perfect number
An positive integer $n$ is called perfect if it is the sum of all positive divisors^{} of $n$ less than $n$ itself. It is not known if there are any odd perfect numbers, but all even perfect numbers have been classified according to the following lemma:
Lemma 1.
An even number^{} is perfect if and only if it equals ${\mathrm{2}}^{k\mathrm{}\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{2}}^{k}\mathrm{}\mathrm{1}\mathrm{)}$ for some integer $k\mathrm{>}\mathrm{1}$ and ${\mathrm{2}}^{k}\mathrm{}\mathrm{1}$ is prime.
Proof.
Let $\sigma $ denote the sum of divisors function. Recall that this function is multiplicative.
Necessity: Let $p={2}^{k}1$ be prime and $n={2}^{k1}p$. We have that
$\sigma (n)$  $=$  $\sigma ({2}^{k1}p)$  
$=$  $\sigma ({2}^{k1})\sigma (p)$  
$=$  $({2}^{k}1)(p+1)$  
$=$  $({2}^{k}1){2}^{k}$  
$=$  $2n,$ 
which shows that $n$ is perfect.
Sufficiency: Assume $n$ is an even perfect number. Write $n={2}^{k1}m$ for some odd $m$ and some $k>1$. Then we have $\mathrm{gcd}({2}^{k1},m)=1$. Thus,
$$\sigma (n)=\sigma ({2}^{k1}m)=\sigma ({2}^{k1})\sigma (m)=({2}^{k}1)\sigma (m).$$ 
Since $n$ is perfect, $\sigma (n)=2n$ by definition. Therefore, $\sigma (n)=2n={2}^{k}m$. Piecing together the two formulas^{} for $\sigma (n)$ yields
$${2}^{k}m=({2}^{k}1)\sigma (m).$$ 
Thus, $({2}^{k}1)\mid {2}^{k}m$, which forces $({2}^{k}1)\mid m$. Write $m=({2}^{k}1)M$. Note that $$. From above, we have:
${2}^{k}m$  $=$  $({2}^{k}1)\sigma (m)$  
${2}^{k}({2}^{k}1)M$  $=$  $({2}^{k}1)\sigma (m)$  
${2}^{k}M$  $=$  $\sigma (m)$ 
Since $m\mid m$ by definition of divides (http://planetmath.org/Divides) and $M\mid m$ by assumption^{}, we have
$${2}^{k}M=\sigma (m)\ge m+M={2}^{k}M,$$ 
which forces $\sigma (m)=m+M$. Therefore, $m$ has only two positive divisors, $m$ and $M$. Hence, $m$ must be prime, $M=1$, and $m=({2}^{k}1)M={2}^{k}1$, from which the result follows. ∎
The lemma can be used to produce examples of (even) perfect numbers:

•
If $k=2$, then ${2}^{k}1={2}^{2}1=3$, which is prime. According to the lemma, ${2}^{k1}({2}^{k}1)={2}^{21}\cdot 3=6$ is perfect. Indeed, $1+2+3=6$.

•
If $k=3$, then ${2}^{k}1={2}^{3}1=7$, which is prime. According to the lemma, ${2}^{k1}({2}^{k}1)={2}^{31}\cdot 7=28$ is perfect. Indeed, $1+2+4+7+14=28$.

•
If $k=5$, then ${2}^{k}1={2}^{5}1=31$, which is prime. According to the lemma, ${2}^{k1}({2}^{k}1)={2}^{51}\cdot 31=496$ is perfect. Indeed, $1+2+4+8+16+31+62+124+248=496$.
Note that $k=4$ yields that ${2}^{k}1={2}^{4}1=15$, which is not prime.
The sequence^{} of known perfect numbers appears in the OEIS as sequence http://www.research.att.com/ njas/sequences/?q=A000396A000396.
Title  perfect number 

Canonical name  PerfectNumber 
Date of creation  20130322 11:45:29 
Last modified on  20130322 11:45:29 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  24 
Author  Wkbj79 (1863) 
Entry type  Definition 
Classification  msc 11A05 
Classification  msc 20D99 
Classification  msc 20D06 
Classification  msc 1800 