Möbius Function is Multiplicative/Corollary

Theorem

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

Then:

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

where $\mu$ denotes the Möbius function.

Proof

Let $\gcd \set {m, n} = k$ where $k > 1$.

Then $m = k r$ and $n = k s$ for some $r, s \in \Z$.

Thus $m n = k^2 r s$.

From Integer is Expressible as Product of Primes there exists $p \in \Z$ such that $p$ is prime and $p \divides k$.

That is:

$\exists t \in \Z: k = p t$

and so:

$m n = p^2 t^2 r s$

That is:

$p^2 \divides m $

Hence the result by definition of the Möbius function.

$\blacksquare$