5
$\begingroup$

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. 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?

  • 1
    Apart from slight wording changes, it is fine. One could also write: As $d$ ranges over the positive divisors of $n$, $\frac{n}{d}$ also ranges over the positive divisors of $n$. Thus $\sigma(n)=\sum_{d|n} \frac{n}{d}$. It follows that $\sigma(n)=2n$ iff $2n=\sum_{d|n} \frac{n}{d}$. Now divide both sides by $n$. (Your proof does exactly the same thing. For concreteness, it is better to use your notation than the condensed notation of this comment.)2012-02-18

1 Answers 1

1

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.

  • 0
    I should point out that if $a = a_1 + \dots + a_n$, then $a$ has $n+1$ positive divisors, the largest of which is $a$ itself.2013-03-09