Searching for a valid proof of the abc Conjecture

[Guest]

Hmm... Of course, the ABC-conjecture is right! Thank Lord God! Amen!

[Guest]

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]

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]

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.

4. [tex]\gamma[/tex] < [tex]2*\prod_{j_2 =1}^{l_2}[/tex][tex]p_{j_2}^{k_{j_2}} * [\prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1} * \prod_{j_2 =1}^{l_2}[/tex][tex]p_{j_2} * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}]^{-\beta_{e}}[/tex] when [tex]\beta_{e}[/tex] > 1.

5. [tex]\gamma[/tex] < [tex]2*\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] when [tex]\beta_{e}[/tex] > 1.

6. [tex]\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] < [tex]\gamma[/tex] < [tex]2*\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] when [tex]\beta_{e}[/tex] > 1.

7. At best, 1 < [tex]\gamma[/tex] < 2 for Case 2 when [tex]\beta_{e}[/tex] > 1. Right?

[Guest]

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]

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.

7. Tentatively, 1 < [tex]\gamma / \alpha < 2[/tex] when [tex]\beta_{e}[/tex] > 1. That's better! But things will probably change soon.

Note:

[tex]A < (A + B)/2 < B[/tex]

or

[tex]A^{2} < (A + B)* A/2 < B^{2}[/tex]

[Guest]

Moreover, [tex]A < (A + B)/2 \le A/\delta_{1} + B/\delta_{2} < B[/tex] for some positive composites, [tex]\delta_{1}[/tex] and [tex]\delta_{2}[/tex], which are products of prime powers according to the Fundamental Theorem of Arithmetic.

The idea is to find how much difference there is between A and B in terms of product of prime powers. And we seek to construct a limiting tool to help us solve our problem.

[Guest]

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]

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].

Moreover, [tex]A < B < A + B[/tex],

therefore [tex]A = \mu^{-1} * (A + B)[/tex] for some [tex]0 < \mu < 1[/tex].

Thus, we have:

[tex]\alpha = \mu^{-1} * \lambda^{-1} * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta_{e}}[/tex].

[Guest]

Hence, we have:

I. [tex]0 < \gamma < 2 * \mu^{-1} * \lambda^{-1} * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta_{e}}[/tex]

for some [tex]0 < \mu < 1[/tex] and for some [tex]0 < \lambda < 1[/tex].

Hmm... The exponents or powers of distinct primes, [tex]p_{j_{3}}[/tex], are constrained by the power or exponent, [tex]-3\beta_{e}[/tex]... .

[Guest]

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 than any further analysis will confirm our conclusion.

Thank Lord God! Amen!

[Guest]

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]

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]

sorr slight correction
1+b = 9^n

[Guest]

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

Does gcd(b = [tex]9^{n}[/tex], c = 1 + [tex]9^{n}[/tex]) = 1 ?

When we speak of ABC-triples in regards to the ABC-conjecture, we assume A, B, and C are coprimes (gcd(A, B) = gcd(A, C) = gcd(B, C) =1) such that A + B = C.

[Guest]

sorr slight correction
1+b = 9^n

Again, does gcd(b, [tex]9^{n}[/tex]) = 1?

[Guest]

Hence, we have:

I. [tex]0 < \gamma < 2 * \mu * \lambda^{-1} * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta_{e}}[/tex]

for some [tex]0 < \mu < 1[/tex] and for some [tex]0 < \lambda < 1[/tex].

Hmm... The exponents or powers of distinct primes, [tex]p_{j_{3}}[/tex], are constrained by the power or exponent, [tex]-3\beta_{e}[/tex]. And rational factors, [tex]\mu[/tex] and [tex]\lambda[/tex], are important too...

Note: [tex]\mu^{-1}[/tex] is wrong factor for statement I and is corrected.

[Guest]

well yes of course. 9^n -1 will never have the any of the same factors as 9^n.

[Guest]

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} - 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?

[Guest]

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?