2020-06-19, 19:49 | #1 |
Mar 2014
South Korea
2^{4} Posts |
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 x^{p}+y^{q}=z^{r} 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=z^{r} >= 2^{10000}, and its merit will easily exceed 48.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. |
2020-08-05, 06:51 | #2 |
Mar 2014
South Korea
2^{4} Posts |
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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