[example: A Circle with Few Integral Points]
The notation $\mathbb{Z}[x,y]$ means the ring of polynomials in the variables $x$ and $y$ with integer coefficients. Let $F(x,y)=x^2+y^2-1\in \mathbb{Z}[x,y]$. Over $\mathbb{R}$, the equation $F(x,y)=0$ is
\begin{align*}
x^2+y^2-1=0,
\end{align*}
equivalently
\begin{align*}
x^2+y^2=1,
\end{align*}
which is the unit circle and has infinitely many real solutions, for example $(\cos \theta,\sin \theta)$ for each $\theta\in\mathbb{R}$.
Over $\mathbb{Z}$, suppose $(x,y)\in\mathbb{Z}^2$ satisfies $x^2+y^2=1$. Since $x^2$ and $y^2$ are nonnegative integers whose sum is $1$, one of them is $1$ and the other is $0$. If $x^2=1$ and $y^2=0$, then $x=1$ or $x=-1$, and $y=0$, giving $(1,0)$ and $(-1,0)$. If $x^2=0$ and $y^2=1$, then $x=0$, and $y=1$ or $y=-1$, giving $(0,1)$ and $(0,-1)$. Substituting these four pairs gives $1^2+0^2=1$, $(-1)^2+0^2=1$, $0^2+1^2=1$, and $0^2+(-1)^2=1$, so the integral solution set is exactly
\begin{align*}
\{(1,0),(-1,0),(0,1),(0,-1)\}.
\end{align*}
The same polynomial therefore defines a continuum of real points but only four integral points.
[/example]