$\newcommand{\spec}{\operatorname{Spec}}\spec\sqrt2=\{\lfloor k\sqrt2\rfloor: k \ge 0\}$.
I have no idea of how I can prove the statement in the question.
Prove that $\spec\sqrt2$ contains infinitely many powers of $2$.
$\newcommand{\spec}{\operatorname{Spec}}\spec\sqrt2=\{\lfloor k\sqrt2\rfloor: k \ge 0\}$.
I have no idea of how I can prove the statement in the question.
Prove that $\spec\sqrt2$ contains infinitely many powers of $2$.
Let $k=\lceil 2^n\sqrt 2\rceil$. Then $2^n\sqrt 2