Find four rational numbers $~\frac{p}{q}~$ with $~\left|\sqrt 2 - \frac{p}{q}\right| \le \frac{1}{q^2}~$
Is there a way to find such numbers?
The only one I can think of is $~p = 1, ~q = 1~$ and I thought maybe if I could convert $~\sqrt 2~$ to it's nearest rational that would help but it didn't.
If $p,q\in \Bbb N$ then $$|p^2-2q^2|=1\implies$$ $$\implies |\frac {p}{q}-\sqrt 2\,|\cdot |\frac {p}{q}+\sqrt 2\,|=|\frac {p^2}{q^2}-2|=\frac {|p^2-2q^2|}{q^2}=\frac {1}{q^2}\implies$$ $$\implies |\frac {p}{q}-\sqrt 2\,|=\frac {1}{q^2}\cdot \frac {1}{|\frac {p}{q}+\sqrt 2\,|}\le\frac {1}{q^2}\frac {1}{\sqrt 2}<\frac {1}{q^2}.$$
E.g. $(p,q)\in \{(1,1),(3,2),(7,5),(17,12)\}.$