Finding the least prime divisor of numbers generated by $X^2-X+q$.

Question

Given that $P_0[m]$ is the least prime factor of m, one can define $$P_0[|f(x)|]:=\min\{P_0[|f(n)|]\mid n\in\mathbb{N}\}$$ for all $f(x)\in\mathbb{Z}[x]$,i.e., the smallest prime that divides a number in the $(|f(n)|)_{n\in\mathbb{N}}$ sequence,

how to prove that $P_0[x^2-x+q]=q$ $\iff$ $q\in\{2,3,5,11,17,41\}$?

I've already shown that $p(dq+r)\equiv_{d} p(r)$ for every integer polynomial, therefore $P_0[x^2-x+q]=\min\{P_0[x²-x+q]\mid x\in\{1,\dots,q\}\}$. Also, I've been able to prove the $\impliedby$ part of the statement above through explicit calculations, but such method gives nothing but the answer.

Answer 1

We first note that, since $x-y|(P(x)-P(y))$ for any integer $x,y$ and polynomial $P$ with integer coefficients,

$$P_0[f(X)]=\min\{P_0[f(n)]\ |\ n \in \{0,\cdots,p-1\}\}.$$

The second main observation is that $\{7,11,19,43,67,163\}$ ($4p-1$ for each prime you gave) are all Heegner numbers. Using the result (found, for example, on the linked Wikipedia page) that if $4p-1$ is a Heegner number, then

$$n^2-n+p$$

is prime for all $0\leq n\leq p-1$, can you finish from here?