# Perfect Numbers ?

#### 1337h4x

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.

#### Drexel28

MHF Hall of Honor
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 $$\displaystyle 2^p-1$$ is prime then $$\displaystyle n=2^{p-1}[2^{p-1}-1]$$ is a prime factorization of it. So, the proper divisors o $$\displaystyle n$$ are $$\displaystyle 1,2,\cdots,2^{p-1},2^p-1,2(2^p-1),\cdots2^{p-2}(2^p-1)$$. So, $$\displaystyle 1+2+\cdots+2^{p-1}+2^p-1+\cdots+2^{p-2}(2^p-1)=\sum_{n=0}^{p-1}2^n+(2^p-1)\sum_{n=0}^{p-2}2^n$$ which using the geometric sum formula is equal to $$\displaystyle 2^p-1+(2^p-1)(2^{p-1}-1)=(2^p-1)(1+2^{p-1}-1)=2^{p-1}(2^p-1)=n$$

• 1337h4x

#### Drexel28

MHF Hall of Honor
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 $$\displaystyle n=p^m$$ then the proper divisors are $$\displaystyle 1,\cdots,p^{m-1}$$ and thus $$\displaystyle 1+\cdots+p^{m-1}=\frac{p^m-1}{p-1}\ne n$$

• 1337h4x

#### 1337h4x

Thank you! That makes good sense!