Guest wrote:There are two cases to consider for [tex]\gamma[/tex]:
Case I : 0 < [tex]\gamma[/tex] < 1
Case 2: [tex]\gamma[/tex] >1
where
1. [tex](A + B) * (A * B)[/tex] = [tex]{( (A + B)/ \gamma )^{1 / \beta_{e}}} * (a * b * c)^{-1}[/tex] when [tex]\beta_{e}[/tex] > 1.
2. [tex]\gamma[/tex] = [tex](A + B) * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}[/tex] when [tex]\beta_{e}[/tex] > 1.
Further analysis will confirm the validity of the ABC-conjecture.
Guest wrote:Guest wrote:There are two cases to consider for [tex]\gamma[/tex]:
Case I : 0 < [tex]\gamma[/tex] < 1
Case 2: [tex]\gamma[/tex] >1
where
1. [tex](A + B) * (A * B)[/tex] = [tex]{( (A + B)/ \gamma )^{1 / \beta_{e}}} * (a * b * c)^{-1}[/tex] when [tex]\beta_{e}[/tex] > 1.
2. [tex]\gamma[/tex] = [tex](A + B) * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}[/tex] when [tex]\beta_{e}[/tex] > 1.
Further analysis will confirm the validity of the ABC-conjecture.
A + B < 2 * B since A < B.
3. [tex]\gamma[/tex] < [tex]2*B * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}[/tex] when [tex]\beta_{e}[/tex] > 1.
Guest wrote:WRONG!!!! 7. At best, 1 < [tex]\gamma[/tex] < 2 for Case 2 when [tex]\beta_{e}[/tex] > 1.
7. At best, 1 < [tex]\gamma / \alpha < 2[/tex] when [tex]\beta_{e}[/tex] > 1.
Note: [tex]\alpha = \prod_{j_2 =1}^{l_2}[/tex][tex]p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1} * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}]^{-\beta_{e}}[/tex]
Hmm...? This analysis needs some fixing.
Guest wrote:We recall [tex]0 < \gamma/\alpha < 2[/tex] and a former note:
[tex]\alpha = \prod_{j_2 =1}^{l_2}[/tex][tex]p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1} * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}]^{-\beta_{e}}[/tex].
But, since [tex]A = \lambda * B[/tex] for some [tex]0 < \lambda < 1[/tex], we have
[tex]\alpha = \lambda^{-1} * \prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1}^{k_{j_1} - 2\beta_{e}} * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{-\beta_{e}}[/tex].
Guest wrote:We can conclude according to the two cases for [tex]\gamma[/tex],
Case 1: There are infinitely many ABC-triples which satisfy the ABC-conjecture when 0 < [tex]\gamma[/tex] < 1.
and
Case 2: There are zero or finitely many ABC-triples which satisfy the ABC-conjecture when [tex]\gamma[/tex] > 1,
that the great ABC-conjecture is true! And we are confident that any further analysis will confirm our conclusion.
Thank Lord God! Amen!
Guest wrote:i personally doubt the abc conjecture is true for one inescapable reason: 1+9^n = c always results in a value for c/ rad function >1
Guest wrote:sorr slight correction
1+b = 9^n
Guest wrote:i personally doubt the abc conjecture is true for one inescapable reason: 1+b = [tex]9^{n}[/tex] always results in a value for c/ rad function >1
Guest wrote:Guest wrote:i personally doubt the ABC-conjecture is true for one inescapable reason: 1+B= [tex]9^{n}[/tex] always results in a value for C/ rad function >1
For what [tex]\beta > 1[/tex], is [tex]9^{n} / rad( 1 * (9^{n} - 1) * 9^{n})^{\beta} > 1?[/tex]
And are there infinitely many ABC-triples in total that satisfy the above question for all appropriate [tex]\beta[/tex] > 1?
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