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perfect number (Definition)

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   An even number is perfect if and only if it equals $ 2^{k-1}(2^k-1)$ for some integer $ k>1$ and $ 2^k-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^{k-1}p$. We have that

$\displaystyle \sigma(n)$ $\displaystyle =$ $\displaystyle \sigma(2^{k-1}p)$  
  $\displaystyle =$ $\displaystyle \sigma(2^{k-1}) \sigma(p)$  
  $\displaystyle =$ $\displaystyle (2^k-1)(p+1)$  
  $\displaystyle =$ $\displaystyle (2^k-1)2^k$  
  $\displaystyle =$ $\displaystyle 2n,$  

which shows that $ n$ is perfect.

Sufficiency: Assume $ n$ is an even perfect number. Write $ n=2^{k-1}m$ for some odd $ m$ and some $ k>1$. Then we have $ \gcd(2^{k-1},m)=1$. Thus,

$\displaystyle \sigma(n)=\sigma(2^{k-1}m)=\sigma(2^{k-1})\sigma(m)=(2^k-1)\sigma(m). $
Since $ n$ is perfect, $ \sigma(n)=2n$ by definition. Therefore, $ \sigma(n)=2n=2^km$. Piecing together the two formulas for $ \sigma(n)$ yields
$\displaystyle 2^km=(2^k-1)\sigma(m). $
Thus, $ (2^k-1)\mid 2^km$, which forces $ (2^k-1)\mid m$. Write $ m=(2^k-1)M$. Note that $ 1\le M<m$. From above, we have:
$\displaystyle 2^km$ $\displaystyle =$ $\displaystyle (2^k-1)\sigma(m)$  
$\displaystyle 2^k(2^k-1)M$ $\displaystyle =$ $\displaystyle (2^k-1)\sigma(m)$  
$\displaystyle 2^kM$ $\displaystyle =$ $\displaystyle \sigma(m)$  

Since $ m\mid m$ by definition of divides and $ M\mid m$ by assumption, we have
$\displaystyle 2^kM=\sigma(m)\geq m+M=2^kM, $
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. $ \qedsymbol$

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^{k-1}(2^k-1)=2^{2-1} \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^{k-1}(2^k-1)=2^{3-1} \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^{k-1}(2^k-1)=2^{5-1} \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 A000396.



"perfect number" is owned by Wkbj79. [ full author list (2) | owner history (1) ]
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Keywords:  number theory

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multiply perfect number (Definition) by CompositeFan
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Cross-references: OEIS, sequence, sufficiency, necessity, multiplicative, function, sum of divisors function, prime, even number, even, odd, divisors, sum, integer, positive
There are 32 references to this entry.

This is version 19 of perfect number, born on 2001-10-15, modified 2007-06-26.
Object id is 206, canonical name is PerfectNumber.
Accessed 9290 times total.

Classification:
AMS MSC: 11A05 (Number theory :: Elementary number theory :: Multiplicative structure; Euclidean algorithm; greatest common divisors)
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"Missing cached output" by CompositeFan on 2007-06-01 16:42:06
For HTML with images mode, this says "Missing cached output, contact an admin," but I can see it just fine in page images mode.
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wat's the 5/n problem called? by Lando47 on 2006-12-04 18:16:54
Or maybe its a conjecture. Taht for n > 3, their are 1 < a < b < c such that 5/n = 1/a + 1/b + 1/c. Do they need to be distinct (ie, can we have 1 <= a <= b <= c)? But most importantely, what is this problem called?
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