Prove that for any positive integers $p,q$ such that $\sqrt{11}>\frac{p}{q}$, the following inequality holds: \[\sqrt{11}-\frac{p}{q}>\frac{1}{2pq}.\]
Problem
Source: Baltic Way 2018, Problem 17
Tags: number theory
Tintarn
06.11.2018 14:26
What I'm referring to as "an old problem" is a very similar question asked with $\sqrt{7}$ instead of $\sqrt{11}$ which seems to originally have appeared in Romania 1979 and later has been posted many, many times on this forum. See also here, here, here, here, here, here, here, here or here. The idea of proof immediately carries over to this problem.
You will want to use that $-1$ is not a quadratic residue modulo $11$.
AnArtist
07.11.2018 14:47
$$11q^2-p^2\ge 1$$
Checking squares modulo 11 we conclude that 1 is not possible. Hence,
$$ 11q^2-p^2 \ge 2$$
$$\Rightarrow \sqrt{11} \ge \frac{\sqrt{p^2+2}}{q}\ge \frac{p}{q}+\frac{1}{2pq}$$