Finite amount of pairs (p,q) problem with $ \mid\sqrt{2}-\frac{p}{q}\mid\leq\frac{1}{q^{3}} $

Question

I have no clue how to solve this one. I tried everything. Im helpfull for any tipps.

Show that their is a finite amount of Pairs (p, q) ∈ $\mathbb{Z}\times \mathbb{N}$ ,so that : $ \mid\sqrt{2}-\frac{p}{q}\mid\leq\frac{1}{q^{3}} $

Answer 1

This is Roth's Theorem with $\epsilon=1$ and $\alpha=\sqrt{2}$. A proof can be found in books on number theory, such as Ireland and Rosen's book "A classical introduction to modern number theory", or in Wolfgang Schmidt's "Diophantine approximation". The question has a long history here on MSE - see for example here and the links given there.

Answer 2

This answer is not intended as a complete proof, but just as a sketch of it.
We may start with the following Lemma, due to Lagrange:

Lemma 1: if $\alpha>0$ is an irrational number and $$\left|\alpha-\frac{p}{q}\right|\leq \frac{1}{2q^2}, $$ $\frac{p}{q}$ is one of the best rational approximations of $\alpha$; in particular, it is a convergent of the continued fraction of $\alpha$.

Lemma 2: the continued fraction of $\sqrt{2}$ has a very simple structure: $$\sqrt{2}=[1;2,2,2,2,2,\ldots].$$ In particular, the convergents of $\sqrt{2}$ are given by ratios of Pell and Pell-Lucas numbers, and by denoting with $\frac{p_n}{q_n}$ the $n$-th convergent of the continued fraction we have: $$ \left|\sqrt{2}-\frac{p_n}{q_n}\right|\color{red}{\geq} \frac{C}{q_n^2} $$ for an absolute constant $C$ close to $\frac{1}{2\sqrt{2}}$.

The second Lemma provides a straightforward proof of the given claim.

Answer 3

Hint: $$\left|\sqrt{2}-\frac{p}{q}\right|=\frac{|\sqrt{2}q-p|}{q}=\frac{|2q^2-p^2|}{q|\sqrt{2}q+p|}\geq\frac{1}{q|\sqrt{2}q+p|}$$