Timeline for The strong twin conjecture can be transformed into the unsolvability of a particular Diophantine equation
Current License: CC BY-SA 4.0
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| Nov 15, 2021 at 17:21 | comment | added | JoshuaZ | @Safwane Right. The way Hilbert's 10th problem is solved by finding a systematic way of making a Diophantine equation which corresponds to any description of a recursively enumerable set en.wikipedia.org/wiki/Computably_enumerable . Since whether certain recursively enumerable sets are in fact empty or not is undecidable, you get unsolvable Diophantine equations out as a bonus. But along the way, you get Diophantine equations which correspond to a whole bunch of sets, including the primes, and nice exponentials. Hence you can apply those parts of the construction. | |
| Nov 15, 2021 at 15:29 | comment | added | Wojowu | No, I can't. A complete explanation is the length of a book, and indeed the book you mention contains such an explanation. | |
| Nov 15, 2021 at 15:15 | comment | added | Safwane | @JoshuaZ: But the problem is to find an equation that is unsolvable. | |
| Nov 15, 2021 at 15:13 | comment | added | Safwane | @Wojowu: Can you explain to me in few words how one can find that equation. | |
| Nov 15, 2021 at 15:04 | comment | added | JoshuaZ | This should be pretty straightforward. There's an explicit Diophantine equation with 26 variables for when a number is prime and getting one for powers of 2 comes pretty easily from the material in the book, and should have 4 variables. You can combine Diophantine equations using the squaring trick. This should give you a Diophantine equation with 26 + 26 + 2+4 = 58 variables. | |
| Nov 15, 2021 at 14:00 | comment | added | Wojowu | I strongly doubt anyone has bothered to write down the equation you are after, but the procedure for generating the equation from the algorithm is, in principle, completely effective. | |
| Nov 15, 2021 at 13:49 | comment | added | Will Sawin | I think the book itself explains how to transform any algorithm with bounded running time into a Diophantine equation which is solvable if and only if that algorithm accepts for some integer $n$. Then take the algorithm which searches the numbers between $n+4$ and $2^n 2^4$ for a twin prime, and accepts if none is found. | |
| Nov 15, 2021 at 13:42 | history | asked | Safwane | CC BY-SA 4.0 |