Elementary Number Theory – Relation Between ?(N), ?(N), and ?(N)

Question

How to prove
$$\sum\limits_{d\mid n} {\sigma (d)\varphi (n/d) = n\tau (n)}$$
and
$$\sum\limits_{d\mid n} {\tau (d)\varphi (n/d) = \sigma (n)}$$
where
${\sigma (N)}$ is the divisor function, ${\tau (N)}$ is the number of positive divisors of $N$, and $\varphi (N)$ is Euler's totient? I am looking for a short proof.

Answer 1

Use Dirichlet convolution: $$ \sum\limits_{d\mid n} \sigma (d)\varphi (n/d) = (\sigma * \varphi) (n) = (u * N * \mu * N)(n) = (N * N)(n) = n \tau(n) $$ where $u(n)=1$ and $N(n)=n$. By definition, $\sigma = u*N$. It's a famous (but easy) theorem that $N=u*\varphi$, from which follows that $\varphi = \mu * N$, since $u$ and $\mu$ are Dirichlet inverses (that's essentially Moebius inversion formula).

The second one follows in the same way: $$ \sum\limits_{d\mid n} \tau (d)\varphi (n/d) = (\tau*\phi)(n) = (u*u*\mu*N)(n) = (u*N)(n)=\sigma (n) $$ since $\tau = u*u$.