Searching for a valid proof of the abc Conjecture

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Sat Sep 01, 2018 10:41 pm

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

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Mon Sep 10, 2018 3:35 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Tue Sep 11, 2018 2:17 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Tue Sep 11, 2018 2:53 pm

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.


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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Tue Sep 11, 2018 3:10 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Wed Sep 12, 2018 1:19 pm

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.


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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Wed Sep 12, 2018 2:47 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 12:25 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 12:44 pm

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].


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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 1:10 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 1:41 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 1:43 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 3:54 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 4:57 pm

sorr slight correction
1+b = 9^n
Guest
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 5:00 pm

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


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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 5:11 pm

Guest wrote:sorr slight correction
1+b = 9^n


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

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 5:29 pm

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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 5:30 pm

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

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 6:11 pm

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} - 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
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 7:04 pm

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?
Guest
 

PreviousNext

Return to Number Theory



Who is online

Users browsing this forum: No registered users and 2 guests