Counter to the abc Conjecture

Feb 2019
5
0
Watertown NY USA
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]
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