Euler Phi Function of Prime Power

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Theorem

Let $p^n$ be a prime power for some prime number $p > 1$.


Then:

$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$

where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function.


Corollary

When $p = 2$, the formula is exceptionally simple:

$\map \phi {2^k} = 2^{k - 1}$


Proof

From Euler Phi Function of Prime:

$\map \phi p = p - 1$


From Prime not Divisor implies Coprime:

$k \perp p^n \iff p \nmid k$

There are $p^{n - 1}$ integers $k$ such that $1 \le k \le p^n$ which are divisible by $p$:

$k \in \set {p, 2 p, 3 p, \ldots, \paren {p^{n - 1} } p}$


Therefore:

$\map \phi {p^n} = p^n - p^{n - 1} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$

$\blacksquare$


Also see


Sources