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.
 A: 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.
A: if $|x- \sqrt 2 | < 1,$  then $x>0 $  and $ x + \sqrt 2 < 1 + 2 \sqrt 2 = 1 + \sqrt 8 $
then
$$ (1 + \sqrt 8) \; |x- \sqrt 2 | > (x+ \sqrt 2 )   |x- \sqrt 2 | > \frac{|p^2 - 2 q^2|}{q^2}  \geq \frac{1}{q^2}   $$
and
$$  |x- \sqrt 2 | > \frac{1}{   (1 + \sqrt 8)     q^2} $$
IF we also demand $$ \frac{1}{q^3}  >   |x- \sqrt 2 | ..... \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc   $$
A: 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 $x=\frac{3}{2}$ ($0< \frac{3}{2} - \sqrt{2} < \frac{1}{10} < \frac{1}{8}$).
A: If $0<|x-\sqrt 2|<1/q^3\le 1$ then $0<q$
and $0<-1+\sqrt 2<x<1+\sqrt 2<3$ so $0< x+\sqrt 2<4.$ So we have $$1\le|p^2-2q^2|=q^2\cdot |x-\sqrt 2|\cdot (x+\sqrt 2)<$$ $$<q^2\cdot q^{-3}\cdot(x+\sqrt 2)<4/q \implies$$ $$\implies 1<4/ q\implies q\in \{1,2,3\}.$$ This narrows the search to those $p/q$ such that $-1+\sqrt 2<p/q=x<1+\sqrt 2$ with $q\in \{1,2,3\}$ and $p\in\Bbb Z^+$.... I leave the rest to you.
