12:32
  • See also http://math.stackexchange.com/questions/521233/what-factor-has-to-be-applied-to-phiab-propto-phia-phib-for-non-coprime. – lhf Nov 05 '13 at 15:28