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Thread: Prove sum of lamda d

  1. #1
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    Prove sum of lamda d

    Prove that $\displaystyle \sum_{d|n}\lambda(d)= \left\{\begin{array}{ll}1,&\mbox{ if } n=m^2 \mbox{ for some m}\\0,&\mbox{ otherwise } \end{array}\right.$

    where $\displaystyle \lambda(n)= (-1)^{\Omega(n)}$ and $\displaystyle \Omega(n)$ is the number of prime divisors of n counting multiplicity.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint : express the sum as a product taken over the prime powers which divide $\displaystyle n$. (Use the fact that Liouville's function is multiplicative.)
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