# Convolution of the Möbius function with itself

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?

• There an exact formula : Let $K$ the number of prime dividing $n$ and whose square does not divide $n$ . Then $(\mu*\mu)(n)=(-2)^K$. The squares can be ignored and the cubes or any higher powers nullify $(\mu*\mu)(n)$ – Jérôme JEAN-CHARLES Aug 4 '18 at 16:48

The easy part is proving that if n is divisible by a cube (greater than 1) then $(f * g)(n) = 0$. Indeed, in this case n is also divisible by the cube of a prime, say $p^3$ (namely the cube of a prime factor of that cube divisor). In that case in the sum that defines $(f * g)(n)$, each term is zero because either d or $n/d$ is divisible by $p^2$.