# Searching for a valid proof of the abc Conjecture

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

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

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

There are two cases to consider for $$\gamma$$:

Case I : 0 < $$\gamma$$ < 1

Case 2: $$\gamma$$ >1

where

1. $$(A + B) * (A * B)$$ = $${( (A + B)/ \gamma )^{1 / \beta_{e}}} * (a * b * c)^{-1}$$ when $$\beta_{e}$$ > 1.

2. $$\gamma$$ = $$(A + B) * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.

Further analysis will confirm the validity of the ABC-conjecture.
Guest

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

Guest wrote:There are two cases to consider for $$\gamma$$:

Case I : 0 < $$\gamma$$ < 1

Case 2: $$\gamma$$ >1

where

1. $$(A + B) * (A * B)$$ = $${( (A + B)/ \gamma )^{1 / \beta_{e}}} * (a * b * c)^{-1}$$ when $$\beta_{e}$$ > 1.

2. $$\gamma$$ = $$(A + B) * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.

Further analysis will confirm the validity of the ABC-conjecture.

A + B < 2 * B since A < B.

3. $$\gamma$$ < $$2*B * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.
Guest

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

Guest wrote:
Guest wrote:There are two cases to consider for $$\gamma$$:

Case I : 0 < $$\gamma$$ < 1

Case 2: $$\gamma$$ >1

where

1. $$(A + B) * (A * B)$$ = $${( (A + B)/ \gamma )^{1 / \beta_{e}}} * (a * b * c)^{-1}$$ when $$\beta_{e}$$ > 1.

2. $$\gamma$$ = $$(A + B) * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.

Further analysis will confirm the validity of the ABC-conjecture.

A + B < 2 * B since A < B.

3. $$\gamma$$ < $$2*B * [(a*b*c)(A+B)(A*B)]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.

4. $$\gamma$$ < $$2*\prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_2 =1}^{l_2}$$$$p_{j_2} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.

5. $$\gamma$$ < $$2*\prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.

6. $$\prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$ < $$\gamma$$ < $$2*\prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$ when $$\beta_{e}$$ > 1.

7. At best, 1 < $$\gamma$$ < 2 for Case 2 when $$\beta_{e}$$ > 1. Right?
Guest

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

WRONG!!!! 7. At best, 1 < $$\gamma$$ < 2 for Case 2 when $$\beta_{e}$$ > 1.

7. At best, 1 < $$\gamma / \alpha < 2$$ when $$\beta_{e}$$ > 1.

Note: $$\alpha = \prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$

Hmm...? This analysis needs some fixing.
Guest

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

Guest wrote:WRONG!!!! 7. At best, 1 < $$\gamma$$ < 2 for Case 2 when $$\beta_{e}$$ > 1.

7. At best, 1 < $$\gamma / \alpha < 2$$ when $$\beta_{e}$$ > 1.

Note: $$\alpha = \prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$

Hmm...? This analysis needs some fixing.

7. Tentatively, 1 < $$\gamma / \alpha < 2$$ when $$\beta_{e}$$ > 1. That's better! But things will probably change soon.

Note:

$$A < (A + B)/2 < B$$

or

$$A^{2} < (A + B)* A/2 < B^{2}$$
Guest

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

Moreover, $$A < (A + B)/2 \le A/\delta_{1} + B/\delta_{2} < B$$ for some positive composites, $$\delta_{1}$$ and $$\delta_{2}$$, 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

We recall $$0 < \gamma/\alpha < 2$$ and a former note:

$$\alpha = \prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$.

But, since $$A = \lambda * B$$ for some $$0 < \lambda < 1$$, we have

$$\alpha = \lambda^{-1} * \prod_{j_1 =1}^{l_1}$$$$p_{j_1}^{k_{j_1} - 2\beta_{e}} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{-\beta_{e}}$$.
Guest

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

Guest wrote:We recall $$0 < \gamma/\alpha < 2$$ and a former note:

$$\alpha = \prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2} - \beta_{e}} * [\prod_{j_1 =1}^{l_1}$$$$p_{j_1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}]^{-\beta_{e}}$$.

But, since $$A = \lambda * B$$ for some $$0 < \lambda < 1$$, we have

$$\alpha = \lambda^{-1} * \prod_{j_1 =1}^{l_1}$$$$p_{j_1}^{k_{j_1} - 2\beta_{e}} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{-\beta_{e}}$$.

Moreover, $$A < B < A + B$$,

therefore $$A = \mu^{-1} * (A + B)$$ for some $$0 < \mu < 1$$.

Thus, we have:

$$\alpha = \mu^{-1} * \lambda^{-1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{k_{j_3} - 3\beta_{e}}$$.
Guest

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

Hence, we have:

I. $$0 < \gamma < 2 * \mu^{-1} * \lambda^{-1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{k_{j_3} - 3\beta_{e}}$$

for some $$0 < \mu < 1$$ and for some $$0 < \lambda < 1$$.

Hmm... The exponents or powers of distinct primes, $$p_{j_{3}}$$, are constrained by the power or exponent, $$-3\beta_{e}$$...
.
Guest

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

We can conclude according to the two cases for $$\gamma$$,

Case 1: There are infinitely many ABC-triples which satisfy the ABC-conjecture when 0 < $$\gamma$$ < 1.

and

Case 2: There are zero or finitely many ABC-triples which satisfy the ABC-conjecture when $$\gamma$$ > 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

Guest wrote:We can conclude according to the two cases for $$\gamma$$,

Case 1: There are infinitely many ABC-triples which satisfy the ABC-conjecture when 0 < $$\gamma$$ < 1.

and

Case 2: There are zero or finitely many ABC-triples which satisfy the ABC-conjecture when $$\gamma$$ > 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

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

sorr slight correction
1+b = 9^n
Guest

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

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 = $$9^{n}$$, c = 1 + $$9^{n}$$) = 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

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

Again, does gcd(b, $$9^{n}$$) = 1?
Guest

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

Hence, we have:

I. $$0 < \gamma < 2 * \mu * \lambda^{-1} * \prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{k_{j_3} - 3\beta_{e}}$$

for some $$0 < \mu < 1$$ and for some $$0 < \lambda < 1$$.

Hmm... The exponents or powers of distinct primes, $$p_{j_{3}}$$, are constrained by the power or exponent, $$-3\beta_{e}$$. And rational factors, $$\mu$$ and $$\lambda$$, are important too...

Note: $$\mu^{-1}$$ is wrong factor for statement I and is corrected.
Guest

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

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

Guest wrote:i personally doubt the abc conjecture is true for one inescapable reason: 1+b = $$9^{n}$$ always results in a value for c/ rad function >1

For what $$\beta > 1$$, is $$9^{n} / rad( 1 * (9^{n -1} - 1) * 9^{n})^{\beta} > 1?$$

And are there infinitely many abc-triples in total that satisfy the above question for all appropriate $$\beta$$ > 1?
Guest

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

Guest wrote:
Guest wrote:i personally doubt the ABC-conjecture is true for one inescapable reason: 1+B= $$9^{n}$$ always results in a value for C/ rad function >1

For what $$\beta > 1$$, is $$9^{n} / rad( 1 * (9^{n} - 1) * 9^{n})^{\beta} > 1?$$

And are there infinitely many ABC-triples in total that satisfy the above question for all appropriate $$\beta$$ > 1?
Guest

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