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We have the arithmetic function $$f(n)=\sum_{d\mid n}\mu (d)\cdot d$$ I want to show that $f\left (p_1^{e_1}\cdots p_k^{e_k}\right )=(-1)^k\cdot (p_1-1)\cdots (p_k-1)$.

We have that $d$ is of the form $p_1^{a_1}\cdots p_k^{a_k}$ with $0\leq a_i\leq e_i$ right?

So $$f\left (p_1^{e_1}\cdots p_k^{e_k}\right )=\sum_{d\mid n}\mu (d)\cdot d=\sum_{0\leq a_i\leq e_i, \forall i}\mu (p_1^{a_1}\cdots p_k^{a_k})\cdot p_1^{a_1}\cdots p_k^{a_k}$$ or not?

How could we continue?

Robert Z
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Mary Star
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    Tip: if $F(n)$ is multiplicative, then so is $\sum_{d\mid n}F(d)$ (if you don't know the proof - try to prove it yourself!) – Wojowu May 30 '18 at 14:37
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    It's been a while, but I recall that the proof for a formula involving something like $$\sum_{d\mid n}\mu(d)\frac nd$$ involved noting that the formula $$\sum_{d\mid n}\mu\left(\frac nd\right)d$$ had all the same terms. The same thing is true here, but I'm not seeing how that could be useful just yet... – abiessu May 30 '18 at 14:37

2 Answers2

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Yes, you are correct so far. Now by the multiplicative property of the Mobius function and its definition, it follows that $$\begin{align*}\sum_{0\leq a_i\leq e_i, \forall i}\mu (p_1^{a_1}\cdots p_k^{a_k})\cdot p_1^{a_1}\cdots p_k^{a_k}&= \sum_{0\leq a_i\leq e_i, \forall i}\mu (p_1^{a_1})\cdots \mu(p_k^{a_k})\cdot p_1^{a_1}\cdots p_k^{a_k}\\&= \sum_{0\leq a_i\leq 1, \forall i}(-1)^{a_1}\cdots (-1)^{a_k}\cdot p_1^{a_1}\cdots p_k^{a_k}\\&= \sum_{0\leq a_1\leq 1}(-p_1)^{a_1}\dots\sum_{0\leq a_k\leq 1}\cdots (-p_k)^{a_k}\\&=(1-p_1)\cdots(1-p_k)\\ &=(-1)^k\cdot (p_1-1)\cdots (p_k-1).\end{align*}$$

Robert Z
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  • After the first sum the upper index is $e_i$ instead of $1$, or not? – Mary Star May 30 '18 at 15:04
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    We can leave $e_i$ at the first step. Then recall that by definition, $\mu (p^{a})=0$ if $a\geq 2$. – Robert Z May 30 '18 at 15:06
  • Thank you for your answer!! I understand it!! Do you maybe have an idea also for my other question: https://math.stackexchange.com/questions/2873939/how-do-we-show-that-the-sum-is-equal-to-0 ? – Mary Star Aug 06 '18 at 14:49
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A little more general:

Theorem: Suppose $f$ is a multiplicative function. Then $$\sum_{d|n} \mu(d) f(d) = \prod_{p|n} (1-f(p)).$$

Proof: Let $$g(n) = \sum_{d|n} \mu(d) f(d).$$ Then $g$ is multiplicative (the product $\mu f$ is obviously multiplicative so the Direchlet convolution $\mu f * u$ where $u(n) = 1$ for every $n$ is multiplicative). This implies that $g$ is completely determined by its behavior on powers of primes.

That is, $$g(p^k) = \sum_{d|p^k} \mu(d) f(d) = \mu(1) f(1) + \mu(p) f(p) = 1-f(p).$$ Therefore, $$g(n) = \prod_{p|n}g(p^k) = \prod(1-f(p)).$$

Note: Try to deal with the prime powers individually rather than dealing with all of them at once. That is, try handling the case where $n = p^k$ first and then apply the multiplicative property to attain the general case.

user328442
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