You've listed quite a few restrictions, but hopefully this example will be "admissible".
Fix an integer $n>0$. Algebraic number theory sheds light on which rational primes are of the form $x^2+ny^2$.
Specifically, consider the order $\mathbb{Z}[\sqrt{-n}]$ of $\mathbb{Q}(\sqrt{-n})$. The relevant theorem is that a prime $p$ is represented by the quadratic form $x^2+ny^2$ if and only if it splits completely in the Ring Class Field of $\mathbb{Z}[\sqrt{-n}]$.
Here is the idea of the proof (for simplicity I'll restrict to the case in which $\mathbb{Z}[\sqrt{-n}]$ is the maximal order of $K=\mathbb{Q}(\sqrt{-n})$:
Direction 1: If $p=x^2 + ny^2$ for a prime $p\nmid 2n$, then $p\mathcal{O}_K=(x + \sqrt{-n}y)(x − \sqrt{-n}y)$. The ideal $(x+\sqrt{-n}y)$ is both prime and principal, so $p$ splits completely in the Hilbert Class Field of $K$.
Direction 2: Suppose conversely that $p$ splits completely in the Hilbert Class Field of $K$. Then we must have $p\mathcal{O}_K=\mathfrak{p}\mathfrak{q}$. By assumption both $\mathfrak{p},\mathfrak{q}$ split completely in the Hilbert Class field of $K$, so they must be principal. Say $\mathfrak{p} =(x+\sqrt{-n} y)\mathcal{O}_K$ and $\mathfrak{q} =(x-\sqrt{-n} y)\mathcal{O}_K$. Then $p=x^2+ny^2$.
Note that it is not difficult, using the Dedekind-Kummer theorem, to show that this is equivalent to the statement:
There is a monic irreducible polynomial $f_n(x)\in \mathbb{Z}[x]$ such that if an odd prime $p$ divides neither n nor the discriminant of $f_n(x)$ then $p$ is represented by $x^2+ny^2$ if and only if $(\frac{-n}{p})=1$ and $f_n(x)\equiv 0\pmod{p}$ has an integral solution.