# Perfect number

• May 25th 2011, 08:20 AM
seit
Perfect number
Prove that, ${2^{(p - 1)}}({2^p} - 1)$ is perfect number if $({2^p} - 1)$ is prime number
• May 25th 2011, 08:32 AM
abhishekkgp
Quote:

Originally Posted by seit
Prove that, ${2^{(p - 1)}}({2^p} - 1)$ is perfect number if $({2^p} - 1)$ is prime number

write the definition of the a perfect number.
list the factors of the number $2^kp$, where $p$ is prime.
then sum of the factors.
• May 25th 2011, 08:35 AM
TheEmptySet
Quote:

Originally Posted by seit
Prove that, ${2^{(p - 1)}}({2^p} - 1)$ is perfect number if $({2^p} - 1)$ is prime number

A number is perfect if the sum of it's divisors are twice the number.

Since by assumption

$2^p-1$ is prime the has only two divisors 1 and itself.

Since the other factor is just powers of 2 its divisors are

$1,2,4,...2^{p-1}$

Since the latter is a geometric series we get that

$\sum_{n=1}^{p}2^{n-1}=2^p-1$

$(2^p-1)+(2^p-1)=2(2^p-1)$
• May 30th 2011, 08:58 AM
Also sprach Zarathustra
Quote:

Originally Posted by seit
Prove that, ${2^{(p - 1)}}({2^p} - 1)$ is perfect number if $({2^p} - 1)$ is prime number

Also, you should know that if $n$ is an even perfect number say $n=2^{k-1}m$ then $m=2^k-1$ and it is a prime number.

Proof:

$m\in \mathbb{N}$ odd, $k\geqslant 2$.

$\text{gcd}(2^{k-1},m)=1$, hence:

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

$n$ is perfect number, therefor:

$\sigma(n)=2n=2^km$

With (1), we get:

$2^km=(2^k-1)\sigma(m)$ (2)

From the above we can deduce that $m=(2^k-1)M$, now if we put this $m$ to (2) we will get:

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

Or:

$2^kM=\sigma(m)$.

$m$ and $M$ are both divisors of m, and therefor:

$2^kM=\sigma(m)\geqslant m+M=2^kM$.

The conclusion is that: $\sigma(m)=m+M$.

From the above follows that $m$ is prime number and $M=1$

Or: $m=(2^k-1)M=2^k-1$