Arithmetical Functions Sum, $\sum\limits_{d|n}\sigma(d)\phi(\frac{n}{d})$ and $\sum\limits_{d|n}\tau(d)\phi(\frac{n}{d})$

Question

$$\sum_{d|n}\sigma(d)\phi\left(\frac{n}{d}\right)=n\tau(n) ,\\ \sum_{d|n}\tau(d)\phi\left(\frac{n}{d}\right)=\sigma(n)$$ The problem (7.4.15) of Burton's Elementary Number Theory has been request to prove the above equalities. In this book Dirichlet multiplication or Riemann's zeta function isn't expressed before this problem. In addition, I know that the first time, Pillai has been proved this equalities but I couldn't find the Pillai's paper on the web. Can you please refer me to a link of Pillai's paper? or give me a proof without the use of Dirichlet multiplication or Riemann's zeta function?

Answer 1

Here I can give you three quick different proofs, first in terms of Dirichlet multiplication we have: $$\sigma*\phi=(\text{id}*1)*\phi=\text{id}*1*\mu*\text{id}=\text{id}*\text{id}=\text{id}\tau$$ $$\tau*\phi=(1*1)*\phi=1*1*\mu*\text{id}=1*\text{id}=\sigma$$

Or note that if $f$ and $g$ are multiplictive with $v_p(n)$ the p-adic order of $n$ then,

$$\sum_{d\mid n}f(d)g\left(\frac{n}{d}\right)=\prod_{p\mid n}\left(\sum_{k=0}^{v_p(n)} f(p^k)g(p^{v_p(n)-k})\right)$$

Thus if $f=\sigma,\tau$ and $g=\phi$ then substituting there values in at prime powers gives both results.

Alternatively if you know their Dirichlet generating functions then multiplying them out yields:

$$\sum_{n=1}^\infty \frac{n\tau(n)}{n^s}=\zeta(s-1)^2=\left(\sum_{n=1}^\infty\frac{\sigma(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{\phi(n)}{n^s}\right)=\sum_{n=1}^\infty \frac{\sum_{d\mid n}\sigma(d)\phi(n/d)}{n^s}$$

$$\sum_{n=1}^\infty \frac{\sigma(n)}{n^s}=\zeta(s)\zeta(s-1)=\left(\sum_{n=1}^\infty\frac{\tau(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{\phi(n)}{n^s}\right)=\sum_{n=1}^\infty \frac{\sum_{d\mid n}\tau(d)\phi(n/d)}{n^s}$$

Therefore we have both $n\tau(n)=\sum_{d\mid n}\tau(d)\phi(n/d)$ and also $\sigma(n)=\sum_{d\mid n}\tau(d)\phi(n/d)$ as required.

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Lastly for future reference I would advise not using $\tau$ to refer to the divisor function, I only did so in this post to reduce any confusion for you. This is because in analytic number theory $\tau$ is often used to refer to Ramanujan's tau function. I would use either of these notations $d(n)=\sigma_0(n)=\sum_{d\mid n}1$.

Answer 2

Prove that both sides are multiplicative functions and that they coincide when $n$ is a prime power.