Searching for a valid proof of the abc Conjecture

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 8:12 pm

Guest wrote:
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?


Example:

The one ABC-triple, [tex](A = 1, B = 9^{5} -1, C = 9^{5})[/tex] satisfy our question when [tex]1 < \beta < 1.32354 (approximately)[/tex] according to the Wolfram Alpha Calculator.

Relevant Reference Links:

http://www.wolframalpha.com/input/?i=factor(9%5E5+-1)

http://www.wolframalpha.com/input/?i=9%5E5%2F(2+*+11+*+61*3)%5E%CE%B2+%3E+1

Please keep searching for more such ABC-triples since infinity is far bigger than any number imaginable.
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Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 9:08 pm

can you not do basic math?

the 9^n -1 will be divisible by 8 at a minimum and 9^n is 3 for the rad function. thus at worst you have 9^n/6*k where k is (9^n -1)/8.
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Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Sep 13, 2018 10:09 pm

Guest wrote:can you not do basic math?

the 9^n -1 will be divisible by 8 at a minimum and 9^n is 3 for the rad function. thus at worst you have 9^n/6*k where k is (9^n -1)/8.
Guest
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Sat Sep 15, 2018 11:32 am

On the ABC-Conjecture:

Please refer to the following excellent link to obtain detailed information about the authors of the great ABC-conjecture and to obtain a detailed explanation of the ABC-conjecture.

https://www.revolvy.com/page/Abc-conjecture?uid=0
Guest
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Mon Oct 01, 2018 3:46 pm

On the ABC-conjecture, here's another excellent reference link:

https://arxiv.org/pdf/1409.2974.pdf
Guest
 

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Tue Oct 02, 2018 3:29 pm

Guest wrote: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]...
.



Since [tex]0 < \mu^{-1} * \lambda^{-1} \le 1[/tex], we have:

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

Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Tue Oct 02, 2018 4:26 pm

Guest wrote:
Guest wrote: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]...
.



Since [tex]0 < \mu * \lambda^{-1} \le 1[/tex], we have:

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

Re: Sea l crching for a valid proof of the abc Conjecture

Postby Guest » Tue Oct 02, 2018 9:53 pm

Guest wrote:
Guest wrote:
Guest wrote:Hence, we have:
I
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]...
.



Since [tex]0 < \mu * \lambda^{-1} < 1[/tex] because A < B , we have:

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

Re: n I a Searching for a valid proof of the abc Conjecture

Postby Guest » Wed Oct 03, 2018 4:17 pm

... Since [tex]1/2 < \mu * \lambda^{-1} < 1[/tex] because A < B and because at maximum, A = B - 1, we have:

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

Re: n I a Searching for a valid proof of the abc Conjecture

Postby Guest » Sat Oct 06, 2018 11:38 am

Guest wrote: ... Since [tex]1/2 < \mu * \lambda^{-1} < 1[/tex] because A < B and because at maximum, A = B - 1, we have:

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


We have considered the two cases for [tex]\gamma[/tex] according to the ABC-conjecture in more detail (please see statement I* above):

Case 1: 0 < [tex]\gamma[/tex] < 1.

Case 2: [tex]\gamma[/tex] > 1.

with:



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.
Guest
 

Re: Searching for a valid proof of the ABC-conjecture

Postby Guest » Sat Oct 06, 2018 6:08 pm

... Since [tex]1/2 < \mu * \lambda^{-1} < 1[/tex] because A < B and because at maximum, A = B - 1, we have:

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


We have considered the two cases for [tex]\gamma[/tex] according to the ABC-conjecture in more detail (please see statement I* above and the two cases below).

And we have concluded 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! :)

Note:

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.
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Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Sun Oct 07, 2018 11:46 am

Notes:

A, B, and C are positive integers such that A < B < C = A + B;

A = [tex]\lambda[/tex] * B where 0 < [tex]\lambda[/tex] < 1;

A = [tex]\mu[/tex] * C where 0 < [tex]\mu[/tex] < 1;

gcd(A, B) = gcd(A, C) = gcd(B, C) = 1;

And according to the Fundamental Theorem of Arithmetic, we have:

A = [tex]\prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1}^{k_{j_1}}[/tex];

B = [tex]\prod_{j_2 =1}^{l_2}[/tex][tex]p_{j_2}^{k_{j_2}}[/tex];

C = [tex]\prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3}}[/tex].

