# Searching for a valid proof of the abc Conjecture

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

Guest wrote:
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?

Example:

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

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.
Guest

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

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

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

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

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

Guest wrote: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, $$-3\beta_{e}$$...
.

Since $$0 < \mu^{-1} * \lambda^{-1} \le 1$$, we have:

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

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

Guest wrote:
Guest wrote: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, $$-3\beta_{e}$$...
.

Since $$0 < \mu * \lambda^{-1} \le 1$$, we have:

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

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

Guest wrote:
Guest wrote:
Guest wrote:Hence, we have:
I
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, $$-3\beta_{e}$$...
.

Since $$0 < \mu * \lambda^{-1} < 1$$ because A < B , we have:

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

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

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

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

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

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

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

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

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

Case 2: $$\gamma$$ > 1.

with:

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.
Guest

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

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

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

We have considered the two cases for $$\gamma$$ 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 $$\gamma$$,

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!

Note:

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.
Guest

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

Notes:

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

A = $$\lambda$$ * B where 0 < $$\lambda$$ < 1;

A = $$\mu$$ * C where 0 < $$\mu$$ < 1;

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

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

A = $$\prod_{j_1 =1}^{l_1}$$$$p_{j_1}^{k_{j_1}}$$;

B = $$\prod_{j_2 =1}^{l_2}$$$$p_{j_2}^{k_{j_2}}$$;

C = $$\prod_{j_3 =1}^{l_3}$$$$p_{j_3}^{k_{j_3}}$$.

Moreover, we have:

a * A = $$\prod_{j_1 =1}^{l_1}$$$$p_{j_1}$$;

b * B = $$\prod_{j_2 =1}^{l_2}$$$$p_{j_2}$$;

c * C = $$\prod_{j_3 =1}^{l_3}$$$$p_{j_3}$$.
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$$(ABC)^{\beta}$$ "
Guest

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

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

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
Guest

### 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:

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

= 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
Guest

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

We shall review our final answer! There's a flaw in the important inequality. It's the $$-3\beta$$ term. What's the correction?
Guest

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

Guest wrote:We shall review our final answer! Opps! There's no obvious flaw in the important inequality...
Guest

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

Guest wrote:
Guest wrote:We shall review our final answer! Opps! There's no obvious flaw in the important inequality...
Guest

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

Guest wrote:We shall review our final answer! There's a flaw in the important inequality. It's the $$-3\beta$$ term. What's the correction?

There are no flaws! Our proof stands! And thus, the ABC-conjecture is true!
Guest

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