This is a homework question and I am to show that $$\sigma(n) = \sum_{d|n} \phi(n) d\left(\frac{n}{d}\right)$$ where $\sigma(n) = \sum_{d|n}d$, $d(n) = \sum_{d|n} 1 $ and $\phi$ is the Euler Phi function.

What I have. Well I know $$\sum_{d|n}\phi(d) = n$$ I also know that for $n\in \mathbb{Z}^n$ it has a certain prime factorization $n = p_1^{a_1} \ldots p_k^{a_k}$ so since $\sigma$ is a multiplicative function, we have $\sigma(n) = \sigma(p_1)\sigma(p_2)...$

I also know the theorem of Möbius Inversion Formula and the fact that if $f$ and $g$ are artihmetic functions, then $$f(n) = \sum_{d|n}g(d)$$ iff $$g(n) = \sum_{d|n}f(d)\mu\left(\frac{n}{d}\right)$$

Please post no solution, only hints. I will post the solution myself for others when I have figured it out.


The convolution product, $f*g$, of two arithmetical functions $f(n)$ and $g(n)$ is defined by $$ (f*g)(n):=\sum_{d\mid n} f(d) g(\frac nd). $$ Now, if ${\bf 1}$ is the function which is always $1$, and $I$ is the identity function, ask yourself:

  • Is $*$ commutative? Why or why not?
  • Is $*$ associative? Why or why not?
  • What is ${\bf 1}*{\bf 1}$?
  • What is $I*{\bf 1}$?
  • What is ${\phi}*{\bf 1}$?
  • $\begingroup$ I got the solution by finding the definitions of the Euler phi function and the divisor function. I never knew those. Finding those made the problem easy. $\endgroup$ – Tyler Hilton Feb 26 '13 at 18:24
  • $\begingroup$ someone post a complete answer please :) $\endgroup$ – emonhossain Oct 26 '18 at 9:47

First hint: verify the formula when $n$ is a power of a prime. Then, prove that the function $n \mapsto \sum_{d \mid n} \phi(n) d(n/d)$ is also multiplicative, so it must coincide with $\sigma$. In fact, prove more generally that if $a_n$ and $b_n$ are multiplicative, then $n\mapsto \sum_{d \mid n}a_db_{n/d}$ is multiplicative.

Second hint (more advanced): If $A(s) = \sum_{n\geq 1}\frac{a_n}{n^{-s}}$ and $B(s) = \sum_{n\geq 1}\frac{b_n}{n^{-s}}$ are Dirichlet series, their product is $$A(s)B(s)=\sum_{n\geq 1}\frac{(a * b)_n}{n^{-s}},$$ where $(a*b)_n = \sum_{d\mid n} a_n b_{n/d}$. Express the Dirichlet series $\sum_{n\geq 1}\frac{\phi(n)}{n^{-s}}$, $\sum_{n\geq 1}\frac{d(n)}{n^{-s}}$ and $\sum_{n\geq 1}\frac{\sigma(n)}{n^{-s}}$ in terms of the Riemann zeta function, and reduce your identity to an identity between them.


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