Dave's Proof of the ABC Conjecture

Dave's Proof of the ABC Conjecture

Postby Guest » Sun Jul 28, 2019 5:40 pm

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.



Notes:

[tex]k_{j_3}[/tex] represent constants (integers [tex]\ge 1[/tex]) while [tex]\beta \ge 1[/tex] is unrestricted (no upper bound) variable.

Case 1: [tex]0 < \gamma < 1[/tex] also implies [tex]k_{j_3} < 3 \beta[/tex] such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} < 1[/tex].

Case 2: [tex]\gamma > 1[/tex] also implies [tex]k_{j_3} \ge 3 \beta[/tex] which forces a upper bound on [tex]\beta[/tex]
such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} > 1[/tex].

And when that upper bound on [tex]{\beta}[/tex] for case 2 is exceeded, there is a solution in case 1.

David Cole.

Relevant Reference Links:

'Searching for a valid proof of the abc Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1793;

'What is the proof of the ABC Conjecture',

https://www.researchgate.net/post/What_is_the_proof_of_the_ABC_Conjecture.
Guest
 

Re: Dave's Proof of the ABC Conjecture

Postby Guest » Sun Jul 28, 2019 6:22 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].

Proof of the ABC Conjecture:

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.



Notes:

The exponent, [tex]k_{j_3}[/tex], represent constants (integers [tex]\ge 1[/tex]) while [tex]\beta \ge 1[/tex] is unrestricted (no upper bound) variable.

Case 1: [tex]0 < \gamma < 1[/tex] also implies [tex]k_{j_3} < 3 \beta[/tex] such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} < 1[/tex].

Case 2: [tex]\gamma > 1[/tex] also implies [tex]k_{j_3} \ge 3 \beta[/tex] which forces a upper bound on [tex]\beta[/tex]
such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} > 1[/tex].

And when that upper bound on [tex]{\beta}[/tex] for case 2 is exceeded, there is a solution in case 1.

David Cole.

Relevant Reference Links:

'Searching for a valid proof of the abc Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1793;

'What is the proof of the ABC Conjecture?',

https://www.researchgate.net/post/What_is_the_proof_of_the_ABC_Conjecture.
Guest
 

Re: Dave's Proof of the ABC Conjecture

Postby Guest » Thu Aug 01, 2019 12:13 am

ABC-Conjecture (Masser-Oesterlé, 1985):

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

Proof of the ABC Conjecture:

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.



Notes:

The exponent, [tex]k_{j_3}[/tex], represent constants (integers [tex]\ge 1[/tex]) while [tex]\beta \ge 1[/tex] is unrestricted (no upper bound) variable.

Case 1: [tex]0 < \gamma < 1[/tex] also implies [tex]k_{j_3} < 3 \beta[/tex] such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} < 1[/tex].

Case 2: [tex]\gamma > 1[/tex] also implies [tex]k_{j_3} \ge 3 \beta[/tex] which forces a upper bound on [tex]\beta[/tex]
such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} > 1[/tex].

And when that upper bound on [tex]{\beta}[/tex] for case 2 is exceeded, there is a solution in case 1.

David Cole.

Relevant Reference Links:

'Searching for a valid proof of the abc Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1793;

'What is the proof of the ABC Conjecture?',

https://www.researchgate.net/post/What_is_the_proof_of_the_ABC_Conjecture.
Guest
 

Re: Dave's Proof of the ABC Conjecture

Postby Guest » Mon Aug 12, 2019 12:24 am

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].

Proof of the ABC Conjecture:

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.



Notes:

The exponent, [tex]k_{j_3}[/tex], represent constants (integers [tex]\ge 1[/tex]) while [tex]\beta > 1[/tex] is unrestricted (no upper bound) variable.

Case 1: [tex]0 < \gamma < 1[/tex] also implies [tex]k_{j_3} < 3 \beta[/tex] such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} < 1[/tex].

Case 2: [tex]\gamma > 1[/tex] also implies [tex]k_{j_3} \ge 3 \beta[/tex] which forces a upper bound on [tex]\beta[/tex]
such that [tex]2 * \prod_{j_3 =1}^{l_3}[/tex][tex]p_{j_3}^{k_{j_3} - 3\beta} > 1[/tex].

And when that upper bound on [tex]{\beta}[/tex] for case 2 is exceeded, there is a solution in case 1.

David Cole.

Relevant Reference Links:

'Searching for a valid proof of the abc Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1793;

'What is the proof of the ABC Conjecture?',

https://www.researchgate.net/post/What_is_the_proof_of_the_ABC_Conjecture.
Guest
 


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