Proof Writing: Sum of divisors of perfect numbers I came across this question in an Olympiad number theory textbook:

$n$ is a perfect number if $\sigma(n) = 2n$ ($\sigma(n)$ stands for the sum of divisors of $n$). If $n$ is a perfect number prove that $\Sigma\{\frac{1}{d} \mid d$ divides $n\} =2$.

I have a pretty good heuristic argument for this question, but I am at a loss as to how I would formally put it in the form of a proof. Further I usually have a great deal of trouble formulating proofs in this manner, and I regularly come across questions where I am unable to mathematically describe what I am thinking despite my intuition usually being correct. In this particular case my argument is as follows:

The sum of the reciprocals of $n$ would have a common denominator in $n$. Once the denominators of all the numbers are made $n$, the resulting sum in the numerator would be equal to the sum of all the divisors, ie. $\frac{\sigma(n)}{n} \Rightarrow \frac{2n}{n} = 2$.

I showed this argument to my instructor who accepted the proof but said that it would not fly in an olympiad environment, where they would demand a more rigorous exercise. I now have two questions:


*

*How would I formalize this proof?

*Is there any method I could follow to translate similar heuristic ideas into formal proofs?

 A: Step back:
Consider the two sets $A = \{d:  d$ divides $n\}$.  And $B = \{\frac nd: d$ divides $n\}$.
Claim: $A = B$.
Pf:  I leave that as an easy excercise.
So $\sum_{d \in A} \frac 1d = \sum_{k \in B} \frac 1k$.
So......
$\sum_{d|n}\frac 1d  =$
$\sum_{d|n}\frac 1{\frac nd} = \sum_{d|n}\frac dn=$
$\frac 1n \sum_{d|n}d$
And as $n$ is perfect we know
$\frac 1n\sum_{d|n}d = \frac 1n (2n) = 2$.
A: Let $n$ be a perfect number. By the definition of $\sigma(n),$ we have $\sigma(n)=\sum_{d|n}d=2n$ (where $d|n$ means $d$ divides $n$.)
Dividing out $n,$ we have $\sum_{d|n}\frac dn=\sum_{d|n}\frac1{n/d}=2.$ Now, notice that $d\mapsto n/d$ is a bijection of the set of divisors of $n,$ hence we have $\sum_{d|n}\frac1{n/d}=\sum_{d|n}\frac1d=2.$
P.S.: proof that $d\mapsto n/d$ is a bijection.
To be exact, this is the map $\{d>0:d|n\}\to \{d>0:d|n\}:d\mapsto n/d.$ Since both sets are finite with the same number of elements, to prove this is a bijection, it is enough to prove it is injective, that is, if $d,d'>0$ are divisors of $n$ such that $n/d=n/d',$ then $d=d'.$ This is trivial.
