I anticipate that the number of lattice points of a special ellipse will be equal to the number of divisors of a number represented by Euler's prime generating polynomial.
Euler's prime generating polynomial: $$f(x)=x^2+x+41 \ \ \ \ \ \ \ \ x\in\mathbb{Z} $$
Special ellipse: $$X^2+163Y^2-2(2x+1)Y-1=0 \ \ \ \ \ \ \ \ \ X,Y\in\mathbb{R}$$
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For example, let $x$ be 40.
Euler's prime generating polynomial:
\begin{eqnarray*} f(40)&=&40^2+40+41\\ &=&1681\\ &=&41^2 \end{eqnarray*}
The number of divisors of $f(40)$ is equal to 3.
Special ellipse:
\begin{eqnarray*} &X^2&+163Y^2-2(2\cdot40+1)Y-1=0\\ &X^2&+163Y^2-162Y-1=0 \end{eqnarray*}
Lattice points of this special ellipse are following. $$(X,Y)=(1,0),(-1,0),(0,1)$$
The number of lattice points is equal to 3.
Please watch this video https://www.youtube.com/watch?v=i5c69-A0cEk.
If you find a counterexample or proof, please let me know.
I assert the following theorem related to this problem.
Theorem 1. $ \forall x, \alpha \in \mathbb{N}, \alpha \neq 1$,
The equation $$x=Yy^2+(Y+1)y+Y\alpha$$ has rational solution $y$ and natural number solution $Y$ $\Rightarrow$ $x^2+x+\alpha$ is a composite number.
Proof. We express the two rational solutions as following: $$y=\frac{n_1}{m_1},\frac{n_2}{m_2},\ \ \ \ where \ m_i\in\mathbb{N},\ n_i\in\mathbb{Z},\ gcd(m_i,n_i)=1,\ (i=1,2)$$ From the factor theorem and $gcd(Y,Y+1)=1$, we can get the following relation. $$(m_1y-n_1)(m_2y-n_2)=Yy^2+(Y+1)y+Y\alpha-x$$ $$m_1m_2y^2-(m_1n_2+m_2n_1)y+n_1n_2=Yy^2+(Y+1)y+Y\alpha-x$$
Hence \begin{eqnarray*} m_1m_2 &=& Y \\ -(m_1n_2+m_2n_1) &=& Y+1 \\ n_1n_2 &=& Y\alpha-x \end{eqnarray*}
So we can get $$x=m_1m_2\alpha-n_1n_2$$ $$m_1n_2+m_2n_1+m_1m_2=-1.$$
We combine the two equation as following: $$x=\frac{n_1n_2-m_1m_2\alpha}{m_1n_2+m_2n_1+m_1m_2}$$
We enter this $x$ into $x^2+x+\alpha$ and calculate the factorization.
We can get $$x^2+x+\alpha = \frac{(n_1^2+m_1n_1+\alpha m_1^2)(n_2^2+m_2n_2+\alpha m_2^2)}{(m_1n_2+m_2n_1+m_1m_2)^2}.$$
Since $\ m_1n_2+m_2n_1+m_1m_2=-1$, $$x^2+x+\alpha = (n_1^2+m_1n_1+\alpha m_1^2)(n_2^2+m_2n_2+\alpha m_2^2).$$
So $x^2+x+\alpha$ is a composite number. $$\tag*{$\square$}$$
Since $y=\frac{-Y-1 \pm \sqrt{(1-4\alpha)Y^2+2(2x+1)Y+1}}{2Y}$, we can get a condition from Theorem 1.
Lemma. $\forall x,\alpha \in \mathbb{N}$,
The ellipse $$X^2 = (1-4\alpha)Y^2+2(2x+1)Y+1, \ \ \ \ \ \ Y>0$$ has lattice points$(X,Y)$. $\Rightarrow$ $y$ is a rational number.
If $Y=0$ is allowed, the ellipse has always $(X,Y) = (\pm 1,0)\ \ $(trivial lattice points).
Hence, the following assertion is correct.
Theorem 2. $\forall x,\alpha \in \mathbb{N}, \alpha \neq 1$,
The ellipse has one or more non-trivial lattice points. $\Rightarrow x^2+x+\alpha$ is a composite number.
The following conjecture is unresolved.
Conjecture.$\forall x \in \mathbb{N} ,\ \forall \alpha \in \{3,5,11,17,41\} $,The ellipse has only trivial lattice points. $\Leftrightarrow x^2+x+\alpha$ is a prime number.
(The ellipse has one or more non-trivial lattice points. $\Leftrightarrow x^2+x+\alpha$ is a composite number.)
If this conjecture is correct, the number of lattice points and the number of divisors are equal.