[QUOTE=Batalov;359404]Exactly.
What proof? What are you talking about?[/QUOTE] Sorry, no more answers. I am insulted all the time and when I defend myself I am moved to Misc. I made last attempt to be here. Batalov, still I wish you good luck. Bye. 
An alternative "proof by dilemma"
[QUOTE=Batalov;359404]What proof? What are you talking about?[/QUOTE]
It seems he's talking about Euler's famous  and untrustworthy, because it relied on Herr Euler's "computation"  proof that 641 divides F5, thus (allegedly and untrustworthily) 'disproving' Fermat's conjecture that all Fn are prime. Allow me to present a simplified version of CRGreathouse's post #2 "untrustworthy alternative computeraided proof"  this one can more easily be done by hand, as it breaks things down into a "powering via repeated doubling and addition of 1 to result" step (Lemma 1) and a "multiplication of one number by another" step (Lemma 2). The mathematically fancypantsy around here may have heard of "proof by contradiction" (as in "you're wrong, thus I'm right.") ... the structure of my proof below is in the form of 2 lemmas  not to be confused with [url=http://en.wikipedia.org/wiki/Lemming]those furry little critters[/url] who famously and counterfactually have a reputation for mass cliffdiving (and which are close relatives of gerbils, as it happens)  thus "proof by dilemma", as it were. I hope that said decomposition will make it easier for the mathematical community to organize the extensive independent doublechecking effort needed to satisfy the OP of the validity of the (allegedly and untrustworthy) alternative proof. [b]Lemma 1:[/b] Let n = 2^7. Then n = 128, and F7 := 2^n+1 = 340282366920938463463374607431768211457. [b]Lemma 2:[/b] F7 = 59649589127497217 x 5704689200685129054721 . QED Is that at all understandable? Perhaps we should organize a special conference  is it too late to propose this as a lastminute addon to this year's WCNTC at Asilomar? 
[QUOTE=Batalov;359404]Exactly.
What proof? What are you talking about?[/QUOTE] I think he's talking about this paragraph from [url]http://en.wikipedia.org/wiki/Fermat_numbers:[/url] [QUOTE] The fact that 641 is a factor of F5 can be easily deduced from the equalities 641 = 2[sup]7[/sup]×5+1 and 641 = 2[sup]4[/sup] + 5[sup]4[/sup]. It follows from the first equality that 2[sup]7[/sup]×5 ≡ −1 (mod 641) and therefore (raising to the fourth power) that 2[sup]28[/sup]×5[sup]4[/sup] ≡ 1 (mod 641). On the other hand, the second equality implies that 5[sup]4[/sup] ≡ −2[sup]4[/sup] (mod 641). These congruences imply that −2[sup]32[/sup] ≡ 1 (mod 641). [/QUOTE] This does show what it's trying to with only arithmetic on numbers no bigger than 641, which is admittedly somewhat clever. However, I don't think this can claim to be any more of a [I]proof[/I] than simply multiplying the two factors. 
[QUOTE=jyb;359432]However, I don't think this can claim to be any more of a [I]proof[/I] than simply multiplying the two factors.[/QUOTE]
Exactly. Similarly, here's a "proof", that 9=3*3. "Proof": 3*3=(1+2)^2=1^2+2*1*2+2^2=9. Alleged added value of the "proof": it operates only with numbers no more than 2. While direct multiplication deals with larger numbers (i.e. 3). 
[QUOTE=ewmayer;359431][b]Lemma 2:[/b] F7 = 59649589127497217 x 5704689200685129054721 .[/QUOTE]
I have through exhaustive labors succeeded in verifying part 2 of my proof by dilemma. We first tabulate multiples of the larger (purported) multiplicand by 09: 5704689200685129054721 x ____ = [code] 0 0 1 5704689200685129054721 2 11409378401370258109442 3 17114067602055387164163 4 22818756802740516218884 5 28523446003425645273605 6 34228135204110774328326 7 39932824404795903383047 8 45637513605481032437768 9 51342202806166161492489[/code] And here is the resulting (no FFTs, DWTs or other suspicious 3letterinitialismdenoted computer flummery allowed here) multiplication rhombus, with blank spaces denoting 0s. Columnwise addition gives  someone please doublecheck my carries!  the indicated sum, matching the desired result: [code]59649589127497217 x 5704689200685129054721: 5: 28523446003425645273605 9: + 51342202806166161492489 6: + 34228135204110774328326 4: + 22818756802740516218884 9: + 51342202806166161492489 5: + 28523446003425645273605 8: + 45637513605481032437768 9: + 51342202806166161492489 1: + 5704689200685129054721 2: + 11409378401370258109442 7: + 39932824404795903383047 4: + 22818756802740516218884 9: + 51342202806166161492489 7: + 39932824404795903383047 2: + 11409378401370258109442 1: + 5704689200685129054721 7: + 39932824404795903383047  Sum= 340282366920938463463374607431768211457 [Nonzero Carries as noted: 11112332345456777777886775664432332 ][/code] Whew! I need a beer. Lemma 1 may need to wait a few days. Hey, even Hercules [or "Herakles" to the OP] needed to rest between his famous labors, I'll bet. 
