20070315, 01:10  #1 
Mar 2007
2_{8} 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^(p1)q is a perfect number. b. Prove that if n is an even perfect number, then n has the form n = 2^(p1) (2^p−1) for some prime of the form 2^p− 1. 
20070315, 07:22  #2 
Feb 2005
2^{2}·3^{2}·7 Posts 

20070315, 09:35  #3 
"Nancy"
Aug 2002
Alexandria
2^{5}×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 20070316 at 09:45 Reason: coected poo splling 
20070315, 19:21  #4 
∂^{2}ω=0
Sep 2002
República de California
2^{5}×307 Posts 
Good username, though. A bit odd, but nearly perfect, given the thread question.

20070315, 20:35  #5 
Jan 2006
JHB, South Africa
157 Posts 

20120609, 09:25  #6  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3·419 Posts 
Quote:
So, we will have to find numbers of the form 2^{k}m, whose sum of all factors including itself is 2^{k+1}m. 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 2^{k}m including itself = 2^{k+1}m. But sum of factors of 2^{k}m including itself = (1 + 2 + 4 + 8 + ... + 2^{k}) × (Sum of factors of m including itself) which implies that, we want numbers of the form 2^{k}m such that, (2^{k+1}1) (m+q) = 2^{k+1}m or, 2^{k+1}m + 2^{k+1}q  m  q = 2^{k+1}m or, m = (2^{k+1}1) q This implies that, q divides m. If 2^{k}m 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 = (2^{k+1}1). But, sum of factors of m including itself = m+q = m+1. which implies that m is prime, or (2^{k+1}1) is prime. And the number, of which we found the condition, to be perfect = 2^{k}m = 2^{k}(2^{k+1}1). Hence, every even perfect number has the form 2^{k}(2^{k+1}1) where 2^{k+1}1 is prime. Or, every even perfect number has the form 2^{k1}(2^{k}1) where 2^{k}1 is prime. 

20120609, 09:32  #7 
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<m, q=1 But when q=m, there is no possibility for any q to exist Does this prove that there are no odd perfect numbers at all? Or what step says that this proof cannot be applied to odd perfect numbers? I want to properly complete my proof with it.^{[SUP][SUP][SUP][SIZE=1][COLOR=White]I'll take from your lack of response for two days that I have PROVED there are no odd perfect numbers at all!!!![/COLOR][/SIZE]}[/SUP][/SUP][/SUP] Last fiddled with by Raman on 20120609 at 09:40 
20120609, 15:06  #8 
Romulan Interpreter
Jun 2011
Thailand
3·13·229 Posts 
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...

20120609, 17:24  #9  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
21704_{8} Posts 
Quote:
You can think anything you want, Raman. That's what distinguishes thoughts from proofs. 

20120609, 18:44  #10  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
2351_{8} Posts 
Quote:
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 = (2^{k+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 

20120609, 20:58  #11  
∂^{2}ω=0
Sep 2002
República de California
23140_{8} Posts 
Quote:
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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