My Math Forum Counter to the abc Conjecture

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 March 11th, 2019, 04:46 PM #1 Newbie   Joined: Feb 2019 From: Watertown NY USA Posts: 5 Thanks: 0 Counter to the abc Conjecture Counter to the abcCounter to the abc C Conjecture (version 14). [*]The abc conjecture states:[*] [*]1.) max (|a|, |b|, |c|) =< C_e PROD_{p|abc} p^(1+e) for any e > 0. [*]2.1) C_m has an upper bound. [*]Result (complete): [*]2.2) C_m has no upper bound nor limit. [*]proof. [*]given: [*]1.) max (|a|, |b|, |c|) =< C_e PROD_{p|abc} p^(1+e) for any e > 0. [*]Let [*]3.1) a = (s_1...s_n)^v where s_i is prime [*]3.2) b = (q_1...q_h)^v where q_j is prime. [*]where v any integer > 0. [*]3.3) a + b = c [*]3.4 ) GCD(a,b) = 1 [*]Let [*]4.1) a > 2[*]4.2) b > a [*](end givens)[*] [*]5.) If p_w | abc then p_w | ab(a+b) since a + b = c. [*]6. from 5.) p_w | [s_1...s_n][q_1...q_h][t_1...t_m ] where [*]where c = t_1^(r_1)...t_m^(r_m) = (s_1...s_n)^v + (q_1...q_h)^v >= t_1...t_m. [*]7. ) c / [t_1^(r_1-1)...t_m^(r_m-1)] = t_1...t_m. [*]8.) c =< max (|a|, |b|, |c|) =< C_e PROD_{p|abc} p^(1 +e). [*]9.) PROD_{p|abc} p^(1 +e) = [(s_1...s_n)(q_1...q_h)(t_1...t_m)]^(1+e) [*]10. from (8. and 9.) ) [c] / [(ab)^(1/v)]^(1+e) =< C_e. [*]11. from 10.) t_1^(r_1)...t_m^(r_m) / [(s_1...s_n)(q_1...q_h)(t_1...t_m)] ^(1+e) =< C_e [*]or [*]12. from 11.) t_1^(r_1)...t_m^(r_m) / [(ab)^(1/v)(t_1...t_m)] < C_e. [*]13. from 12.) c / [(ab)^(1/v)(t_1...t_m)] < C_e. [*]As v increases without bound then [*](ab)^(1/v) tends to 1 and c increases without bound nor limit. [*]14 from 13.) c /[(t_1...t_m)] < C_e. [*]Recall please c = t_1^(r_1)...t_m^(r_m). [*]If c increases without bound then c /[(t_1...t_m)] also increases w/out bound and therefore C_e hasn't a boundry or BOUND. [*]the proof complete. [*]Simon C. Robe[/LEFT][/LEFT][/LIST][/SIZE][/LIST][/LEFT]

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