Question

Is the Möbius function completely multiplicative? Please provide full solution to answer.

Answer 1

μ(n)= 1 ; n=1

=0 ; n is not square free

=(-1)^k ; n is square-free with k prime factors.

suppose that (n,m)=1. If m or n is not squarefree, then mn is not squarefree, and μ(mn)=0 and either one of μ(n) or μ(m) is 0, so equality holds. We may assume thus that m and n are squarefree. Then so is mn, and since (m,n)=1; the prime factors of m and n are different Thus if m has k (different) prime factors and n has (different) l prime factors, mn has k+l (different) prime factors and μ(n)μ(m)=(−1)k(−1)l=(−1)k+l=μ(nm).

Proved. Thanks

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