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 2007-03-15, 01:10 #1 toilet   Mar 2007 28 Posts perfect number proof An integer n is called a perfect number if it is equal to the sum of its proper divisors. Thus 6 = 3 + 2 + 1 is a perfect number. a. If q = 2^p− 1 is a (Mersenne-)prime, prove that 2^(p-1)q is a perfect number. b. Prove that if n is an even perfect number, then n has the form n = 2^(p-1) (2^p−1) for some prime of the form 2^p− 1.
 2007-03-15, 07:22 #2 maxal     Feb 2005 22·32·7 Posts
 2007-03-15, 09:35 #3 akruppa     "Nancy" Aug 2002 Alexandria 25×7×11 Posts Can you please give us the email address of your teaching assistant? Thank you very much, Alex Last fiddled with by akruppa on 2007-03-16 at 09:45 Reason: coected poo splling
 2007-03-15, 19:21 #4 ewmayer ∂2ω=0     Sep 2002 República de California 25×307 Posts Good username, though. A bit odd, but nearly perfect, given the thread question.
2007-03-15, 20:35   #5
Patrick123

Jan 2006
JHB, South Africa

157 Posts

Quote:
 Originally Posted by akruppa Can you please give us the email address of you teaching assistant? Thank you very much, Alex
We'll FLUSH it out of him, 1, 2, 3 or 6 times over.

2012-06-09, 09:25   #6
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

3·419 Posts

Quote:
 Originally Posted by toilet An integer n is called a perfect number if it is equal to the sum of its proper divisors. Thus 6 = 3 + 2 + 1 is a perfect number. a. If q = 2^p− 1 is a (Mersenne-)prime, prove that 2^(p-1)q is a perfect number. b. Prove that if n is an even perfect number, then n has the form n = 2^(p-1) (2^p−1) for some prime of the form 2^p− 1.
It can be proved that every even perfect number has the form 2k-1(2k-1) where 2k-1 is a prime. Every number can be expressed in the form, 2km, where m is an odd number, and for an even number k > 0.
So, we will have to find numbers of the form 2km, whose sum of all factors including itself is 2k+1m.
Let the sum of factors of m, including itself be m+q, where q is the sum of factors of m, except m.

We now have,
Sum of factors of 2km including itself = 2k+1m.

But sum of factors of 2km including itself =
(1 + 2 + 4 + 8 + ... + 2k) × (Sum of factors of m including itself)

which implies that, we want numbers of the form 2km such that,
(2k+1-1) (m+q) = 2k+1m
or, 2k+1m + 2k+1q - m - q = 2k+1m
or, m = (2k+1-1) q

This implies that, q divides m.

If 2km is odd, then k = 0, we forget that case here.
If k > 0, q < m.
Also, q divides m.

But, q is sum of all factors of m, that are less than m.

which implies that, q is sum of factors of m, including q.

which implies that, q cannot be composite, since q will have additional factors.

Also, q cannot be prime, since every prime number also has 1 as its factor.

which implies that, q = 1.

Thus, m = (2k+1-1).
But, sum of factors of m including itself = m+q = m+1.
which implies that m is prime, or (2k+1-1) is prime.

And the number, of which we found the condition, to be perfect = 2km = 2k(2k+1-1).
Hence, every even perfect number has the form 2k(2k+1-1) where 2k+1-1 is prime.
Or, every even perfect number has the form 2k-1(2k-1) where 2k-1 is prime.

 2012-06-09, 09:32 #7 Raman Noodles     "Mr. Tuch" Dec 2007 Chennai, India 3×419 Posts Have a look at my proof on perfect numbers I have written in the penultimate paragraph the case when k>0 But for odd numbers k=0 In that case, q=m But also q divides m q is the sum of factors less than m q is sum of factors of m including q If q
2012-06-09, 15:06   #8
LaurV
Romulan Interpreter

Jun 2011
Thailand

3·13·229 Posts

Quote:
 Originally Posted by Raman I'll take from your lack of response for two days that I have PROVED there are no odd perfect numbers at all!!!!
Or that your question is so dull everyone died laughing and no one can reply anymore. I did not die yet because I am a bit stupid and did not understand it yet... If k=0 you start with sigma(m)=sigma(m) and of course you can prove anything you like from this...

2012-06-09, 17:24   #9
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

217048 Posts

Quote:
 Originally Posted by Raman I'll take from your lack of response for two days that I have PROVED there are no odd perfect numbers at all!!!!
I take it that you use while color for your thoughts? We cannot read neither white color, nor your thoughts. Maybe it's a good thing.

You can think anything you want, Raman. That's what distinguishes thoughts from proofs.

2012-06-09, 18:44   #10
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

23518 Posts

Quote:
 Originally Posted by Batalov I take it that you use while color for your thoughts? We cannot read neither white color, nor your thoughts. Maybe it's a good thing. You can think anything you want, Raman. That's what distinguishes thoughts from proofs.
Whatever that I had posted within the white color font, it was being certainly an old-style post.

What happens when the condition k = 0 is being imposed together?
We will get that q = m, but that q = σ(m) - m
q = the sum of all factors for m except m
since m = (2k+1-1) q, q divides m

q is being the sum of all factors for m including q
Excluding the case for q = 1, σ(1) - 1 = 0
if m is being composite, then q is being a factor for m
Atleast three factors for m exist then, namely 1, q, (m/q)
whose sum > q. Then, we will have that certainly σ(m) > m + q.
σ(m) > 2m. Even if m is being prime, then m = q,
σ(m) = 1 + q > q = m.

Is it being possible for such a 'q' value to exist at all?
We have got that σ(1) - 1 = 0; σ(prime) - prime = 1

2012-06-09, 20:58   #11
ewmayer
2ω=0

Sep 2002
República de California

231408 Posts

Quote:
 Originally Posted by Batalov I take it that you use while color for your thoughts? We cannot read neither white color, nor your thoughts. Maybe it's a good thing. You can think anything you want, Raman. That's what distinguishes thoughts from proofs.
By analogy with the Einsteinian/German term for "thought experiment", Is Gedankenproof a term-in-common-usage?

Or perhaps it's meant in the same sense as 'foolproof', that is in the sense of 'immune to": I have a thoughtproof solution to the problem.

I take it from the lack of objections to the above while I composed it that you all agree. Excellent!

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