Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Prove it Bulgaria

Let $f_1 \in \mathbb{R}[x]$ be a quadratic polynomial with positive leading coefficient. Set $f_{n+1} = f_1 \circ f_n$ for $n \ge 1$ . It is known that the polynomial $f_2$ has four non-positive different zeroes. Prove that the polynomial $f_n$ has $2^n$ different real zeroes.

Solution

Note that if $x_1, \dots, x_{2^n}$ are the zeroes of $f_n$ , then the zeroes of $f_{n+1}$ are the solutions of the equations $f_1(x) = x_k$ , $1 \le k \le 2^n$ . Moreover, the equation $f_1(x) = a$ has two different real roots if and only if $a > m := \min f_1$ .

Assume that $f_2$ has four non-positive different zeroes. Then it is easy to see that $f_1$ has two zeroes $x_1 < x_2 \le 0$ and $x_1 > m$ . Using the argument from the beginning, it follows by induction on $n$ that all the zeroes of $f_{n+1}$ are different and belong to the interval $(x_1, x_2]$ .

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