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!
Best Answer
$$\sum_{d\mid n} \frac{n}d \sigma(d) =\sum_{d_1\mid n}\frac{n}{d_1}\sum_{d_2\mid d_1} d_2 = \sum_{d_2\mid d_1\mid n}\frac{n}{d_1/d_2}$$
$$\sum_{d\mid n} d\tau(d)= \sum_{d_3\mid n} d_3\sum_{d_4\mid d_3}1 = \sum_{d_4\mid d_3\mid n} d_3$$
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 $f(m,n)=m$.