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Old 2007-03-15, 01:10   #1
toilet
 
Mar 2007

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Default 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^p1 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^p1) for some prime of the form 2^p1.

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Old 2007-03-15, 07:22   #2
maxal
 
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See http://primes.utm.edu/notes/proofs/EvenPerfect.html
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Old 2007-03-15, 09:35   #3
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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
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Old 2007-03-15, 19:21   #4
ewmayer
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Good username, though. A bit odd, but nearly perfect, given the thread question.
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Old 2007-03-15, 20:35   #5
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Quote:
Originally Posted by akruppa View Post
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.
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Old 2012-06-09, 09:25   #6
Raman
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Quote:
Originally Posted by toilet View Post
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^p1 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^p1) for some prime of the form 2^p1.

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.
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Old 2012-06-09, 09:32   #7
Raman
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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 2012-06-09 at 09:40
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Old 2012-06-09, 15:06   #8
LaurV
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Quote:
Originally Posted by Raman View Post
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...
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Old 2012-06-09, 17:24   #9
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Quote:
Originally Posted by Raman View Post
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.
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Old 2012-06-09, 18:44   #10
Raman
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Quote:
Originally Posted by Batalov View Post
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
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Old 2012-06-09, 20:58   #11
ewmayer
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Quote:
Originally Posted by Batalov View Post
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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