Approximating the square root of two with fractions I would like to prove that there exist only finitely many $m, n \in \mathbb{N}$ satisfying $$\left | \sqrt{2} - \frac{m}{n} \right | < \frac{1}{4n^2}.$$
Any thoughts?
Thank you for your help.

$$|\sqrt{2} - \frac{m}{n}| < \frac{1}{4n^2}\\\implies
|\sqrt{2}n-m|< \frac{1}{4n}\\\implies
\sqrt{2}n-\frac{1}{4n}<m<\sqrt{2}n+\frac{1}{4n}\\\implies
0<|(\sqrt{2}n-m)(\sqrt{2}n+m)|<\frac{1}{4n}(2\sqrt{2}n+\frac{1}{4n})=\frac{\sqrt{2}}{2}+\frac{1}{16n^2}<1.$$
But $|(\sqrt{2}n-m)(\sqrt{2}n+m)|=|2n^2-m^2|$ is an integer, a contradiction.
