# Perfect Numbers ?

• May 21st 2010, 06:22 PM
1337h4x
Perfect Numbers ?
Okay, I have a few questions pertaining to perfect numbers.

Q1:
The theorem in my book states that a prime power is never a perfect number. Therefore if p is a prime, then p^n cannot be a perfect number. Can anyone explain why this is?

Q2:
Suppose that p is a prime and that (2^p)-1 is also prime (in fact a Mersenne prime) . It can be prove that :

n = [2^(p-1)]*[(2^p)-1] is a perfect number. How can this be proven? I've tried like 5 times and can't get anywhere with this.
• May 21st 2010, 06:42 PM
Drexel28
Quote:

Originally Posted by 1337h4x
Q2:
Suppose that p is a prime and that (2^p)-1 is also prime (in fact a Mersenne prime) . It can be prove that :

n = [2^(p-1)]*[(2^p)-1] is a perfect number. How can this be proven? I've tried like 5 times and can't get anywhere with this.

If $2^p-1$ is prime then $n=2^{p-1}[2^{p-1}-1]$ is a prime factorization of it. So, the proper divisors o $n$ are $1,2,\cdots,2^{p-1},2^p-1,2(2^p-1),\cdots2^{p-2}(2^p-1)$. So, $1+2+\cdots+2^{p-1}+2^p-1+\cdots+2^{p-2}(2^p-1)=\sum_{n=0}^{p-1}2^n+(2^p-1)\sum_{n=0}^{p-2}2^n$ which using the geometric sum formula is equal to $2^p-1+(2^p-1)(2^{p-1}-1)=(2^p-1)(1+2^{p-1}-1)=2^{p-1}(2^p-1)=n$
• May 21st 2010, 06:43 PM
Drexel28
Quote:

Originally Posted by 1337h4x
Q1:
The theorem in my book states that a prime power is never a perfect number. Therefore if p is a prime, then p^n cannot be a perfect number. Can anyone explain why this is?

Suppose that $n=p^m$ then the proper divisors are $1,\cdots,p^{m-1}$ and thus $1+\cdots+p^{m-1}=\frac{p^m-1}{p-1}\ne n$
• May 21st 2010, 07:27 PM
1337h4x
Thank you! That makes good sense!