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

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.

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.

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

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

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 the way Brad, what is your proof?

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

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:

I've been able to deduce it to

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

Did you get it down to the form:

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)×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 s(m).


Regards,
Olof

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

s(n)=2n (definition of perfect number)

Notice that I define s(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

s(n)=s(2^k)s(m)=(2^k+1-1)s(m).

Equating our two expressions for s(n), we get

(2^k+1-1)s(m)=2^k+1m

so we have

m=(2^k+1-1)L, s(m)=2^k+1L, some positive integer L.

If L > 1, then L|m. So s(m) ³ m+L+1. But L+m=2^k+1L=s(m), contradiction.

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

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

Kerwin

s(m)×(2^k-1)=2^k×m

Therefore 2^k-1|m, so m=(2^k-1)×n

So s(m)×(2^k-1)=2^k×(2^k-1)×n

s(m)=2^k×n

n and m both divide m therefore:

s(m) ³ n+m=2^k n

(Since m=(2^k-1)×n=2^kn-nÞ m+n=2^kn).

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