Along the lines of Polynomial representing all nonnegative integers, but likely well-known question:
is there a polynomial $f \in \mathbb Q[x_1, \dots, x_n]$ such that $f(\mathbb Z\times\mathbb Z\times\dots\times\mathbb Z) = P$, the set of primes?
MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.
Sign up to join this communityAlong the lines of Polynomial representing all nonnegative integers, but likely well-known question:
is there a polynomial $f \in \mathbb Q[x_1, \dots, x_n]$ such that $f(\mathbb Z\times\mathbb Z\times\dots\times\mathbb Z) = P$, the set of primes?
No. Any such polynomial would have the property that any of its restrictions $f(x)$ to one variable consist only of primes, but this is easily seen to be impossible, since if $p(a)$ is prime then $p(k p(a) + a)$ is divisible by $p(a)$. (Even accounting for the coefficients in $\mathbb{Q}$ is straightforward by multiplying by the common denominator and using CRT; in fact, we can show that given an integer polynomial $q(x)$ and a positive integer $n$ there exists $x_n$ such that $q(x_n)$ is divisible by $n$ distinct primes.)
However, there do exist multivariate polynomials with the property that their positive integer outputs consist of the set of primes. See the Wikipedia article.