If $\nu(n)=$ Number of positive divisors of $n,$ $\mu$ is the Möbius function and $\sigma(n)$ is the sum of positive divisors.
show that;
$\sum\limits_{d|n} \mu(n/d)\nu(d)=1$ for all $n.$
$\sum\limits_{d|n} \mu(n/d)\sigma(d)=n$ for all $n.$
If $\nu(n)=$ Number of positive divisors of $n,$ $\mu$ is the Möbius function and $\sigma(n)$ is the sum of positive divisors.
show that;
$\sum\limits_{d|n} \mu(n/d)\nu(d)=1$ for all $n.$
$\sum\limits_{d|n} \mu(n/d)\sigma(d)=n$ for all $n.$