XIX OBM

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 $x^2-|x|=|x|(|x|-1)>(|c|+1)|c|\ge |c|⟹x^2-|c|>|x|$, so $|f(x)|=|x^2+c|\ge x^2-|c|>|x|$, that is, $|f^{(n+1)}(x)|>|f^{(n)}(x)|>\cdots >|x|$ 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 $$\frac{u}{v}={\left(\frac{y}{z}\right)}^2+\frac{r}{s}⟺y^{2}sv=z^2(us-rv)$$ Since $\gcd (y,z)=1$, $z^2$ divides $sv$, so $sv\ge z^2⟺v\ge \frac{z^2}{s}$. If $\frac{z^2}{s}>z⟺z>s$ 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.