Since there has been considerable work done to answer this famous and important question, we expect a clear and definitive proof of the ABC Conjecture within 48 months.
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On simple forms, we shall also consider the following mathematical statements:
e * f = a + b and e / f = a - b.
Example: We can consider the following Fermat's Equations (1, 2, and 3):
1. xn + yn = zn; 2. x(n + ε)/2 = zn/2 + yn/2; and 3. x(n - ε)/2 = zn/2 - yn/2 where
e = xn/2, f = xε/2, a = zn/2, and b = yn/2 with the control parameter, ε, such that
0
Conjecture is sort of unproved statements. But in this case, it is claimed by Shinichi Mochizuki with more than 500 pages. I don't think it is still completely approved in the number theory community
Dear David
You can see Emzari Papava's page on ABC theorem
https://www.facebook.com/photo.php?fbid=1016304698512989&set=pcb.1046780248760397&type=3&theater
Sincerely Yuri
Hmm... We recall: max( |a|, |b|, |c| ) ≤ β(λ) * rad(a*b*c)1+λ.
We can first explore β(λ) as a non-constant function of λ for all values,
λ > 0. And hopefully, with tools/resources/research/insight, we'll reduce β(λ) to a constant function or a constant for which the abc Conjecture is true for all values, λ > 0.
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Conjecture I: k * |a + b | ≤ β(λ) * rad(a*b*(a + b) )1+λ
where 1 ≤ k ≤ 3 and a 0.
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Conjecture III: β(λ) = 3 (?) for λ > 0.
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Keywords: Geometry of Numbers and Optimization.
What is the true value of β(λ) for which the abc Conjecture holds?
The answer is β(λ) = 3 (?) since a, b, and c = a + b are the three (3) crucial components of the abc Conjecture. If we chose β(λ) = 1, then c = a
or c = b which are false statements. And if we chose β(λ) = 2, we have considered only two of the three crucial components, a, b, and c. And of course, β(λ) > 3 is obviously invalid (?)
Moreover, | a + b | ≤ |a| + |b|. And if we assume | a + b | < |a| + |b| since
3 * (|a| + |b|) > 3 * | a + b | ≤ β(λ) * rad(a * b * c)(1+λ) for λ > 0 , then we
discover 3 * | a + b | is the minimum relative to 3 * (|a| + |b|) for which the conjecture is valid , and it is also the perimeter of a equilateral triangle whose side length is |a + b| where min(|a|, |b|) 0.
Thus, we conclude the abc Conjecture is correct when β(λ) = 3 (?) for λ > 0.
Reference link:
https://plus.google.com/u/0/115467095957398372179/posts/dCTYHHAyY5e.
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Praise Lord GOD! Thank Lord GOD! Love Lord GOD!
Amen! :-)
*****
Oops! Could it be that β(λ) = π = 3.14159... (Conjecture IV) is best according to the works of Riemann, Lagarias, and of other top experts?
*****
Yum... Yum. I like apple pie! :-)
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ABC Conjecture, Data Analysis, and Relevant Prime Factors of a, b,
or c, and the Relation Between Sum of Primes and the Product of Primes:
Keywords: Prime Decomposition and Knot Theory, Synchronization and Entanglement of quantum states (ak or bk) over field, Z2, Asymmetrical Parity between two knots (connected data states of p and of q).
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We shall also keep in mind the following equation,
1 / p +/- 1/p = (q +/- p) / (p *q) for large prime numbers, p and q such that q > p.
Given n such that n = p*q.
Example: Let n = 1513, and we assume n = p*q. Find distinct primes p and q.
Check Sqrt(n) = prime? Otherwise, the complete the algorithm below.
Start simply: Let prime, p = 10*a_1 + a_0 where a_1 ∈ {1, 2, 3, 4,5,6,7,9} and a_1 ∈ {1, 3,7,9}.
Let prime, q = 10*b_1 + b_0 where b_1 ∈ {1, 2, 3, 4,5,6,7,9}, and
b_1 ∈ {1, 3,7,9}.
Compute p * q = n = 1513.
What are a_0, a_1, b_1 and b_0?
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Let a = 3 and b = 5, then c = 8.
We have max( |a|, |b|, |c| ) = 8, and rad(a * b * c) = 3 * 5 * 2 = 30.
Clearly, 8 ≤ 3 * 301+λ1 when β(λ1) = 3 (?) for λ1 > 0.
Therefore, 23 / 3 = 2.666... ≤ (2 * 3 * 5)1+λ1.
While the equations,
3 * x1 + 5 * y1 = 1,
3 * x2 + 8* y2 = 1, and
5 * x3 + 8 * y3 = 1,
imply (x1 = 2, y1 = -1), (x2 = 3, y2 = -1), and (x3 = -3, y3 = 2).
Therefore,
sqrt( (x1 - x2)2 + (y1 - y2)2 ) + sqrt( (x1 - x3)2 + (y1 - y3)2 ) +
sqrt( (x2 - x3)2 + (y2 - y3)2 )
= sqrt( (2 - 3)2 + (-1 - (-1))2 ) + sqrt( (2 - (-3))2 + (-1 - 2)2 ) +
sqrt( (3 - (-3))2 + (-1 - 2)2 ) = 1 + sqrt(34) + sqrt(45) ≈ 13.53916.
Therefore, 13.53916 / 3 = 4.513052 ≤ (2 * 3 * 5)1+λ2.
And clearly, 0 < λ1 ≤ λ2.
Notes: NextPrime(2.666...) = 3 and NextPrime(4.513052) = 5 which are prime factors of a and b, respectively.
2.666... < avg = (2 + 3 + 5)/ 3 = 10/3 = 3.333... < 4.513052.
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Let a = 3 and b = -5, then c = -2.
We have max( |a|, |b|, |c| ) = 5, and rad(a * b * c) = 3 * 5 * 2 = 30.
Clearly, 5 ≤ 3 * 301+λ1 when β(λ1) = 3 for λ1 > 0.
Therefore, 5/ 3 = 1.666... ≤ 301+λ1.
While the equations,
3 * x1 - 5 * y1 = 1,
3 * x2 - 2 * y2 = 1, and
-5 * x3 - 2 * y3 = 1,
imply (x1 =2, y1 = 1), (x2 = 1, y2 = 1), and (x3 = -1, y3 = 2).
Therefore,
sqrt( (2-1)2 +(1-1)2 ) + sqrt( (2- (-1))2 +(1-2)2 ) + sqrt( (1-(-1))2 +(1-2)2 )
= 1 + sqrt(10) + 1 = 2 + sqrt(10) ≈ 5.162278.
Therefore, 5.162278 / 3 = 1.720759 ≤ 301+λ1.
And clearly, 0 < λ1 ≤ λ2.
Notes: NextPrime(1.666...) = NextPrime(1.720759) = 2 which is a prime factor of c.
1.666... < 1.720759 < avg = (2 + 3 + 5)/ 3 = 10/3 = 3.333...
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Let a = 15 and b = 14, then c = 29.
We have max( |a|, |b|, |c| ) = 29, and rad(a * b * c) = 3*5*2*7*29 = 6090.
Clearly, 29 / 3 = 9.666... ≤ 60901+λ1 when β(λ1) = 3 for λ1 > 0.
While the equations,
15 * x1 + 14 * y1 = 1,
15 * x2 + 29* y2 = 1, and
14 * x3 + 29* y3 = 1,
imply (x1 = 1, y1 = -1), (x2 = 2, y2 = -1), and (x3 = --2, y3 = 1).
Therefore,
sqrt( (1-2)2 +(-1-(-1))2 ) + sqrt( (1+2)2 +(-1-1)2 ) + sqrt( (2+2)2 +(-1-1)2 )
= 1 + sqrt(13) + sqrt(20) ≈ 9.07768.
Therefore, 9.07768 / 3 ≈ 3.02589 ≤ 60901+λ2.
And clearly, 0 < λ2 ≤ λ1.
Notes: Nearest Prime(9.666...) = 11, and NearestPrime(3.02589) = 3 which is a prime factor of a.
3.02589
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