Follow up to "Modular Arithmetic Question $p(x)=x^2−x+41$"

Question

This is the original question. Aside from simply calculating the values of $p(0)$ through $p(40)$, can anyone suggest how modular arithmetic might be used to show that $p(x)$ returns primes for all values of $x$ in the set $\{0, 1, 2, ..., 40\}$? For context, this question comes from a relatively introductory math text, though it is marked as a challenging question.

Answer 1

I prefer to write $x^2 + x + 41.$ Just habit. Same values, note $$ (x+1)^2 - (x+1) + p = x^2 + x + p $$

The statement is this: given a (positive) prime number $p,$ the unary polynomial $$ x^2 + x + p $$ takes prime values for all integer $x$ with $0 \leq x \leq p-2$ if, and only if, the (positive) binary quadratic form $$ x^2 + xy + p y^2 $$ is the only form (up to $SL_2 \mathbb Z$ equivalence) of its discriminant. Checking class number of a discriminant $\Delta$ is a finite computation, we need only consider Gauss reduced forms, meaning $$ a x^2 + b xy + c y^2 $$ with $$ b^2 - 4ac = \Delta \; , \; $$ $$ \gcd(a,b,c) = 1 \; , \; $$ $$ |b| \leq a \leq c $$ IF $a=c,$ then $$ b \geq 0 $$ IF $|b| = a,$ then $$ b = a $$