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Jan 24, 2023 at 8:52 comment added Wlod AA Perhaps we would have $\ r_0\ $ and $\ R_0\ $ since you're avoiding prime $\ 2$.
Jan 24, 2023 at 8:45 comment added Wlod AA Somehow, your notation is hard on my eyes. Perhaps, we can have notation for the set-radix $\ R\ $ and the standard integer-radix $\ r\ $ so that $\ r(x)\ :=\ \prod R(x). $ (Thus, your $\ P(x)\ $ would become $\ R(x)$).
Jan 24, 2023 at 8:26 comment added Stefan Kohl @R.vanDobbendeBruyn Thanks! -- I edited out the redundant factor of 2 from the example.
Jan 24, 2023 at 8:25 history edited Stefan Kohl CC BY-SA 4.0
Edited out the redundant factor of 2 from the example.
Jan 23, 2023 at 20:18 comment added R. van Dobben de Bruyn Ah, I'll have a look! (My comment about the factors of 2 is just that without loss of generality, you can assume $a$, $b$, and $c$ have no common factors, so in particular two of them are odd and the last one even. I think it's customary to get rid of common factors when solving Diophantine equations, but it's a free world!)
Jan 23, 2023 at 17:15 comment added Stefan Kohl @R.vanDobbendeBruyn To prove that a given triple does not to satisfy the condition, one can use the algorithm described in: B. M. M. de Weger, Solving exponential Diophantine equations using lattice basis reduction algorithms, J. Number Theory 26 (1987), no. 3, 325–367. An example is $\{\{5\},\{13\},\{17\}\}$.
Jan 23, 2023 at 16:01 comment added Stefan Kohl @R.vanDobbendeBruyn Well -- it is clear that not all numbers can be odd, as the sum of two odd numbers is even. On the other hand, not all numbers need to be even. E.g. we have $\{\emptyset, \{3\}, \{5\}\} \in \Delta$ since $2 + 3 = 5$.
Jan 23, 2023 at 12:42 comment added R. van Dobben de Bruyn All of your examples have a common factor $2$; is this intentional? It doesn't show up in the formulation of the problem. In addition, do you have examples of triples that provably do not satisfy the condition?
Jan 23, 2023 at 12:18 history asked Stefan Kohl CC BY-SA 4.0