20110507, 17:43  #1 
Apr 2009
2 Posts 
oddperfect numbers don't exist
Hello, Readers.
I have proven that oddperfect numbers don't exist; it's a 1page proof that combines Euler's work with that of Jacques Touchard's 1953paper. Anyone with the knowledge of basic algebra can enjoy it. The answer has been right under our noses!; please visit the root page of my website... www.oddperfectnumbers.com to see for yourself; no additional math is needed to construct this very valid proof. I've also included proofs of both Zarankiewicz's and Richard K. Guy's crossing number formulas on the page Other Short Proofs; just click on the heading to view them. Best Regards, Bill Bouris 
20110507, 20:16  #2 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
9559_{10} Posts 
This is was posted in the wrong part of the forum.
That makes me think that the proof may have problems too. 
20110507, 20:43  #3 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20110507, 21:21  #4 
Aug 2006
3^{2}×5×7×19 Posts 
Paragraph 5 seems to contain the first mistake: "the first factor of the sum is always equal to onethird the size of the number being compared". The previous paragraphs have no mathematical content relating to odd perfect numbers.

20110507, 22:33  #5 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
I have only skimmed the web page. Would you say this is Misc. Math. material?

20110507, 22:39  #6 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 

20110507, 22:47  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
okay Q=3 allows it Q=6n+1 or 6n1 means Q^2 is of the form 6n+1 to make it divisible by three (4x+1)^(4y+1) must be 0 mod 3: 1,5,9 = 1,2,0 mod 3 so the base doesn't always allow 0 mod 3 so next to the exponents so to ensure the biggest factor is 1/3 of the values of the number 1^(4x+1) ,2^(4y+1) , and 0^(4y+1) , must all be 0 mod 3 , and they aren't. even replacing Q=2 doesn't fix that 1^(4y+1) will always be one and hence with Q=2 it will end up as four ( and yes I know that 4 is already not included). of course I'm not familiar to all requirement of the equation you talk of. Last fiddled with by science_man_88 on 20110507 at 22:52 

20110507, 22:53  #8 
Aug 2006
3^{2}·5·7·19 Posts 

20110507, 23:07  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20110508, 03:26  #10 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Moved to Misc. Math.

20110508, 03:27  #11 
Aug 2006
3^{2}×5×7×19 Posts 
Science man, I'm not sure what you're trying to show. Since you're talking about an odd perfect number Q must be odd, and hence Q ≠ 2. It's know that Q is divisible by at least eight different primes so in particular Q ≠ 3. 4x+1 is prime, so (4x+1)^(4y+1) is not 0 mod 3.

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