Fermat’s Last Theorem: A one-operation proof

ramsey2879

It is possible to choose a "d" where ad = n^s-1 but r < k+2. If this is the case, then, for instance if r=2 then d must be multiplied by 10101...01 where the number of zeros between the 1's = r-1 to get a new R. It thus is possible to chose a "d" such that R > is any multiple of r > k+2.


statement 2 is false
let a= 251, b = 326502, c=53 base 7
Then a+b-c=330000; thus d = 2020202 and k=4, r=8
A= 2020202*a=540404032, B=2020202*b= 666266660004, C=2*c=140404036

You state that for k<i<r [i.e. 4<i<8] that A_i + B_i + C_i =0, but this is false.
You should use variables consistently and watch your statements. Also please do not make us have to refer to your earlier proof to understand what you are talking about in the current proof.

moshek

[QUOTE=ramsey2879]


Hi ramsey2879

thank you for your complicate example
You probably mean A_i+B_i - C_i = 0
let's see Victor replay to your contra example

But Isn't Victor FLT contradiction (?) appear only at the base n
which is the power of the original equation a^n+b^n=c^n.



Moshe

Victor Sorokine

Dear ramsey2879, Hurkyl, robert Ihnot, learningphysics and moshek,
there is a question: is it THE END:

Start-situation:
Let a^n + b^n – c^n = 0 (1°)
u = a + b – c with k-zero-ending (k > 0);
r – maximum rank of c^n (1a°);

PROOF

1. Let’s multiply the equation (1°) by dn , where d = 1 + gn^r, in order to transform the digit
u_{r + k + 1} into 1 (2°). (The digits of u_(r + k) do not change.)
2. Now from
[a_(r + k) + a_{r + k + 1}]^n + [b_(r + k) + b_{r + k + 1}]^n + [c_(r + k) + c_{r + k + 1}]^n = 0 we have:
{[a_(r + k)] _(r + k + 1) + [b_(r + k)] _(r + k + 1) – [c_(r + k)] _(r + k + 1)} +
+ (n^(r + k + 2))(a_{r + k + 1} x 1 + b_{r + k + 1} x 1 – c_{r + k + 1} x 1) + (n^(r + k + 3))P = 0, where
{[a_(r + k)] _(r + k + 1) + [b_(r + k)] _(r + k + 1) – [c_(r + k)] _(r + k + 1)} = 0 (cf. 1a°) and
(a_{r + k + 1} x 1 + b_{r + k + 1} x 1 – c_{r + k + 1} x 1)_1 = u_{r + k + 1}.
From here u_{r + k + 1} = 0, that contradicts to (2°).
The proof is done.

Victor Sorokine

Hurkyl

So? All you need to do is to (correctly) derive one contradiction in order to prove that one of your original assumptions must be false.

moshek

Do you think, that Victor prove for FLT is wrong ?

Moshe Klein

moshek

hi victor

well, even A.willes had a mistak in FLT untill he fix it ( I hope)
Are all these calculation are in base n ?


Thank you
Moshe Klein

moshek

ramsey2879 hi

I am still study victor work on FLT , I know that it is Organic and very nice but I dont have a final opinion if it is really correct with no mistakes.

Does post 71 answer to your claim that statement 2 is palse ?

Thank you
Moshe Klein

Victor Sorokine

Dear Moshe,
thank you and excuse my mistake.
There is interesting correction.

It is obvious:

If a^n + b^n – c^n = 0 (1°), then there is such s, that
/(a + n^(–s))^n + (a + n^(–s))^n – (a + n^(–s))^n/ < n^(–sn) (2°), where s > r + 1 and r is maximum rank of the number c, and
/(an^s + 1)^n + (bn^s + 1)^n – (cn^s + 1)^n/ < 1 (3°).
From here: n^sn(a^n + b^n – c^n) = 0, nn^[s(n – 1)](a^(n – 1) + b^n(n – 1) – c^n(n – 1)) = 0
{or a^(n – 1) + b^(n – 1) – c^(n – 1) = 0}, … a + b – c = n^(–s) 0, that is impossible.
V.S.

moshek

Dear Victor,

I am really like your non conventual's way of thinking. for the first time I find real interest in FLT because of your organic attitude that can explain ( I hope so ) why Fermat wrote what he wrote about this problem ! if your prove is right [ I don't know that yet ] you should present it at the icm exactly 365 day's from today .

Please look at http://www.icm2006.org/

I hope to have my final opinion about your nice work at October
but since you correct it already few times in this thread ( and that's fine )can you sent here the most update version of your work.


Thank you
Moshe

Victor Sorokine

Counter-examples are done for such u where u'_{k + 1} = 0.
But in my published proof u'_{k + 1} =/ 0.(Not infrequently, to find the error in a counter-example it is more difficult then to solve the same problem.)
For 10 months 70 mathematicians (members of AMS) and
2000 amateurs could not find an error of principle in the demonstration.
I terminate the participation in the discussion about all following proofs.
So the discussion about the original proof continues.

