Brazilian Mathematical Olympiad

For sake of simplicity, call $x$ a periodic number if $f^{(0)}(x)$, $f^{(1)}(x)$, $f^{(2)}(x)$, $\dots$ takes finitely many values.

If $|x| > |c| + 1$ then $x^2 - |x| = |x|(|x| - 1) > (|c| + 1)|c| \ge |c| \implies x^2 - |c| > |x|$, so $|f(x)| = |x^2 + c| \ge x^2 - |c| > |x|$, that is, $|f^{(n+1)}(x)| > |f^{(n)}(x)| > \dots > |x|$ and $f^{(0)}(x)$, $f^{(1)}(x)$, $f^{(2)}(x)$, $\dots$ takes infinitely many values. So if $x$ is periodic then $|x| \le |c| + 1$, that is, all periodic numbers lie in the interval $[-(|c| + 1), |c| + 1]$.

Let $c = \frac{r}{s}$, $x = \frac{y}{z}$ and $f(x) = \frac{u}{v}$, $\gcd(r, s) = \gcd(y, z) = \gcd(u, v) = 1$, $s, z, v > 0$. So
$$\frac{u}{v} = \left(\frac{y}{z}\right)^2 + \frac{r}{s} \iff y^2 s v = z^2 (u s - r v)$$
Since $\gcd(y, z) = 1$, $z^2$ divides $s v$, so $s v \ge z^2 \iff v \ge \frac{z^2}{s}$. If $\frac{z^2}{s} > z \iff z > s$ then the denominator of $x$ is less than the denominator of $f(x)$ and consequently the denominator of $f^{(n)}(x)$ is less than the denominator $f^{(n+1)}(x)$, so $x$ is not a periodic point.

So all rational periodic points of $f$ lie in the interval $[-(|c|+1), |c|+1]$ and have denominator not greater than the denominator of $c$. Thus the number of rational periodic points of $f$ is finite.