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P: 894
 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.