I am trying to prove the following theorem:
Theorem. A number is perfect iff the sum of the reciprocal of its divisors, excluding $1$, is $1$.
Thus far, this is the proof that I have managed to sew:
Proof. Let $n$ be perfect. Then $2n=1+a_1+a_2+\cdots+a_m+n$, where each $a_j$ is a divisor of $n$. Moreover, let $1<a_j<a_{j+1}<n$. It follows that $2=\frac{1}{n}+\frac{a_1}{n}+\frac{a_2}{n}+\cdots+\frac{a_m}{n}+1\Longrightarrow$ $1=\frac{1}{n}+\frac{1}{a_m}+\frac{1}{a_{m-1}}+\cdots+\frac{1}{a_1}$, as required. Conversely, let $1=\frac{1}{n}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_m}$. It follows that $1=\frac{1}{n}+\frac{a_m}{n}+\frac{a_{m-1}}{n}+\cdots+\frac{a_1}{n}\Longrightarrow$ $n=1+a_1+a_2+\cdots+a_m$. Therefore, $n$ is perfect. $\square$
Nevertheless, I do not feel very confident about it. What do you guys think?
Best Answer
Let $a$ be a perfect number, My '$a$' has $n$ factors.
According to perfect number,
$a=a_1+a_2+a_3+...a_n$
$a_1=1$
$\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+...\frac{1}{a_n}$
Notice that
$a_1.a=a$
$a_2.a_n=a$
.
.
.
$a_n.a_2=a$
You can surely take on from here.