# An application of Mobius Inversion $\sum_{d \mid n} \mu(\frac{n}{d})\nu(d) = 1$

Show that $\sum_{d \mid n} \mu(\frac{n}{d})\nu(d) = 1$, for any positive integer n. Where $\mu$ denotes the Mobius function defined by $\mu(n)=(-1)^{s}$ if $n=p_{1} \dotsc p_{s}$ for distinct primes $p_{1} \dotsc p_{s}$ and $\mu(n)=0$ otherwise, and $\nu(n)$ denotes the number of divisors of $n$.

I think I have to apply the Mobius Inversion Theorem somehow or use properties of the Dirichlet product, but I'm not sure. Any help would be greatly appreciated.

• Can you prove it's true for primes? prime powers? Can you prove it's a multiplicative function? – Gerry Myerson Oct 9 '16 at 10:06
• OK I have managed to prove all three things and after a bit of work the result should follow? – jackwo Oct 9 '16 at 11:53
• Well, that's the point of multiplicative functions – once you can evaluate them at prime powers, you can evaluate them everywhere. – Gerry Myerson Oct 9 '16 at 11:59
• One formula for $\nu$ is $\nu(n) = \sum_{d|n} 1$. What does Mobius inversion do to that? – B. Goddard Oct 9 '16 at 12:56
• This is simply $$\frac{1}{\zeta(s)}\times \zeta(s)^2 = \zeta(s).$$ – Marko Riedel Oct 9 '16 at 20:21

$\nu(n)= \sum_{d|n} 1$ => According to the Moebius Inversion Formula $\sum_{d|n} μ(n/d)\nu(d) = 1$