# 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 $2^{k-1}(2^{k}-1)$ 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,

 $\sigma(n)=\sigma(2^{k-1}m)=\sigma(2^{k-1})\sigma(m)=(2^{k}-1)\sigma(m).$

Since $n$ is perfect, $\sigma(n)=2n$ by definition. Therefore, $\sigma(n)=2n=2^{k}m$. 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 $1\leq M. From above, we have:

 $\displaystyle 2^{k}m$ $\displaystyle=$ $\displaystyle(2^{k}-1)\sigma(m)$ $\displaystyle 2^{k}(2^{k}-1)M$ $\displaystyle=$ $\displaystyle(2^{k}-1)\sigma(m)$ $\displaystyle 2^{k}M$ $\displaystyle=$ $\displaystyle\sigma(m)$

Since $m\mid m$ by definition of divides (http://planetmath.org/Divides) and $M\mid m$ by assumption  , we have

 $2^{k}M=\sigma(m)\geq m+M=2^{k}M,$

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, $2^{k-1}(2^{k}-1)=2^{2-1}\cdot 3=6$ 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 http://www.research.att.com/ njas/sequences/?q=A000396A000396.

Title perfect number PerfectNumber 2013-03-22 11:45:29 2013-03-22 11:45:29 Wkbj79 (1863) Wkbj79 (1863) 24 Wkbj79 (1863) Definition msc 11A05 msc 20D99 msc 20D06 msc 18-00