On pages 956-957 of [this paper](http://arxiv.org/pdf/math/9811185.pdf), it is established that for any two $v\_1, v\_2$ satisfying $v\_1^2 + 1 \equiv 0\operatorname{(mod} d\_1), v\_2^2 + 1\equiv 0\operatorname{(mod} d\_2)$, $$\left\lVert \frac{v\_1}{d\_1} - \frac{v\_2}{d\_2}\right\rVert > \frac{1}{4\sqrt{d\_1d\_2}}.$$ It is unclear how one might generalize this to general quadratic congruences, since in general, their roots do not correspond to representing the modulus as some quadratic form that depends on the congruence. How might one modify the arguments in the paper to yield a similar result for other quadratic congruences?


Edit: As pointed out in one of the comments, similar results do not actually hold for all quadratic congruences. However, how might one show that this in the cases where it actually holds?



Well-spacing of the roots of a quadratic congruence
What you have written is not true:
$8^2 + 1 \equiv 0\operatorname{(mod} 13), 3^2 + 1\equiv 0\operatorname{(mod} 5)$ yet $$\left\lVert \frac{8}{13} - \frac35\right\rVert=\frac1{5\cdot 13}=\frac{1}{\sqrt{5 \cdot 13}\sqrt{5 \cdot 13}} <\frac{1}{8\sqrt{5 \cdot 13}}< \frac{1}{4\sqrt{5 \cdot 13}}.$$


The well written paper you cite has two further conditions for the inequality to hold. It also remarks that the situation is exceptional. 


One might inspect $v^2+k \equiv 0\operatorname{(mod} d)$ perhaps for $k=2$ or $k=-1$ and, if something similar seems to happen, attempt to imitate the proof.


