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Apr 12 at 20:02 vote accept The T
Apr 12 at 12:38 comment added Marzio De Biasi @TheT: I think it's hard to find a smaller counterexample (I just looked at the KNOWN 6-th power diophantine equations, and it is very small). And be aware to play with larger exponents, because number-theory is full of UNKNOWN/UNPROVED stuff even for "smaller" objects so you risk getting nowhere. Even for your actual question, it is unknown if you can represent a 3-element subset uniquely with the sum of 3 6-th prime powers without any collision (so even if there had been no counterexample to the entire reduction, you would have been stuck)
Apr 12 at 10:08 comment added The T @MarzioDeBiasi So what about counterexamples significantly smaller or is it impossible for a counter example to be smaller than that? Because then I can just raise the exponent to see if I could learn something about the numbers.
Apr 12 at 7:28 history edited Marzio De Biasi CC BY-SA 4.0
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Apr 12 at 7:20 history edited Marzio De Biasi CC BY-SA 4.0
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Apr 12 at 7:18 comment added Marzio De Biasi @TheT: ... and don't give up! perhaps something interesting could come from the probabilities of getting an overlapping sequence?!?
Apr 12 at 7:12 comment added Marzio De Biasi @TheT: $S$ contains at least 2*(4097+4097+46657+46657+1746657 +46657+11390627+46657)+8 distinct elements (each one mapped to a $p_i^6$) in order to build enough 3 element subsets that sum up to 251^6; plus a bunch of other elements to fill the gaps and make the total count divisible by 3 (7^6,19^6, 27^6, ..., 241^6, 251^6, 257^6) ... something like N=13230648 elements .
Apr 12 at 1:19 comment added The T Just curious how large was S and C?
Apr 12 at 1:04 comment added The T Its been a hurtin' for me. I'm impressed. Of course I will continue my research into seeing what larger exponents would do.
Apr 12 at 1:02 comment added Tayfun Pay You actually went through it and gave an example. Nice !
Apr 12 at 0:37 history answered Marzio De Biasi CC BY-SA 4.0