where the sum is taken over all positive divisors of . Here is a counting argument for a proof.
Let . For each divisor of , let us count the number of elements with . When , this is just . What about when ? If is odd then clearly there is no such . Otherwise, implies that and must both be even. Moreover, this happens iff . Hence, as before, the number of such is .
This works in general: for any , the number of elements such that is . So summing over all , which is the same as summing over all , gives the total number of elements of , which is . This proves the desired result.
This begs for a generalisation. I will add it here when I come up with one. (I really need to sleep now!)