Show that set of pre-periodic points is finite

Solution 1:

Let us fix $c = \frac{c_1}{c_2}$ with $\gcd(c_1,\ c_2)=1$. For each prime $p$, let $p^s$ be the largest power of $p$ dividing $c_2$. Let us write this as $p^s\Vert c_2$

For some $x_0 \in \mathbb{Q}$ let us consider $x_1 = f(x_0)$ and more generally, $x_{k+1} = f(x_k)$ for some $$f(x) = x^2 + \frac{c_1}{c_2}$$ Let us write $x_k = \frac{m_k}{n_k}$ with $\gcd(m_k,\ n_k)=1$.

Lemma 1: Suppose that $p^s\Vert c_2$. If $p^{t+1}\mid n_k$ for some $t \ge s$ then $p^{t+2}\mid n_{k+1}$.

Proof: Consider $x_{k+1}$ given by $$f(x_k) = \frac{m_k^2}{n_k^2} + \frac{c_1}{c_2} = \frac{m_k^2c_2 + n_k^2c_1}{c_2n_k^2}$$ Clearly $p^{2t+2}\mid c_2n_k^2$. We will show that $p^{t+1}$ does not divide the numerator. For otherwise, we have $$p^{t+1}\mid m_k^2c_2$$ But $m_k$ and $n_k$ are coprime, so $p^{t+1}$ is coprime to $m_k^2$. It follows that $p^{t+1}\mid c_2$ which is a contradiction to the fact that $p^s\Vert c_2$.

It follows that $p^t$ is the largest possible power of $p$ which can divide the numerator. Therefore the smallest power of $p$ possible to divide $n_{k+1}$ is $p^{(2t + 2) - t} = p^{t+2}$. $\square$

Lemma 2: Let $x_0 = \frac{m_0}{n_0}$. Suppose that $p^s\Vert c_2$. If $p^{s+1}\mid n_0$ then $x_0$ is not a pre-periodic point for $f$.

Proof: We show that the power of $p$ dividing each denominator $n_k$ is strictly larger than any before. Letting $t_k$ be the largest power of $p$ dividing each $n_k$, we wish to show that $$t_0 < t_1 < t_2 < t_3 < \cdots $$ Indeed if $p^{t_k}\Vert n_k$ for some $t_k \ge s+1$ then by lemma 1 we have $p^{t_k + 1}\mid n_{k+1}$. It follows that $t_{k+1} > t_k$.

Since the powers of $p$ dividing the denominator is larger than every previous iterate, it follows that the point cannot be pre-periodic. $\square$

Finally, we use a well known condition for the boundedness of the function: If the iterates of $f$ ever exceed $2$ in magnitude then the orbit will escape to infinity (this is a very well known condition for the Mandelbrot set, I will not prove this here but I'm sure you can find a proof online somewhere). Therefore we need the requirement $|x_k| \le 2$ for each $k\ge 1$.

We now have all the tools we need. The only possible candidates for pre-periodic orbits are $x\in \mathbb{Q}$ with denominators which divide $c_2$ so the denominator is bounded above by $|c_2|$. In addition, the numerator is bounded above by the restriction $$|x^2| = |x^2 + c - c| \le |x^2 + c| + |c| \le 2 + |c|$$ It follows that the set of candidates are finite.