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.

Relevant Reference Links:

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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