Moreover, we have:

a * A = [tex]\prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1}[/tex];

b * B = [tex]\prod_{j_2 =1}^{l_2}[/tex][tex]p_{j_2}[/tex];

c * C = [tex]\prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}[/tex].
Guest
 

Re: Searching for a valid proof of the ABC-conjecture

Postby Guest » Sun Oct 07, 2018 8:36 pm

"ABC-Conjecture (Masser-Oesterlé, 1985):

Let [tex]\beta[/tex] > 1. Then, with finitely many exceptions, we have C < rad[tex](ABC)^{\beta}[/tex] "
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Re: Searching for a valid proof of the abc Conjecture

Postby Guest » Thu Oct 11, 2018 5:58 am

The abc-conjecture has been proved, since 2013, 5 years ago. The whole math community knew that.

Here is the proof: [Page:13]
Riemann Shatters The Gordian Knot.2018.
https://hal.archives-ouvertes.fr/hal-01852157v1

or here: [Page:2-4]
Stealth Elliptic Curves and The Quantum Fields.2013.
https://hal.archives-ouvertes.fr/hal-01513663v2

The question which remains is are we honest enough to recognize that publicly?
Guest
 

Re: Searching for a valid proof of the ABC-conjecture

Postby Guest » Sun Oct 14, 2018 6:21 pm

Guest wrote:"ABC-Conjecture (Masser-Oesterlé, 1985):

Let [tex]\beta[/tex] > 1. Then, with finitely many exceptions, we have C < rad[tex](ABC)^{\beta}[/tex] "


In a 'nutshell', for positive integers, A, B, and C we have have:

[tex]\prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1}[/tex] [tex]\prod_{j_2 =1}^{l_2}[/tex] [tex]p_{j_2}[/tex] [tex]\prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}[/tex]

= rad(ABC) = [tex](\frac{C}{\gamma})^{1/{\beta}}[/tex] = [tex](\frac{\prod_{j_3 =1}^{l_3}p_{j_3}^{k_{j_3}}}{\gamma})^{1/{\beta}}[/tex]

for some [tex]{\beta} > 1[/tex] such that:

Case 1: [tex]0 < \gamma < 1[/tex] implies an infinite set of triples, (A, B, C);

Case 2: [tex]\gamma > 1[/tex] implies an empty or finite set of triples, (A, B, C);

according to [tex]0 < \gamma < 2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta}[/tex]

such that gcd(A, B) = gcd(A, C) = gcd(B, C) = 1 with A < B < C = A + B.

-- Dave, https://www.researchgate.net/profile/David_Cole29
Guest
 

Re: Searching for a valid proof of the ABC-conjecture

Postby Guest » Sun Oct 14, 2018 7:03 pm

Guest wrote:
Guest wrote:"ABC-Conjecture (Masser-Oesterlé, 1985):

Let [tex]\beta[/tex] > 1. Then, with finitely many exceptions, we have C < rad[tex](ABC)^{\beta}[/tex] "


In a 'nutshell', for positive integers, A, B, and C, we have:

[tex]\prod_{j_1 =1}^{l_1}[/tex][tex]p_{j_1}[/tex] [tex]\prod_{j_2 =1}^{l_2}[/tex] [tex]p_{j_2}[/tex] [tex]\prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}[/tex]

= rad(ABC) = [tex](\frac{C}{\gamma})^{1/{\beta}}[/tex] = [tex](\frac{\prod_{j_3 =1}^{l_3}p_{j_3}^{k_{j_3}}}{\gamma})^{1/{\beta}}[/tex]

for some [tex]{\beta} > 1[/tex] such that:

Case 1: [tex]0 < \gamma < 1[/tex] implies an infinite set of triples, (A, B, C);

Case 2: [tex]\gamma > 1[/tex] implies an empty or finite set of triples, (A, B, C);

according to [tex]0 < \gamma < 2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta}[/tex]

such that gcd(A, B) = gcd(A, C) = gcd(B, C) = 1 with A < B < C = A + B.

-- Dave, https://www.researchgate.net/profile/David_Cole29
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