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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?

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  • $\begingroup$ 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 $(\mu*\mu)(n)$ $\endgroup$ – Jérôme JEAN-CHARLES Aug 4 '18 at 16:48
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HINT: The convolution of two multiplicative functions is multiplicative, therefore check the claim for prime powers and extend the result from there.

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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 $.

As for the part that the function is nonzero for other numbers, use Tib's suggestion above.

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