# How to prove that $n\sum_{d\mid n}\frac{|\mu(d)|}{d}=\sum_{d^2\mid n}\mu(d)\sigma\left(\frac{n}{d^2}\right)$?

This is problem 11 part b in chapter 3 of Tom M. Apostol's "Introduction to Analytic Number Theory". A variation on Euler's totient function is defined as $$\varphi_1(n) = n \sum_{d \mid n} \frac{|\mu(d)|}{d}$$ The question asks to show that $$\varphi_1(n) = \sum_{d^2 \mid n} \mu(d) \sigma\left( \frac{n}{d^2} \right)$$ My attempt so far: I have proved in part (a) of the same question that $$\varphi_1(n) = n \prod_{p \mid n}\left(1 + \frac{1}{p} \right)$$ And so in an attempt to equate these two expressions I write \begin{eqnarray} \varphi_1(n) &=& n \prod_{p \mid n}\frac{p + 1}{p} \\ &=& n \left(\prod_{\substack{p \mid n \\ p^2 \mid n}}\frac{p + 1}{p}\right)\left(\prod_{\substack{p \mid n \\ p^2 \nmid n}}\frac{p + 1}{p}\right) \end{eqnarray} Let $s = \prod_{\substack{p \mid n \\ p^2 \mid n}} p$ and $r = \prod_{\substack{p \mid n \\ p^2 \nmid n}} p$. Then \begin{eqnarray} \varphi_1(n) &=& n \frac{\sigma(s)}{s} \frac{\sigma(r)}{r} \\ &=& \sigma(s) \sigma(r) \frac{n}{sr} \end{eqnarray} Using $N = \mu * \sigma$ where $*$ is the Dirichlet convolution and $N(n) = n$ we obtain \begin{eqnarray} \varphi_1(n) &=& \sigma(s) \sigma(r) \sum_{d \mid \frac{n}{sr}}\mu(d) \sigma\left( \frac{n}{srd} \right) \end{eqnarray} Because $(r, \frac{n}{srd}) = 1$ we can simplify the expression: \begin{eqnarray} \varphi_1(n) &=& \sigma(s) \sum_{d \mid \frac{n}{sr}}\mu(d) \sigma\left( \frac{n}{sd} \right) \end{eqnarray} We know that $\frac{n}{r}$ is square, and that all $d$ that contribute (a non zero value) to the above sum are squarefree, because otherwise $\mu(d) = 0$, and so the sum over $d$ such that $d \mid \frac{n}{sr}$ is the same as the sum over $d$ such that $d^2 \mid n$. This brings us to where I am stuck: \begin{eqnarray} \varphi_1(n) &=& \sigma(s) \sum_{d^2 \mid n}\mu(d) \sigma\left( \frac{n}{sd} \right) \end{eqnarray} I cannot see how to proceed from here. Hints or answers for how to proceed from where I currently am, or how to show what is required using a different approach would be appreciated.

## 1 Answer

Since $\varphi_1$ is multiplicative it suffices to show this for prime powers $p^k$. The product definition in part (a) yields $$\varphi_1(p^k)=p^k\prod_{p|n}(1+p^{-1})=p^k+p^{k-1}$$ The formula in part (b) yields $$\sum_{d^2|n}\mu(d)\sigma(\frac{n}{d^2})=\sigma(p^k)-\sigma(p^{k-2})=\frac{(p^{k+1}-1)-(p^{k-1}-1)}{p-1}=p^k+p^{k-1}.$$ so that the two definitions are the same.

• Thank you. I knew I was over complicating it! – Ross Pure Jan 8 '14 at 9:42