Please check [my post](https://math.stackexchange.com/questions/3485228/the-condition-that-eulers-prime-generating-polynomial-is-a-composite-number).

It indicates that the number of divisors of $x^2+x+41$ is equal to the number of lattice points of $X^2+163Y^2-2(2x+1)Y-1=0$.

This formula is transformed in this way.

$$163X^2+163^2Y^2-2\cdot163(2x+1)Y=163$$
$$163X^2+\{163Y-(2x+1)\}^2-(2x+1)^2=163$$
$$163X^2+(163Y-2x-1)^2=4x^2+4x+164$$


$X':=163Y-2x-1,\ Y':=X$ and we divide both sides by 4,

$$\frac{{X'}^2+{163Y'}^2}{4}=x^2+x+41=P(x)$$
>$$N\left(\frac{X'+Y'\sqrt{-163}}{2}\right)=P(x)$$

$\frac{X'+Y'\sqrt{-163}}{2}$ is an element of $\mathbb Q(\sqrt{-163})$.

**This formula indicates that the elements of $\mathbb Q(\sqrt{-163})$ with norm $P(x)$ is linked to the product pattern of two integers of $P(x)$.**


The phenomenon you are interested in is based on this fact.


For example, let $x$ be $76$.

\begin{eqnarray*}
    P(x)&=&76^2+76+41\\
    &=&1\cdot5893\\
    &=&71\cdot83
\end{eqnarray*}

The number of product pattern of $P(x)$ is $2$.

On the other hand, the elements of $\mathbb Q(\sqrt{-163})$ with norm $P(x)$ (ignore sign) are

$$\frac{153+\sqrt{-163}}{2},\ 5+6\sqrt{-163}\ .$$

The number is $2$.

The left one is a trivial element $\frac{(2x+1)+\sqrt{-163}}{2}$.The other one is a non-trivial element that indicates that $P(x)$ is a composite number.