Möbius Function is Multiplicative

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

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.