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## Fermat’s Last Theorem: A one-operation proof

 Quote by Victor Sorokine The proof contains 3 tools only: 1) Scale of notation in base prime; 2) Lemma: for number a (there last digit a_1 =/ 0) there are such d and s, that ad = n^s – 1 (corollary from Little Fermat's theorem). After multiplication u (= a + b – c) by p {sic"d"} in new Fermat's equation the number u has the form (in base 7): 666…666000…000 with r digits equal to n – 1 or "9" (with k zeros in the end; r > k + 2).
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.

 Quote by {continued} 3) The number h = (c – u)/u > 0. From c > a and c > b we have: 2c > a + b, c > a + b – c, c – u > 0, h > 0. THE PROOF of FLT 1. Let a^n + b^n – c^n = 0 (1°). 2. For any rank i, where k < i < r, it fulfils the equality: a_i + b_i – c_i = 0, and for rank r this equality has a form: (a_r + b_r – c_r)_1 = n – 1 = «9».

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.

[QUOTE=ramsey2879]

 statement 2 is false You state that for k

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
 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

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 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.
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.

 Quote by 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.

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

Moshe Klein

 Quote by 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

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

 Quote by 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
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
 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.
 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
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

 Quote by 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.
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 )

 The Clay Mathematics Institute included the conjecture on its list of \$1 million prize problems. In April 2002, M. J. Dunwoody produced a five-page paper that purports to prove the conjecture. However, Dunwoody's manuscript was quickly found to be fundamentally flawed (Weisstein 2002). A much more promising result has been reported by Perelman (2002, 2003; Robinson 2003). Perelman's work appears to establish a more general result known as the Thurston's geometrization conjecture, from which the Poincaré conjecture immediately follows (Weisstein 2003). Mathematicians familiar with Perelman's work describe it as well thought-out and expect that it will be difficult to locate any substantial mistakes (Robinson 2003, Collins 2004). In fact, Collins (2004) goes so far as to state, "everyone expects [that] Perelman's proof is correct."

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

 ..Finally, on April 18, 2002, Mihailescu sent a manuscript purporting to prove the entire conjecture to several mathematicians. The paper was apparently also accompanied by an expository paper by colleague Yuri Bilu that analyzes and summarizes Mihailescu's argument (van der Poorten 2002).

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
 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.

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 Don't forget the story of Ramanujan who could not prove his 4000 amassing results.
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.

 please read the strange situation Perelman paper on Poincaré conjecture : ( was taken from www.mathworld.com )
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 also from there about the strange situation Mihailescu paper on the Catalan problem:
And I have no idea what point you're trying to make with this one.
 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
 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
 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

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