We know --- as is in essence the result of Matiyasevich's proof of Hilbert's tenth problem --- that the language, call it $L$, of all $\langle p(\cdot) \rangle$ where $p$ is a multivariate polynomial with integer coefficients s.t. some assignment of its variables with positive integers evaluates to zero, is undecidable.
Now define the language $X$ similarly, but instead of requiring the $p(\cdot)$s to evaluate to zero, we require that it evaluates to some prime $q$.
I'm trying to prove that I can reduce finding a Turing Machine deciding $L$ to finding a Turing Machine deciding $X$. Showing that $X$ is undecidable thus. I'm completely stuck however. Without spoiling the entire proof (which I figure is rather easy, I just am not thinking in the right direction) could someone hint me in the right direction?
Given a polynomial $p$, you want to find another polynomial $p'$ (computable from $p$) such that $p(x_1,\dots,x_n)=0$ iff $p'(x_1,\dots,x_n)$ is prime. There are several ways you could do this; here's a hint towards one. Try to construct $p'$ from $p$ such that $p'$ is always even, and will be equal to $2$ iff $p$ is $0$.