XXVI OBM

Let the orbit $\mathrm{orb}(P)$ of $P=(x_0,y_0)$ be the least $n>0$ such that $f_{a,b}^n(P)=P$.

Also, let $f_{a,b}^k(P)=(x_k,y_k)$. We have $(x_{k+1},y_{k+1})=(a-b{y}_k-x_k^2,x_k)$, that is, $y_{k+1}=x_k$ and, consequently, $x_{k+1}=a-b{x}_{k-1}-x_k^2$.

Sum this relation over $k$, $k$ from $0$ to $\mathrm{orb}(P)-1$. Setting $\mathrm{orb}(P)=m$, $S_1=x_0+x_1+\cdots +x_{m-1}$ and $S_2=x_0^2+x_1^2+\cdots +x_{m-1}^2$ for simplicity, we obtain $S_1=ma-b{S}_1-S_2$ which implies $$S_2=ma-(b-1)S_1.$$ But it is known by Cauchy, for example, that $S_2\ge S_1^2/m$.

So $$ma-(b-1)S_1\ge \frac{S_1^2}{m}⟺\frac{S_1^2}{m}+(b-1)S_1-ma\le 0,$$ which has solution iff $$(b-1{)}^2-4\cdot \frac{1}{m}\cdot (-ma)\ge 0⟺a\ge -\frac{(b-1{)}^2}{4}.$$

Thus $A_b=\left[-\frac{(b-1{)}^2}{4},+\infty \right[$.