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March 11th, 2019, 04:46 PM   #1
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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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