How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$ Find all positive real number $\beta$,there are infinitely many relatively prime integers $(p,q)$ such that 
$$\left|\dfrac{p}{q}-\sqrt{2}\right|<\dfrac{\beta}{q^2}$$
maybe this problem background is Hurwitz's theorem:
$$\left|\sqrt{2}-\dfrac{p}{q}\right|<\dfrac{1}{\sqrt{5}q^2}$$
so I guess my problem ?  $\beta\ge\dfrac{1}{\sqrt{5}}$
and this problem is Germany National Olympiad 2013 last problem (1), see: http://www.mathematik-olympiaden.de/aufgaben/52/4/A52124b.pdf

Proof sketch. First write
$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|}$$
The Pell equation $$p^2 - 2q^2 = 1$$ 
is known to have infinitely many solutions (should be proven) so
$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|} = \frac{1}{q^2}\frac{1}{\left|\sqrt{2 + \frac{1}{q^2}} + \sqrt{2}\right|}$$
for infinitely many $p,q$ so all $\beta \geq \frac{1}{\sqrt{8}}$ are possible. Now try to show that it fails for $\beta < \frac{1}{\sqrt{8}}$. Let $p^2 - 2q^2 = k$ then $p/q = \sqrt{2 + k/q^2}$ and by inserting this into the inequality above show that it cannot hold for large enough $q$.
