Prove that $\sum \limits_{d|n}(n/d)\sigma(d) = \sum \limits_{d|n}d\tau(d)$

Question

How can I prove:

$$\sum \limits_{d|n}(n/d)\sigma(d) = \sum \limits_{d|n}d\tau(d)?$$

Few observations : Left side is a sum function and the right side is a number of divisors function. Both the sides are multiplicative. I don't want to start expanding like this. Appreciate any help on how to interpret the sums!

Answer 1

$$\sum_{d\mid n} \frac{n}d \sigma(d) =\sum_{d_1\mid n}\frac{n}{d_1}\sum_{d_2\mid d_1} d_2 = \sum_{d_2\mid d_1\mid n}\frac{n}{d_1/d_2}$$

$$\sum_{d\mid n} d\tau(d)= \sum_{d_3\mid n} d_3\sum_{d_4\mid d_3}1 = \sum_{d_4\mid d_3\mid n} d_3$$

Now, map $(d_1,d_2)$ to $(d_3,d_4)=(nd_2/d_1,n/d_1)$ and we see we have the same sums.

So, more generally, if $S_n=\{(d_1,d_2): d_2\mid d_1\mid n\}$ then the map $S_n\to S_n$ defined by $(d_1,d_2)\to\left(\frac{nd_2}{d_1},\frac n{d_1}\right)$ is a bijection. Thus, for any function $f(m,n)$ of two natural numbers, we have that:

$$\sum_{(d_1,d_2)\in S_n} f(d_1,d_2)=\sum_{(d_1,d_2)\in S_n} f\left(\frac{nd_2}{d_1},\frac{n}{d_1}\right)$$

The above is just the case of $f(m,n)=m$.

Answer 2

First we note that by definition we have

$$\sigma(d) = \sum_{d_1|d}d$$

$$\tau(d) = \sum_{d_1|d}1$$

Then by substituting these into each side of the equation we get

$$\sum_{d|n}\frac{n}{d}\sigma(d) = \sum_{d|n}\frac{n}{d} \sum_{d_1|d} d_1 = \sum_{d|n}\sum_{d_1|d} \frac{n}{d/d_1}= \sum_{\substack{n=de\\d=d_1d_2}}\frac{n}{d/d_1} = \sum_{n=d_1d_2e}\frac{n}{d_2} = \sum_{n = d_1d_2e} d_1e$$

$$\sum_{d|n}d\tau(d) = \sum_{d|n}d\sum_{d_1|d} 1 = \sum_{d|n} \sum_{d_1|d} d = \sum_{\substack{n = de\\ d = d_1d_2}} d = \sum_{n = d_1d_2e} d_1d_2$$

And we can see the last two sums are the same.

Answer 3

Note that $\sigma=N*1$ and $\tau=1*1$ where "*" is the Dirichlet convolution and $N$ is the identical fucntion in positive integers. We have $$\sum_{d\mid n}{\frac{n}{d}\sigma\left(d\right)=N*\sigma}=N*\left(N*1\right)=\left(N*N\right)*1=\left(\sum_{d\mid n}{d\cdot\frac{n}{d}}\right)*1=\left(\sum_{d\mid n}n\right)*1$$ $$=\left(N\tau\right)*1=\sum_{d\mid n}{d\cdot\tau\left(d\right).}$$