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Brazilian Mathematical Olympiad

Algebra Difficulty 6.5 National Olympiad Prove it Brazil

Let $a$ and $b$ be real numbers. Define $f_{a,b}: \mathbb{R}^2 \to \mathbb{R}^2$ by $f_{a,b}(x, y) = (a - b y - x^2, x)$ . If $P = (x, y) \in \mathbb{R}^2$ , define $f_{a,b}^0(P) = P$ and $f_{a,b}^{k+1}(P) = f_{a,b}(f_{a,b}^k(P))$ for all nonnegative integers $k$ .

The set $\operatorname{per}(a, b)$ of the periodic points of $f_{a,b}$ is the set of points $P \in \mathbb{R}^2$ such that $f_{a,b}^n(P) = P$ for some positive integer $n$ .

Fix $b$ . Prove that the set $A_b = \{a \in \mathbb{R} \mid \operatorname{per}(a, b) \neq \emptyset\}$ admits a minimum. Find this minimum.

Solution

Let the orbit $\operatorname{orb}(P)$ 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 $\operatorname{orb}(P) - 1$ . Setting $\operatorname{orb}(P) = m$ , $S_1 = x_0 + x_1 + \cdots + x_{m-1}$ and $S_2 = x_0^2 + x_1^2 + \cdots + x_{m-1}^2$ for simplicity, we obtain $S_1 = m a - b S_1 - S_2$ which implies
$S_2 = m a - (b-1) S_1.$
But it is known by Cauchy, for example, that $S_2 \geq S_1^2/m$ .

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

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

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