Euler Phi Function of Integer

Theorem
Let $n \in \Z_{>0}$, that is, a (strictly) positive integer.

Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$-function.

Then for any $n \in \Z_{>0}$, we have:


 * $\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \ldots \paren {1 - \dfrac 1 {p_r} }$

where $p_1, p_2, \ldots, p_r$ are the distinct primes dividing $n$.

Or, more compactly:
 * $\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$

where $p \divides n$ denotes the primes which divide $n$.

Proof
If $n = 1$ the result holds by inspection.

Let $n \ge 2$.

We express $n$ in its prime decomposition:


 * $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}, p_1 < p_2 < \ldots < p_r$

as it is always possible to do.

By definition, all primes are coprime to each other.

Hence from Euler Phi Function is Multiplicative:


 * $\map \phi n = \map \phi {p_1^{k_1} } \map \phi {p_2^{k_2} } \cdots \map \phi {p_r^{k_r} }$

and from Euler Phi Function of Prime Power:
 * $\displaystyle \map \phi {p^k} = p^k \paren {1 - \frac 1 p}$

So:
 * $\map \phi n = p_1^{k_1} \paren {1 - \dfrac 1 {p_1} } p_2^{k_2} \paren {1 - \frac 1 {p_2} } \cdots p_r^{k_r} \paren {1 - \dfrac 1 {p_r} }$

and the result follows directly from $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$.