...and if you did it in octal, you would have been done already, because calculation of 2^128+1 in octal doesn't need a separate calculation (no Lemma 1!).
[CODE]First, prepare multiples of 3237257607274243001: 3237257607274243001 (x1) 6476537416570506002 (x2) 11736017226064751003 (x3) 15175277035361214004 (x4) 20434556644655457005 (x5) 23674036454151722006 (x6) 27133316263446165007 (x7) Now, add them in a staircase 3237257607274243001 1152401672664431414535001 * _________________________ 3237257607274243001 20434556644655457005 11736017226064751003 20434556644655457005 15175277035361214004 3237257607274243001 15175277035361214004 3237257607274243001 11736017226064751003 15175277035361214004 15175277035361214004 23674036454151722006 23674036454151722006 6476537416570506002 27133316263446165007 23674036454151722006 3237257607274243001 15175277035361214004 6476537416570506002 20434556644655457005 3237257607274243001 3237257607274243001 _______________________________________________ 4000000000000000000000000000000000000000001[/CODE] 
There you go, invoking mysterious and highly untrustworthy "computational magic" again. :)

Your posts are real eyes opener for me. In fact I learned today much more about people of this forum than in the past 3 years.
Batalov, you quoted me wrongly. You wrote 255 instead of 2^55. Someone could think that I made a mistake. Of course you cannot correct it, but be careful next time. BTW. No matter how much you write these proofs (i.e. about F5, F6, and F7) will stay my proofs and the only thing left for you it will be to write how you can divide 2 numbers or to multiply 2 numbers. 
Of course, they will. They will stay a monument to how one can scratch one's [B]left[/B] ear not simply with the right hand, but more elegantly  with the toe on one's right foot.
On to the same Herculean task for F8, then? Is it already in the plans? 
Here's how one "scratches left ear with the left hand".
(All you need is a pencil and one sheet of paper for up to F7 and if your handwriting is neat enough, with space to spare for the F8.) [CODE]Lemma 5A. 641 divides 2^32+1. Proof: 2^8 = 256. Let's square this value two more times modulo 641, and compare to 6411. (256^2)%641 = 154 (154^2)%641 = 640. QED. Lemma 6A. 274177 divides 2^64+1. Proof: 2^16 = 65536. Let's square this value two more times modulo 274177, and compare to 2741771. (65536^2)%274177= 258768. (258768^2)%274177= 274176. QED. Lemma 7A and so on. Same thing over and over again.[/CODE] 
[QUOTE=Batalov;359450]Of course, they will. They will stay a monument to how one can scratch one's [B]left[/B] ear not simply with the right hand, but more elegantly  with the toe on one's right foot.
On to the same Herculean task for F8, then? Is it already in the plans?[/QUOTE] I wrote "no more answers". I can make exception this time. For a while I thought I had similar proof for largest known factor of F12. You might notice that proofs for F6 and F7 are based on the same concept. I noticed some regularities. Since there are high degree polynomials let me introduce some abbreviations: Instead of a polynomial (3)*x^2+4*x5 I will write (3)(4)(5). Take 2 polynomials {1){0}(1)(1)(1)(1)(0)(1)(0)(1)(0)(2)(2)(2)(1)(1)(1)(1)(0)(1)(0)(0)(0)(1) and (1)(0)(1)(1)(0)(1)(0)(0)(0)(1) Product of these polynomials is (1)(0)(0)(0)(0)(0)(0)(0)(0)(0)(0)(0)(1)(5)(5)(5)(5)(5)(5)(4)(0)(0)(0)(0)(0)(0)(0)(0)(0)(0)(0)(1) The value of the last polynomial for the point x=4 is F6. Hence values of first 2 polynomials must be factors of F6. What nice in this it is that last polynomial is nice looking almost symmetric polynomial. 
All times are UTC. The time now is 07:28. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2022, Jelsoft Enterprises Ltd.