Perfect Numbers

By Brad Rodgers (P1930) on Monday, April 30, 2001 - 11:35 pm :

I know how to prove that if (2^k -1) is a prime then 2^k-1 (2^k -1) is a perfect number , but how does one prove that all even perfect numbers are o2316.html f this form? Also, has any work ever been done on finding an odd perfect number?

Brad

By William Astle (Wja24) on Monday, April 30, 2001 - 11:57 pm :

I haven't thought about this much but if all perfect numbers are of the form 2^k-1 (2^k -1) where 2^k -1 is prime then there can't be any odd perfect numbers since for k> 1 2^k-1 (2^k -1) is even and no other case gives a non trivial natural number.

By Tim Martin (Tam31) on Tuesday, May 1, 2001 - 12:07 am :

I imagine there has probably been some work done on the subject of odd perfect numbers, but to date nobody has proved whether or not they exist.

By Olof Sisask (P3033) on Tuesday, May 1, 2001 - 05:55 pm :

I think someone has shown that there are no odd perfect numbers less than 10³⁰⁰ , or something along those lines. I think it was Euler who demonstrated several properties that an odd perfect number must have, if one exists, but I can't remember offhand what it was he showed. I'll have a look for the information and post it if I find it.

Olof

By Tim Martin (Tam31) on Tuesday, May 1, 2001 - 06:06 pm :

Any odd perfect number will have to satisfy the following criteria:

A perfect square multiplied by an odd power of a single prime
Have at least 8 distinct prime factors
Have at least 29 prime factors (not necessarily distinct)
Have a prime factor greater than 10⁶
Have its second largest prime factor greater than 10²
Be divisible by a prime component greater than 10²⁰

By Olof Sisask (P3033) on Tuesday, May 1, 2001 - 07:02 pm :

Also, an odd number perfect number must be of the form

(4n+1)^4k+1 b²

where 4n+1 is prime, according to Euler.

By Olof Sisask (P3033) on Tuesday, May 1, 2001 - 09:56 pm :

By the way Brad, what is your proof?

By Brad Rodgers (P1930) on Wednesday, May 2, 2001 - 12:01 am :

Note that the factors of 2^k-1 (2^k -1) if the second term is prime are

1,2,4,...,2^k-1 ,2^k -1,2(2^k -1),4(2^k -1),...,2^k-2 (2^k -1)

So the addition of all these will be

$begin{displaymath}sum\_{n=0}^{k-1} 2^n + sum\_{n=0}^{k-2} 2^n(2^k-1) = (2^k-1) + (2^{k-1}-1)(2^k-1) = 2^{k-1}(2^k-1) end{displaymath}$

which is what we want.

Brad

By Olof Sisask (P3033) on Thursday, May 3, 2001 - 09:52 pm :

The proof I know of the converse is similar. I'm not sure if you want the entire proof or just a general direction, so just in case here's the direction:

If you let

n

be an even perfect number, then it must be of the form

2^{k - 1} \times m

, where

k

\geq 2

and

m

is odd. Compare the sum of the factors of this expression with the sum of the factors of

n (= 2 n = 2^{k} \times m)

.

Hope that makes some sort of sense to you.

Regards,
Olof

By Brad Rodgers (P1930) on Friday, May 4, 2001 - 12:27 am :

I've been able to deduce it to

if there exists an even perfect number, then there is a number

m

and a number

k

such that

(σ (m) - 1) + (2^{k} - 1) = m

I haven't been able to show yet that this implies $m = (2^{k} - 1)$ and is prime. Where do I go from here?

Brad

By Olof Sisask (P3033) on Friday, May 4, 2001 - 07:47 am :

Did you get it down to the form:

σ (m) \times (2^{k} - 1) = 2^{k} \times m

in getting to what you have above? If so, you can deduce from this point that $2^{k} - 1$ divides $m$ , i.e. $m = (2^{k} - 1) \times n$ . Try substituting this expression for $m$ into the RHS, and then consider that we know that both $m$ and $n$ divide $m$ , and therefore can create a lower bound for $σ (m)$ .

Regards,
Olof

By Kerwin Hui (Kwkh2) on Friday, May 4, 2001 - 09:51 am :

I'll quickly outline a proof:

If $n = 2^{k} m$ , where $m$ is odd, then

$σ (n) = 2 n$ (definition of perfect number)

Notice that I define $σ (n)$ to be the sum of divisors of $n$ from $1$ up to $n$ . This way, $s$ will be multiplicative (Prove it!). Now from the multiplicative property, we get

$σ (n) = σ (2^{k}) σ (m) = (2^{k + 1} - 1) σ (m)$ .

Equating our two expressions for $σ (n)$ , we get

$(2^{k + 1} - 1) σ (m) = 2^{k + 1} m$

so we have

$m = (2^{k + 1} - 1) L$ , $σ (m) = 2^{k + 1} L$ , some positive integer $L$ .

If $L > 1$ , then $L | m$ . So $σ (m) \geq m + L + 1$ . But $L + m = 2^{k + 1} L = σ (m)$ , contradiction.

so $L = 1$ . $σ (m) = m + 1$ . Hence $m$ is prime.

Now just work through $σ (n) = (2^{k + 1} - 1) (m + 1) = 2^{k + 1} m$ to show $m$ is the Mersenne prime $2^{k + 1} - 1$ .

Kerwin

By Olof Sisask (P3033) on Friday, May 4, 2001 - 01:09 pm :

Virtually identical:

$σ (m) \times (2^{k} - 1) = 2^{k} \times m$

Therefore $2^{k} - 1 | m$ , so $m = (2^{k} - 1) \times n$

So $σ (m) \times (2^{k} - 1) = 2^{k} \times (2^{k} - 1) \times n$

$σ (m) = 2^{k} \times n$

$n$ and $m$ both divide $m$ therefore:

$σ (m) \geq n + m = 2^{k} n$

(Since $m = (2^{k} - 1) \times n = 2^{k} n - n \Rightarrow m + n = 2^{k} n$ ).

This is what we had above, therefore $n$ and $m$ are the only factors of $σ (m)$ , so $m$ is prime and $n = 1$ .