For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$

Solution 1:

Define the functions $N$ and $1$ by $N(n)=n$ and $1(n)=1$ for all $n \in \mathbb{N}.$

An equivalent, but easier to write, solution is to note that the function $\sum_{d|n}\sigma(d)$ is the Dirichlet convolution $\sigma\ast 1$ and thus the definition $\sigma=N*1$ implies that $$\sigma*1=(N*1)*1=N*(1*1).$$ Now notice that $1*1$ is the classical divisor function $\tau$ and we therefore get $$ \sigma*1=N*\tau,$$ which provides the required answer.

Solution 2:

You can reduce to prime powers to get the job done, but that doesn't get at the heart of the equality.

$$\sum_{d\mid n}\sigma(d)=\sum_{d\mid n}\sum_{r\mid d}r=\sum_{r\mid d\mid n}r=\cdots$$

Can you continue? For each $r\mid n$, how many times is $r$ a summand of the above sum?