Does there exist an algorithm which to each Diophantine equation assigns an integer which is greater than the modulus of integer solutions, if these solutions form a finite set? (Q2843817)
From MaRDI portal
| This is the item page for this Wikibase entity, intended for internal use and editing purposes. Please use this page instead for the normal view: Does there exist an algorithm which to each Diophantine equation assigns an integer which is greater than the modulus of integer solutions, if these solutions form a finite set? |
scientific article; zbMATH DE number 6201404
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Does there exist an algorithm which to each Diophantine equation assigns an integer which is greater than the modulus of integer solutions, if these solutions form a finite set? |
scientific article; zbMATH DE number 6201404 |
Statements
26 August 2013
0 references
Davis-Putnam-Robinson-Matiyasevich theorem
0 references
Matiyasevich's conjecture on finite-fold Diophantine representations
0 references
Does there exist an algorithm which to each Diophantine equation assigns an integer which is greater than the modulus of integer solutions, if these solutions form a finite set? (English)
0 references
The solution of Hilbert's Tenth Problem by Martin Davis, Hilary Putnam, Julia Robinson and Yuri Matyasevich also contained a proof that every recursively enumerable set is Diophantine. In other words, given an r.e. subset of \(\mathbb Z\), there exists a polynomial \(P(t,\bar x)\) with integer coefficients such that \(t\in A\) if and only if \(P(t,\bar x)=0\) has solutions in \(\mathbb Z\). A finite-fold conjecture for rational integers asserts that for each r.e. set \(A\) one could find a polynomial \(P(t,\bar x)\) as above so that for each \(t \in A\), the number of tuples \(\bar x\) in \(\mathbb Z\) solving \(P(t,\bar x)=0\) is finite. The conjecture is still open.NEWLINENEWLINEThe author of the paper under review makes conjectures about solutions to systems of equations in a particularly simple form and shows that if his conjectures are true, then the finite-fold conjecture is false.
0 references