How to prove $\sum_{d|n} \mu(d) \sigma(n / d) = n$? How can I show that for all $n \in \mathbb{N}$;$$\sum_{d \mid n} \mu(d)\sigma(n/d) = n$$ 
Where $\sigma$ is the divisor function and $\mu$ is the Möbius function.
I am confused because isn't $\sum_{d \mid n}\mu(d)= 1$ if $n=1$ and 0 otherwise?

The functions you are finding the convolution of are multiplicative. So their convolution is.
Thus you need only verify the equality for prime powers, and the result will follow for all $n$.
If a number $k$ is divisible by a square greater than $1$, then $\mu(k)=0$. So for prime powers the sum is easy to compute.
Remark: If $m$ is the product of $d$ distinct primes, then $\mu(m)=(-1)^d$. If $m$ is divisible by a square $\gt 1$, then $\mu(m)=0$.  For details, google the Möbius function.
