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.

Notice that your problem is equivalent to show that:
$$-\frac{1}{q^3}<\sqrt{2}-\frac{p}{q}< \frac{1}{q^3}$$
And this is equivalent to solve the following system of inequalities over $\mathbb{Z}$:
$$\sqrt{2}q^3-pq^2-1<0$$
$$\sqrt{2}q^3-pq^2+1>0$$
where $q\neq0$. From this we can split into two cases: $q>0$ or $q<0$
If $q>0$ then $q\geq1$ and so $\sqrt{2}q^3-pq^2=q^2\left(\sqrt{2}q-p\right)\geq \sqrt{2}-p$. Thus:
$$\sqrt{2}-p-1\leq\sqrt{2}q^3-pq^2-1<0\longrightarrow p>\sqrt{2}-1$$
$$\sqrt{2}q^3-pq^2+1>\sqrt{2}-p+1>0 \longrightarrow p<\sqrt{2}+1$$
Therefore $\sqrt{2}+1>p>\sqrt{2}-1$ which implies $p\in\lbrace0,1,2\rbrace$. Checking the three possible values for $p$ one can easily get that $(p,q)\in\lbrace{(1,1),(2,1)\rbrace}$
In a similar way with $q<0$ one gets $(p,q)\in\lbrace{(-1,-1),(-2,-1)\rbrace}$.
Since we do not care about the specific values of $p$ and $q$,but on the quotient $\frac{p}{q} $we get that $\frac{p}{q}\in\lbrace1,2\rbrace$ and so $A=\lbrace 1,2\rbrace$.
Hope you can fill the details.
