Constant to Power of Number of Distinct Prime Divisors is Multiplicative Function

Theorem
Let $c \in \R$ be a constant.

Let $f: \Z \to \Z$ denotes the mapping defined as:
 * $\forall n \in \Z: f \left({n}\right) = c^k$

where $k$ is number of distinct primes that divide $n$.

Then $f$ is multiplicative.

Proof
Let $r, s \in \Z$ such that $r \perp s$.

Let $r$ be composed of $p$ distinct primes:
 * $r_1, r_2, \ldots r_p$

Thus:
 * $f \left({r}\right) = c^p$

Let $s$ be composed of $q$ distinct primes:
 * $s_1, s_2, \ldots s_q$

Thus:
 * $f \left({s}\right) = c^q$

As $r \perp s$, all the $r_k$ and $s_k$ are distinct.

Thus $r s$ is composed of:
 * the $p$ distinct primes $r_1, r_2, \ldots r_p$

and:
 * the $q$ distinct primes $s_1, s_2, \ldots s_q$

which is a total of $p + q$ distinct primes.

Thus:

Hence the result.

Also see

 * Möbius Function is Multiplicative, a similar result.