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 $n=2^{k-1}p$. We have that

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

$=$$\sigma (2^{k-1})\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 $n=2^{k-1}m$ 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 . 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, $2^{k-1}(2^k-1)=2^{2-1}\cdot 3=6$ 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.