Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

HINT: The convolution of two multiplicative functions is multiplicative, therefore check the claim for prime powers and extend the result from there.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.