בתשובה לדורון שדמי, 29/08/05 16:22
פרסום (?) למתמטיקה מונדית 326480
האם הוא פרסם את עבודתך בלי ששלחת אותה אליו?
פרסום (?) למתמטיקה מונדית 326488
בוודאי שלא.

ראה נא את תגובתי הקודמת אליך.

תודה,

דורון
ועוד חדשות על השערת פרמה 326796
http://www.physicsforums.com/showthread.php?t=82541&...

Fantastic idea for my friends

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Quote:
Originally Posted by Victor Sorokine
Condition at present

Fantastic idea for my friends

Right contradiction: the number u is infinite

(1°) Let a^n + b^n – c^n = 0,
(2°) where for integers a, b, c the number u = a + b – c > 0, where (a_1b_1c_1)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the digit u_{k+1} into 1.

(4°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(4a°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 0, then a_{k+1} + a_{k+1} – a_{k+1} = 1,
U"_{k+2} = a_{k+1} + a_{k+1} – a_{k+1} = 1 and the number U' contains only one non-zero digit (U'_{k+2} = 1).
Or: u is even, but a^n + b^n – c^n is odd, that is impossible.
(4b°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 1, then a_{k+1} + a_{k+1} – a_{k+1} = 0,
U"_{k+2} = 0 and U'_{k+2} = 1. Or: u is odd, but a^n + b^n – c^n even is, that is impossible.
Therefore there exists second non-zero digit in the number u: u_s.

(5°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(5a°) if ((a_(s) + b_(s) – c_(s))_{k+1} is odd, then u is even, but U"_{s+1} (and a^n + b^n – c^n) is odd, that is impossible.
(5b°) if ((a_(s) + b_(s) – c_(s))_{k+1} is even, then u is odd, but U"_{s+1} (and a^n + b^n – c^n) is even, that is impossible.
Therefore there exists third non-zero digit in the number u: u_r.

(6°) Let's assume…
AND SO AD INFINITUM

Victor Sorokine

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