Prove that, $\displaystyle {2^{(p - 1)}}({2^p} - 1)$ isperfect numberif $\displaystyle ({2^p} - 1)$ isprime number

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- May 25th 2011, 07:20 AMseitPerfect number
Prove that, $\displaystyle {2^{(p - 1)}}({2^p} - 1)$ is

*perfect number*if $\displaystyle ({2^p} - 1)$ is*prime number* - May 25th 2011, 07:32 AMabhishekkgp
- May 25th 2011, 07:35 AMTheEmptySet
A number is perfect if the sum of it's divisors are twice the number.

Since by assumption

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

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

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

Since the latter is a geometric series we get that

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

Since the has already added the divisor 1 we just add other divisor

$\displaystyle (2^p-1)+(2^p-1)=2(2^p-1)$ - May 30th 2011, 07:58 AMAlso sprach Zarathustra
Also, you should know that if $\displaystyle n$ is an even perfect number say $\displaystyle n=2^{k-1}m$ then $\displaystyle m=2^k-1$ and it is a prime number.

Proof:

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

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

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

$\displaystyle n$ is perfect number, therefor:

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

With (1), we get:

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

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

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

Or:

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

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

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

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

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

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