Euler Phi Function of Product with Prime

Theorem
Let $p$ be prime and $n \in \Z: n \ge 1$.

Then:
 * $\map \phi {p n} = \begin{cases}

\paren {p - 1} \map \phi n & : p \nmid n \\ p \map \phi n & : p \divides n \end{cases}$ where:
 * $\map \phi n$ denotes the Euler $\phi$ function of $n$
 * $\divides$ denotes divisibility.

Thus for all $n \ge 1$ and for any prime $p$, we have that $\map \phi n$ divides $\map \phi {p n}$.

Proof
First suppose that $p \nmid n$.

Then by Prime not Divisor implies Coprime, $p \perp n$.

So by Euler Phi Function is Multiplicative, $\map \phi {p n} = \map \phi p \map \phi n$.

It follows from Euler Phi Function of Prime that $\map \phi {p n} = \paren {p - 1} \map \phi n$.

Now suppose that $p \divides n$.

Then $n = p^k m$ for some $k, m \in \Z: k, m \ge 1$ such that $p \perp m$.

Then:

At the same time: