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Old 2020-06-19, 19:49   #1
didgogns
 
Mar 2014
South Korea

1610 Posts
Default Can we prove Beal conjecture assuming ABC conjecture?

It is known that Beal's conjecture has finite number of counterexamples assuming ABC conjecture.

If we use the explicit form of ABC conjecture such that the merit (relevant paper) (known ABC triples with highest merit) does not exceed 48, only "reasonable" number of candidates remain.

Let's assume xp+yq=zr is counterexample. Its quality as ABC triple is at least 1/(1/p+1/q+1/r). In the following I'm assuming you are familiar with the partial results presented in the wikipedia.

i) {p, q, r} contains 3:
a) {p, q, r} contains another 3: The other number is bigger than 10,000. C=zr >= 210000, and its merit will easily exceed 48.
b) {p, q, r} contains 4: Recently it was shown that remaining number is at least 13. Therefore searching up to exp(72) is enough, and there are ~17B {4th power, 13th or more power} pairs to check. That {2, 3, 11} proof reduced ~350B pairs.
c) {p, q, r} contains 5:
I) {p, q, r} = {3, 5, 7}: We need to search up to exp(79) and it contains ~580B pairs.
II) {p, q, r} != {3, 5, 7}: The other power is at least 11, search range is only exp(59) and there are only ~28M pairs.
d) {p, q, r} does not contain number less than 7: Again search range is less than exp(60) and there are less than 20M pairs to check.
ii) {p, q, r} does not contain 3: You may continue similarly, and will find that the number of pairs does not exceed 1B.

It looks doable if we collect some computational power to calculate sum and difference of 600 billion pair of large numbers and compute their cubic root.
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Old 2020-08-05, 06:51   #2
didgogns
 
Mar 2014
South Korea

100002 Posts
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With ~48 hours of work on my computer, I could show that there are no counterexample in Beal conjecture with 48 or less merit.

Of the 10 known counterexample to Fermat-Catalan conjecture, the triple with highest merit has only ~1.26 merit, so chance that there is a counterexample to Beal conjecture is really small.
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