Skip to main content
7 events
when toggle format what by license comment
Feb 10, 2021 at 17:26 comment added Will Sawin The easiest way to make a polynomial that represents all primes is to make a polynomial that represents all nonnegative numbers. This can be done with a one-variable polynomial, for instance.
Feb 10, 2021 at 12:01 comment added Gerry Myerson There have been papers which reduce the number of variables from $26$, at the cost of incresing the degree of the polynomial.
Feb 10, 2021 at 11:40 review Close votes
Feb 25, 2021 at 3:04
Feb 10, 2021 at 11:23 comment added Jan-Christoph Schlage-Puchta The existence of such a polynomial follows from the negative solution of Hilbert's tenth problem. Robinson and Matijasevic showed that every enumerable set is diophantine, that is, if for a set of integers $A$ there exists a computer program, that produces all elements of $A$, not necessarily in the correct order, then there exists a polynomial $P(x, y_1, \ldots, y_k)$, such that the equation $P(n, y_1, \ldots, y_k)=0$ is solvable in integers if and only if $n\in A$. The set of primes is enumerable, and from the polynomial $P$ it is easy to construct a prime producing polynomial.
Feb 10, 2021 at 9:51 comment added gmvh Check e.g. primes.utm.edu/glossary/page.php?sort=MatijasevicPoly and references therein.
Feb 10, 2021 at 9:50 review First posts
Feb 10, 2021 at 11:23
Feb 10, 2021 at 9:45 history asked W. Wongcharoenbhorn CC BY-SA 4.0