Sum Over Divisors Equals Sum Over Quotients

Theorem
Let $$n$$ be a positive integer.

Let $$f: \Z^*_+ \to \Z^*_+$$ be a function on the positive integers.

Let $$\sum_{d \backslash n} f \left({d}\right)$$ be the sum of $f \left({d}\right)$ over the divisors of $n$.

Then:
 * $$\sum_{d \backslash n} f \left({d}\right) = \sum_{d \backslash n} f \left({\frac n d}\right)$$.

Proof
If $$d$$ is a divisor of $$n$$ then $$d \times \frac n d = n$$ and so $$\frac n d$$ is also a divisor of $$n$$.

Therefore if $$d_1, d_2, \ldots, d_r$$ are all the divisors of $$n$$, then so are $$\frac n {d_1}, \frac n {d_2}, \ldots, \frac n {d_r}$$ except in a different order.

Hence:

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