20131114, 03:24  #1 
Mar 2010
C0_{16} Posts 
Number 59649589127497217 is a factor of Fermat number F7
Recently I have written a page http://www.literka.addr.com/mathcoun...mth/proof4.htm with a proof that59649589127497217 is a factor of Fermat number F7. It is something similar to my previous work about F5 and F6 (see my posts in this section "Factoring" or visit my description page http://www.literka.addr.com/mathcountry.
Unfortunately factors of F7 are large numbers, hence some computation had to be with numbers of same size. Still these numbers are incomparably smaller than F7. Last fiddled with by literka on 20131114 at 03:26 
20131115, 01:37  #2 
Aug 2006
3×1,993 Posts 
Code:
Mod(2,59649589127497217)^2^7+1 
20131115, 08:42  #3 
Mar 2010
2^{6}·3 Posts 

20131115, 14:16  #4 
Aug 2010
Kansas
547 Posts 
He was giving a simple code to accomplish the same thing as your proof. Except his version goes straight to (11) in yours, negating the need for, well, your entire "proof". Might I ask why you are "proving" known Fermat divisors? Do you think this will give some unknown insight into Fermat divisors to allow one to predict what unknown divisors are?

20131115, 14:47  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×3×1,093 Posts 
Even one my my lessor minions can do this longhand
Code:
5704689200685129054721  59649589127497217)340282366920938463463374607431768211457 298247945637486085  420344212834523784 417547123892480519  279708894204326563 238598356509988868  411105376943376953 357897534764983302  532078421783936517 477196713019977736  548817087639587814 536846302147474953  119707854921128616 119299178254994434  408676666134182074 357897534764983302  507791313691987723 477196713019977736  305946006720099871 298247945637486085  76980610826137867 59649589127497217  173310216986406506 119299178254994434  540110387314120728 536846302147474953  326408516664577521 298247945637486085  281605710270914361 238598356509988868  430073537609254934 417547123892480519  125264137167744155 119299178254994434  59649589127497217 59649589127497217  0 
20131115, 15:10  #6  
Mar 2010
2^{6}×3 Posts 
Quote:
It is proof not a "proof". The purpose is obvious, so obvious that I did not mention if: to find a proof that something is a factor with no help of a computer and avoid timeconsuming computations as presented by Retina. In mathematics something computed by a computer is not regarded as a proof. This problem arose with a "proof" of Four Colors Theorem. Many mathematicians regarded this as not a proof because of intense using of computers. I suspect that my proof of F7 is not best possible i.e. there must be a proof more straightforward, using less computations. Last fiddled with by literka on 20131115 at 15:12 

20131115, 15:22  #7 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6558_{10} Posts 

20131115, 15:27  #8 
Mar 2010
2^{6}×3 Posts 

20131115, 15:44  #9 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}·3·311 Posts 

20131115, 16:09  #10 
Mar 2010
192_{10} Posts 
I don't want to ague with you about it, it is not a subject of this post. No statistics is made. I heard the story, but it is only a story, I am not sure that it is a true story. And the story is this: Long time ago it was announced that a proof of Four Colors Theorem was found. It was presented in Finland and I talked to mathematicians, who were there. They did not accept this proof because of use of computers. They even told that the method was known long before. Presenters just wrote a program to verify what was known before. This is everything I know about it. I heard this from a second hand so I cannot be sure that it is true. 
20131115, 16:11  #11  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
4278_{10} Posts 
Quote:
Since you're already working from the prior knowledge that the factor exists, why not take both factors and multiply them together? That'd be a much less difficult feat. Last fiddled with by MiniGeek on 20131115 at 16:16 

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