Euler Phi Function of Prime Power

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

  • 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.28)$
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 25$
  • 1976: Tom M. Apostol: Introduction to Analytic Number Theory ... (previous) ... (next): $2.5$: A product formula for $\varphi$
  • 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.3$: Congruences
  • 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $27$
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