Möbius Function is Multiplicative

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Theorem

The Möbius function $\mu$ is a multiplicative function:

$m \perp n \implies \map \mu {m n} = \map \mu m \map \mu n$

where $m, n \in \Z_{>0}$.


Corollary

Let $\gcd \left\{{m, n}\right\} > 1$.

Then:

$\mu \left({m n}\right) = 0$


Proof

First note that we have $\map \mu 1 = 1$, which agrees with Value of Multiplicative Function at One.


Let $m, n \in \Z_{>0}$ such that $m \perp n$.


First, suppose that either $\map \mu m = 0$ or $\map \mu n = 0$.

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 $\map \mu {m n} = 0$.

So the result holds when $\map \mu m = 0$ or $\map \mu n = 0$.


Now suppose that $\map \mu m \ne 0$ and $\map \mu n \ne 0$.

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.

Then:

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

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 $\map \mu {m n} \ne 0$.

So:

$\map \mu {m n} = \paren {-1}^{r + s} = \paren {-1}^r \paren {-1}^s = \map \mu m \map \mu n$


Hence the result.

$\blacksquare$


Sources