Möbius Function is Multiplicative

Theorem
The Moebius function $\mu$ is a multiplicative function:
 * $m \perp n \implies \mu \left({m n}\right) = \mu \left({m}\right) \mu \left({n}\right)$

where $m, n \in \Z^*_+$.

Corollary
If $\gcd \left\{{m, n}\right\} > 1$ then $\mu \left({m n}\right) = 0$.

Proof
First note that we have $\mu \left({1}\right) = 1$, which agrees with that result from Basic Properties of Multiplicative Function.

Let $m, n \in \Z^*_+$ such that $m \perp n$.


 * First, suppose that either $\mu \left({m}\right) = 0$ or $\mu \left({n}\right) = 0$.

Then either $m$ or $n$ has a factor $p^2$ where $p$ is prime.

Thus it will follow that $m n$ will also have a factor $p^2$ and hence $\mu \left({m n}\right) = 0$.

So the result holds when $\mu \left({m}\right) = 0$ or $\mu \left({n}\right) = 0$.


 * Now, suppose that $\mu \left({m}\right) \ne 0$ and $\mu \left({n}\right) \ne 0$.

Let $m = p_1 p_2 \ldots p_r$, $n = q_1 q_2 \ldots q_s$ where all the $p_i, q_j$ are prime.

So $m n = p_1 p_2 \ldots p_r q_1 q_2 \ldots q_s$.

As $m \perp n$ it follows that $\forall i, j: p_i \ne q_j$.

Hence there is no prime in $m n$ whose power is higher than $1$, which means that $\mu \left({m n}\right) \ne 0$.

So $\mu \left({m n}\right) = \left({-1}\right)^{r+s} = \left({-1}\right)^{r} \left({-1}\right)^{s} = \mu \left({m}\right) \mu \left({n}\right)$.

Hence the result.