***
A key of the proof:
For a and b:
if a + b = (a + b)_(k) = 0, then (a^n + b^n)_(k + 1) = 0,
if a + b = (a + b)_(k) =/ 0, then (a^n + b^n)_(k + 1) =/ 0 (cf. ADDENDUM).
For a, b, c:
if a + b – c = (a + b – c)_(k) = 0, then (a^n + b^n – c^n)_(k + 1) = 0 (cf. counter-examples),
if a + b – c = (a + b – c)_(k) =/ 0, then (a^n + b^n – c^n)_(k + 1) =/ 0 (cf. my proof).

Publications of the proof:
doc: Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm
pdf: http://fox.ivlim.ru/docs/sorokine/flt.pdf

Correction in (9°) (cf. the proof):
(9°) u''_{k+2} = [v + (a_(k+1) + b_(k+1) – c_(k+1))_ {k+2}]_1.
Cause of the error: erroneous copying.
Victor

Hurkyl

Don't confuse "could not find an error" with "didn't bother looking for an error". Most people don't bother trying to fill in the gaps of an argument for which they have little to no expectation of validity.

The problem is that you don't actually write a proof: you write a bunch of equations, occasionally with a brief comment, and very few, if any, are self-evident.

moshek

Don't forget the story of Ramanujan who could not prove his 4000 amassing results. [ few of them are wrong ] Victor have a new interesting vision about FLT and he may need the help and the giddiness of the 'expert' to expand / check / correct / or maybe unfortunately cancel his attitude.

please read the strange situation Perelman paper on Poincaré conjecture : ( was taken from www.mathworld.com )


And also from there about the strange situation Mihailescu paper on the Catalan problem:


So I think that Victor ask for meangfull interaction with his work, I will try my best to have my final opinion on his work on FLT until October.

Moshe

Victor Sorokine

The simplest case for analysis: u_{k + 1} = u'_{k + 1} + u''_{k + 1} = 1

If u'_{k + 1} = 1, then u''_{k + 1} = 0, U'_{k + 2} =/ 0 (?), U''_{k + 2} = 0 and U' + U'' =/ 0;
if u''_{k + 1} = 1, then u'_{k + 1} = 0, U'_{k + 2} = 0 (?), U''_{k + 2} =/ 0 and U' + U'' =/ 0.

Hurkyl

I highly doubt that: as I told Victor, don't confuse "did not prove" with "cannot prove". Remember that many of his 4000 amazing results were in his private notebooks, not in papers he was trying to publish, so there wasn't any real reason to give a rigorous proof.


And mathematicians have not described Victor's paper as "well thought out", and have not stated they expect it to be correct, so I don't see your point. It sounds like Perelman actually tried to write a proof, rather than an outline he claims to work, as Victor has done.


And I have no idea what point you're trying to make with this one.

Victor Sorokine

Specially for MocheK

(1°) Let (an^t)^n + (bn^t)^n – (cn^t)^n = 0, where t >> maximum rank R(c) = r.
From here
(2°) n^r > (an^t)^(n – 1) + (bn^t)^ (n – 1) – (cn^t)^ (n – 1) > 0,
and
(3°) (an^t + 1)^n + (bn^t + 1)^n – (cn^t + 1)^n > 0 and hence
(4°) (an^t + e)^n + (bn^t + 1)^n – (cn^t + 1)^n = 0, where e < 1.
From (4°) we have:
(5°) n^t(a^n + b^n – c^n) + n^[t(n – 1) + 1][a^(n – 1) + b^(n – 1) – c^(n – 1)e] + P = 0,
where P < n^[t(n – 2) + r] << n^[t(n – 1) + 1].
From (5°) we have: [a^(n – 1) + b^(n – 1)]_(t) = [c^(n – 1)e]_(t), or
(6°) a^(n – 1) + b^(n – 1) = c^(n – 1)e.
But it's easy to show that it is impossible!!!

Victor

Victor Sorokine

Condition at present

1. The last (from 1 to {k+s}) and first (from {r – k – s + 2}) digits of the number u do not show the contradiction.
2. Counter-examples are right.
3. My error: the digit u_{k+2} shows error.
4. The right solution: the digit u_{k + s + 1} shows error (k + s is maximum rank of the number u).

Deciding Lemma:
If U_(t + 1) = 0, then U*' _{t + 1} = U_{t + 1} (corollary from 21° - 25° in the Proof).

Idea of the new proof:
After transformation of the digit u_{k + s + 1} into 1 (with multiplication by 1 + n^(s+1)), we see that
U* _{ k + s + 1} = U*'' _{ k + s + 1} = 1 and hence U*'_{ k + s + 1} = –1.
Therefore up to multiplication U' _{ k + s + 1} = –1 and hence U _{ k + s + 1} = –1 =/ 0.

Let's come back to the discussion of the original proof in new version.
Dear Hurkyl, moshek, Robert Ihnot and Ramsay2879,
I would like to thank you all for your active participation and for your help.
Victor

moshek

Dear Victor,

Please call me Moshe. Since 1980 my investigation is about the Organic unity of mathematics, and it's hidden connection to physics ( Please read the end of Hilbert lecture at Paris 1900).

I would like now to start and studies carefully all the details your interesting Organic work about FLT at your web-site :

www.fmatem.moldnet.md/1_(v_sor_05).htm

Did you correct there already your mistake?

Another very interesting attitude to FLT I found yesterday here:

http://noticingnumbers.net/FLTsummary.htm

Yours
Moshe