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-   -   Number 59649589127497217 is a factor of Fermat number F7 (https://www.mersenneforum.org/showthread.php?t=18877)

 literka 2013-11-15 20:22

[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.

 ewmayer 2013-11-15 23:08

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 computer-aided 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 fancy-pantsy 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 cliff-diving (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 double-checking 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 last-minute add-on to this year's WCNTC at Asilomar?

 jyb 2013-11-15 23:21

[QUOTE=Batalov;359404]Exactly.

What proof? What are you talking about?[/QUOTE]

[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.

 Batalov 2013-11-15 23:31

[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).

 ewmayer 2013-11-16 00:49

[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 0-9: 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 3-letter-initialism-denoted computer flummery allowed here) multiplication rhombus, with blank spaces denoting 0s. Columnwise addition gives - someone please double-check 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.

 Batalov 2013-11-16 00:57

...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]

 ewmayer 2013-11-16 01:01

There you go, invoking mysterious and highly untrustworthy "computational magic" again. :)

 literka 2013-11-16 01:28

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.

 Batalov 2013-11-16 01:42

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?

 Batalov 2013-11-16 02:09

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 641-1.
(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 274177-1.
(65536^2)%274177= 258768.
(258768^2)%274177= 274176. QED.

Lemma 7A and so on. Same thing over and over again.[/CODE]

 literka 2013-11-16 02:18

[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*x-5 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.

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