# Dave's Proof of the ABC Conjecture

### Dave's Proof of the ABC Conjecture

ABC-Conjecture (Masser-Oesterlé, 1985):

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

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

$$\prod_{j_1 =1}^{l_1}$$$$p_{j_1}$$ $$\prod_{j_2 =1}^{l_2}$$ $$p_{j_2}$$ $$\prod_{j_3 =1}^{l_3}$$$$p_{j_3}$$

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

for some $${\beta} > 1$$ such that:

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

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

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

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

Notes:

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

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

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

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

David Cole.

'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

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

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

Proof of the ABC Conjecture:

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

$$\prod_{j_1 =1}^{l_1}$$$$p_{j_1}$$ $$\prod_{j_2 =1}^{l_2}$$ $$p_{j_2}$$ $$\prod_{j_3 =1}^{l_3}$$$$p_{j_3}$$

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

for some $${\beta} > 1$$ such that:

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

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

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

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

Notes:

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

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

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

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

David Cole.

'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

ABC-Conjecture (Masser-Oesterlé, 1985):

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

Proof of the ABC Conjecture:

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

$$\prod_{j_1 =1}^{l_1}$$$$p_{j_1}$$ $$\prod_{j_2 =1}^{l_2}$$ $$p_{j_2}$$ $$\prod_{j_3 =1}^{l_3}$$$$p_{j_3}$$

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

for some $${\beta} > 1$$ such that:

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

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

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

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

Notes:

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

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

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

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

David Cole.

'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

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

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

Proof of the ABC Conjecture:

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

$$\prod_{j_1 =1}^{l_1}$$$$p_{j_1}$$ $$\prod_{j_2 =1}^{l_2}$$ $$p_{j_2}$$ $$\prod_{j_3 =1}^{l_3}$$$$p_{j_3}$$

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

for some $${\beta} > 1$$ such that:

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

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

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

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

Notes:

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

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

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

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

David Cole.

'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

Why was the first post copied three times without comment?
Guest

### Re: Dave's Proof of the ABC Conjecture

Guest wrote:Why was the first post copied three times without comment?

It is human to error... However, those copies are all updates with minor or important changes.
Guest