Sum Over Divisors Equals Sum Over Quotients

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Theorem

Let $n$ be a positive integer.

Let $f: \Z_{>0} \to \Z_{>0}$ be a mapping on the positive integers.


Let $\displaystyle \sum_{d \mathop \divides n} \map f d$ be the sum of $\map f d$ over the divisors of $n$.

Then:

$\displaystyle \sum_{d \mathop \divides n} \map f d = \sum_{d \mathop \divides n} \map f {\frac n d}$.


Proof

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

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

Hence:

\(\displaystyle \sum_{d \mathop \divides n} \map f {\frac n d}\) \(=\) \(\displaystyle \map f {\frac n {d_1} } + \map f {\frac n {d_2} } + \cdots + \map f {\frac n {d_r} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {d_1} + \map f {d_2} + \cdots + \map f {d_r}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{d \mathop \divides n} \map f d\)

$\blacksquare$