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