Cardinal of the set $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$ Let $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$
We want to prove that $A$ is finite and find its cardinal? we can prove that it's finite by using directly Roth’s theorem which is a generalized theorem of Lionville's theorem. But, I am stuck on finding  the number of its elements. Any help, and thanks in advance.

Note that  $|2 - \frac{p^2}{q^2}| \ge \frac{1}{q^2}$, and $\sqrt{2}+\frac{p}{q}< \sqrt{2} + (\sqrt{2} + \frac{1}{q^3})$  therefore
$$\frac{1}{q^3} >|\sqrt{2} - \frac{p}{q}| = \frac{|2 - \frac{p^2}{q^2}|}{\sqrt{2}+\frac{p}{q}}> \frac{\frac{1}{q^2}}{2 \sqrt{2}+\frac{1}{q^3}} $$
and so
$$2\sqrt{2}+\frac{1}{q^3}> q$$
We conclude that $q\le 2$, and so $x=1$, $x=2$, or $x=\frac{3}{2}$ ($0< \frac{3}{2} - \sqrt{2} < \frac{1}{10} < \frac{1}{8}$).
