Let k be the number of prime factors other than a positive integer n. Prove that $$\sum_{d|n}{|\mu(d)|} = 2^{k}$$ I'm not sure how to approach this problem. Can anyone give me a hint about how to start/approach this proof?
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2If $n = \prod_{j=1}^k p_j$ then what are its divisors ? Does it change something when $n= \prod_{j=1}^k p_j^{e_j}$ – reuns Jan 16 '20 at 02:15
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... where $k$ is the number of distinct primes that divide $n$ ... & $\mu$ is the Mobius function ... How far have you got ? – Donald Splutterwit Jan 16 '20 at 02:15
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If $a(n)$ is a multiplicative function then so is $\sum_{d|n}a(d)$. – Somos Jan 16 '20 at 02:48
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I'm sorry, I've tried and I can't understand ... – R.R Jan 16 '20 at 02:51
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$\mu(m)=1$ if $m$ is square-free and has an even number of prime factors, and $-1$ if $m$ is square-free and has an odd number of prime factors. Otherwise it is 0.
So suppose $n$ has exactly $k$ prime factors. The only divisors $d$ of $n$ which have $\mu(d)\ne0$ are those which are products of distinct prime factors of $n$. There are exactly $2^k$ such factors, because each prime factor can either divide $d$ or not. For example, if $n=2\cdot3\cdot5$, then there are 8 divisors with $\mu(d)\ne0$, namely 1, 2, 3, 5, 6, 10, 15, 30. If $\mu(d)\ne0$ then $|\mu(d)|=1$, so $$\sum_{d|n}|\mu(d)|=2^k$$
almagest
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