[Rational Points on Elliptic Curves- Joseph H. Silverman, ex 3.1]
Given a rational number $x=\frac mn$ in its lowest terms, define $$H(x):=\max\left\{|m|,|n|\right\}.$$ Let $R(\kappa)$ be the number of rational numbers $x$ with $H(x)$ less than $\kappa$. First prove that $R(\kappa)\le 2\kappa^2+\kappa$ and then show that $$\lim_{\kappa \to \infty} \frac{R(\kappa)}{\kappa^2}=\frac{12}{\pi^2}$$
The first part is quite easy since the number of choices for the numerator is clearly bounded by $2\kappa+1$ and the denominator is bounded by $\kappa$.
But, I have no idea how to proceed with the second part.
Hints (with the risk of raising further questions). It's not too difficult to derive the recursive formula ($|...|$ - means cardinality in this case, not absolute value): $$R(k+1)=R(k)+\left|\left\{x \in\mathbb{Q} | H(x)=k\right\}\right|=\\ R(k)+4\cdot\varphi(k)$$ where $\varphi(n)$ is the Euler's totient function. That's similar to A171503.
As a result: $$\frac{R(k+1)}{(k+1)^2}= \frac{R(1)}{(k+1)^2}+\frac{4}{(k+1)^2}\left(\sum_{n=1}^{k}\varphi(n)\right)$$ Now, according to a result of Arnold Walfisz (search within the Wikipedia article): $${\displaystyle \sum_{n=1}^{k}\varphi (n)= {\frac {3k^{2}}{\pi ^{2}}}+O\left(k(\log k)^{\frac {2}{3}}(\log \log k)^{\frac {4}{3}}\right)\quad {\text{as }}k\rightarrow \infty ,}$$ and the result follows.