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

There is a request for Pillai's paper or proof. The paper is a bit hard to find and quite old, so I scanned it. It's called "On An Arithmetic Function" and was published at Annamalai University.

I've made the scan available at my website. [Presumably I'll host it somewhere else sometime - I do not know of another online copy]. The first desired equation is the combination of his Theorem II and Theorem III, for instance. The second desired equation isn't an explicit result, but instead is a byproduct of his other proofs and results.

Answer 3

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