# Searching for a valid proof of the abc Conjecture

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

Guest wrote:
Guest wrote:"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.

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

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 $$\beta$$.
Guest

### Re: Searching for a valid 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.

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

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 $$2 * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{k_{j_3} - 3\beta} \ge 1$$.
Guest

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

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

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

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.

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

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 $$2 * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{k_{j_3} - 3\beta} > 1$$.
Guest

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

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

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

An Update:

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) real 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: Searching for a valid proof of the abc Conjecture

FYI: 'Does ABC (conjecture) implies Fermat's last theorem?'

https://math.stackexchange.com/questions/1157932/does-abc-implies-fermats-last-theorem.

Since the abc Conjecture is true, then the following conjectures or theorems are also confirmed:

Modified Szpiro conjecture;

Beal conjecture ("I am still waiting for that million-dollar award for proving the Beal Conjecture..." -- David Cole);

Fermat–Catalan conjecture;

Roth's theorem;

Tijdeman's theorem.

'The abc Conjecture',

https://en.wikipedia.org/wiki/Abc_conjecture.
Guest

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

On the Beal Conjecture:

I am still waiting for that million-dollar award for proving the Beal Conjecture directly or via the proof of the abc Conjecture.

Sincerely,

David Cole,

openmind123omega@gmail.com

https://www.researchgate.net/profile/David_Cole29

P.S. Please do not spam me. Thank you!
Guest

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

Guest wrote:On the Beal Conjecture:

I am still waiting for that million-dollar award for proving the Beal Conjecture directly or via the proof of the abc Conjecture.

Sincerely,

David Cole,

openmind123omega@gmail.com

https://www.researchgate.net/profile/David_Cole29

P.S. Please do not spam me. Thank you!

I, David Cole, have made some significant contributions to the mathematical sciences here and elsewhere, and I have not received any significant recognition nor any significant award for my works thus far. And therefore, I shall stop sharing my work...

May Lord God bless me and help me in this very miserable so-called human world,

David Cole.
Guest

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