The Möbius function $\mu(n)$ is defined as:
- $μ(n) = 1$ if $n$ is a square-free positive integer with an even number of prime factors.
- $μ(n) = −1$ if $n$ is a square-free positive integer with an odd number of prime factors.
- $μ(n) = 0$ if $n$ is not square-free.
We want to prove that $\lambda (n)=(\mu * \mu)(n)$ equals 0 if and only if n is divisible by some cube. The convolution is defined as $(f \, * \, g)(n) = \sum_{d|n} f(d) \, g \left( \frac{n}{d} \right)$
Anyone has an idea how to handle this?