Perfect Numbers

**BlakeRobertsonMD** · October 19th 2008, 08:33 AM

Hey I was having a little trouble with this one problem and was wondering if you all could help me.

I have to explain why:

2 ^(k-1)(2 ^(k)-1) is perfect if 2 ^(k)-1 is prime.

Any ideas on how I would start this?

**ThePerfectHacker** · October 19th 2008, 08:40 AM

Originally Posted by BlakeRobertsonMD

Hey I was having a little trouble with this one problem and was wondering if you all could help me.

I have to explain why:

2 ^(k-1)(2 ^(k-1)) is perfect if 2 ^(k-1) is prime.

Any ideas on how I would start this?

Let $\sigma (n)$ be the sum of divisors of $n$ .
It is easy to show $\sigma (2^r) = 2^{r+1} - 1$ .
Also $\sigma (p) = p+1$ if $p$ is prime.
Finally $\sigma (ab) = \sigma (a ) \sigma (b)$ if $\gcd(a,b)=1$ .

A number of perfect if and only if $\sigma (n) = 2n$ .
Set $n=2^{k-1} (2^k -1)$ and use the above results.

**BlakeRobertsonMD** · October 19th 2008, 09:54 AM

Thanks. I written up a proof of induction and I think I got the problem to work well. Can any of you tell me if this sounds about correct?

Proof: 2 ^ (k – 1) ( 2 ^ k – 1) is always perfect when 2 ^ k - 1 is prime.
We will use induction for this proof.
Let P(k) be the predicate: 2^(k-1) (2 ^(k) -1).
Base Case: Let P(2).
2^(2-1) (2 ^ (2) -1) = 2 * 3 = 6
As you see k = 2 lets 2 ^ k – 1 become 3 which is the lowest prime number possible for this equation.
Our answer to the equation is 6 which is a perfect number because:
6 = 1 + 2 + 3.
Inductive Step: Assume that P(k) is true, where k >= 2. Then P(k + 1) is also true, because:
Using assumption we add (k+1) to the equation:
2 ^ ((k+1)-1) * (2 ^ (k+1) – 1)
As the equation shows P(k) implies P(k+1) for every natural number k >=2.
By Principle of induction, 2^(k – 1)(2^(k)-1) is a perfect number if 2^(k) -1 is a prime number for every natural number k>=2, which proves the claim.
How does it sound?

**o_O** · October 19th 2008, 10:10 AM

That doesn't work. The statement doesn't say that it works for every k greater than or equal to 2. For example, take k = 6. $2^{6} - 1 = 63$ which isn't prime. I don't quite see how you got your inductive step to work. It seems you just said 'P(k) implies P(k+1)' without actually showing it.

Use the hints ThePerfectHacker gave you.

$\begin{array}{rcll}\sigma (n) & = & \sigma \! \bigg(2^{k-1} \left(2^k - 1\right) \bigg) & \\ & = & \sigma (2^{k-1}) \cdot \sigma (2^k - 1) & \quad \text{Since: } \text{gcd}\!\left(2^{k}-1, 2^{k-1}\right) = 1 \\ & = & (2^k - 1) \cdot 2^k & \quad \text{Since: } \sigma (2^r) = 2^{r+1} - 1 \ \text{and} \ \sigma (p) = p + 1 \\ & \vdots & & \end{array}$

**BlakeRobertsonMD** · October 19th 2008, 10:12 AM

So is n just any natural number or what? Not really sure where n came from.

Also I made a mistake in the first posting, 2^(k) -1 is prime, not 2^(k-1). sorry ><

**o_O** · October 19th 2008, 10:13 AM

No. As TPH gave you, $n = 2^{k-1} \left(2^k - 1\right)$ .

We have to prove that $\sigma (n) = 2n$ since this is what it means for $n$ to be perfect. Read through TPH's hints then the post I gave.

**BlakeRobertsonMD** · October 19th 2008, 10:41 AM

Okay personally I'm lost. I was never good at discrete math and I was never really good at induction either.

I've never used the divisor function nor was I ever taught it. Usually I would think that the sum of all the divisors of a number would equal the number if it was truely perfect, not 2 * the number as shown with 2n.

Sorry but after the hints you and the perfect hacker gave me, i'm still lost. Such as how did you go from o(2^(k-1)) * o(2^(k) - 1) to just 2^(k) -1 * 2 ^(k).

I'm sorry if I seem slow with this.

**o_O** · October 19th 2008, 11:01 AM

A number exist iff it is equal to the sum of its positive divisors excluding itself, i.e. $\underbrace{\sigma (n)}_{\text{the sum of its divisors}} - \underbrace{n}_{\text{excluding itself}} = n$ .

Simply moving n to the other side gives us $\sigma (n) = 2n$ . So if n is perfect, then $\sigma (n) = 2n$

To answer your questions:

1. Why $\sigma (2^{k-1}) = 2^k - 1$ .

What are the divisors of $2^r$ ? Well, there's $1, 2, 2^2, 2^3, \hdots, 2^{r-1}, 2^r$ . For example, $2^8$ can be divided by 1, 2, 4, 8, etc. since $\frac{2^8}{2^2} = 2^6, \ \frac{2^8}{2^3} = 2^5, \ \hdots$

Now, TPH said that the sum of the divisors of $2^r$ is $2^{r+1} - 1$ . In other words: $1 + 2 + 2^2 + \hdots + 2^r = 2^{r+1} - 1$

To prove this, let $S = 1 + 2 + 2^2 + \hdots + 2^r$ and then consider $S = 2S - S$ and you will get the desired result.

Thus, $\sigma(2^r) = 1 + 2 + 2^2 + \hdots + 2^{r-1} + 2^r = 2^{r+1} - 1$ . Now imagine $r = k-1$ and you're set.
__________________________________________________ ____________________________________

2. Why does $\sigma (2^k - 1) = 2^k$ ?

We were given that $2^k - 1$ is prime. For any prime, its divisors are 1 and itself. So, the sum of its divisors is just itself plus 1, i.e. $\sigma (p) = p + 1$ where $p$ is prime.

Now put (1) and (2) together and you get what you needed.

**BlakeRobertsonMD** · October 19th 2008, 01:33 PM

Okay so the overall equation is 2^(2k)-(2^(k))

**o_O** · October 19th 2008, 03:00 PM

No. Should be $2^{k+1} - 2^k$ . But that isn't the way to go.

Factor out a 2 from $2^k$ and you should notice that: $2^{k} \left( 2^{k} - 1\right) = 2^1 \cdot \underbrace{ 2^{k-1}\left( 2^{k} - 1\right)}_{n} = 2n$

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