Let $A$ be an Apollonian gasket generated by the quadruple $(k_1, k_2, k_3, k_4)$. The curvatures of the circles that form the gasket satisfy Descartes Theorem:
$$ (k_1+k_2+k_3+k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2) \tag{1} $$
An Apollonian gasket generated by (-10, 18, 23, 27). From Wikipedia
As we can see in the above example, from the initial quadruple of curvatures $(-1, 18, 23, 27)$ we can generate ever-increasing integer curvatures (and smaller and smaller circles) by taking a circle that is tangential to three other. For example, we have $35, 47, 62, 63, 78, 83, \cdots$.
Suppose $k_1 = -1$ (i.e., the largest enclosing circle) and $k_2 = N$ where $N = p_1^{e_1} \cdot p_2 ^{e_2} \cdots p_r^{e_r}$ is an integer with unknown factorization.
We need at least $3$ curvatures in order to derive the fourth curvature through the equation
$$k_4 = k_1 + k_2 + k_3 \pm \sqrt{k_1k_2 + k_2k_3 + k_3k_1} \tag{2}$$
Question: How should we choose $(k_1, k_2, k_3, k_4)$ such that at least one divisor $d|N$ appears in the set of integer curvatures of $A$ keeping in mind that the $p_i$ are unknown? What we know is that the term under the square root in Eqn. $(2)$ must be a perfect square for $k_4$ to be an integer.
When is a divisor $d$ of $N$ guaranteed to appear as one of the curvatures of the gasket i.e. $k_1, k_2, k_3, k_4, k_5, k_6, \cdots, k_r, \cdots, k_s (\le \beta)$ where $\beta$ is a suitable bound such as $N$ or $\sqrt{N}$?